Category: Class 12

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Physics Guide Pdf Chapter 7 Dual Nature of Radiation and Matter Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Physics Solutions

Chapter 7 Dual Nature of Radiation and Matter

12th Physics Guide Dual Nature of Radiation and Matter Text Book Back Questions and Answers

Part – 1:

Text Book Evaluation:

I. Multiple Choice Questions:

Question 1.
The Wavelength λ_e of an electron and λ_p of a photon of same energy E are related
by
a) λ_p α λ_e
b) λ_p α \(\sqrt{\lambda_{e}}\)
c) λ_p α \(\frac{1}{\sqrt{\lambda_{e}}}\)
d) λ_p α λ_e²
Answer:
d) λ_p α λ_e²
Solution:

Question 2.
In an electron microscope, the electrons are accelerated by a voltage of 14 kV. If the voltage is changed to 224 kV, then the de Brogue wavelength associated with the electrons would
a) increase by 2 times
b) decrease by 2 times
c) decrease by 4 times
d) increase by 4 times
Answer:
c) decrease by 4 times
Solution:

Question 3.
A particle of mass 3 × 10^-6 g has the same wavelength as an electron moving with a velocity 6 × 10⁶ m s^-1. The velocity of the particle is
a) 1.82 × 10^-18 ms^-1
b) 9 × 10^-2 ms^-1
c) 3 × 10^-31 ms^-1
d) 1.82 × 10^-15 ms^-1
Answer:
d) 1.82 × 10^-15 ms^-1
Solution:

Question 4.
When a metallic surface is illuminated with radiation of wavelength λ, the stopping potential is V. If the same surface is illuminated with radiation of wavelength 2λ, the stopping potential is V/4. The threshold wavelength for the metallic surface is
a) 4λ
b) 5λ
c) \(\frac{5}{2}\)λ
d) 3λ
Answer:
d) 3λ
Solution:
hν – hν₀ = eV
\(\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_{\mathrm{o}}}\) = eV ………(1)
\(\frac{\mathrm{hc}}{2 \lambda}-\frac{\mathrm{hc}}{\lambda_{\mathrm{a}}}\) = \(\frac{\mathrm{eV}}{4}\) …………(2)
From (1) and (2)

Question 5.
If light of wavelength 330 nm is incident on metal with work function 3.55 eV, the electrons are emitted. Then the wavelength of the emitted electron is (Take h = 6.6 × 10^-34 Js)
a) < 2.75 × 10^-9 m
b) ≥ 2.75 × 10^-9 m
c) ≤ 2.75 × 10^-12 m
d) < 2.5 × 10^-10 m
Answer:
b) ≥ 2.75 × 10^-9 m
Solution:
hv – Φ₀ = \(\frac{1}{2}\) mv² = K.E

Question 6.
A photoelectric surface is illuminated successively by monochromatic light of wavelength λ and λ/2. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function at the surface of material is
a) \(\frac{\mathrm{hc}}{\lambda}\)
b) \(\frac{2 h c}{\lambda}\)
c) \(\frac{\mathrm{hc}}{3 \lambda}\)
d) \(\frac{\mathrm{hc}}{2 \lambda}\)
Answer:
d) \(\frac{\mathrm{hc}}{2 \lambda}\)
Solution:

Question 7.
In photoelectric emission, radiation whose frequency is 4 times the threshold frequency of a certain metal is incident on the metal. Then the maximum possible velocity of the emitted electron will be
a) \(\sqrt{\frac{\mathrm{hv}_{0}}{\mathrm{~m}}}\)
b) \(\sqrt{\frac{6 h v_{0}}{m}}\)
c) \(2 \sqrt{\frac{\mathrm{hv}_{0}}{\mathrm{~m}}}\)
d) \(\sqrt{\frac{\mathrm{hv}_{0}}{2 \mathrm{~m}}}\)
Answer:
b) \(\sqrt{\frac{6 h v_{0}}{m}}\)
Solution:
ν = 4ν₀
ν = ?
hν – hν₀ = \(\frac{1}{2}\) mv²

4hν₀ – hν₀ = \(\frac{1}{2}\) mv²

3hν₀ = \(\frac{1}{2}\) mv²

ν² = \(\frac{6 \mathrm{~h} v_{\mathrm{o}}}{\mathrm{m}}\)

⇒ ν = \(\sqrt{\frac{6 h v_{0}}{m}}\)

Question 8.
Two radiations with photon energies 0.9 eV and 3.3 eV respectively are falling on a metallic surface successively. If the work function of the metal is 0.6 eV, then the ratio of maximum speeds of emitted electrons will be
a) 1 : 4
b) 1 : 3
c) 1 : 1
d) 1 : 9
Answer:
b) 1 : 3
Solution:
\(\frac{1}{2}\) mv² = \(\frac{\mathrm{hc}}{\lambda}\)
\(\frac{1}{2}\) mv₁² = 0.9 – 0.6 = 0.3 ……………….(1)
\(\frac{1}{2}\) mv₂² = 3.3 – 0.6 = 2.7 ……………….(2)
eqn (1) / (2)
\(\frac{v_{1}^{2}}{v_{2}^{2}}\) = \(\frac{0.3}{2.7}=\frac{1}{9}\)

\(\frac{v_{1}}{v_{2}}=\sqrt{\frac{1}{9}}=\frac{1}{3}\)
ν₁ : ν₂ = 1 : 3

Question 9.
A light source of wavelength 520 nm emits 1.04 × 10¹⁵ photons per second while the second source of 460 nm produces 1.38 × 10¹⁵ photons per second. Then the ratio of power of second source to that of first source is
a) 1.00
b) 1.02
c) 1.5
d) 0.98
Answer:
c) 1.5
Solution:

Question 10.
The mean wavelength of light from the sun is taken to be 550 nm and its mean power is 3.8 × 10²⁶ W. The number of photons received by the human eye per second on average from sunlight is of the order of
a) 10⁴⁵
b) 10⁴²
c) 10⁵⁴
d) 10⁵¹
Answer:
a) 10⁴⁵
Solution:
Power P = En = \(\frac{\mathrm{hc}}{\lambda}\).n
n = \(\frac{P \lambda}{h c}\)
= \(\frac{3.8 \times 10^{26} \times 550 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^{8}}\)
= 105.5 × 10⁴³
n = 1.055 × 10⁴⁵

Question 11.
The threshold wavelength for a metal surface whose photoelectric work function is 3.313 eV is
a) 4125 Å
b) 3750 Å
c) 6000 Å
d) 2062.5 Å
Answer:
b) 3750 Å
Solution:
Φ = hν₀ = \(\frac{\mathrm{hc}}{\lambda_{\mathrm{o}}}\)

λ₀ = \(\frac{\mathrm{hc}}{\phi}\)
= \(\frac{6.64 \times 10^{-34} \times 3 \times 10^{8}}{500 \times 10^{-9}}\)
λ₀ = 3.757 × 10^-7
λ₀ = 3750 Å

Question 12.
Light of wavelength 500 nm is incident on a sensitive plate of photoelectric work function 1.235 eV. The kinetic energy of the photoelectrons emitted is (Take h = 6.6 × 10^-34 Js)
a) 0.58 eV
b) 2.48 eV
c) 1.24 eV
d) 1.16 eV
Answer:
c)1.24 eV
Solution:
λ = 500 nm;
φ = 1.235 eV
hν = \(\frac{\mathrm{hc}}{\lambda}\)

hν = \(\frac{6.64 \times 10^{-34} \times 3 \times 10^{8}}{500 \times 10^{-9}}\)
= 0.0396 × 10^-17
hν = 3.96 × 10^-19 J
hν = \(\frac{3.96 \times 10^{-19}}{1.6 \times 10^{-19}}\) = 2.475 evacuated
K.E = hν – φ
= 2.475 – 1.235
= 1.24 eV

Question 13.
Photons of wavelength λ are incident on a metal. The most energetic electrons ejected from the metal are bent into a circular arc of radius R by a perpendicular magnetic field having magnitude B. The work function of the metal is
a) \(\frac{\mathrm{hc}}{\lambda}-\mathrm{m}_{\mathrm{e}}+\frac{\mathrm{e}^{2} \mathrm{~B}^{2} \mathrm{R}^{2}}{2 \mathrm{~m}_{\mathrm{e}}}\)

b) \(\frac{\mathrm{hc}}{\lambda}+2 \mathrm{me}\left[\frac{\mathrm{eBR}}{2 \mathrm{~m}_{\mathrm{e}}}\right]^{2}\)

c) \(\frac{\mathrm{hc}}{\lambda}-\mathrm{m}_{\mathrm{e}} \mathrm{C}^{2} \frac{\mathrm{e}^{2} \mathrm{~B}^{2} \mathrm{R}^{2}}{2 \mathrm{~m}_{\mathrm{e}}}\)

d) \(\frac{\mathrm{hc}}{\lambda}-2 \mathrm{~m}_{\mathrm{e}}\left[\frac{\mathrm{eBR}}{2 \mathrm{~m}_{\mathrm{e}}}\right]^{2}\)
Answer:
d) \(\frac{\mathrm{hc}}{\lambda}-2 \mathrm{~m}_{\mathrm{e}}\left[\frac{\mathrm{eBR}}{2 \mathrm{~m}_{\mathrm{e}}}\right]^{2}\)
Solution:

Question 14.
The work functions for metals A, B and C are 1.92 eV, 2.0 eV and 5.0 eV respectively. The metals which will emit photoelectrons for radiation of wavelength 4100 Å is/are
a) A Only
b) both A and B
c) all these metals
d) none
Answer:
b) both A and B
Solution:
Energy in eV = \(\frac{12375}{\lambda}\)
\(\frac{12375}{4100}\) =3.01 eV
Work functions at A and B are less than 3.01 eV so both A and B emit.

Question 15.
Emission of electrons by the absorption of heat energy is called ________ emission
a) photoelectric
b) field
c) thermionic
d) secondary
Answer:
c) thermionic

II. Short Answer Questions:

Question 1.
Why do metals have a large number of free electrons?
Answer:
In metals, the electrons in the outermost shells are loosely bound to the nucleus. Even at room temperature, there are a large number of free electrons which are moving inside the metal in a random manner.

Question 2.
Define the work function of a metal. Give its unit.
Answer:

• The minimum energy needed for an electron to escape from the metal surface is called the work function of a metal.
• It is denoted by Φ₀ and its unit is eV

Question 3.
What is the photoelectric effect?
Answer:
The ejection of electrons from a metal plate when illuminated by light or any other electromagnetic radiation of suitable wavelength (or) frequency.

Question 4.
How does photocurrent vary with the intensity of the incident light?
Answer:
Photocurrent – the number of electrons emitted per second is directly proportional to the intensity of the incident light.

Question 5.
Give the definition of intensity of light and its unit.
Answer:

• The intensity is the power of light commonly referred to brightness.
• Its unit is candela (cd).

Question 6.
How will you define threshold frequency?
Answer:
For a given metallic surface, the emission of photoelectrons takes place only if the frequency of the incident light is greater than a certain minimum frequency called the threshold frequency.

Question 7.
What is a photocell? Mention the different types of photocells.
Answer:
photocells: Photoelectric cell or photocell is a device which converts light energy into electrical energy. It works on the principle of the photoelectric effect.
Types:

• Photo emissive cell
• Photovoltaic cell
• Photoconductive cell

Question 8.
Write the expression for the de Broglie wavelength associated with a charged particle of charge q and mass m, when it is accelerated through a potential V
Answer:

• De Broglie wavelength λ = \(\frac{\mathrm{h}}{\sqrt{2 \mathrm{qmv}}}\)
• Where q = ne, h= Planck’s constant

Question 9.
State de Broglie hypothesis.
Answer:
De Broglie’s hypothesis, all matter particles like electrons, protons, neutrons in motion are associated with waves.

Question 10.
Why we do not see the wave properties of a baseball?
Answer:
Since the momentum of a baseball is very low and the wavelength is of the order of 10^-34 m the wave properties cannot be seen in a baseball.

Question 11.
A proton and an electron have the same kinetic energy. Which one has a greater de Brogue wavelength? Justify.
Answer:
de-Broglie wavelength of the particle is λ = \(\frac { h }{ p }\) = \(\frac { h }{ \sqrt { 2mK } } \)
i.e. λ ∝ \(\frac { h }{ \sqrt { m } } \)
As m_e << m_p, so λ_e >> λ_p
Hence protons have greater de-Broglie wavelength.

Question 12.
Write the relationship of de Broglie wavelength \ associated with a particle of mass m in terms of its kinetic energy K.
1. De Broglie wavelength h h
λ = \(\frac{\mathrm{h}}{\mathrm{m} \cdot \mathrm{v}}\)
= \(\frac{\mathrm{h}}{\sqrt{2 \mathrm{emV}}}\)

2. Since kinetic energy of the electron K = eV then the de Broglie wavelength is given by
λ = \(\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}\)

Question 13.
Name an experiment which shows wave nature of the electron. Which phenomenon was observed in this experiment using an electron beam?
Answer:

• Davisson – Germer experiment confirmed the wave nature of electrons.
• They demonstrated that electron beams are diffracted when they fall on crystalline solids.

Question 14.
An electron and an alpha particle have same kinetic energy. How are the de Broglie wavelengths associated with them related?
Answer:
1. De Broglie wavelength is given by
λ = \(\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}\)
2. Since m_α > m_e
λ α \(\frac{1}{\sqrt{\mathrm{m}}}\) then
λ_α < λ_e
3. An α particle having a greater mass than electron so it is having lesser de Broglie wavelength.

III. Long Answer Questions:

Question 1.
What do you mean by electron emission? Explain briefly various methods of electron emission.
Answer:
The liberation of electrons from any surface of a metallic substance is called electron emission.
There are mainly four types of electron emission.
i) Thermionic emission:
1. When a metal is heated to a high temperature, the free electrons on the surface of the metal get sufficient thermal energy so that they are emitted from the metallic surface.
2. The intensity of the thermionic emission depends on the metal used and its temperature.
(Examples: Cathode ray tubes, electrons microscopes, X-ray tubes, etc.)

Electrons in the (a) metal (b) heated metal

ii) Field emission:
1. Electron field emission occurs when a very strong electric field is applied across the metal.
2. This strong field pulls the free electrons and helps them to overcome the surface barrier of the metal.
Examples:
Field emission scanning electron microscopes, Field emission display, etc….

Field emission

iii) Photoelectric emission :
1. When electromagnetic radiation of suitable frequency is incident on the surface of the metal, the energy is transferred from the radiation to the free electrons.
2. Hence the free electrons get sufficient energy to cross the surface barrier and the photoelectric emission takes place.
3. The number of electrons emitted depends on the intensity of the incident radiation.
Examples: Photodiodes, Photoelectric cells, etc.

Photoelectric emission

iv) Secondary emission:
1. When a beam of fast-moving electrons strikes the metal surface, the kinetic energy of the striking electrons is transferred to the free electrons on the metal surface.
2. Thus the free electrons get sufficient kinetic energy so that the secondary emission occurs.
Example: Image intensifiers, Photomultiplier tubes, etc..

Secondary emission of electrons

Question 2.
Briefly discuss the observations of Hertz, Hallwachs, and Lenard.
Answer:
Hertz Observation:
1. In 1887, Hertz generated and detected electromagnetic waves with his high voltage induction coil to cause a spark discharge between two metallic spheres.
2. When a spark is formed, the charges will oscillate back and forth rapidly and the electromagnetic waves are
produced.
3. The electromagnetic waves thus produced were detected by a detector that has a copper wire bent in the shape of a circle
4. But the tiny spark produced in the detecter cannot be explained by hertz.

Hallwach Observation:

Irradiation of ultraviolet light on

• uncharged zinc plate
• negatively charged plate
• positively charged plate

1. In 1888, Hallwach confirmed that the strange behaviour of the spark is due to the action of ultraviolet light.
2. A clean circular plate of zinc is mounted on an insulating stand and is attached to a gold leaf electroscope by a wire.
3. When the uncharged zinc plate is irradiated by ultraviolet light from an arc lamp, it becomes positively charged and the leaves will open.
4. If the negatively charged zinc plate is exposed to ultraviolet light, the leaves will close as the charges leaked away quickly.

Lenard’s observation:

1. In 1902, Lenard constructed an apparatus consists of two metallic plates A and C placed in an evacuated quartz bulb.
2. The Galvanometer G and battery B are connected in the circuit.
3. When ultraviolet light is incident there is a deflection in the galvanometer.
4. If anode A is exposed to ultraviolet light, no current is observed.
5. The electron emission observed when ultraviolet light falling on the negative plate called photoelectric emission.

Experimental setup for the study of the photoelectric effect

Question 3.
Explain the effect of potential difference on photoelectric current.
Answer:
1. The effect of potential difference on photoelectric current can be studied by keeping the frequency and the intensity of the incident light constant.

Variation of photocurrent with a potential difference

2. If the cathode is irradiated the potential of A is increased thereby increase in the photocurrent.
3. When all the photoelectrons from C are collected by A the photocurrent reaches a maximum called saturation current.
4. When a negative (retarding) potential is applied to A with respect to C the photoelectrons start to decrease because more and more photoelectrons are being repelled away.
5. The photocurrent becomes zero at a particular negative potential V₀ called stopping or cut off potential.
6. The negative potential applied to anode A which is just sufficient to stop the most energetic photoelectrons to make photocurrent zero is called stopping potential.
7. The kinetic energy of the fastest electron (K_max) is equal to the work done by the stopping potential to stop it (eV₀)
K_max = \(\frac{1}{2}\) mv_max² = eV₀ …………..(1)
where ν_max is the maximum speed of the photo electrons
ν_max = \(\sqrt{\frac{2 \mathrm{eV}_{\mathrm{o}}}{\mathrm{m}}}\)
= \(\sqrt{\frac{2 \times 1.602 \times 10^{-19}}{9.1 \times 10^{-31}} \times V_{\mathrm{O}}}\)
ν_max = 5.93 × 10⁵ \(\sqrt{\mathrm{V}_{\mathrm{O}}}\) …………….(2)
K_max = eV₀ (in Joule) or K_max = V₀ (in eV)
8. The maximum kinetic energy of the photoelectrons is independent of the intensity of the incident light.

Question 4.
Explain how the frequency of incident light varies with stopping potential.
Answer:
1. The effect of frequency of incident light on stopping potential can be studied by keeping the intensity of the incident light constant
2. The electrode potential varies for different frequencies of the incident light.
3. As the frequency is increased the photoelectrons are emitted with greater kinetic energies so that the retarding potential needed to stop the photoelectrons is also greater.

variation or photocurrent with corrector electrode potential for different frequencies of the incident radiation

4. From the graph between frequency and stopping potential, the stopping potential varies linearly with the frequency of the incident light.
5. The stopping potential is zero when no electrons are emitted below a certain frequency called threshold frequency.

Variation of stopping potential with frequency of the incident radiation for two metals

Question 5.
List out the laws of the photoelectric effect.
Answer:

1. For a given frequency of incident light, the number of photoelectrons emitted is directly proportional to the intensity of the incident light.
2. The maximum kinetic energy of the photoelectrons is independent of the intensity of the incident light.
3. The maximum kinetic energy of the photoelectrons from a given metal is directly proportional to the frequency of incident light.
4. For a given surface, the emission of photoelectrons takes place only if the frequency of incident light is greater than a certain minimum frequency called the threshold frequency.
5. There is no time lag between the incidence of light and emission of photoelectrons.

Question 6.
Explain Why the photoelectric effect cannot be explained on the basis of the wave nature of light.
Answer:
Failures of classical wave theory:
From Maxwell’s theory, light is an electromagnetic wave consisting of coupled electric and magnetic oscillations that move with the speed of light and exhibit typical wave behaviour. Let us try to explain the experimental observations of the photoelectric effect using wave picture of light.

1. When light is incident on the target, there is a continuous supply of energy to the electrons. According to wave theory, light of greater intensity should impart greater kinetic energy to the liberated electrons (Here, the Intensity of light is the energy delivered per unit area per unit time). But this does not happen. The experiments show that the maximum kinetic energy of the photoelectrons does not depend on the intensity of the incident light.

2. According to wave theory, if a sufficiently intense beam of light is incident on the surface, electrons will be liberated from the surface of the target, however low the frequency of the radiation is. From the experiments, we know that photoelectric emission is not possible below a certain minimum frequency. Therefore, the wave theory fails to explain the existence of threshold frequency.

3. Since the energy of light is spread across the wavefront, the electrons which receive energy from it are large in number. Each electron needs considerable amount of time (a few hours) to get energy sufficient to overcome the work function and to get liberated from the surface. But experiments show that photoelectric emission is almost instantaneous process (the time lag is less than 10“9 s after the surface is illuminated) which could not be explained by wave theory.

Question 7.
Explain the quantum concept of light.
Answer:
Planck’s concepts of quantum :

1. Max Planck proposed the quantum concept in 1900 in order to explain the thermal radiations emitted by a black body and the shape of the radiation curves.
2. If an oscillator vibrates with frequency ν, its energy can have only certain discrete values.
   E = nhν n = 1,2,3…………
3. Where h is Planck’s constant.
4. The oscillators emit or absorb energy in small packets or quanta and the energy of each quantum is E = hν
5. Energy is quantized and it is not continuous as believed in the wave nature. This is called the quantization of energy.

Einstein’s concept of quantum:

1. Einstein extended Planck’s quantum concept to explain the photoelectric effect in 1905.
2. According to Einstein, the energy in light is not spread out over wavefronts but is concentrated in small packets called energy quanta.
3. The light of frequency ν from any source having energy E = hν
4. The light quanta having the magnitude of linear momentum
   p = \(\frac{\mathrm{hv}}{\mathrm{c}}\)

Question 8.
Obtain Einstein’s photoelectric equation with the necessary explanation.
Answer:
When a photon of energy hν is incident on a metal surface, it is completely absorbed by a single electron and the electron is ejected. In this process, a part of the photon energy is used for the ejection of the electrons from the metal surface (photo electric work function Φ₀) and the remaining energy is imparted as kinetic energy of the ejected electron.
hν = Φ₀ + \(\frac{1}{2}\) mv² ………………….(1)
Where m is the mass and v is the velocity.
At threshold frequency, ν₀
There is no ejection of electrons hence zero kinetic energy then,
hν₀ = Φ₀ …………………(2)
eqn (1) => hν = hν₀ + \(\frac{1}{2}\) mv² ……………(3)
This is Einsteins photo electric equation.
If the electron does not lose energy by internal collisions, then it is emitted with maximum
kinetic energy K_max. Then
K_max = \(\frac{1}{2}\) mv_max²
Where ν_max maximum velocity of the electron ejected
K_max = hν – Φ₀ ……………..(4)

Emission of photoelectrons

A graph between K_max maximum kinetic energy of a photoelectron and frequency ν of the incident light is a straight line.

K_max vs ν graph

Question 9.
Explain experimentally observed facts of the photoelectric effect with the help of Einstein’s explanation.
Answer:

1. The experimentally observed facts of the photoelectric effect can be explained with the help of Einstein’s photoelectric equation.
2. As each incident photon liberates one electron, then the increase of intensity of the light (the number of per unit area per unit time) increases the number of electrons emitted thereby increasing the photocurrent.
3. From K_max = hν – Φ₀ it is evident that K_max is proportional to the frequency and independent of the intensity of the light.
4. From hν = hν₀ + \(\frac{1}{2}\) mv², there must be minimum energy equal to the work function of the metal to liberate electrons from the metal surface. Below which the emission of electrons is not possible.
5. There exists a minimum frequency called threshold frequency below which there is no photoelectric emission.
6. The transfer of photon energy to the electrons is instantaneous so that no time lag between the incidence of photons and ejection of electrons.
7. Thus the photoelectric effect is explained on the basis of the quantum concept of light.

Question 10.
Give the construction and working of a photoemissive cell.
Answer:
Photo emissive cell:
Construction:

Construction of photo cell

1. It consists of an evacuated glass or quartz bulb in which two metallic electrodes, cathode and anode are fixed.
2. The cathode C is Semi cylindrical in shape and is coated with a photo-sensitive material.
3. anode A is a thin rod or wire kept along the axis of the semi-cylindrical cathode.
4. A potential difference is applied between the anode and the cathode through a galvanometer G.

Working:
When cathode is illuminated, electrons are emitted from it. These electrons are attracted by anode and hence a current is produced which is measured by the galvanometer. For a given cathode, the magnitude of the current depends on (i) the intensity of incident radiation and (ii) the potential difference between anode and cathode.

Question 11.
Derive an expression for de Broglie wavelength of electrons.
Answer:
An electron of mass m is accelerated through a potential difference of V volt. The kinetic energy aquired by the electron is given by
\(\frac{1}{2}\) mv² = eV
Therefore, the speed ν of the electron is
ν = \(\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}\) …………..(1)
From de Broglie equation
λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\) = \(\sqrt{\frac{\mathrm{h}}{2 \mathrm{emV}}}\) ……………(2)
Substituting the known values

When electron is accelerated with 100 V, then
λ = \(\frac{12.27}{\sqrt{100}}\) = 1.227 Å
Since the kinetic energy of the electron, K = eV, then the de Broglie wavelength associated with electron is given by
λ = \(\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}\) …………..(4)

Question 12.
Briefly explain the principle and working of an electron microscope.
Answer:
Principle:

(a) Optical Microscope
(b) Electron Microscope

1. The wave nature of the electron is used in the construction of an electron microscope.
2. The resolving power of a microscope is inversely proportional to the wavelength of the radiation used.
3. Since the de Broglie wavelength of the electron is much smaller than the visible light, the resolving power and the magnification are high.
4. Electron microscopes giving magnification more than 2,00,000 times than the optical microscope.

Working:

1. The construction and working of an electron microscope are similar to the optical microscope.
2. Focussing of electron beam is done by the electrostatic or magnetic lenses.
3. The divergence and convergence can be done by adjusting electric and magnetic fields.
4. The accelerated electrons from the source with high potential are made parallel by a magnetic condenser lens.
5. With the help of a magnetic objective lens and magnetic projector lens system, the magnified image is obtained.
6. These electron microscopes are being used in almost all branches of science.

Question 13.
Describe briefly Davisson – Germer experiment which demonstrated the wave nature of electrons.
Answer:

Experimental set up of Davisson – Germer experiment

1. De Broglie hypothesis of matter waves was experimentally confirmed by Clinton Davisson and Lester Germer in 1927.
2. They demonstrated that electron beams are diffracted when they fall on crystalline solids.
3. Since crystals can act as a three-dimensional diffraction grating for matter waves, the electron waves incident on crystals are diffracted off in certain specific directions.
4. The filament F is heated by a low tension battery, electrons are emitted from the hot filament by thermionic emission.
5. Electrons are accelerated due to-the potential between the filament and the anode aluminium cylinder by a high tension battery.
6. The electron beam is collimated and it is allowed to strike a single crystal of nickel by two thin aluminum diaphragms.
7. The electron detector measures the intensity of the scattered electron beam.
8. The intensity of the scattered beam is measured for various incident angles θ.
9. For accelerating voltage of 54 V, the scattered wave shows a peak at an angle of 50° to the incident beam.
10. The constructive interference of diffracted electrons for various atomic layers obtained known as interplanar spacing of Nickel.
11. The de-Broglie wavelength for V = 54 V is given by,

Variation of intensity of diffracted electron beam with the angle
λ = \(\frac{12.27}{\sqrt{\mathrm{V}}}\) Å
λ = \(\frac{12.27}{\sqrt{54}}\)
= 1.67 Å

IV. Numerical Problems:

Question 1.
How many photons per second emanate from a 50 mW laser of 640 nm?
Answer:
Given:
P = 50 × 10^-3 W
λ = 640 × 10^-9 m
N = ?
N => Number of photons emitted per second.

Question 2.
Calculate the maximum kinetic energy and maximum velocity of the photoelectrons emitted when the stopping potential is 81 V for the photoelectric emission experiment.
Answer:
Given:
V = 81 V
E_K = ?
E_K maximum kinetic energy
v_K maximum velocity
\(\frac{1}{2}\) mv² = eV
E_K = 1.6 × 10^-19 × 81
= 129.6 × 10^-19
E_K = 1.3 × 10^-17 J

Question 3.
Calculate the energies of the photons associated with the following radiation
(i) violet light of 413 nm,
(ii) X-rays of 0.1 nm,
(iii) radio waves of 10 m
Answer:
E = hν
E = \(\frac{\mathrm{hc}}{\lambda}\) in Joule

E = \(\frac{\mathrm{hc}}{\lambda \mathrm{e}}\) in eV
(i) Violet light of 413 nm
E = \(\frac{6.625 \times 10^{-34} \times 3 \times 10^{8}}{413 \times 10^{-9} \times 1.6 \times 10^{-19}}\)
= \(\frac{19.875}{660.8}\) × 10^-26 × 10²⁸
= 0.032 × 10²
E = 3 eV

(ii) X-rays of 0.1 nm
= \(\frac{6.625 \times 10^{-34} \times 3 \times 10^{8}}{0.1 \times 10^{-9} \times 1.6 \times 10^{-19}}\)

= \(\frac{19.875 \times 10^{-26}}{0.16 \times 10^{-28}}\)
= 124.24 × 10²
E = 12424 eV

(iii) radio waves of 10 m
E = \(\frac{6.625 \times 10^{-34} \times 3 \times 10^{8}}{10 \times 1.6 \times 10^{-19}}\)
= \(\frac{19.875}{1.6}\) × 10^-26 × 10^-1 × 10¹⁹
E = 1.24 × 10^-7 eV

Question 4.
A 150 W lamp emits light of mean wavelength of 5500 A. If the efficiency is 12%, find out the number of photons emitted by the lamp in one second.
Answer:
P = 150 W
λ = 5500 × 10^-10 m
Eff = 12%
N = \(\frac{\mathrm{P} \lambda}{\mathrm{hc}}\)

= \(\frac{150 \times 5500 \times 10^{-10}}{6.626 \times 10^{-34} \times 3 \times 10^{8}}\)

= \(\frac{825 \times 10^{-7}}{19.878 \times 10^{-26}}\)
= 41.5 × 10¹⁹
Number of photons emitted is with 12% efficiency = 41.5 × 10¹⁹ × \(\frac{12}{100}\)
N = 4.98 × 10¹⁹

Question 5.
How many photons of frequency 10¹⁴ Hz will make up 19.86 J of energy?
Answer:
E = 19.86 J
ν = 10¹⁴ Hz
E = nhν
n = \(\frac{E}{h v}\)
= \(\frac{19.86}{6.626 \times 10^{-34} \times 10^{14}}\)
= 2.997 × 10²⁰
Number of Photons N = 3 × 10²⁰

Question 6.
What should be the velocity of the electron so that its momentum equals that of 4000 Å wavelength photon.
Answer:
Given:
λ = 4000 × 10^-10 m

Question 7.
When a light of frequency 9 × 10¹⁴ Hz is incident on a metal surface, photoelectrons are emitted with a maximum speed of 8 × 10⁵ ms^-1. Determine the threshold frequency of the surface.
Answer:
Given:
ν = 9 × 10¹⁴ Hz
ν = 8 × 10⁵ms^-1
From Einstein’s photo electric equation
hν = hν₀ – \(\frac{1}{2}\) mv²
hν₀ = hν – \(\frac{1}{2}\) mv²
ν₀ = \(\frac{\mathrm{hv}-1 / 2 \mathrm{mv}^{2}}{\mathrm{~h}}\)
ν₀ = ν – \(\frac{\mathrm{mv}^{2}}{2 \mathrm{~h}}\)
= 9 × 10¹⁴ – \(\frac{9.1 \times 10^{-31} \times 64 \times 10^{10}}{2 \times 6.626 \times 10^{-34}}\)
= 9 × 10¹⁴ – \(\frac{582.4 \times 10^{-21}}{13.252 \times 10^{-34}}\)
= 9 × 10¹⁴ – 4.39 × 10¹⁴
= (9 – 4.39) × 10¹⁴
ν₀ = 4.61 × 10¹⁴ Hz

Question 8.
When a 6000 A light falls on the cathode of a photocell and produced photoemission. if a stopping potential of 0.8 V is required to stop the emission of an electron, then determine the
(i) frequency of the light
(ii) the energy of the incident photon
(iii) the work function of the cathode material
(iv) threshold frequency and
(v) the net energy of the electron after it leaves the surface.
Answer:
Given:
λ = 6000 × 10^-10 m;
eV₀ = 0.8 eV
i) ν = \(\frac{C}{\lambda}\)
= \(\frac{3 \times 10^{8}}{6 \times 10^{-7}}\)
= 0.5 × 10¹⁵
ν = 5 × 10¹⁴ Hz

(ii) E = hν
= \(\frac{6.626 \times 10^{-34} \times 5 \times 10^{14}}{1.6 \times 10^{-19}}\)
= \(\frac{53.13}{1.6}\) × 10^-20 × 10¹⁹
E = 2.07 eV

(iii) eν₀ = \(\frac{1}{2}\) mv² = 0.8 eV
W = hν – \(\frac{1}{2}\) mv²
= (2.07 – 0.8) eV
W = 1.27 eV

(iv) W = hν₀
ν₀ = \(\frac{1.27 \times 1.6 \times 10^{-19}}{6.626 \times 10^{-34}}\)
= \(\frac{2.032}{6.626}\) × 10¹⁵
ν₀ = 3.06 × 10¹⁴ Hz

(v) Net Energy = E = hν – hν₀
E = (2.07 – 1.27) ev = 0.8 eV
E = 0.8 eV

Question 9.
A 3310 Å photon liberates an electron from a material with energy 3 × 10^-19 J while another 5000 Å photon ejects an electron with energy 0.972 × 10^-19 J from the same material. Determine the value of Planck’s constant and the threshold wavelength of the material.
Answer:
Given:
λ₁ = 3310 × 10^-10 m
λ₂ = 5000 × 10¹⁰ m
K.E₁ = 3 × 10^-19 J
K.E₂ = 0.972 × 10^-19 J
i) ν₁ = \(\frac{C}{\lambda_{1}}\)
= \(\frac{3 \times 10^{8}}{3310 \times 10^{-10}}\)
= 9.06 × 10¹⁴ Hz
ν₂ = \(\frac{C}{\lambda_{2}}\)
= \(\frac{3 \times 10^{8}}{5000 \times 10^{-10}}\)
= 6 × 10¹⁴ Hz
hν₁ = W + K.E₁ …………….(1)
hν₁ = W + K.E₂ …………….(2)
eqn (1) – (2)
h(ν₁ – ν₂) = K.E₁ – K.E₂
h(9.06 – 6) × 10¹⁴ = 3 × 10^-19 – 0.972 × 10^-19
h(3.06 × 10¹⁴) = 2.028 × 10^-19
h = \(\frac{2.028 \times 10^{-19}}{3.06 \times 10^{14}}\)
= 0.662 × 10^-33
h = 6.62 × 10^-34 Js

ii)

Question 10.
At the given point of time, the earth receives energy from the sun at 4 cal cm min^-1. Determine the number of photons received on the surface of the Earth per cm^-2 per minute. (Given: Mean wavelength of sunlight = 5500 Å)
Answer:
Given:
E = 4 cal cm^-2 min^-1
λ = 5500 × 10^-10m
We know that 1 cal = 4.2 J
E = hnν

Number of photons n = 4.652 × 10¹⁹ per cm² per minute

Question 11.
UV light of wavelength 1800 Å is incident on a lithium surface whose threshold wavelength 4965 Å. Determine the maximum energy of the electron emitted.
Answer:
Given:
λ = 1800 × 10^-10 m
λ₀ = 4965 × 10^-10 m
hν = hν₀ +K.E
h\(\frac{c}{\lambda}\) =h\(\frac{c}{\lambda_{o}}\) + K. E
K.E = \(\frac{\mathrm{hc}}{\lambda}\) – \(\frac{\mathrm{hc}}{\lambda_{\mathrm{o}}}\)
hc = 6.625 × 10^-34 × 3 × 10⁸
hc = 19.86 × 10^-24
K.E = \(\frac{19.86 \times 10^{-26}}{1800 \times 10^{-10}}\) – \(\frac{19.86 \times 10^{-24}}{4965 \times 10^{-10}}\)
= 0.0110 × 10^-16 – 0.004 × 10^-16
K.E in eV = \(\frac{7 \times 10^{-19}}{1.6 \times 10^{-19}}\) = 4.375 eV
K.E = 4.4 eV

Question 12.
Calculate the de Brogue wavelength of a proton whose kinetic energy is equal to 81.9 x 1O’ J. (Given: mass of proton is 1836 times that of electron).
Answer:
Given:
K.E = 81.9 × 10^-15 J
m_p = 836 m_e

Question 13.
A deuteron and an alpha particle are accelerated with the same potential. Which one of the two has
i) greater value of de Broglie wavelength associated with it and
ii) less kinetic energy? Explain.
Answer:
Given:
deutron => e → 1e \(\left({ }_{1} \mathrm{H}^{2}\right)\)
m → 2m
alpha => e → 2e (2He⁴)
m → 4m
i)

ii) \(\frac{1}{2}\) mv² = eV
K_d = eV
K_α = 2 eV
K_α = 2 K_d
K_d = \(\frac{\mathrm{K}_{\alpha}}{2}\)

Question 14.
An electron is accelerated thorugh a potential difference of 81 V. What is the de Brogue wavelength associated with it? To which part of electromagnetic spectrum does this wavelength correspond?
Answer:
Given:
V = 81 V
λ = \(\frac{12.27}{\sqrt{\mathrm{V}}}\) Å
λ = \(\frac{12.27}{\sqrt{81}}\) ⇒ λ = \(\frac{12.27}{9}\)
λ = 1.36 Å

Question 15.
The ratio between the de Brogue wavelengths associated with protons. accelerated through a potential of 512 V and that of alpha particles accelerated through a potential of X volts is found to be one. Find the value of X.
Answer:
Given:
For proton charge = e
mass = m
For charge = 2e
mass = 4m

Part II:

12th Physics Guide Dual Nature of Radiation and Matter Additional Questions and Answers

I. Match the following:

Question 1.

Answer:
A. b
B. d
C. a
D. c

Question 2.

Answer:
A. b
B. c
C. d
D. a

Question 3.

Answer:
A. c
B. a
C. d
D. b

Question 4.

Answer:
A. d
B. c
C. b
D. a

II. Fill in the blanks:

Question 1.
The liberation of electrons from any surface of a substance is called _______.
Answer:
electron emission

Question 2.
1 eV is equal to ___________ Joule.
Answer:
1.602 × 10^-19

Question 3.
The stopping potential is independent of ________ of the incident light.
Answer:
intensity

Question 4.
The quality of X-rays is measured in terms of their _______.
Answer:
penetrating power

Question 5.
The graph between maximum kinetic energy of the photoelectron and frequency of the incident light is ________.
Answer:
straight line

Question 6.
Einstein’s photoelectric equation was experimentally confirmed by ________.
Answer:
Millikan

III. Choose the odd man out:

Question 1.
a) Field emission display
b) Photodiodes
c) Photomultiplier tubes
d) Electron microscope
Answer:
d) Electron MicroScope

Question 2.
a) Work function
b) Discharge tube
c) Threshold frequency
d) Stopping potential
Answer:
b) Discharge tube

Question 3.
a) mass
b) frequency
c) Intensity
d) wavelength
Answer:
a) mass

Question 4.
a) neutrons
b) tungsten
c) particles
d) x-rays
Answer:
a) tungsten

Question 5.
a) Photon
b) electron
c) proton
d) neutron
Answer:
a) photon

IV. Choose the incorrect pair:

Question 1.

Answer:
D) Intensity of X-rays – Constant for all substances

Question 2.

Answer:
D) Emission of photoelectrons – does not depend upon the potential of the electrodes

Question 3.

Answer:
B) Stopping potential – maximum potential between the electrodes

V. Choose the correct pair:

Question 1.

Answer:
B) de – Broglie wavelength – 1.67 Å

Question 2.

Answer:
D) V = \(\frac{\sqrt{2 \mathrm{em}}}{\mathrm{V}}\)

VI. Assertion and Reason:

Question 1.
Assertion:
The resolving power of the electron microscope is high.
Reason:
The wavelength of electrons is very much lesser than the visible light.
i) Assertion is true but reason is false
ii) The assertion and reason both are false
iii) Both assertion and reason are true and the reason is the correct explanation of the assertion.
iv) Both assertion and reason are true and the reason is not the correct explanation of the assertion.
Answer:
(iii) Both assertion and reason are true and the reason is the correct explanation of the assertion.

Question 2.
Assertion:
The oscillators emit or absorb energy in small packets or quanta.
Reason:
Energy is not discrete, the energy posses wave nature.
i) The assertion and reason both are false
i) The assertion is true but reason is false
iii) The assertion is false but reason is true
iv) Both the assertion and reason are true
Answer:
(ii) The assertion is true but the reason is false

VII. Choose the correct statement:
Question 1.
a) Particle cannot be localised in space and time
b) Wave can be localised in space and time
c) Black body radiation can be explained by wave nature.
d) The minimum energy needed for an electron to escape from the metal surface is called the work function.
Answer:
d) The minimum energy needed for an electron to escape from the metal surface is called the work function.

Question 2.
a) When the momentum of the particle increases de-Broglie wavelength also increases.
b) A photon of energy 2E is an incident on a photosensitive surface of photoelectric work function E. The maximum kinetic energy of photoelectron emitted is 2E.
c) The number of de-Broglie waves of an electron in the nth orbit of an atom is n.
d) The resolving power of the electron microscope is 100 times greater than an optical microscope.
Answer:
b) A photon of energy 2E is an incident on a photosensitive surface of photoelectric work function E. The maximum kinetic energy of the photoelectron emitted is 2E.

VIII. Choose the incorrect statement:

Question 1.
a) Wave mechanical concept of the atom was based on the de-Broglie hypothesis
b) Maximum kinetic energy of the photoelectrons varies linearly with the frequency of incident radiation.
c) The stopping potential of a metal surface is independent of the intensity of the incident radiation.
d) The graph drawn by taking the frequency of the radiation along the x-axis and stopping potential along the y-axis for a photo-sensitive metal is a parabola.
Answer:
d) The graph drawn by taking the frequency of the radiation along the x-axis and stopping potential along the y-axis for a photo-sensitive metal is a parabola.

Question 2.
a) Photoelectric emission is an instantaneous process.
b) Maximum kinetic energy of photoelectrons is directly proportional to the frequency of the incident radiation.
c) Maximum kinetic energy of photoelectrons is indirectly proportional to the intensity of the incident radiation.
d) Photoelectron emission is not possible below a minimum frequency of the incident radiation.
Answer:
c) Maximum kinetic energy of photoelectrons is indirectly proportional to the intensity of incident radiation.

IX. Choose the best answer:

Question 1.
The maximum kinetic energy of photoelectrons emitted from a surface when photons of
energy 3 eV fall on it is 4 eV. The stopping potential, in volt, is
(a) 2
(b) 4
(c) 6
(d) 10
Answer:
(b) 4
Hint:
Stopping potential, V₀ = \(\frac { { K }_{ max } }{ e } \) = \(\frac { 4eV }{ e }\) = 4v

Question 2.
The threshold frequency is constant with respect to _______.
a) type of the metal
b) Velocity of electrons
c) Voltage applied
d) intensity of the incident light.
Answer:
a) type of the metal

Question 3.
In a photoemissive cell, the anode is made up of _______.
a) copper
b) gold
c) platinum
d) zinc
Answer:
c) platinum

Question 4.
The work function of a substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from this substance is approximately
(a) 540 nm
(b) 400 nm
(c) 310 nm
(d) 220 nm
Answer:
(c) 310 nm
Hint:
λ₀ = \(\frac { hc }{ W }\) = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{4.0 \times 1.6 \times 10^{-19}}\) = m = 310 x 10^-9 m = 310 nm

Question 5.
Duane-Hunt law is ________.
a) λ = \(\frac{12400}{\mathrm{v}}\) m
b) λ = \(\frac{\mathrm{hc}}{\mathrm{V}}\) m
c) λ = \(\frac{\mathrm{hc}}{\sqrt{\mathrm{E}_{\mathrm{k}}}}\) m
d) none of the above
Answer:
a) λ = \(\frac{12400}{\mathrm{v}}\) m

Question 6.
The resolving power of the electron microscope is _____ times greater than the resolving power of the optical
microscope
a) 10²
b) 10⁴
c) 10⁵
d) 10³
Answer:
c) c) 10⁵

Question 7.
4 eV is the energy of the incident photon and the work function is 2 eV. The stopping potential will be
(a) 2V
(b) 4V
(c) 6V
(d) 2√2V
Answer:
(d) 2√2V
Hint:
eV₀= hv- W₀ = 4eV – 2eV = 2eV
∴ V₀= \(\frac { 2ev }{ e }\) = 2v

Question 8.
Which of the following graph represents the variation of the particle momentum and associated de Broglie wavelength.
a)

b)

c)

d)
Answer:
c)

Question 9.
The radiation produced from decelerating electron is called ________.
a) reverse radiation
b) Photon emission
c) black body radiation
d) breaking radiation
Answer:
d) breaking radiation

Question 10.
If the striking electrons knocks of an electron in n = 2 state, the resulting transition give ________.
a) K-lines
b) L-lines
c) M-lines
d) N-lines
Answer:
b) L-lines

Question 11.
When a proton is accelerated through IV, then its kinetic energy will be
(a) 1 eV
(b) 13.6 eV
(c) 1840 eV
(d) 0.54 eV
Answer:
(a) 1 eV
Hint:
K = qV = e x 1V= 1 eV

Question 12.
The work function of a photoelectric material is 3.3 eV. The threshold frequency will be equal to
a) 8 × 10¹⁴ Hz
b) 8 × 10¹⁰ Hz
c) 5 × 10²⁰ Hz
d) 4 × 10¹⁴ Hz
Answer:
a) 8 × 10¹⁴ Hz
Solution:

Question 13.
The momentum of the electron having wavelength of 4 Å is
a) 1.6 × 10²⁴ kg ms^-1
b) 1.65 × 10^-24 kg ms^-1
c) 3.3 × 10²⁴ kg ms^-1
d) 3.3 × 10^-24 kg ms^-1
Answer:
b) 1.65 × 10^-24 kg ms^-1
Solution:
λ = \(\frac{\mathrm{h}}{\mathrm{p}}\)
p = \(\frac{h}{\lambda}\)
= \(\frac{6.6 \times 10^{-34}}{4 \times 10^{-10}}\)
p = 1.65 × 10^-24 kgms^-1

Question 15.
The number of photo-electrons emitted for the light of a frequency υ (higher than the threshold frequency υ₀) is proportional to
(a) Threshold frequency (υ₀)
(b) Intensity of light
(c) Frequency of light (υ)
(d) υ – υ₀
Answer:
(b) Intensity of light
Hint:
The photoelectric current of Intensity of incident light

Question 15.
The maximum kinetic energy of photoelectrons emitted in a photoelectric effect phenomenon is 3.2eV What is the value of stopping potential to stop the electrons not to reach the anode?
a) 1.6 eV
b) -3.2 eV
c) -1.6 eV
d) -6.4 eV
Answer:
b) -3.2 eV
Solution:
K.E_max = \(\frac{1}{2}\) mv²
V₀ = \(\frac{\mathrm{K} . \mathrm{E}}{e}\)
= \(\frac{3.2 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-19}}\)
V₀ = -3.2 eV

Question 16.
Radiation of energy 6.2 eV is incident on the metal surface of work function 4.7eV. Find the kinetic energy of the
electrons emitted.
a) 3 × 10^-19 J
b) 1.6 × 10^-19 J
c) 2.4 × 10^-19 J
d) 3.2 × 10^-19 J
Answer:
c) 2.4 × 10^-19 J
Solution:
hν = 6.2 eV
W = 4.7 eV
\(\frac{1}{2}\) mv² = hν – W
= (6.2 – 4.7) eV
= 1.5 ev
= 1.5 × 1.6 × 10^-19
\(\frac{1}{2}\) mv² = 2.4 × 10^-19 J
K.E = 2.4 × 10^-19 J

Question 17.
The de-Broglie wavelength of electrons moving with a speed of 500 km/s
a) 1.45 nm
b) 1.45 Å
c) 1.45 m
d) 4.6 × 10²⁴ HZ
Answer:
a) 1.45 nm
Solution:
λ = \(\frac{\mathrm{h}}{\mathrm{mv}}\)
= \(\frac{6.6 \times 10^{-34}}{9.1 \times 10^{-31} \times 500 \times 10^{3}}\)
λ = 1.45 × 10^-19 ;
λ = 1.45 nm

Question 18.
Electron volt is a unit of
(a) Energy
(b) potential
(c) current
(d) charge
Answer:
(a) Energy
Hint:
An electron volt is a unit of energy

Question 19.
The work function of iron is 4.7 V. calculate the cut-off wavelength for this metal.
a) 2633 Å
b) 26.4 × 10^-7 m
c) 13.2 × 10^-10 m
d) 1320 Å
Answer:
a) 2633 Å
Solution:
hν₀ = W;
h\(\frac{\mathrm{C}}{\lambda_{0}}\) = W;
λ₀ = \(\frac{\mathrm{hc}}{\mathrm{W}}\)
= \(\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{4.7 \times 1.6 \times 10^{-19}}\)
= 2633 × 10^-10
λ₀ = 2633 Å

Question 20.
The time taken by a photoelectron to come out after photon strikes is approximately
(a) 10^-14 s
(b) 10^-10 s
(c) 10^-16 s
(d) 10^-1 s
Answer:
(b) 10^-10 s
Hint:
The time lag between the incident of photon and the emission of photoelectrons is 10^-10 s approximately.

Question 21.
A Coolidge tube operates at 37200 V maximum frequency of x rays emitted is _______.
a) 9 × 10²⁸ Hz
b) 100 Hz
c) 9 × 10¹⁸ Hz
d) 3 × 10¹⁸ Hz
Answer:
c) 9 × 10¹⁸ Hz
Solution:

Question 22.
De – Broglie wavelength of an electron accelerating by 100 V
a) 1.227 Å
b) 122.7 μm
c) 0.2227 m
d) 12.27 Å
Answer:
a) 1.227 Å
Solution:
λ = \(\frac{12.27}{\sqrt{\mathrm{V}}}\)
= \(\frac{12.27}{\sqrt{100}}\)
= \(\frac{12.27}{10}\)
λ = 1.227 Å

Question 23.
The energy of photon of wavelength λ is
(a) \(\frac { hc }{ λ }\)
(b) hλc
(c) \(\frac { λ }{ hc }\)
(d) \(\frac { hλ }{ c }\)
Answer:
(a) \(\frac { hc }{ λ }\)
Hint:
E = hυ = \(\frac { hc }{ λ }\)

X. Two Mark Questions:

Question 1.
Why electron is preferred over X-ray in the microscope?
Answer:

1. The de-Broglie wavelength of an electron is very less compared to X-rays.
2. We can build a high resolving power microscope using electrons.
3. Resolving power of a microscope inversely proportional to the wavelength
   r₀ α \(\frac{1}{\lambda}\)

Question 2.
What are photoelectrons?
Answer:
These are the electrons emitted from a metal surface when it is exposed to electromagnetic radiations of a suitable frequency.

Question 3.
What are matter waves?
Answer:
The moving elementary particles behave as waves under suitable conditions. These waves associated with particles are called matter waves or de-Broglie waves.

Question 4.
What are continuous x rays?
Answer:
Continuous x-ray spectrum consists of radiations of all possible wavelengths from a certain lower limit to higher values continuously as in the case of visible light.

Question 5.
Why is a photo-cell also called an electric eye?
Answer:
Like an eye, a photo-cell can distinguish between weak and intense light. But a photocell gives a measure of light intensity in terms of photoelectric current. So it is also called an electric eye.

Question 6.
What is a photoemissive cell?
Answer:
Its working depends on the electron emission from a metal cathode due to irradiation of light and other radiation.

Question 7.
What are X-ray spectra?
Answer:
X-rays are produced when fast-moving electrons strike the metal target. The intensity of the X-rays when plotted against its wavelength gives a curve called X-ray spectrum.

Question 8.
What are photoconductive cells?
Answer:
The resistance of the semiconductor changes in accordance with the radiation energy incident on it.

Question 9.
Define resolving power of the microscope.
Answer:
The resolving power of a microscope is inversely proportional to the wavelength of the radiation used.

Question 10.
Who invented the particle nature of light?
Answer:
Hertz confirmed that the light is an electromagnetic wave. But the same experiment also produced the first evidence for particle nature of light.

Question 11.
Define intensity of light.
Answer:
Intensity denotes power of light means brightness.

Question 12.
What is diffraction?
Answer:
The bending of light waves round the edges of obstacles is called diffraction.

Question 13.
When light behaves like waves and matter?
Answer:
Light behaves as a wave during its propagation and behaves as a particle during its interaction with matter.

XI. Three Marks Questions:

Question 1.
What is nature of light?
Answer:
The wave nature of light explains phenomena such as interference, diffraction, and polarisation. Certain phenomenon like black body radiation and photoelectric effect light behaves as particles. Thus light posses dual nature, that is particle and wave.

Question 2.
Define one electron volt.
Answer:
One electron volt is defined as the kinetic energy gained by an electron when accelerated by a potential difference of 1 V.
1 eV = K.E gained by the electron or workdone by the electric field
1 eV = qv = 1.602 × 10^-19 C × 1 V
1 eV = 1.602 × 10^-19 J

Question 3.
Differentiate particle and wave.
Answer:
Particle is a material object which is considered as a tiny concentration of matter localized in space and time whereas wave is a broad distribution of energy not localized in space and time.

XII. Five Mark Questions:

Question 1.
What are the applications of the photocells?
Answer:

1. Photocells have many applications especially switches and sensors.
2. Automatic switching ON and OFF ordinary lights as well as street lights
3. They are used for the reproduction of sound in motion pictures.
4. Photocells are used as timers to measure the speeds of athletes during a race.
5. Photocells are used to measure the intensity of the given light and to calculate the exact time of exposure.

Question 2.
Write down the characteristics of photons.
Answer:
Characteristics of photons:
According to the particle nature of light, photons are the basic constituents of any radiation and possess the following characteristic properties:

• The photons of light of frequency v and wavelength λ will have energy, given by E = hυ = \(\frac { hc }{ λ }\).
• The energy of a photon is determined by the frequency of the radiation and not by its intensity and the intensity has no relation with the energy of the individual photons in the beam.
• The photons travel with the velocity of light and its momentum is given by p
• Since photons are electrically neutral, they are unaffected by electric and magnetic fields.
• When a photon interacts with matter (photon-electron collision), the total energy, total linear momentum, and angular momentum are conserved. Since photon may be absorbed or a new photon may be produced in such interactions, the number of photons may not be conserved

Question 3.
Write a note on characteristic X-rays. Give the applications of X-rays in medical therapy and industry.
Answer:
Characteristic X-ray Spectra:
1. X-ray spectra snow some narrow peaks at some, well-defined wavelengths when the target is hit by fast electrons. The line spectrum showing these peaks is called the characteristic X-ray spectrum.
2. When an energetic electron removes electrons from K-shell electrons then the electrons from the outer orbit jump to fill the vacancy.
3. Energy difference between the levels is given out as X-ray photon of definite wavelength.

Origin of characteristics X-ray Spectra

K-Series (K_α and K_β): This line is due to electronic transitions from L, M, and N shells to K-level.
L-Series (L_α and L_β): It arises due to electronic transitions from M, N, and O shells.

Uses of X-rays:

1. Medical diagnosis: used to detect fractures, foreign bodies, diseased organs, etc.
2. Medical therapy: To kill diseased tissues, they are employed to cure skin diseases, malignant tumors.
3. Industry: To check for flaws in welded joints, motor tires, tennis balls, and wood. At the customs post, they are used for the detection of contraband goods.
4. Scientific research: To study crystalline structure i.e. arrangement of atoms and molecules in crystals.

XIII. Conceptual Questions:

Question 1.
Are other particles, other than electrons having a wave nature?
Answer:
The particles like neutrons and alpha particles are also associated with waves. They undergo diffraction when they scattered by suitable crystals.

Question 2.
Why diffraction effects of ordinary light is very small?
Answer:
The wavelength of light is high compared to the spacing between the lattice planes so that the diffraction pattern of ordinary light is less pronounced.

Question 3.
Why crystals are used for three-dimensional grating?
Answer:
The wavelength of X-rays are in the order of 10^-10 m which is comparable to the spacing between the crystal lattice planes. Since the crystals can serve as three-dimensional grating.

XIV. Additional Problems:

Question 1.
Calculate the momentum and the de- Broglie wavelength of an electron with kinetic energy 25 eV.
Solution
i) Momentum of the electron:
p = \(\sqrt{2 \mathrm{mk}}\)
= \(\sqrt{2 \times 9.1 \times 10^{-31} \times 25 \times 1.6 \times 10^{-19}}\)
= \(\sqrt{728 \times 10^{-50}}\)
= 26.98 × 10^-25 kgms^-1

ii) de Broglie wavelength:
λ = \(\frac{\mathrm{h}}{\mathrm{p}}\)
= \(\frac{6.626 \times 10^{-34}}{26.98 \times 10^{-25}}\)
= 0.2455 × 10^-34 × 10²⁵
= 0.2455 ×10⁹
λ = 2.455 Å

Question 2.
Monochromatic light of frequency 6 x 10¹⁴ Hz is produced by a laser. The power emitted is 2 x 10^-3w.
(i) What is the energy of each photon in the light?
(ii) How many photons per second, on average, are emitted by the source?
Solution:
(i) Energy of each photon,
E = hυ = 6.6 x 10^-34 x 6 x 10¹⁴
E = 3.98 x 10^-19J
(ii) If N is the number of photons emitted per second by the source, then
Power transmitted in the beam = N x energy of each photon
P = N
N = \(\frac { P }{ E }\) = \(\frac{2 \times 10^{-3}}{3.98 \times 10^{-19}}\)
N = 5 x 10¹⁵ Photons per second.

Question 3.
A proton is moving at a speed of 0.900 times the velocity of light. Find the kinetic energy in Joules and Mev.
Answer:
Given:
ν = 0.9 × 3 × 10⁸ ms^-1
Mass of proton m = 1.6 × 10^-27 kg
K.E = \(\frac{1}{2}\) mv²
= \(\frac{1}{2}\) × 1.673 × 10^-27 × 2.7 × 2.7 × 10¹⁶
= \(\frac{6.098 \times 10^{-11}}{1.6 \times 10^{-19}}\)
= 3.811 × 10⁸ eV
K.E = 3.81 MeV

Question 4.
Calculate the momentum of an electron with kinetic energy 2 eV.
Answer:
Given:
K.E = 2 eV
p = \(\sqrt{2 \mathrm{mk}}\)
p = \(\sqrt{2 \times 9.1 \times 10^{-31} \times 2 \times 1.6 \times 10^{-19}}\)
p = 7.63 × 10^-25 kgms^-1

Question 5.
Calculate the cut-off wavelength and cut off frequency of X-rays from an X-ray tube of accelerating potential 20,000 V.
Answer:
The cut-off wavelength of the characteristic X-rays is
λ₀ = \(\frac{12400}{\mathrm{v}}\) Å = \(\frac{12400}{20000}\) Å
= 0.62 Å
The corresponding frequency is
ν₀ = \(\frac{c}{\lambda_{0}}\)
= \(\frac{3 \times 10^{8}}{0.62 \times 10^{-10}}\)
= 4.84 × 10¹⁸ Hz

Question 6.
What is the (a) momentum, (b) speed, and (c) de-Broglie wavelength of an electron with the kinetic energy of 120 eV.
Solution:
Kinetic energy, K.E = 120 eV = 120 x 1.6 x 10^-19
K = K.E = 1.92 x 10^-17 J
(a) Momentum of an electron, P = \(\sqrt { 2mK } \)
P = \(\sqrt{2 \times 9.1 \times 10^{-31} \times 1.92 \times 10^{-17}}\)
P = 5.91 x 10^-24 kg ms^-1
(b) Speed of an electron,
v = \(\frac { p }{ m }\) = \(\frac{5.91 \times 10^{-24}}{9.1 \times 10^{-31}}\) = 6.5 x 10⁶ kg ms^-1
(c) de-Broglie wavelength,
λ = \(\frac { h }{ p }\) = \(\frac{6.6 \times 10^{-34}}{5.91 \times 10^{-24}}\) = 1.117 x 10^-10 = 0.112 x 10^-9 m
λ = 0.112 nm

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Computer Applications Guide Pdf Chapter 2 An Introduction to Adobe Pagemaker Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Computer Applications Solutions Chapter 2 An Introduction to Adobe Pagemaker

12th Computer Applications Guide An Introduction to Adobe Pagemaker Text Book Questions and Answers

Part I

Choose The Correct Answers

Question 1.
DTP stands for ……………….
(a) Desktop Publishing
(b) Desktop Publication
(c) Doctor To Patient
(d) Desktop Printer
Answer:
(a) Desktop Publishing

Question 2.
________ is a DTP software,
(a) Lotus 1-2-3
(b) PageMaker
(c) Maya
(d) Flash
Answer:
(b) PageMaker

Question 3.
Which menu contains the Newoption?
(a) File menu
(b) Edit menu
(c) Layout menu
(d) Type menu
Answer:
(a) File menu

Question 4.
In PageMaker Window, the areaoutside of the dark border is referredto as …………….
(a) page
(b) pasteboard
(c) blackboard
(d) dashboard
Answer:
(b) pasteboard

Question 5.
Shortcut to close a document in PageMaker is …………..
(a) Ctrl + A
(b) Ctrl + B
(c) Ctrl + C
(d) Ctrl + W
Answer:
(d) Ctrl + W

Question 6.
A …………………… tool is used for magnifying the particular portion of the area.
(a) Text tool
(b) Line tool
(c) Zoom tool
(d) Hand tool
Answer:
(c) Zoom tool

Question 7.
…………………. tool is used for drawing boxes.
(a) Line
(b) Ellipse
(c) Rectangle
(d) Text
Answer:
(c) Rectangle

Question 8.
Place option is present in ………………….menu.
(a) File
(b) Edit
(c) Layout
(d) Window
Answer:
(a) File

Question 9.
To select an entire document using the keyboard, press …………….
(a) Ctrl + A
(b) Ctrl + B
(c) Ctrl + C
(d) Ctrl + D
Answer:
(a) Ctrl + A

Question 10.
Character formatting consfstsof which of the following textproperties?
(a) Bold
(b) Italic
(c) Underline
(d) All of these
Answer:
(d) All of these

Question 11.
Which tool lets you edit text?
(a) Text tool
(b) Type tool
(c) Crop tool
(d) Hand tool
Answer:
(a) Text tool

Question 12.
Shortcut to print a document in Pagemaker is ……………….
(a) Ctrl + A
(b) Ctrl + P
(c) Ctrl + C
(d) Ctrl + V
Answer:
(b) Ctrl + P

Question 13.
Adobe PageMaker is a ware………………… software.
Answer:
Page Layout

Question 14.
…………….. Bar is the topmost part of the PageMaker window.
Answer:
Title

Question 15.
…………… is the process of movingup and down or left and right through the document window.
Answer:
Scrolling

Question 16.
___________ tool is used to draw acircle.
Answer:
Ellipse

Question 17.
The Insert pages option is available on clicking the………………menu.
Answer:
Layout

Question 18.
Match the following.
Cut – (i) Ctrl + Z
Copy – (ii) Ctrl + V
Paste – (iii) Ctrl + X
Undo – (iv) Ctrl + C
Answer:
Match: iii, iv, ii, i

Question 19.
Choose the odd man out.
i) Adobe PageMaker, QuarkXPress, Adobe InDesign, Audacity
ii) File, Edit, Layout, Type, Zip
iii) Pointer Tool, Line Tool, Hide Tool, Hand Tool
iv) Bold, Italic, Portrait, Underline
Answer:
(i) Audacity, (ii) Zip, (iii) Hide Tool, (iv) Portrait

Question 20.
Choose the correct statement.
i. (a) Text can be selected using mouse only.
(b) Text can be selected using mouseor the keyboard.
ii. (a) DTP is an abbreviation for Desktop publishing, (b) DTP is an abbreviation for Desktop publication.
Answer:
(i)-b, (ii)-a

Question 21.
Choose the correct pair
(a) Edit and Cut
(b) Edit and New
(c) Undo and Copy
(d) Undo and Redo
Answer:
(a) Edit and Cut

Part II

Short Answers

Question 1.
What is desktop publishing?
Answer:
Desktop publishing (abbreviated DTP) is the creation of page layouts for documents using DTP software.

Question 2.
Give some examples of DTP software.
Answer:
Some of the popular DTP software are Adobe PageMaker, Adobe InDesign, QuarkXPress, etc.

Question 3.
Write the steps to open PageMaker.
Answer:
In the Windows 7 operating system, we can open Adobe PageMaker using the command sequence
Start →All Programs → Adobe → Pagemaker 7.0 → Adobe PageMaker 7.0.

Question 4.
How do you create a New document in PageMaker?
Answer:
To create a new document in. PageMaker:

• Choose File > New in the menu bar. (or) Press Ctrl + N in the keyboard. Document Setup dialog box appears.
• Enter the appropriate settings for your new doc¬ument in the Document Setup dialog box.
• Click on OK.

Question 5.
What is a Pasteboard in PageMaker?
Answer:
A document page is displayed within a dark border. The area outside of the dark border is referred to as the pasteboard. Anything that is placed completely in the pasteboard is not visible when you print the document.

Question 6.
Write about the Menu bar of PageMaker.
Answer:
It contains the following menus File, Edit, Layout, Type, Element, Utilities, View, Window, Help. When you click on a menu item, a pull down menu appears. There may be sub-menus under certain options in the pull-down menus.

Question 7.
Differentiate Ellipse tool from Ellipse frame tool.
Answer:

Question 8.
What is text editing?
Answer:
Editing encompasses many tasks, such as inserting and deleting words and phrases, correcting errors, and moving and copying text to different places in the document.

Question 9.
What is text block?
Answer:
A text block contains text you type, paste, or import. You cannot see the borders of a text block until you select it with the pointer tool.

Question 10.
What is threading text blocks?
Answer:
A Text block can be connected to other text blocks so that the text in one text block can flow into another text block. Text blocks that are connected in this way are threaded. The process of connecting text among Text blocks is called threading text.

Question 11.
What is threading text?
Answer:
The process of connecting text among Text blocks is called threading text.

Question 12.
How do you insert a page in PageMaker?
Answer:
To insert pages

1. Go to the page immediately before the page you want to insert.
2. Choose Layout > Insert Pages in themenu bar. The Insert Pages dialog box appears.
3. Type the number of pages you want to insert.
4. To insert pages after the current page, choose ‘after’ from the pop-up menu.
5. Click on Insert.

Part III

Explain In Brief Answer

Question 1.
What is PageMaker? Explain its uses,
Answer:
Adobe PageMaker is a page layout software. It is used to design and produce documents that can be printed. You can create anything from a simple business card to a large book.

Page layout software includes tools that allow you to easily position text and graphics on document pages. For example, using PageMaker, you could create a newsletter that includes articles and pictures on each page. You can place pictures and text next to each other, on top of each other, or beside each other wherever you want them to go.

Question 2.
Mention three tools in PageMaker and write their keyboard shortcuts.
Answer:

1. Pointer Tool F9
2. Rotating Tool Shift + F2
3. Line Tool Shift + F3

Question 3.
Write the use of arty three tools in PageMaker along with symbols.
Answer:

Question 4.
How do you rejoin split blocks?
Answer:
Rejoining split blocks
To rejoin the two text blocks

1. Place the cursor on the bottom handle of the second text.block, click and drag the bottom handle up to the top.
2. Then place the cursor on the bottom handle of the first text block, and click and drag the bottom handle down if necessary.

Question 5.
How do you Sink frames containing text?
Answer:

• Draw a second frame with the Frame tool of your choice.
• Click the first frame to select it.
• Click on the red triangle to load the text icon.
• Click the second frame. PageMaker flows the text into the second frame.

Question 6.
What is the use of Master Page?
Answer:
Any text or object that you place on the master page will appear on the entire document pages to which the master is applied. It shortens the amount of time because you don’t have to create the same objects repeatedly on subsequent pages. Master Pages commonly contain repeating logos, page numbers, headers, and footers. They also contain non-printing layout guides, such as column guides, ruler guides, and margin guides.

Question 7.
How to you insert page numbers in Master pages?
Answer:

• Click on Master Pages icon.
• Then click on Text Tool. Now the cursor changes to I – beam.
• Then Click on the left Master page where you want to put the page number.

Part IV

Explain In Details

Question 1.
Explain the tools in PageMaker toolbox,
Answer:

Question 2.
Write the steps to place the text in a frame.
Answer:
Placing Text in a Frame
You can also use frames to hold the text in place of using text blocks.
To place text in a Frame

1. Click on one of a Frame tool from the Toolbox.
2. Draw a frame with one of PageMaker’s Frame tools (Rectangle frame tool or Ellipse Frame Tool or Polygon frame Tool). Make sure the object remains selected.
3. Click on File. The File menu will appear.
4. Click on Place. The Place dialog box will appear.
5. Locate the document that contains the text you want to place, select it.
6. Click on Open.
7. Click in a frame to place the text in it. The text will be placed in the frame.

Question 3.
How can you convert text in a text block to a frame?
Answer:
After created text in a text block, if you want to convert it to a frame. You can do this by using these steps.

• Draw the frame of your choice using one of the PageMaker’s Frame tool.
• Select the text block you want to insert in the frame.
• Click the frame while pressing the Shift key. Now both elements will be selected.
• Choose Element > Frame > Attach Content on the Menu bar.
• Now the text appears in the frame.

Question 4.
Write the steps to draw a star using polygon tool?
Answer:
Drawing a Star using Polygon tool
To draw a Star

1. Click on the Polygon tool from the toolbox. The cursor changes to a crosshair.
2. Click and drag anywhere on the screen. As you drag, a Polygon appears.
3. Release the mouse button when the Polygon is of the desired size.
4. Choose Element > Polygon Settings in the menu bar. Now Polygon Settings dialogue box appears.
5. Type 5 in the Number of the sides text box.
6. Type 50% in Star inset textbox.
7. Click OK. Now the required star appears on the screen.

12th Computer Applications Guide An Introduction to Adobe Pagemaker Additional Important Questions and Answers

Part A

Choose The Correct Answers:

Question 1.
Which of the following is DTP software?
a) page maker
b) in design
c) quark x press
d) all of the above.
Answer:
d) all of the above.

Question 2.
The main components of the page maker windows are
a) title bar, the menu bar
b) toolbar, ruler
c) scroll bars and text area
d) all of the above.
Answer:
d) all of the above.

Question 3.
…………………. is the topmost part of the windows.
a) title bar
b) menu bar
c) toolbar
d) toolbox.
Answer:
a) title bar

Question 4.
How many control buttons are present in the title bar?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Question 5.
There are ………………… ruler bar.
a) 1
b) 2
c) 3
d) 4
Answer:
b) 2

Question 6.
When you move the mouse pointer on a button in the toolbar, a short text that appears is called ……………………………..
(a) Tooltip
(b) Text
(c) Show text
(d) Short text
Answer:
(a) Tooltip

Question 7.
To select a paragraph, press ………….. with I – Beam.
a) select click
b) right-click
c) double click
d) triple-click.
Answer:
d) triple-click.

Question 8.
………………… tool is used to select .move and resize text objects and graphics.
a) pointer tool
b) text tool
c) rotating tool
d) cropping tool.
Answer:
a) pointer tool

Question 9.
The ………………… key is used to press down and
the movements keys.
a) Ctrl
b) shift
c) alt
d) tab
Answer:
b) shift

Question 10.
In page maker, the text of the document can type inside a …………………
a) text tool
b) text area
c) text block
d) text box
Answer:
c) text block

Question 11.
How many scroll bars are there?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 12.
Text can be contained …………………
a) text blocks
b) text frame
c) either a (or) b
d) both a and b
Answer:
c) either a (or) b

Question 13.
Master pages commonly contain …………………
a) repeating logos, page numbers
b) headers and footers
c) both a and b
d) none of these.
Answer:
c) both a and b

Question 14.
A new document in page maker is called…………………
a) Untitled-1
b) Document-1
c) New Document
d) None of these
Answer:
a) Untitled-1

Question 15.
The flashing verticle bar is called ……………………………
(a) scroll bar
(b) ruler
(c) Footer
(d) Insertion point
Answer:
(d) Insertion point

Fill in The Blanks:

1. DTP expands
Answer:
Desktop publishing,

2. Adobe page Is a layout software,
Answer:
page

3. To make an adobe page make in windows is
Answer:

4. The area outside of the border is referred to as the
Answer:
pasteboard

5. In the toolbar a short text will appear as its description called
Answer:
Tooltip

6. The command is used to reverse the action of the last command.
Answer:
undo

7. The short cut key for undo is
Answer:
ctrl + z

8. A contains the text you type, paste, or import in Pagemaker.
Answer:
text block

9. The two handles are seen above and below of the text, the block is called _________
Answer:
window shades

10. Generate a new page by selecting in the menu bar.
Answer:
layout → insert pages.

11. All text in page maker resides inside containers called
Answer:
text blocks.

12. In page maker, text and graphics that you draw or import is called
Answer:
objects.

13. The process of connecting text among text blocks is called text
Answer:
threading.

14. Text the flows through one or more .threaded blocks is called a
Answer:
story,

15. The palette is especially useful when you are doing lot of formatting
Answer:
control palette.

16. Page maker has a line tool
Answer:
2

17. As the characters are typed, the flashing vertical bar called the
Answer:
insertion point

18. is the process of changing the general arrangement of text.
Answer:
Formatting

19. means making changes to the text.
Answer:
Editing

20. Reversing the Undo command is known as.
Answer:
Redo

Short Cut Keys

1. Ctrl+N – Create New Document
2. F9 – Pointer Tool
3. Shift+F2 – Rotating Tool
4. Shift+F3 – Line Tool
5. Shift*F4 – Rectangle Too!
6, Shift+F5 – Ellipse Too!
7. Shift+F6 – Polygon Too!
8. Shift+Alt +Drag Left Mouse Button- Hand Tool
9. Shift +Ait+ FI – Text Tool
10. Shift +Ait +F2 – Cropping Too!
11. Shift+Alt +F3 – Constrained Line Tool
12, Shift+Alt +F4 – Rectangle Frame Too!
13. Shift + <– – One Character To Left
14. Shift +→ – One Character To The Light
15. Shift +^ – One Line Up
16. Shift* – One Line Down
17. Shift +End “ To End Of The Current Line
18. Shift+Home – To The Beginning Of The Current Line
19. Ctrl* A – Select Entire Document
20. Ctrl+Z – Undo
21. Ctrl+X – Cut
22. Ctri+V – Paste
23. Ctri+C – Copy
24. Ctri+S – Saving A Document
25. Ctri+W – Closing A Document
26. Left Arrow (←) – One Character To The Left
27. Right Arrow (→) – One Character To The Right
28. One Word To The Left – Ctrl +Left Arrow
29. One Word To The Right – Ctrl + Right Arrow
30. Up Arrow – Up One Une
31. Down Arrow- Down One Line
32. End – ToThe End Of A Line
33. Home – To The Beginning Of A Line
34. Ctrl + Up Arrow – Up One Paragraph
35. Ctrl + Down – Down One Paragraph
36. Ctrl + Q – Opening An Existing Document
37. Ctrl + Space Bar – to zoom in
38. Ctrl + Alt + Space Bar-ar- To Zoom Out
39. Ctrl+ T – Character Formatting
40. Ctrl+ – Control Palette
41. Alt+ Ctrl+ G – Going To Specific
42. Ctrl4- Alt+ P – Page Number Displays
43. Ctri+ P – Print

Assertion And Reason

Question 1.
Assertion (A): Adobe PageMaker is a page layout software.
Reason (R): It is used to design and produce documents that can be printed,
a) Both (A) and (R) are correct and (R) is the correct explanation of (A)
b) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)
c) (A) is true and (R) is false d) (A) is false and (R) is true
Answer:
a) Both (A) and (R) are correct and (R) is the correct explanation of (A)

Question 2.
Assertion (A): A document pageis displayed within a dark border.
Reason (R): The area outside of the dark border is referred to as the Margin.
a) Both (A) and (R) are correct and (R) is the correct explanation of (A)
b) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)
c) (A) is true and (R) is false d) (A) is false and (R) is true
Answer:
c) (A) is true and (R) is false

Question 3.
Assertion (A): Menu bar ¡s the topmost part of the window.
Reason (R): Title bar shows the name of the software and the name ofthe document at the left, and the control buttons (Minimize, Maximize and Close) at theright.
a) Both (A) and (R) are correct and (R) is the correct explanation of (A)
b) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)
c) (A) is true and (R) is false
d) (A) is false and (R) Is true
Answer:
d) (A) is false and (R) Is true

Question 4.
Assertion (A): Editing means creating the text. Reason (R); Editing encompasses many tasks, such as inserting and deleting words and phrases, correcting errors, and moving and copying text to different places in the document. a) Both (A) and (R) are correct and (R) is the correct explanation of (A)
b) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)
c) (A) is true and (R) is false d) (A) is false and (R) is true Answer: d) (A) is false and (R) is true

Question 5.
Assertion (A): Text blocks that are connected the way are threaded.
Reason (R): A Text block can be connected to another text block so that the text in one text block can flow into another text block.
a) Both (A) and (R) are correct and (R) is the correct explanation of (A)
b) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)
c) (A) is true and (R) is false
d) (A) is false and (R) is true
Answer:
a) Both (A) and (R) are correct and (R) is the correct explanation of (A)

Match The Following

1. One character to the left a) Ctrl + Left Arrow
2. One character to the right b) Ctrl + Right Arrow
3. One word to the left c) Left Arrow
4. One word to the right d) Right Arrow
5. Up one line e) Ctrl +Up Arrow
6. Down one line f) Ctrl +Down Arrow
7. To the end of a line g) End
8. To the beginning of a line h) Home
9. Up one paragraph i) Up Arrow
10. Down one paragraph j) Down Arrow
Answer:
1. c 2. d 3. a 4. b 5. I 6. j 7. g 8. h 9. e 10. f

Find The Odd One On The Following

1. (a) Page Maker
(b) Indesign
(c) QuarkXpress
(d) Ubuntu
Answer:
(d) Ubuntu

2. (a) File
(b) Tool Tip
(c) Elements
(d) Utilities
Answer:
(b) Tool Tip

3. (a) Type
(b) Paste
(c) Import
(d) Print
Answer:
(d) Print

4. (a) Text Block
(b) Ruler
(c) Text Tool
(d) InsertionPoint
Answer:
(b) Ruler

5. (a) Ctrl+Z
(b) Ctrl+Y
(c) Ctrl+T
(d) Ctrl+C
Answer:
(c) Ctrl+T

6. (a) Text Tool
(b) Line Tool
(c) HandTool
(d) Window
Answer:
(d) Window

7. (a) Type
(b) Select
(c) zoom
(d) edit Answer:
(c) zoom

8. (a) Shift+end
(b) Shift+home
(c) Shift +→
(d) shift+F1
Answer:
(d) shift+F1

9. (a) Threaded
(b) Text
(c) Threading Text
(d) story
Answer:
(b) Text

10. (a) Save
(b) Element
(c) Frame
(d) Delete
Answer:
(a) Save

Choose The In Correct Pair

1. (a) Edit and Paste
(b) Layout and Go to Page
(c) Window and Hide tools
(d) Element and Cascade
Answer:
(d) Element and Cascade

2. a) File → Print and Ctrl + P
b) Centre Alignment and Ctrl+C
c) Window→ Show colors and Press Ctrl + J
d) Type → Character and Press Ctrl + T
Answer:
b) Centre Alignment and Ctrl+C

3. a) File →New and Edit → Paste
b) Ctrl + X → to Cut and Ctrl + V → to Paste
c) Ctrl + C → to Copy and Ctrl + V → to Paste
d) Up one line, Press Up Arrow and Down one line, press Down Arrow
Answer:
a) File →New and Edit → Paste

4. a) Draw Star; Polygon tool
b) Draw Rounded corner, Rectangle tool
c) Draw Dotted line, the Pointer tool
d) Draw Rectangle, Ellipse tool
Answer:
c) Draw Dotted line, the Pointer tool

5. a) Zoom tool, Magnify
b) Hand tool, Scroll
c) Rotating tool. Trim
d) Ellipse tool, Circles
Answer:
c) Rotating tool. Trim

Part B

Short Answers

Question 1.
What is the purpose of the page layout tool?
Answer:
Page layout software includes tools that allow you to easily position text and graphics on document pages.

Question 2.
Write a note on scroll bars?
Answer:
Scrolling is the process of moving up and down or left and right through the document window. There are two scroll bars namely Vertical and Horizontal scroll bars for scrolling the document vertically or horizontally.

Question 3.
What is the title bar?
Answer:

• It is the topmost part of the window.
• It shows the name of the software and the name of the document at the left, and the control buttons (Minimize, Maximize and Close) at the right.

Question 4.
What is the use of entering key?
Answer:
The Enter key should be pressed only at the end of a paragraph or when a blank line is to be inserted.

Question 5.
What is a tooltip?
Answer:
If you place the mouse pointer on a button in the Toolbar, a short text will appear as its description called ‘Tool Tip’

Question 6.
Write the steps to show toolbox in page maker,
Answer:

1. Click on Window. The Window menu will appear.
2. Click on Show tools.

Question 7.
What are the two ways of creating text blocks?
Answer:

1. Click or drag the text tool on the page or pasteboard, and then type
2. Click a loaded text icon in an empty column or page.

Question 8.
Write the steps to show ruler in page maker.
Answer:

1. Click on View. The View menu will appear.
2. Click on Show Rulers. Rulers appear along the top and left sides of the document window.

Question 9.
Write the steps to hide ruler in page maker.
Answer:

1. Click on View, The View menu will appear.
2. Click on Hide Rulers to hide the rulers.

Question 10.
What is the purpose of Undo command?
Answer:

• The Undo command is used to reverse the action of the last command.
• To reverse the last command, click on Edit Undo in the menu bar
  (or)
• Press Ctrl + Z on the keyboard.

Question 11.
What are the two-line tools in page maker?
Answer:
PageMaker has two Line tools. The first one creates a straight line at any orientation. The second is a constrained Line tool that draws only at increments of 45 degrees.

Question 12.
Write a short note on the Text block in the page maker.
Answer:
A text block contains the text you type, paste, or im¬port. You can’t see the borders of a text block until you select it with the pointer tool.
You create text blocks in two ways:

1. Click or drag the text tool on the page or pasteboard, and then type.
2. Click a loaded text icon in an empty column or page.

Question 13.
How will you rejoin the split blocks?
To rejoin the two blocks

• Place the cursor on the bottom handle of the sec¬ond text block, click and drag the bottom handle up to the top.
• Then place the cursor on the bottom handle of the first text block, and ciick and drag the bottom handle down if necessary.

Question 14.
What do you mean by threading text?
Answer:

• Text blocks that are connected in this way are threaded.
• The process of connecting text among Text blocks is called threading text.

Question 15.
How will you close a document?
Answer:
The document can be closed using the File > Close command in the menu bar (or) Ctrl +W in the keyboard.

Question 16.
How will you create a text block with a text tool?
Answer:
To create a text block with the text tool:
(i) Select the text tool (T) from the toolbox. The pointer turns into an I-beam.

(ii) On an empty area of the page or pasteboard, do one of the following:
Click the I-beam where you want to insert text. This creates a text block the width of the column or page. By default, the insertion point jumps to the left side of the text block.

(iii) Type the text you want.
Unlike with a text frame, you do not see the borders of a text block until you click the text with the pointer tool.

Question 17.
How to Hide the Master Items?
Answer:
To make the master items invisible on a particular page, switch to the appropriate page, then choose View> Display Masteritems (which is usually ticked).

Part C

Explain In Brief Answer

Question 1.
Write the steps to resize a text block.
Answer:

• Click on the Pointer tool.
• Click either the left or right corner handle on the bottom of the text block and drag. When you release the mouse button, the text in the text: block will reflow to fit the new size of the text block.
• A red triangle in the bottom windowshade means there is more text In the text block than is visi¬ble on the page. Drag the window shade handle down to show more text.

Question 2.
Write the steps to Seles® Text using the mouse
Answer:
To select text using a mouse, follow these steps :

• Place the insertion point to the left of the first character to be selected.
• Press the left mouse button and drag the mouse to a position where you want to stop selecting.
• Release the mouse button.
• The selected text gets highlighted.

Question 3.
Differentiate copying and moving the text
Answer:

Question 4.
Write the steps to delete a character or word or block of text
Answer:
Deleting Text
You can easily delete a character, or word, or block of text.
To delete a character, do the following :

1. Position the insertion point to the left of the character to be deleted.
2. Press the Delete key on the keyboard.
   (or)
3. Position the insertion point to the right of the character to be deleted.
4. Press the Backspace key on the keyboard.

To delete a block of text, do the following :

1. Select the text to be deleted.
2. Press Delete or Backspace in the keyboard (or) Edit → Clear command.

Question 5.
How will you create a text box with the text tool?
Answer:
To create a text block with the text tool:
1. Select the text tool (T) from the toolbox. The pointer turns into an I-beam.

2. On an empty area of the page or pasteboard, do one of the following:

• Click the I-beam where you want to insert text.
• This creates a text block the width of the column or page. By default, the insertion point jumps to the left side of the text block.

3. Type the text you want.
Unlike with a text frame, you do not see the borders of a text block until you click the text with the pointer tool.

Question 6.
Write the steps to resize a text block.
Answer:

1. Click on the Pointer tool.
2. Click either the left or right corner handle on the bottom of the text block and drag.
3. When you release the mouse button, the text in the text block will reflow to fit the new size of the text block.
4. A red triangle in the bottom window shade means there is more text in the text block than is visible on the page.
5. Drag the window shade handle down to show more text.

Question 7.
Write the steps to split a textbox into two.
Answer:
To split a text block into two

1. Place the cursor on the bottom handle, click and drag upwards. When you release the bottom handle will contain a red triangle.
2. Click once on this, and the cursor changes to a loaded text icon.
3. Position this where the second part of the text is to be, and click.

Question 8.
Write the steps to Select Text Using the Keyboard
Answer:
To select text using a keyboard, follow these steps:

1. Place the insertion point to the left of the first character you wish to select.
2. The Shift key is pressed down and the movement keys are used to highlight the required text.
3. When the Shift key is released, the text is selected

Question 9.
Write the steps to split a textbox into two.
Answer:
To split a text block into two

1. Place the cursor on the bottom handle, click and drag upwards. When you release the bottom handle will contain a red triangle.
2. Click once on this, and the cursor changes to a loaded text icon.
3. Position this where the second part of the text is to be, and click.

Question 10.
Write the steps to import the text.
Answer:

1. Choose File → Place. The Place dialog box will appear.
2. Locate the document that contains the text you want to place and select it.
3. Click on the Open in the Place dialog box. The pointer changes to the loaded text icon.
4. Make a text block to place the text. (Or) Click on the page to place the text. The text will be placed on the page.
5. If the text to be placed is too big to fit on one page, PageMaker allows you to place it on several pages. This can be done manually or automatically.

Question 11.
What are the various options to save a document?
Answer:

1. Choose File > Save in the menu bar. (or)
2. The file name is given in the File name list box.
3. Then click on the Save button to save the document.
4. The document is now saved and a file name appears in the title bar.

Once a file is saved under a name, to save it again the name need not be entered again. The file can be saved simply by selecting the File > Save command or by clicking the Save button (or) clicking Ctrl + S on the keyboard.

Question 12.
Write the steps to save a document with a new name or in a different location
Answer:
You can save a document with a new name or in a different location using Save AS command. Save AS command creates a new copy of the document. So, two versions of the document ex¬ist. The versions are completely separate, and the work you do on one document has no effect on the other.

To save a document with a new name or in a different location:

1. Choose File > Save As in the menu bar. (or) Press Shift + Ctrl + S on the keyboard. Now Save Publication dialog box will appear.
2. Type a new name or specify a new location.
3. Click the Save button.

Question 13.
How will you open an existing document?
Answer:

1. Choose File > Open in the menu bar (or)Click on the Open icon () in the Tool bar (or) Press Ctrl + 0 in the Keyboard. An open Publication dialog box as shown that appears on the screen.
2. The file name is given in the File name list box. The name of the file to be opened can be chosen from the list, which is displayed.
3. Then click on the Open button. Now the required file is opened.

Question 14.
Write the procedure to scroll the document.
Answer:
The scrolling procedure is as follows:

1. To scroll left and right the left and right arrow respectively should be clicked.
2. To scroll up and down the up and down arrow respectively should be clicked.
3. To scroll a relative distance in the document the scroll box should be drawn up or down.

Question 15.
Write the steps to draw a line.
Answer:

1. Select the Line tool from the toolbox. The cursor changes to a crosshair.
2. Click and drag on the screen to draw your line. As you drag, a line appears. 44
3. Release the mouse button and the line will be drawn and selected, with sizing handles on either end. Resize the line by clicking and dragging the handles, if necessary.

Question 16.
Write the steps to Draw Rectangles or Ellipses.
Answer:
You can also draw rectangles and ellipses shapes by using the same technique as used in line drawing.

To draw a rectangle or ellipse:

1. Click on the Rectangle or Ellipse tool from the toolbox. The cursor changes to a crosshair.
2. Click and drag anywhere on the screen. As you drag, a rectangle or ellipse appears.
3. Release the mouse button when the rectangle or ellipse is of the desired size.
4. Press the Shift key while you’re drawing to constrain the shape to a square or circle.

Question 17.
Write the steps to Draw Polygon
Answer:
To draw a Polygon

1. Click on the Polygon tool from the toolbox. The cursor changes to a crosshair.
2. Click and drag anywhere on the screen. As you drag, a Polygon appears.
3. Release the mouse button when the Polygon is of the desired size.

Question 18.
Write the steps to Draw a star with given number of sides and required inset.
Answer:

1. The value of ‘Star inset’ is 50% The number of sides is 15
2. The value of ‘Star inset’ is 25% The number of sides is 25
3. The value of ‘Star inset’ is 35% The number of sides is 70

Question 19.
Write a short note on Master Page.
Answer:

• Master Pages commonly contain repeating logos, page numbers, headers, and footers. They also contain nonprinting layout guides, such as column guides, ruler guides, and margin guides.
• A master item cannot be selected on a document page.
• You can create, modify, and delete objects on master pages just like any other objects, but you must do so from the master pages themselves.

Question 20.
Write the steps to create a new Master Page
Answer:

1. Click the New Master Page icon in the Master Pages palette. The New Master Page dialog box appears.
2. Enter the name of the new master page in the Name field.
3. Make the appropriate changes in the Margins and Column Guides fields.
4. Click on OK. A new Master Page appears in the Master Pages palette.

Part D

Explain In Detail.

Question 1.
What are the different ways of selecting the text?
Answer:
Selecting Text:
Text can be selected using the mouse or the keyboard.
Selecting Text using the mouse:
To select text using a mouse, follow these steps :

1. Place the insertion point to the left of the first character to be selected.
2. Press the left mouse button and drag the mouse to a position where you want to stop selecting.
3. Release the mouse button.
4. The selected text gets highlighted.

To Select Press:

1. A Word Double-click with I-beam
2. A Paragraph Triple-click with I-beam

Selecting Text using the Keyboard:
To select text using a keyboard, follow these steps :

1. Place the Insertion point to the left of the first character you wish to select.
2. The Shift key is pressed down and the movement keys are used to highlight the required text.
3. When the Shift key is released, the text is selected.

To Select – Press
One character to the left – Shift + ←
One character to the right – Shift + →
One line up – Shift + ↑
One line down – Shift + ↓
To the end of the current line – Shift +End
To the beginning of the current line – Shift + Home,
Entire Document – Ctrl + A

Question 2.
Write the steps to place text in a frame
Answer:

1. Click on one of a Frame tool from the Toolbox.
2. Draw a frame with one of PageMaker’s Frame tools (Rectangle frame tool or Ellipse Frame Tool or Polygon frame Tool). Make sure the object remains selected.
3. Click on File. The File menu will appear.
4. Click on Place. The Place dialog box will appear.
5. Locate the document that contains the text you want to place, select it.
6. Click on Open.
7. Click in a frame to place the text in it. The text will be placed in the frame.

Question 3.
Explain the magnifying and reducing with the zoom tool.
Answer:
Use the zoom tool to magnify or reduce the display of any area in your publication

To magnify or reduce with the zoom tool:
1. Select the zoom tool. The pointer becomes a magnifying glass with a plus sign in its center, in¬dicating that the zoom tool will magnify your view of the image. To toggle between magnification and reduction, press the Ctrl key.

2. Position the magnifying glass at the center of the area you want to magnify or reduce, and then click to zoom in or out. Continue clicking until the publication is at the magnification level you want. When the publication has reached its maximum magnification or reduction level, the center of the magnifying glass appears blank.

To magnify part of a page by dragging:

1. Select the zoom tool.
2. Drag to draw a marquee around the area you want to magnify.

To zoom in or out while using another tool:
Press Ctrl+Spacebar to zoom in. Press Ctrl+Alt+Spacebar to zoom out.

Question 4.
Explain how will you draw a dotted line?
Answer:
To draw a Dotted line:

1. Double click the Line tool from the toolbox. A Custom Stroke dialogue box appears.
2. Select the required Stroke style in the drop-down list box.
3. Then click the OK button. Now the cursor changes to a crosshair.
4. Click and drag on the screen to draw your dotted line. As you drag, the line appears.
5. Release the mouse button and the line will be drawn and selected, with sizing handles on either end.
   Resize the line by clicking and dragging the handles, if necessary.

Question 5.
Write the steps To draw a Dotted line
Answer:

1. Double click the Line tool from the toolbox. A Custom Stroke dialogue box appears.
2. Select the required Stroke style in the drop-down list box.
3. Then click the OK button. Now the cursor changes to a crosshair.
4. Click and drag on the screen to draw your dotted line. As you drag, the line appears.
5. Release the mouse button and the line will be drawn and selected, with sizing handles on either end.
6. Resize the line by clicking and dragging the handles, if necessary.

Question 6.
Write the steps to Drawing a Rounded Corner Rectangle
Answer:
To draw a rounded-corner rectangle:

1. Double-click the Rectangle tool in the toolbox. The Rounded Corners dialog box appears.
2. Choose a corner setting from the preset shapes.
3. Click on OK. The cursor changes to a crosshair.
4. Click and drag anywhere on the screen.
5. Release the mouse button when the rectangle is the desired size.
6. Press the Shift key as you draw to constrain the shape to a rounded corner square.

Question 7.
Write the steps to Fill Shapes with Colors and Patterns
Answer:
Filling Rectangle with colour:

1. Draw a rectangle using the Rectangle tool.
2. Select the rectangle.
3. Choose Window → Show colors in the menu bar (or) Press Ctrl + J. Now Colors palette appears.
4. Click on the required colour from the Colors Palette.
5. The rectangle has been filled with colour.

Question 8.
Write all the methods to go to a specific page.
Answer:
Pagemaker provides several methods for navigating the pages in your publication.

Method 1:
You can move from one page to another by using the Page up and Page down keys on your key- i 10. board. These is probably the navigation methods you will use most often.
Method 2;
You can move from one page to another by using the page icons at the left bottom of the screen.
Click on the page icon that corresponds to the page that you want to view. The page is displayed,

Method 3:
Using the Go to Page dialog box.
To go to a specific page in a document

1. Choose Layout → Go to Page in the menu bar (or) Press Alt + Ctrl + G in the keyboard. Now the Go to Page dialog box appears.
2. In the dialog box, type the page number that you want to view
3. Then click on OK, The required page is displayed on the screen.

Question 9.
What are the various methods to the inset page numbers in PageMaker Software?
Answer:
To make page numbers appear on every page

1. Click on the Master Pages icon.
2. Then dick on Text Tool. Now the cursor changes to I – beam.
3. Then Click on the left Master page where you want to put the page number,
4. Press Ctrl + Ait + P.
5. The page number displays as ’LM’ on the left master page.
6. Similarly, click on the right Master page where you want to put the page number.
7. Press Ctrl + Alt + P.
8. The page number displays as ‘RM’ on the right master page, but will appear correctly on the actual pages.

Question 10.
Write the steps to print a document.
Answer:
1. Choose File → Print in the menu bar (or) Press Ctrl + P on the keyboard.
The Print Document dialog box appears.
2. Choose the settings in the Print Document dialog box as

• Select the printer from the Printer drop-down list box,
• Choose the pages to be printed in the Pages group box by selecting one of the following available options

All This option prints the whole document.
Ranges: This option prints individual pages by the page number or a range of pages.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Economics Guide Pdf Chapter 11 Economics of Development and Planning Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Economics Solutions Chapter 11 Economics of Development and Planning

12th Economics Guide Economics of Development and Planning Text Book Back Questions and Answers

PART – A

Multiple Choice questions

Question 1.
“Redistribution with Growth” became popular slogan under which approach?
a) Traditional approach
b) New welfare oriented approach
c) Industrial approach
d) None of the above
Answer:
b) New welfare oriented approach

Question 2.
Which is not the feature of economic growth?
a) Concerned with developed nations
b) Gradual change
c) Concerned with quantitative aspect
d) Wider concept
Answer:
d) Wider concept

Question 3.
Which among the following is a characteristic of underdevelopment?
a) Vicious circle of poverty
b) Rising mass consumption
c) Growth of Industries
d) High rate of urbanization
Answer:
a) Vicious circle of poverty

Question 4.
The non-economic determinant of economic development
a) Natural resources
b) Human resource
c) Capital formation
d) Foreign trade
Answer:
b) Human resource

Question 5.
Economic growth measures the ……………………
a) Growth of productivity
b) Increase in nominal income
c) Increase in output
d) None of the above
Answer:
c) Increase in output

Question 6.
The supply side vicious circle of poverty suggests that poor nations remain poor because
a) Saving remains low
b) Investment remains low
c) There is a lack of effective government
d) a and b above
Answer:
d) a and b above

Question 7.
Which of the following plan has focussed on the agriculture and rural economy?
a) People’s Plan
b) Bombay Plan
c) Gandhian Plan
d) Vishveshwarya Plan
Answer:
c) Gandhian Plan

Question 8.
Arrange following plans in correct chronological order
a) People’s Plan
b) Bombay Plan
c) Jawaharlal Nehru Plan
d) Vishveshwarya Plan
Answer choices
a) (i) (ii) (iii) (iv)
b) (iv) (hi) (ii) (i)
c) (i) (ii) (iv) (ii)
d) (ii) (i) (iv) (iii)
Answer:
b) Bombay Plan

Question 9.
M. N. Roy was associated with …………..
a) Congress Plan
b) People’s Plan
c) Bombay Plan
d) None of the above
Answer:
b) People’s Plan

Question 10.
Which of the following country adopts indicative planning?
a) France
b) Germany
c) Italy
d) Russia
Answer:
b) Germany

Question 11.
Short-term plan is also known as ……………………
a) Controlling Plans
b) De-controlling Plans
c) Rolling Plans
d) De-rolling Plans
Answer:
a) Controlling Plans

Question 12.
Long-term plan is also known as …………
a) Progressive Plans
b) Non-progressive Plan
c) Perspective Plans
d) Non-perspective Plans
Answer:
c) Perspective Plans

Question 13.
The basic philosophy behind long – term planning is to bring ………………. changes in the economy?
a) Financial
b) Agricultural
c) Industrial
d) Structural
Answer:
c) Industrial

Question 14.
Sarvodaya Plan was advocated by ……………..
a) Mahatma Gandhi
b) J.P. Narayan
c) S. N Agarwal
d) M. N. Roy
Answer:
b) J.P. Narayan

Question 15.
Planning Commission was set up in the year ………….
a) 1950
b) 1951
c) 1947
d) 1948
Answer:
a) 1950

Question 16.
Who wrote the book ‘The Road to Serfdom’?
a) Friedrich Hayek
b) H. R. Hicks
c) David Ricardo
d) Thomas Robert Malthus
Answer:
a) Friedrich Hayek

Question 17.
Perspective plan is also known as ………….
a) Short-term plan
b) Medium-term plan
c) Long-term plan
d) None of the above
Answer:
c) Long-term plan

Question 18.
NITI Aayog is formed through ……………
a) Presidential Ordinance ‘
b) Allocation of business rules by President of India
c) Cabinet resolution
d) None of the above
Answer:
c) Cabinet resolution

Question 19.
Expansion of NITI Aayog?
a) National Institute to Transform India
b) National Institute for Transforming India
c) National Institution to Transform India
d) National Institution for Transforming India
Answer:
d) National Institution for Transforming India

Question 20.
The Chair Person of NITI Aayog is
a) Prime Minister
b) President
c) Vice – President
d) Finance Minister
Answer:
a) Prime Minister

PART – B

Answer the following questions in one or two sentences.

Question 21.
Define economic development
Answer:

1. Economic development is regarded as a process whereby there is an increase in the consumption of goods and services by individuals.
2. From the welfare perspective, economic development is defined as a sustained improvement in health, literacy and standard of living.

Question 22.
Mention the indicators of development.
Answer:

• Gross National product (GNP)
• GNP Per capita
• Welfare
• Social Indicators

Question 23.
Distinguish between economic growth and development
Answer:

Question 24.
What is GNP?
Answer:
Gross National Product (GNP):

1. GNP is the total market value of all final goods and services produced within a nation in a particular year, plus income earned by its citizens (including income of those located abroad), minus income of non – residents located in that country.
2. GNP is one measure of the economic condition of a country, under the assumption that a higher GNP leads to a higher quality of living, all other things being equal.

Question 25.
Define economic planning.
Answer:
Economic planning is “collective contral or suppression of private activities of production and exchange”. – Robbins.

Question 26.
What are the social indicators of economic development?
Answer:
Social Indicators:

1. Social indicators are normally referred to as the basic and collective needs of the people.
2. The direct provision of basic needs such as health, education, food, water, sanitation, and housing facilities check social backwardness.

Question 27.
Write a short note on NITI Aayog.
Answer:

1. NITI Aayog (National Institution for Transforming India) was formed on January 1, 2015, through a Union Cabinet resolution.
2. NITI Aayog is a policy think-tank of the Government of India. It replaced the Planning Commission from 13th August 2014.
3. The Prime Minister is the Chairperson of NITI Aayog and Union Ministers will be Ex – officio members.
4. The Vice-Chairman of the NITI Aayog is the functional head and the first Vice-Chairman was Arvind Panangariya.

PART-C

Answer the following questions in one paragraph.

Question 28.
Elucidate major causes of vicious circle of poverty with diagram
Answer:
The vicious circle of Poverty

A country is poor because of low capital income. The vicious circle of poverty operates both on the demand side and the supply side.
It is associated with low rate of saving and investment on the supply side and low level of income leads to low level of demand on the demand side.

Question 29.
What are the non-economic factors determining development?
Answer:

1. Human Resource
2. Technical Know-how
3. Political Freedom
4. Social Organization
5. Corruption free administration
6. Desire for Development
7. Moral, ethical and social values
8. Casino Capitalism
9. Patrimonial Capitalism

Question 30.
How would you break the vicious circle of poverty?
Answer:

• In the demand side Nurkse suggested the strategy of balanced growth to break vicious circle of poverty.
• The balanced growth (i.e.,) simultaneous investment in large number of industries creates mutual demand.
• Thus, through the strategy of balanced growth, vicious circle of poverty operating on the demand side of capital formation can be broken.
• In the supply side to break the circle a country has to keep its marginal rate of savings higher than its average rate of savings.

Question 31.
Trace the evolution of economic planning in India.
Answer:
The evolution of planning in India is stated below:

(I) Sir M. Vishveshwarya (1934):
A prominent engineer and politician made his first attempt in laying the foundation for economic planning in India in 1934 through his book, “Planned Economy of India”. It was a 10-year plan.

(II) Jawaharlal Nehru (1938):
Set-up “National Planning Commission” by a committee but due to the changes in the political era and Second World War, it did not materialize.

(III) Bombay Plan (1940):
The 8 leading industrialists of Bombay presented the “Bombay Plan”. It was a 15 Year Investment Plan.

(IV) S.N Agarwal (1944):
Gave the “Gandhian Plan” focusing on the agricultural and rural economy.

(V) M.N. Roy (1945):
Drafted ‘People’s Plan”. It was aiming at the mechanization of agricultural production and distribution by the state only.

(VI) J.P. Narayan (1950):
Advocated, “Sarvodaya Plan” which was inspired by the Gandhian Plan and with the idea of Vinoba Bhave. It gave importance not only for agriculture but encouraged small and cottage industries in the plan.

Question 32.
Describe the case for planning.
Answer:

The economic planning is justified on the following grounds:-

• To accelerate and strengthen market mechanism.
• To remove unemployment ‘
• To achieve balanced development
• To remove poverty and inequalities.
  Arthur lewis says, ‘planning is more necessary in backward countries to devise ways and means and to make concentrated efforts to raise national income.

Question 33.
Distinguish between functional and structural planning.
Answer:
Functional planning refers to that planning which seeks to remove economic difficulties by directing all the planning activities within the existing economic and social structure.
The structural planning refers to a good deal of changes in the socioeconomic framework of the country.
This type of planning is adopted mostly in under developed countries.

Question 34.
What are the functions of NITI Aayog?
Answer:
Functions of NITI Aayog:
(I) Cooperative and Competitive Federalism:
To enable the States to have active participation in the formulation of national policy.

(II) Shared National Agenda:
To evolve a shared vision of national development priorities and strategies with the active involvement of States.

(III) Decentralized Planning:
To restructure the planning process into a bottom-up model.

(IV) Vision and Scenario Planning:
To design medium and long – term strategic frameworks towards India’s future.

(V) Network of Expertise:
To mainstream external ideas and expertise into government policies and programmes through collective participation.

(VI) Harmonization:
To facilitate harmonization of actions across different layers of government, especially when involving cross-cutting and overlapping issues across multiple sectors; through communication, coordination,
collaboration, and convergence amongst all the stakeholders.

(VII) Conflict Resolution:
To provide a platform for mutual consensus to inter-sectoral, interdepartmental, interstate as well as center-state issues for all speedy execution of the government programmes.

(VIII) Coordinating Interface with the World:
It will act nodal point to harness global expertise and resources coming from International organizations for India’s developmental process.

(IX) Internal Consultancy:
It provides internal consultancy to Central and State governments on policy and programmes.

(X) Capacity Building:
It enables to provide capacity building and technology up-gradation across government, benchmarking with latest global trends and providing managerial and technical know-how.

(XI) Monitoring and Evaluation:
It will monitor the implementation of policies and progammes and evaluate the impacts.

PART – D

Answer the following questions in about a page

Question 35.
Discuss the.economic determinants of economic development.
Answer:
Economic factors:

1. Natural Resource:-
The existence of natural resources in abundance is essential for development.
A country deficient in natural resources may not be in a position to develop rapidly.

2. Capital Formation:-
Capital formation refers to the net addition to the exist¬ing stock of capital goods which are either tangible like plants and machinery or intangible like health, education and research.

3 Size of the Market:-
Large size of the market would stimulate production,increase employment and raise the National per capita income.

4. Structural change:-
Structural change refers to change in the occupational structure of the economy.

5. Financial System:-
Financial system implies the existence of an efficient and organized banking system in the country.

6. Marketable Surplus :-
Marketable surplus refers to the total amount of farm output cultivated by farmers over and above their family consumption needs. This is a surplus that can be sold in the market for earning income.

7. Foreign Trade:-
The country which enjoys favourable balance of trade is always developed. It has huge forex reserves and stable exchange rate.

8. Economic system:-
The countries which adopt free market mechanism enjoy better growth rate compared to controlled economics.

Question 36.
Describe different types of Planning.
Answer:
(I) Democratic Vs Totalitarian:
A form of rule in which the government attempts to maintain ‘total’ control over society, including all aspects of the public and private lives of its citizens.

(II) Centralized Vs Decentralized:

1. Under centralized planning, the entire planning process in a country is under a central planning authority.
2. This authority formulates a central plan, fixes objectives, targets and priorities for every sector of the economy.
3. In other words, it is called ‘planning from above’.

(III) Planning by Direction Vs Inducement:
Under planning by direction, there is a central authority which plans, directs and orders the execution of the plan in accordance with pre-determined targets and priorities.

(IV) Indicative Vs Imperative Planning:

1. Indicative planning is peculiar to the mixed economies. It has been in practice in France since the Monnet Plan of 1947-50.
2. In a mixed economy, the private sector and the public sector work together.
3. Under this plan, the outline of plan is prepared by the Government.
4. Then it is discussed with the representatives of private management, trade unions, consumer groups, finance institutions and other experts.

(V) Short, Medium and Long term Planning:

1. Short-term plans are also known as ‘controlling plans’.
2. They encompass the period of one year, therefore, they are also known as ‘annual plans’

(VI) Financial Vs Physical Planning:
Financial planning refers to the technique of planning in which resources are allocated in terms of money while physical planning pertains to the allocation of resources in terms of men, materials and machinery.

(VII) Functional Vs Structural Planning:
Functional planning refers to that planning seeks to remove economic difficulties by directing all the planning activities within the existing economic and social structure.

(VIII) Comprehensive Vs Partial Planning:
General planning which concerns itself with the major issues for the whole economy is known as comprehensive planning whereas partial planning is to consider only a few important sectors of the economy.

Question 37.
Bring out the arguments against planning.
Answer:
The prime goals of economic planning are stabilization in developed countries and growth in LDCS But the economic planning also is not free from limitations.
The arguments against planning are,

1. Loss of Freedom:
The absence of freedom in decision making may acts as an obstacle for economic growth. Under planning, the crucial decisions are made by the central planning Authority. The consumers, producers and the workers enjoy no freedom of choice. Therefore, Hayek explains, that centralized planning leads to loss of personal freedom and ends in economic stagnation. The decisions by the Government are not always rational Freedom to private producers will be misused, Profit will be given top priority, welfare will be relegated.

2. Elimination of Initiative:
Under centralized planning, there will be no incentive for initiatives and innovations. Planning follows routine procedure and may cause stagnation in growth.

• The absence of private ownership and profit motive discourages entrepreneurs from taking bold decisions and risk taking.
• Equal reward to all discourages interested in undertaking new and risky ventures.
• The bureaucracy and red tapism which are the features of planned economy.

3. High cost of Management:
The cost of management of the economic affairs outweights the benefits of planning. Inadequate data, faulty estimations and improper implementation of plans result in wastage of resources and cause either surplus or shortages.

Difficulty in advance calculations :
Advance calculation in a precise manner are impossible to make decisions regarding the consumption and production. It is also very difficult to put the calculations into practice under planning.

12th Economics Guide Economics of Development and Planning Additional Important Questions and Answers

I. Match the following:
Question 1.
A) Development- 1) Narrow concept
B) Todaro – 2) Wider concept
C) Economic Growth – 3) Structural change
D) Economic Development – 4) Development

Answer:
b) 3 4 2 1

Question 2.
a) Innovation – 1) Dadabai Naoroji
b) Poverty and un British Rule- in India – 2) Douglas c. North
c) Moral and social values – 3) Thomas Piketty
d) Casino capitalism – 4) Schumpeter

Answer:
d) 4 1 2 3

II. Choose the correct pair

Question 1.
a) Patrimonial capitalism – Douglas c. North
b) Vicious circle of poverty – Thomas Piketty
c) Great Depression – 1930
d) Planning commission – 1951
Answer:
c) Great Depression – 1930

Question 2.
a) M.N. Roy – People’s plan
b) J.P. Narayan – Gandhian plan
c) S. N. Agarwal – Sarvodaya plan
d) Jawaharlal Nehru – Bombay plan
Answer:
a) M.N. Roy – People’s plan

III. Choose the incorrect pair

Question 1.
a) Planning commission – March 15,1950
b) Plan era – April 1,1951
c) First five-year plan – 1951-56
d) Planning – Gandhi
Answer:
d) Planning – Gandhi

Question 2.
a) Underdeveloped country – high percapita Income
b) Low-Income countries – $ 906 and below
c) Middle Income countries – Between $ 906 and $ 11,115
d) High Income countries – $ 11,116 or more
Answer:
a) Underdeveloped country – high percapita Income

Choose the correct statement

Question 1.
a) The UDCS are characterized by the predominance of the tertiary sector.
b) The Prime minister is the functional head of NITI Aayog.
c) The first vice-chairman of NITI Aayog was Arvind panangariya.
d) NITI Aayog assist planning commission
Answer:
c) The first vice-chairman of NITI Aayog was Arvind panangariya.

Question 2.
a) Centralised planning is called planning from below.
b) Decentralised planning is called planning from above.
c) Long-term plans are operational plans.
d) Medium-term planning is tactical planning.
Answer:
d) Medium-term planning is tactical planning.

V. Choose the incorrect statement

Question 1.
a) Short-term plans are for the period up to 1 year.
b) Medium-term plans last for 5-8 years
c) Long term plans last for the period of 10 – 30 years
d) Short term plans are also known as ‘controlling plans’
Answer:
b) Medium-term plans last for 5-8 years

Question 2.
a) NITI Aayog was formed on January 1, 2015
b) NITI Aayog replaced the planning commission on 13th August 2014
c) Ayushmaan Bharat’s approach was towards forest conservation.
d) National medical commission was replaced by the medical council of India.
Answer:
c) Ayushmaan Bharat’s approach was towards forest conservation.

VI. Pick the odd one out:

Question 1.
a) Economic planning
b) Financial Planning
c) Physical planning
d) Perspective Planning
Answer:
a) Economic planning

Question 2.
a) Low productivity
b) High Investment
c) Low percapita Income
d) Low saving
Answer:
b) High Investment

Analyse the reason:

Question 1.
Assertion (A): Structural changes refers to change in the occupational structure of the Economy.
Reason (R): A country is generally divided into primary, secondary, and tertiary sectors.
a) Both (A) and (R) are true, (R) is the correct explanation of (A)
b) Both (A) and (R) are true, (R) is not the correct explanation of (A)
c) (A) is true; but (R) is false
d) Both (A) and (R) are false.
Answer:
b) Both (A) and (R) are true, (R) is not the correct explanation of (A)

Question 2.
Assertion(A): Economic development depends on Economic, social, political, and religious factors.
Reasons (R): Being a wider concept economic development includes all the as¬pects of the country.
a) Both (A) and (R) are true, (R) is the correct explanation of (A)
b) Both (A) and (R) are true, (R) is not the correct explanation of (A)
c) Both (A) and (R) is false, d) (A) is true (R) is false
Answer:
a) Both (A) and (R) are true, (R) is the correct explanation of (A)

VIII. Fill in the blanks with the correct option.

Question 1.
Gross National Product is ……………………..
a) GDP
b) GNP
c) GNI
d) None
Answer:
b) GNP

Question 2.
The countries which adopt free-market mechanism are
a) Socialist
b) Mixed economy
c) Laissez-faire
d) Communist
Answer:
c) Laissez faire

Question 3.
Capital is a necessary but not a sufficient condition of progress is a statement of …………………….
a) Thomas piketty
b) Douglas c. North
c) Schumpeter
d) Ragnar Nurkse
Answer:
d) Ragnar Nurkse

IX. Choose the best answer.

Question 1.
The first chairman of the planning commission was
a) J. P. Narayan
b) S. N. Agarwal
c) Jawaharlal Nehru
d) Sri. M. Vishveshwarya
Answer:
c) Jawaharlal Nehru

Question 2.
A declaration of industrial policy was announced in
a) 1950
b) 1948
c) 1947
d)1951
Answer:
b) 1948

Question 3.
The better we try to plan, the more planners we need’ is the statement of ………………………..
a) Thomas piketty
b) Douglas c. North
c) Ragnar Nurkse
d) Arthur Lewis
Answer:
d) Arthur Lewis

X. Answer the following in one or two sentences.

Question 1.
Define “Traditional Approach”?
Answer:
Traditional Approach:

1. The traditional approach defines development strictly in economic terms.
2. The increase in GNP is accompanied by a decline in the share of agriculture in output and employment while those of manufacturing and service sectors increase.

Question 2.
State the meaning of Under development.
Answer:
The term underdevelopment refers to that state of an economy where levels of living of masses are extremely low due to very low levels of per capita income, resulting from low levels of productivity and high growth rate of population.

Question 3.
Write “UDC” characteristics?
Answer:
The UDCs are characterized by the predominance of primary sector i.e. agriculture, low per capita income, widespread poverty, wide inequality in the distribution of income and wealth, overpopulation, low rate of capital formation, high rate of unemployment, technological backwardness, dualism, etc.

Question 4.
What are the pillars of NITI Aayog?
Answer:

• pro-people
• pro – Activity.
• Participation
• Empowering
• Inclusion of all
• Equality
• Transparency

Question 5.
Define “Financial system.”?
Answer:
Financial System:

1. A financial system implies the existence of an efficient and organized banking system in the country.
2. There should be an organized money market to facilitate the easy availability of capital.

Question 6.
What is GNP Per Capita?
Answer:
It relates to an increase in per capita real income of the economy over a long period.

Question 7.
Define “Sarvodaya plan”?
Answer:
“Sarvodaya Plan” was inspired by the Gandhian Plan and with the idea of Vinoba Bhave.
It gave importance not only to agriculture but encouraged small and cottage industries in the plan.

XI. Answer the following questions in paragraph

Question 1.
Briefly explain the Measurement of Economic Development?
Answer:
Measurement of Economic Development:
Economic development is measured on the basis of four criteria

(I) Gross National Product (GNP):
1. GNP is the total market value of all final goods and services produced within a nation in a particular year, plus income earned by its citizens (including income of those located abroad), minus income of non-residents located in that country.

2. GNP is one measure of the economic condition of a country, under the assumption that a higher GNP leads to a higher quality of living, all other things being equal.

(II) GNP per capita:

1. This relates to increasing in the per capita real income of the economy over the long period.
2. This indicator of economic growth emphasizes that for economic development the rate of increase in real per capita income should be higher than the growth rate of the population.

(III) Welfare:

1. Economic development is regarded as a process whereby there is an increase in the consumption of goods and services by individuals.
2. From the welfare perspective, economic development is defined as a sustained improvement in health, literacy and standard of living.

(IV) Social Indicators:

1. Social indicators are normally referred to as the basic and collective needs of the people.
2. The direct provision of basic needs such as health, education, food, water, sanitation, and housing facilities check social backwardness.

Question 2.
Distinguish Between Economic Growth and Economic Development.
Answer:

Question 3.
Explain Thomas Piketty’s non – Economic factors contributing to Economic development
(or)
Explain
1) casino capitalism
2) Patrimonial capitalism
Answer:

1. Casino capitalism: If people spend a larger proportion of their income and time on the proportion of their income and time on entertainment liquor and other illegal activities, productive activities may Suffer.
2. Patrimonial capitalism: If the assets are simply passed on to children from their parents, the children would not work hard, because the children do not know the value of the assets Hence productivity will below.
   These two ideas were contributed by Thomas Piketty.

Question 4.
What is Crony capitalism?
Answer:
Social Organization:

1. People show interest in the development activity only when they feel that the fruits of development will be fairly distributed.
2. Mass participation in development programs is a pre-condition for accelerating the development process.
3. Whenever the defective social organization allows some groups to appropriate the benefits of growth.
4. The majority of the poor people do not participate in the process of development.
5. This is called crony capitalism.

Question 5.
Explain the concepts of NITI Aayog.
Answer:
Initiatives like the Atal Innovation mission, the Ayushmaan Bharat approach towards water conservation measures, and the draft bill to establish the National Medical Commission to replace the medical council of India have all been conceptualized in NITI Aayog.

Question 6.
Explain the price mechanism?
Answer:

1. Price mechanism provides for the automatic adjustment among price, demand, and supply in a Laissez-Faire economy.
2. The producers and consumers adjust their supply and demand based on price changes.
3. There is no such mechanism in a planned economy.
4. Advance calculations in a precise manner are impossible to make decisions regarding consumption and production.
5. It is also very difficult to put the calculations into practice under planning.
6. Excess supply and excess demand can also happen in the market-oriented economy.
7. In fact, it has happened in many capitalistic economies, including the US.

Question 7.
Explain the supply side of the vicious circle of poverty.
Answer:
On the supply side, the low level of real income means low savings. The low level of savings leads to low investment and to deficiency of capital. The deficiency of capital leads to a low level of productivity and back to low income. Thus the vicious circle is complete from the supply side.

XII. Answer the following questions

Question 1.
Explain the Economic planning in India?
Answer:
Economic Planning in India:

• Consists of economic decisions, schemes formed to meet certain pre-determined economic objectives, and a road map of directions to achieve specific goals within a specific period of time.
• The idea of economic planning was strengthened during the Great Depression in the 1930s.
• The outbreak of World War II also required adequate and suitable planning of economic resources for effective management after the effects of the post-war economy.
• After Independence, in 1948, a declaration of industrial policy was announced.
• The policy suggested the creation of a National Planning Commission and the elaboration of the policy of a mixed economic system.
• On January 26, 1950, the Constitution came into force.
• In logical order, the Planning Commission was created on March 15, 1950, and the plan era began on April 1, 1951, with the launch of the first five-year plan (1951-56). The evolution of planning in India.

Question 2.
Explain the arguments in favour of economic planning.
Answer:
The economic planning is justified on the following grounds.

1. To accelerate and strengthen market mechanism:
The market mechanism works imperfectly in underdeveloped countries because of ignorance and unfamiliarity with it. A large part of the economy comprises the non – monetized sector. The product, factor, money, and capital markets are not organized properly. Therefore the planned economy will be a better substitute for a free economy.

2. To remove unemployment
The need for planning in underdeveloped countries is further stressed by the necessity of removing widespread unemployment and disguised unemployment in such economies.

3. To achieve balanced development:
In the absence of sufficient enterprise and initiative, the planning authority is the only institution for planning the balanced development of the economy.

For rapid economic development:

• The development of Agriculture and Industrial sectors.
• The development of Infrastructure
• The development of money and capital markets is necessary.

To remove poverty and inequalities:
Planning is the only path open to underdeveloped countries, for raising national and per capita income, reducing inequalities and poverty, and increasing employment opportunities.

Question 3.
Briefly explain Indicative and Imperative planning?
Answer:
Indicative Vs Imperative Planning:

1. Indicative planning is peculiar to the mixed economies.
2. It has been in practice in France since the Monnet Plan of 1947-50.
3. In a mixed economy, the private sector and the public sector work together.
4. Under this plan, the outline of the plan is prepared by the Government.
5. Then it is discussed with the representatives of private management, trade unions, consumer groups, financial institutions, and other experts.
6. The essential function of planning is the coordination of different economic units.
7. The state provides all types of facilities to the private sector.
8. The private sector is expected to fulfill the targets and priorities.
9. The state does not force the private sector but just indicates the areas of operation and targets to be fulfilled.
10. In short, the planning procedure is soft and flexible.
11. Under imperative planning, the state is all-powerful in the preparation and implementation of the plan.
12. Once a plan is drawn up, its implementation is a matter of enforcement.
13. The USSR President Stalin used to say, ‘Our plans are our instructions’.
14. There is complete control over the entire resources by the state.
15. There is no consumer sovereignty.
16. The Government policies and procedures are rigid.
17. China and Russia follow imperative planning.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Economics Guide Pdf Chapter 10 Environmental Economics Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Economics Solutions Chapter 10 Environmental Economics

12th Economics Guide Environmental Economics Text Book Back Questions and Answers

PART- A

Multiple Choice questions

Question 1.
The term environment has been derived from a French word ………………..
a) Environ
b) Environs
c) Environia
d) Envir
Answer:
c) Environia

Question 2.
The word biotic means environment
a) living
b) non – living
c) physical
d) None of the above
Answer:
a) living

Question 3.
Ecosystem is smallest unit of
a) Ionosphere
b) Lithosphere
c) Biosphere
d) Mesosphere
Answer:
c) Biosphere

Question 4.
Who developed Material Balance Models?
a) Thomas and Picardy
b) Alenkneese and R.V.Ayres
c) Joan Robinson and J.M.Keynes
d) Joseph Stiglitz and Edward Chamberlin
Answer:
d) Joseph Stiglitz and Edward Chamberlin

Question 5.
Environmental goods are ……………
a) Market goods
b) Non – market goods
c) Both
d) None of the above
Answer:
b) Non – market goods

Question 6.
In a pure public good, consumption is ……………..
a) Rival
b) Non – rival
c) Both
d) None of the above
Answer:
a) Rival

Question 7.
One of the most important market failures is caused by …………………………..
a) Positive externalities
b) Negative externalities
c) Both
d) None of the above
Answer:
b) Negative externalities

Question 8.
The common source of outdoor air pollution is caused by combustion processes from the following ……………..
a) Heating and cooking
b) Traditional stoves
c) Motor vehicles
d) All the above
Answer:
b) Traditional stoves

Question 9.
The major contributor of Carbon monoxide is
a) Automobiles
b) Industrial process
c) Stationary fuel combustion
d) None of the above
Answer:
a) Automobiles

Question 10.
Which one of the following causes of global warming?
a) Earth gravitation force
b) Oxygen
c) Centripetal force
d) Increasing temperature
Answer:
d) Increasing temperature

Question 11.
Which of. the following is responsible for protecting humans from harmful ultraviólet rays? .
a) UV-A
b) UV-C .
c) Ozone layer
d) None of the above
Answer:
c) Ozone layer

Question 12.
Global warming also refers to as
a) Ecological change
b) Climate Change
c) Atmosphere changé
d) None of the above
Answer:
d) None of the above

Question 13.
Which of the following is the anticipated effect of Global Warming?
a) Rising sea levels
b) Changing precipitation
c) Expansion of deserts
d) All of the above
Answer:
b) Changing precipitation

Question 14.
The process of nutrient enrichment is termed as
a) Eutrophication
b) Limiting nutrients
c) Enrichment
d) Schistosomoasis
Answer:
b) Limiting nutrients

Question 15.
Primary cause of Soil pollution is ……………..
a) Pest control measures
b) Land reclamation
c) Agricultural run off
d) Chemical fertilizer
Answer:
d) Chemical fertilizer

Question 16.
Which of the following is main cause for deforestation?
a) Timber harvesting industry
b) Natural afforestation
c) Soil stabilization
d) Climate stabilization
Answer:
a) Timber harvesting industry

Question 17.
Electronic waste is commonly referred as ……………….
a) Solid waste
b) Composite waste
c) e- waste
d) Hospital waste
Answer:
c) e- waste

Question 18.
Acid rain is one of the consequences of ………………………………… Air pollution
a) Water Pollution
b) Land pollution
c) Noise pollution
Answer:
a) Water Pollution

Question 19.
Sustainable Development Goals and targets are to be achieved by …………………….
a) 2020
b) 2025
c) 2030
d) 2050
Answer:
c) 2030

Question 20.
Alkali soils are predominantly located in the ……………………… plains?
a) Indus-Ganga
b) North-Indian
c) Gangetic plains
d) All the above
Answer:
d) All the above

PART -B

Answer the following questions in one or two sentences.

Question 21.
State the meaning of environment.
Answer:

1. The meaning of Environmental Economics is a different branch of economics that recognizes the value of both the environment and economic activity and makes choices based on those values.
2. The goal is to balance the economic activity and the environmental impacts by taking into account all the costs and benefits.
3. In short, Environmental Economics is an area of economics that studies the financial impact of environmental issues and policies.
4. Environmental Economics involves theoretical and empirical studies of the economic effects of national or local environmental policies around the world.

Question 22.
What do you mean by ecosystem?
Answer:
The ecosystem is the interacting system of a biological community and its non living environmental surroundings.

Question 23.
Mention the countries where per capita carbon dioxide emission is the highest in the world.
Answer:

• United States of America – (USA)
• Europian Union – (EU)
• Japan
• Russian Federation
• United Arab Emirates (UAE)
• Saudi Arabia
• China

Question 24.
What are environmental goods? Give examples.
Answer:
Environmental goods are typically non – market goods.
Eg: Rivers, Forests, Mountains

Question 25.
What are the remedial measures to control noise pollution?
Answer:
Remedial measures to control Noise Pollution

1. Use of noise barriers
2. Newer roadway for surface transport
3. Traffic control
4. Regulating times for heavy vehicles
5. Installations of noise barriers in the workplace
6. Regulation of Loudspeakers

Question 26.
Define Global warming.
Answer:
Global warming is the increase in temperature of the Earth’s surface , due to greenhouse gases.

Question 27.
Specify the meaning of seed ball.
Answer:

1. A seed ball (or seed bomb) is a seed that has been wrapped in soil materials, usually a mixture of clay and compost, and then dried.
2. Essentially, the seed is ‘pre-planted’ and can be sown by depositing the seed ball anywhere suitable for the species, keeping the seed safely until the proper germination window arises.
3. Seed balls are an easy and sustainable way to cultivate plants that provide a larger window of time when the sowing can occur.

PART -C

Answer the following questions in one paragraph

Question 28.
Brief the linkage between economy and environment.
Answer:

• Man’s life is interconnected with various other living and non-living things. The life also depends on social, political, ethical, philosophical and other aspects of the economic system.
• In fact, the life of human beings is shaped by his living environment. The relationship between the economy and the environment is explained in the form of a ‘Material Balance Model’.

Question 29.
Specify the meaning of material balance principle.
Answer:

• The Material Balance Model was developed by Alenkneese and R.V.Ayres. It explains the relationship between the economy and the environment. The model considers the total economic process as a physically balanced flow between inputs and outputs.
• The first law of thermodynamics emphasizes that in any production system ” what goes in must come out” and this is known as the Material Balance Approach.

Question 30.
Explain different types of air pollution.
Answer:
Types of Air pollution:
1. Indoor Air Pollution:
It refers to toxic contaminants that we encounter in our daily lives in our homes, schools and workplaces. For example, cooking and heating with solid fuels on open fires or traditional stoves results in high levels of indoor air pollution.

2. Outdoor Air Pollution:
It refers to ambient air. The common sources of outdoor air pollution are caused by combustion processes from motor vehicles, solid fuel burning and industry.

Question 31.
What are the causes of water pollution?
Answer:
Water pollution is caused due to several reasons. Few are:

1. Discharge of sewage and wastewater.
2. Dumping of solid wastes.
3. Discharge of industrial wastes.
4. Oil spill
5. Acid rain
6. Global warming
7. Eutrophication.

Question 32.
State the meaning of e-waste.
Answer:
1. Electronic waste which is commonly referred as “e-waste” is the new byproduct of the Info-Tech society.

2. It is a physical waste in the form of old discarded, end of life electronics.

3. It includes a broad and growing range of electronic devices from large household appliances such as refrigerators, air conditioners, cellular phones, computers and other electronic goods”.

4. Similarly, e-waste can be defined as the result when consumer, business and household devices are disposed or sent for re-cycling (example, television, computers, audio-equipments, VCR, DVD, telephone, Fax, Xerox machines, wireless devices, video games, other households electronic equipment).

Question 33.
What is land pollution? Mention the causes of land pollution.
Answer:
Land pollution is defined as,” the degradation of land because of the disposal of waste on the land” Any substance that is discharged, emitted or deposited in the environment in such a way that it alters the environment causes land pollution.

Causes:

• Deforestation and Soil erosion
• Agricultural Activities
• Mining Activities
• Land fills
• Industrialization
• Construction activities
• Nuclear waste.

Question 34.
Write a note on a) Climate change and b) Acid rain Climate Change:
Answer:
(a) Climate Change:
1. climate change refers to seasonal changes over a long period with respect to the growing accumulation of greenhouse gases in the atmosphere.

2. Recent studies have shown that human activities since the beginning of the industrial revolution, have contributed to an increase in the concentration of carbon dioxide in the atmosphere by as much as 40%, from about 280 parts per million in the pre-industrial period, to 402 parts per million in 2016, which in turn has led to global warming.

3. Several parts of the world have already experienced the warming of coastal waters, high temperatures, a marked change in rainfall patterns, and increased intensity and frequency of storms. Sea levels and temperatures are expected to be rising.

(b) Acid Rain:

1. Acid rain is one of the consequences of air pollution.
2. It occurs when emissions from factories, cars or heating boilers contact with the water in the atmosphere.
3. These emissions contain nitrogen oxides, sulphur dioxide and sulphur trioxide which when mixed with water become sulfurous acid, nitric acid and sulfuric acid.
4. This process also occurs by nature through volcanic eruptions.
5. It can have harmful effects on plants, aquatic animals and infrastructure.

PART -D

Answer the following questions in about a page.

Question 35.
Briefly explain the relationship between GDP growth and the quality of environment.
Answer:

• Environmental quality is a set of properties and characteristics of the environment either generalized or local, as they impinge on human beings and other organism.
• Environmental quality has been continuously declining due to capitalistic mode of functioning.
• Environment is a pure public good that can be consumed simultaneously by everyone and from which no one can be excluded.
• Pure public goods pose a free rider problem. As a result, resources are depleted.
• The contribution of the nature to GDP as well as depletion of natural resources are not accounted in the present system of National Income Enumeration.
• In Environmental Economics one of the most important market failures is caused by negative externalities arising from production and consumption of goods and services.
• Externality may be defined as ” the cost or benefit imposed by the consumption and production activities of the individuals in these activity and towards which no payment is made.
• Beneficial externalities are called “Positive externalities” and adverse ones are called ” negative externalities”.
• As these externalities occur outside of the market ie) they affect people not directly involved in the production and consumption of a good or service.
• These externalities both positive and negative are not included in GDP. If proper calculation is made including positive and negative externalities the correlation between GDP growth and environmental quality can be assessed.

Question 36.
Explain the concepts of externality and its classification Externality:
Answer:
Externalities refer to external effects or spillover effects resulting from the act of production or consumption on the third parties. Externalities arise due to interdependence between economic units.

Positive Consumption Externality:
When some residents of a locality hire a private security agency to patrol their area, the other residents of the area also benefit from better security without bearing cost.

Negative Consumption Externality:
A person smoking cigarette gets may gives satisfaction to that person, but this act causes hardship (dissatisfaction) to the non – smokers who are driven to passive smoking.

Positive Production Externality:

1. The ideal location for beehives is orchards (first growing fields).
2. While bees make honey, they also help in the pollination of apple blossoms.
3. The benefits accrue to both producers (honey as well as apple). This is called reciprocal untraded interdependency.
4. Suppose training is given for the workers in a company. If those trained workers leave the. company to join some other company, the later company gets the benefit of skilled workers without incurring the cost of training.

Question 37.
Explain the importance of sustainable development and its goals.
Answer:
Sustainable Development Goals (SDGs):
1. It is crucial to harmonize three core elements such as economic growth, social inclusion, and environmental protection.

2. A set of 17 goals for the World’s future can be achieved before 2030 with three unanimous principles fixed by United Nations such as Universality, Integration, and Transformation.

1. End Poverty in all its forms everywhere
2. End hunger, achieve food security and improved nutrition and promote sustainable agriculture
3. Ensure healthy lives and promote well-being for all at all ages
4. Ensure inclusive and quality education for all and promote lifelong learning
5. Achieve gender equality and empower women and girls
6. Ensure access to water and sanitation for all
7. Ensure access to affordable, reliable, sustainable, and modem energy for all.
8. Promote inclusive and sustainable economic growth, employment, and decent work for all.
9. Build resilient infrastructure, promote sustainable industrialization, and foster innovation.
10. Reduce inequality within and among countries
11. Make cities inclusive, safe, resilient, and sustainable
12. Ensure sustainable consumption and production pattern
13. Take urgent action to combat climate change and its impacts
14. Conserve and sustainably use the oceans, seas, and marine resources
15. Sustainably manage forests, combat desertification, halt and reverse land degradation, halt biodiversity loss
16. Promote just, peaceful and inclusive societies
17. Revitalize the global partnership for sustainable development.

12th Economics Guide Environmental Economics Additional Important Questions and Answers

I. Choose the best Answer

Question 1.
“All the conditions, circumstances, and influences surrounding and affecting the development of an organism or group of organisms” is called ………………………..
(a) Environment
(b) Economics
(c) Eco system
(d) Biosphere
Answer:
(a) Environment

Question 2.
………………… refers to external effects resulting from the act of production or consumption on the third parties.
a) Externalities
b) Ecosystem
c) Pollution
d) Environmental quality
Answer:
a) Externalities

Question 3.
…………………….. Economics involves theoretical and empirical studies of the economic effects.
(a) Biosphere
(b) Political
(c) Environment
(d) Philosophical
Answer:
(c) Environment

Question 4.
……………………… is the introduction of contaminants into the natural environment.
a) Externalities
b) Pollution
c) Ecosystem
d) Spillover effects
Answer:
b) Pollution

Question 5.
…………………………. is the current increase in temperature of the Earth’s surface as well as its atmosphere. .
a) Global warming
b) Climate change
c) Pollution
d) None of the above
Answer :
a) Global warming

Question 6.
Vehicles smoke happens to release high amounts of ……………………..
(a) Carbon – di – oxide
(b) Carbon – monoxide
(c) Carbon
(d) Oxygen
Answer:
(b) Carbon – monoxide

Question 7.
Negative Production externalities are caused by ………………………….
a) Industries
b) Agriculture
c) Transport
d) Computer
Answer:
a) Industries

Question 8.
……………………. is unwanted or excessive sound that can have deleterious effects on human health and environmental quality.
a) Sound pollution
b) Water pollution
c) Noise pollution
d) Land pollution
Answer:
a) Sound pollution

Question 9.
…………………….. is unwanted or excessive sound that can have deleterious effects on human health and environmental quality.
(a) Air pollution
(b) Water pollution
(c) Noise pollution
(d) Land pollution
Answer:
(c) Noise pollution

Question 10.
………………………… is a system of agricultural production which avoids the use of synthetic fertilizer, pesticides, and livestock additives.
a) Organic farming
b) Green Revolution
c) Sustainable Development
d) Green Initiatives
Answer:
a) Organic Farming

Question 11.
…………………….. is an increased level of nutrients in water bodies.
(a) Eutrophication
(b) Global warming
(c) Acid rain
(d) Oil spill
Answer:
(a) Eutrophication

Question 12.
…………………….. is the supplier of all forms of resources like renewable and non – renewable.
(a) Environment
(b) Environmental goods
(c) Environmental quality
(d) Environmental wastes
Answer:
(a) Environment

III Choose the correct pair:

Question 1.
a) Environmental Goods – Industries
b)Environmental Quality – Properties and characteristics of the Environment
c) The Air Prevention and Control of Pollution Act – 1985
d) Acid Rain – Sound pollution
Answer:
b) Environmental Quality – Properties and characteristics of the Environment

Question 2.
a) Concentration of carbon dioxide in the pre-industrial – 380 ppm
b) Concentration Co2 in 2016 – 400ppm
c) Externalities – spillover effects
d) Organic farming – Chemical Fertilizers
Answer:
c) Externalities – spillover effects

Question 3.
a) Alkali soils – Indo – Gangetic plains
b) E – wastes – Environmental wastes
c) Ozone Layer – Infrared rays
d) Non – conventional – Petrol
Answer:
a) Alkali soils – Indo – Gangetic plains

IV.Choose the incorrect pair:

Question 1.
a) Environmental goods – Mountains, Rivers
b) Ecosystem – the foundation of the Biosphere
c) Spillover effects – Externalities
d) Negative Production Externality – beehives
Answer:
d) Negative Production Externality – beehives

Question 2.
a) Non – conventional fuels – Biogas, CNG, LPG
b) Eutrophication – the bloom of algae in the water
c) Indoor air pollution – cooking with solid fuels on open fires

II. Match the following:
Question 1.
A) Environia – 1) R. V Ayres
B) Material Balance Approach – 2) To surround
C) Non – conventional fuel – 3) Electronic wastes
D) E – waste – 4) Biogas

Answer:
c) 2 1 4 3

Question 2.
a) Acid Rain – 1) Sound pollution
b) Eutrophication – 2) Water pollution
c) Oil spills – 3) Air pollution
d) Hearing loss – 4) Depletion of oxygen on water

Answer:
c) 4 3 2 1

Question 3.
a) Production sector – 1) F = W₂
b) House hold sector – 2) R = W₁+ W₂
c) Input = Output – 3) Final product
d) F – 4) R = F + W₁

answer:
b) 4 1 2 3

III. Choose the correct pair:

Question 1.
a) Environmental Goods – Industries
b) Environmental Quality – Properties and characteristics of the Environment
c) The Air Prevention and Control of Pollution Act – 1985
d) Acid Rain – Sound pollution
Answer:
b) Environmental Quality – Properties and characteristics of the Environment.

Question 2.
a) Concentration of carbon dioxide in the pre-industrial – 380 ppm
b) Concentration CO₂ in 2016 – 400ppm
c) Externalities – spillover effects
d) Organic farming – Chemical Fertilizers
Answer:
c) Externalities – spillover effects

Question 3.
a) Alkali soils – Indo – Gangetic plains
b) E – wastes – Environmental wastes
c) Ozone Layer – Infrared rays
d) Non – conventional – Petrol
Answer:
a) Alkali soils – Indo – Gangetic plains

IV.Choose the incorrect pair:

Question 1.
a) Environmental goods – Mountains, Rivers
b) Ecosystem – the foundation of the Biosphere
c) Spillover effects – Externalities
d) Negative Production Externality – beehives
Answer:
d) Negative Production Externality – beehives

Question 2.
a) Concentration of carbon dioxide in the pre-industrial – 38Oppm
b) Concentration CO2 in 2016 – 400ppm
c) Externalities – spillover effects
d) Organic farming – Chemical Fertilizers
Answer:
c) Externalities – spillover effects

Question 3.
a) Alkali soils – Indo – Gangetic plains
b) E – wastes – Environmental wastes
c) Ozone Layer – Infrared rays
d) Non – conventional – Petrol
Answer:
Answer:
a) Alkali soils – Indo – Gangetic plains

IV.Choose the incorrect pair:

Question 1.
a) Environmental goods – Mountains, Rivers
b) Ecosystem – the foundation of the Biosphere
c) Spillover effects – Externalities
d) Negative Production Externality – beehives.
Answer:
d) Negative Production Externality – beehives.

Question 2.
a) Non – conventiõnal fuels – Biogas, CNG, LPG
b) Eutrophication – the bloom of algae in the water
c) Indoor air pollution – cooking with solid fuels on open fires
d) Industrial wastes – Noise pollution.
Answer:
d) Industrial wastes – Noise pollution

Question 3.
a) Oil spills – Seawater gets polluted
b) Trees – CO₂ emission
c) Sustainable Development – economic growth, social inclusion, and environmental protection.
d) Organic farming – crop rotation
Answer:
b) Trees – CO₂ emission

V. Choose the correct Statement:

Question 1.
a) Average temperatures around the world have risen by 1°c over the last 100 years.
b) Global warming reduces the level of greenhouse gases.
c) Vehicles smoke happens to release high amounts of carbon monoxide.
d) Trees absorb oxygen from the air and release carbon dioxide.
Answer:
c) Vehicles smoke happens to release high amounts of carbon monoxide.

Question 2.
a) Water pollution increases the oxygen level in the water
b) Pollution is the introduction of contaminants into the natural environment.
c) Beneficial externalities are called “positive externalities”.
d) Environment is a private good.
Answer:
b) Pollution is the introduction of contaminants into the natural environment.

Question 3.
a) Environmental goods are typically non – market goods.
b) Positive production externalities include pollution generated by a factory that ’ imposes costs on others.
c) The term environment has been derived from the Greek word “Environia” which means to surround.
d) The relationship between the economy and pollution is explained in the form of a “Material Balance Models”.
Answer:
a) Environmental goods are typically non-market goods.

VI. Choose the incorrect statement

Question 1.
a) Acid rain is one of the consequences of Air pollution.
b) Sustainable Development includes economic growth, social inclusion, and environmental protection.
c) Trees release oxygen and pollutes the air.
d) Chronic exposure to noise may cause noise-induced hearing loss.
Answer:
c) Trees release oxygen and pollutes the air

Question 2.
a) Increasing temperature in the atmosphere leads to global warming.
b) Ozone layer is responsible for protecting humans from harmful ultraviolet rays.
c) Earth’s ozone layer is depleting due to the presence of chlorofluorocarbons and hydrochlorofluorocarbons in the atmosphere.
d) Atmospheric pollution increases the level of ozone.
Answer:
d) Atmospheric pollution increases the level of ozone.

Question 3.
a) Surface water includes natural water found on the earth’s surface, like rivers, lakes, lagoons, and oceans.
b) Eutrophication is an increased level of oxygen in water bodies.
c) CO₂ is the most important of the greenhouse gases contributing to 50% of global warming.
d) ‘Problem soils’ exist mainly in arid and semi-arid regions.
Answer:
b) Eutrophication is an increased level of oxygen in water bodies.

VII. Pick the odd one out:

Question 1.
a) Sea pollution
b) Air pollution
c) Land pollution
d) Water pollution
Answer:
a) Sea pollution

Question 2.
a) BlOGAS
b) Petrol
c) CNG
d) LPG
Answer:
b) Petrol

VIII. Analyze the Reason:

Question 1.
Assertion (A): “E-waste ” is the new by-product of the Info-Tech society.
Reason (R): E-waste can be generated as the result when consumer, business, and household devices are disposed or sent for recycling.
Answer:
a) Assertion (A) and (R) both are true, and (R) is the correct explanation of (A).

Question 2.
Assertion (A): Solid waste is basically the discharge of useless and unwarranted materials as a result of human activity.
Reason (R): Solid waste consists of the discards of households, hospital refuse, dead animals, debris from a construction site, etc.,
Answer:
b) Assertion (A) and Reason (R) both are true, but (R) is not the correct explanation of (A).

Question 3.
Assertion (A): Global warming is the current increase in temperature of the Earth’s surface as well as its atmosphere.
Reason (R): The increase in the number of greenhouse gases warms the earth’s surface.
Options:
a) Assertion (A) and (R) both are true, and (R) is the correct explanation of (A).
b) Assertion (A) and Reason (R) both are true, but (R) is not the correct explanation of (A).
c) (A) is true but (R) is false.
d)Both (A) and (R) are false.
Answer:
a) Assertion (A) and (R) both are true, and (R) is the correct explanation of (A).

IX. 2 Marks Questions

Question 1.
Define “Solid wastes”?
Answer:
Solid Wastes:

1. Non-liquid, non-soluble materials, ranging from municipal garbage to industrial wastes that contain complex, and hazardous, substances.
2. Solid wastes include sewage sludge, agricultural refuse, demolition wastes, and mining residues.

Question 2.
Draw the flow diagram for Material Balance Approach.
Answer:

Question 3.
Define “Sustainable Development”?
Answer:

1. Sustainable Development is a development that meets the needs of the present generation without compromising the ability of future generations to meet their own needs.
2. “The alternative approach (to sustainable development) is to focus on natural capital assets and suggest that they should not decline through time.”

Question 4.
What is pollution?
Answer:
Pollution is the introduction of contaminants into the natural environment that causes an adverse change in the form of life, toxicity, of environmental damage to the ecosystem and aesthetics of our surroundings.

Question 5.
Define “Deforestation”?
Answer:

1. Humans depend on trees for many things including life.
2. Trees absorb carbon dioxide from the air and release Oxygen, which is needed for life.
3. Forest helps replenish soils and helps retain nutrients being washed away.
4. Deforestation is led to land pollution.

Question 6.
What are the causes of Air pollution?
Answer:

• Vehicle exhaust smoke
• Fossil fuel-based power plants
• Exhaust from industrial plants and factories
• Construction and Agricultural Activities
• Natural causes
• Household activities.

Question 7.
What are the effects of Air pollution?
Answer:

• Respiratory and heart problems
• Acid rain
• Eutrophication
• Effect on wildlife
• Depletion of the ozone layer
• Human Health
•  Global warming

Question 8.
What are the remedial measures to control Air pollution?
Answer:

• Establishment of industries away from the towns and cities.
• Increasing the length of the chimneys in Industries.
• Growing more plants and trees.
• Use of non – conventional fuels like Biogas, CNG, and LPG.
• Use of Mass Transit system.

Question 9.
What are the types of water pollution?
Answer:

1. Surface water pollution
2. Groundwater pollution
3. Microbiological pollution.
4. Oxygen depletion pollution

Question 10.
What are the remedial measures to control water pollution?
Answer:

• Comprehensive water management plan.
• Construction of proper storm drains and settling ponds.
• Maintenance of drain life.
• Effluent and sewage treatment plant.
• Regular monitoring of water and wastewater.
• Stringent actions towards illegal dumping of waste into the water bodies.

Question 11.
Define Noise Pollution.
Answer:
Noise pollution is unwanted or excessive sound that can have deleterious effects on human health and environmental quality. Noise pollution is commonly generated by many factories. It also comes from the highway, railway, and airplane traffic and from outdoor construction activity.

Question 12.
Name the types of Noise Pollution.
Answer:

1. Atmospheric Noise
2. Industrial Noise
3. Man-made Noise

Question 13.
What are the causes of Noise Pollution?
Answer:

• Poor urban planning
• Sounds from motor vehicles
• Crackers
• Factory machinery

Question 14.
What are the effects of Noise Pollution?
Answer:

• Hearing Loss
• Damage physiological and psychological health.
• Cardiovascular effects.
• Detrimental effect on animals and aquatic life.
• Effects on wildlife and aquatic animals.

X. 3 Mark Questions

Question 1.
Explain the types of Noise pollution?
Answer:
Types of Noise Pollution:
(I) Atmospheric Noise:
Atmospheric noise or static is caused by lightning discharges in thunderstorms and other natural electrical disturbances occurring in the atmosphere.

(II) Industrial Noise:

1. Industrial noise refers to noise that is created in factories.
2. When sound becomes noise it becomes unwanted.
3. Heavy industries like shipbuilding, iron, and steel have long been associated with Noise-Induced Hearing Loss (NIHL).

(III) Man-made Noise:
The main sources of man-made noise pollution are ships, aircraft, seismic exploration, marine construction, drilling, and motorboats.

Question 2.
State the Effect of Land Pollution.
Answer:

• Soil pollution
• Healthy Impact
• Cause for Air pollution
• Effect on wildlife.

Question 3.
State the remedial measures to control Land Pollution.
Answer:

• Making people aware of the concept of a Reduce, Recycle, and Reuse.
• Buying biodegradable products.
• Minimizing the usage of pesticides
• Shifting cultivation.
• Disposing of unwanted garbage properly either by burning or by burying it under the soil.
• Minimizing the usage of plastics.

Question 4.
Define sustainable development.
Answer:
“Sustainable development is a development that meets the needs of the present without compromising the ability of future generations to meet their own needs” World commission of Environment and Development, 1987.

Question 5.
What is alkali soil?
The occurrence of accumulation of excess salt/ acid in the root zone results in a partial or complete loss of soil productivity and such soil is defined as ‘Problem (alkali, saline, acid) soils’.
This soil exists mainly in arid and semi-arid regions.

Question 6.
What is Acid – rain?
Answer:

• Acid rain is one of the consequences of air pollution.
• The emissions from factories, cars, or heating boilers contain nitrogen oxides, sulphur dioxide, and sulphur trioxide.
• These oxides when mixed with rainwater become sulfurous acid, nitric acid, and sulfuric acid and come down to earth as acid rain.

Question 7.
What is organic farming?
Answer:

• Organic farming is a system of agricultural production which relies on animal manure, organic waste, crop rotation, legumes, and biological pest control.
• It avoids the use of synthetic fertilizer, pesticides, and livestock additives.
• Organic inputs have certain benefits, such as enriching the soil for microbes.

XI. 5 Mark Questions

Question 1.
Name the sources of E-waste.
Answer:

Question 2.
Write a note on solid waste.
Answer:

• Solid waste is basically the discharge of useless and unwarranted materials as a result of human activity.
• Most commonly, they are composed of solids, semisolids, or liquids.
• Solid wastes consist of the discards of households, hospital refuse, dead animals, debris from a construction site, ashes, agricultural wastes, and industrial wastes, etc.
• When waste is not removed from the streets and public places in time it poses severe public – health and hygiene hazards.

Question 3.
Explain the causes of Air pollution?
Answer:
Causes of Air Pollution:
(I) Vehicle exhaust smoke:

1. Vehicles smoke happens to release high amounts of Carbon monoxide.
2. Millions of vehicles are operated every day in cities, each one leaving behind its own carbon footprint on the environment.

(II) Fossil fuel-based power plants:

1. Fossil fuels also present a wider-scale problem when they are burned for energy in power plants.
2. Chemicals like sulphur dioxide are released during the burning process, which travels straight into the atmosphere.
3. These types of pollutants react with water molecules to yield something known as acid rain.

(III) Exhaust from Industrial Plants and Factories:
Heavy machinery located inside big factories and industrial plants also emits pollutants into the air.

(IV) Construction and Agricultural activities:

1. Potential impacts arising from the construction debris would include dust particles and gaseous emissions from the construction sites.
2. Likewise, using ammonia for agriculture is a frequent byproduct that happens to be one of the most dangerous gases affecting the air.

(V) Natural Causes:

1. Earth is one of the biggest polluters itself, through volcanoes, forest fires, and dust storms.
2. They are nature-borne events that dump massive amounts of air pollution into the atmosphere.

(VI) Household activities:
Household activities like cooking, heating, and lighting, use of various forms of mosquito repellents, pesticides and chemicals for cleaning at home, and use of artificial fragrances are some of the sources that contribute to air pollution.

Question 4.
What are the general principles of organic farming?
Answer:

1. Protect the environment, minimize soil degradation and erosion, decrease pollution, optimize biological productivity and promote a sound state of health.
2. Maintain long-term soil fertility by optimizing conditions for biological activity within the soil.
3. Maintain biological diversity within the system.
4. Recycle materials and resources to the greatest extent possible within the enterprise.
5. Provide attentive care that promotes health and meets the behavioural needs of livestock.
6. Prepare organic products, emphasizing careful processing and handling methods in order to maintain the organic integrity and vital qualities of the products at all stages of production.
7. Rely on Renewable resources in the locally organized agricultural system.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Computer Applications Guide Pdf Chapter 18 Electronic Data Interchange – EDI Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Computer Applications Solutions Chapter 18 Electronic Data Interchange – EDI

12th Computer Applications Guide Electronic Data Interchange – EDI Text Book Questions and Answers

Part I

Choose The Correct Answers

Question 1.
EDI stands for
a) Electronic Details Information
b) Electronic Data Information
c) Electronic Data Interchange
d) Electronic Details Interchange
Answer:
a) Electronic Details Information

Question 2.
Which of the following is an internationally recognized standard format for trade, transportation, insurance, banking and customs?
a) TSLFACT
b) SETFACT
c) FTPFACT
d) EDIFACT
Answer:
d) EDIFACT

Question 3.
Which is the first industry-specific EDI standard?
a) TDCC
b) VISA
c) Master
d) ANSI
Answer:
a) TDCC

Question 4.
UNSM stands for
a) Universal Natural Standard Message
b) Universal Notations for Simple Message
c) United Nations Standard Message
d) United Nations Service Message
Answer:
c) United Nations Standard Message

Question 5.
Which of the following is a type of EDI?
a) Direct EDI
b) Indirect EDI
c) Collective EDI
d) Unique EDI
Answer:
a) Direct EDI

Question 6.
Who is called the father of EDI?
a) Charles Babbage
b) Ed Guilbert
c) Pascal
d) None of the above
Answer:
b) Ed Guilbert

Question 7.
EDI interchanges start with ……………. and end with ……………
a) UNA, UNZ
b) UNB, UNZ
c) UNA, UNT
d) UNB, UNT
Answer:
b) UNB, UNZ

Question 8.
EDIFACT stands for
a) EDI for Admissible Commercial Transport
b) EDI for Advisory Committee and transport
c) EDI for Administration, Commerce, and Transport
d) EDI for Admissible Commerce and Trade
Answer:
c) EDI for Administration, Commerce, and Transport

Question 9.
The versions of EDIFACT are also called as
a) Message types
b) Subsets
c) Directories
d) Folders
Answer:
c) Directories

Question 10.
Number of characters in a single EDIFACT messages
a) 5
b) 6
c) 4
d) 3
Answer:
b) 6

Part II

Short Answers

Question 1.
Define EDI.
Answer:
The Electronic Data Interchange (EDI) is the exchange of business documents between one trade partner and another electronically. It is transferred through a dedicated channel or – through the Internet in a predefined format without much human intervention.

Question 2.
List few types of business documents that are transmitted through EDI.
Answer:

1. Delivery notes
2. Invoices
3. Purchase orders
4. Advance ship notice
5. Functional acknowledgments etc.

Question 3.
What are the 4 major components of EDI?
Answer:
There are four major components of EDI. They are:

1. Standard document format
2. Translator and Mapper
3. Communication software
4. Communication network

Question 4.
What is meant by directories inEDIFACT?
Answer:

• The versions of EDIFACT are also called as directories.
• These EDIFACT directories will he revised twice a year.

Question 5.
Write a note on EDIFACT subsets.
Answer:
Due to the complexity, branch-specific subsets of EDIFACT have developed. These subsets of EDIFACT include only the functions relevant to specific user groups.
Example:

• CEFIC – Chemical industry
• EDIFURN – furniture industry
• EDIGAS – gas business

Part III

Explain In Brief Answer

Question 1.
Write a short note on EDI.

• The Electronic Data Interchange (EDI)is the exchange of business documents between one trade partner and another electronically,
• It is transferred through a dedicated channel or through the Internet in a predefined format without much human intervention,
• It is used to transfer documents such as delivery notes, invoices, purchase orders, advance ship notices, functional acknowledgments, etc.

Question 2.
List the various layers of EDI.
Answer:
Electronic data interchange architecture specifies four different layers namely

1. Semantic layer
2. Standa, us translation layer
3. Transport layer
4. Physical layer

These EDI layers describe how data flows from one computer to another.

Question 3.
Write a note on UN/EDIFACT.
Answer:

• United Nations / Electronic Data Interchange for Administration, Commerce, and Transport
• (UN / EDIFACT) is an international EDI – a standard developed under the supervision of the United Nations.
• In 1987, the UN / EDIFACT syntax rules were approved as ISO: IS09735 standard by the International Organization for Standardization.
• EDIFACT includes a set of internationally agreed standards, catalogs, and guidelines for the electronic exchange of structured data between independent computer systems.

Question 4.
Write a note on the EDIFACT message.
Answer:

• The basic standardization concept of EDIFACT is that there are uniform message types called United Nations Standard Message (UNSM).
• In so-called subsets, the message types can be specified deeper in their characteristics depending on the sector.
• The message types, all of which always have exactly one nickname consisting of six uppercase English alphabets.
• The message begins with UNH and ends with UNT.

Question 5.
Write about EDIFACT separators
Answer:
EDIFACT has the following punctuation marks that are used as standard separators.
Character Uses

Part IV

Explain In Detail

Question 1.
Briefly explain various types of EDI.
Answer:
The types of EDI were constructed based on how EDI communication connections and the conversion were organized. Thus based on the medium used for transmitting EDI documents the following are the major EDI types.

1. Direct EDI
2. EDI via VAN
3. EDI via-FTP/VPN, SFTP, FTPS
4. Web EDI
5. Mobile EDI
6. Direct EDI/Point-to-Point

It is also called as Point-to-Point EDI. It establishes a direct connection between various business stakeholders and partners individually. This type of EDI suits to larger businesses with a lot of day to day business transactions.

EDI via VAN:
EDI via VAN (Value Added Network) is where EDI documents are transferred with the support of third-party network service providers. Many businesses prefer this network model to protect them from the updating ongoing complexities of network technologies.

EDI via FTP/VPN, SFTP, FTPS:
When protocols like FTP/VPN, SFTP, and FTPS are used for the exchange of EDI-based documents through the Internet or Intranet it is called EDI via FTP/VPN, SFTP, FTPS.

Web EDI:
Web-based EDI conducts EDI using a web browser via the Internet. Here the businesses are allowed to use any browser to transfer data to their business partners. Web-based EDI is easy and convenient for small and medium organizations.

Mobile EDI:
When smartphones or other such handheld devices are used to transfer EDI documents it is called mobile EDI. Mobile EDI applications considerably increase the speed of EDI transactions.

Question 2.
What are the advantages of EDI?
Answer:

• EDI was developed to solve the problems inherent in paper-based transaction processing and in other forms of electronic communication.
• Implementing an EDI system offers a company greater control over its supply chain and allow it to trade more effectively. It also increases productivity and promotes operational efficiency.

The following are the other advantages of EDI.

• Improving service to end-users
• Increasing productivity
• Minimizing errors
• Slashing response times
• Automation of operations
• Cutting costs
• Integrating all business and trading partners

Question 3.
Write about the structure of EDIFACT.
Answer:

• EDIFACT is a hierarchical structure where the top level is referred to as an interchange, and lower levels contain multiple messages.
• The messages consist of segments, which in turn consist of composites.
• The final iteration is a data element.

Segment Tables

• The segment table lists the message tags.
• It contains the tags, tag names, requirements designator, and repetition field.
• The requirement designator may be mandatory (M) or conditional (C).
• The (M) denotes that the segment must appear at least once. The (C) denotes that the segment may be used if needed.
• Example: CIO indicates repetitions of a segment or group between 0 and 10.

EDI Interchange

• Interchange is also called an envelope.
• The top-level of the EDIFACT structure is Interchange.
• An interchange may contain multiple messages. It starts with UNB and ends with UNZ

EDIFACT message

• The basic standardization concept of EDIFACT is that there are uniform message types called United Nations Standard Message (UNSM).
• In so-called subsets, the message types can be specified deeper in their characteristics depending on the sector.
• The message types, all of which always have exactly one nickname consisting of six uppercase English alphabets.
• The message begins with UNH and ends with UNT

Service messages

• To confirm/reject a message, CONTRL and APERAK messages are sent.
• CONTRL- Syntax Check and Confirmation of Arrival of Message
• APERAK – Technical error messages and acknowledgment

Data exchange

• CREMUL – multiple credit advice
• DELFOR- Delivery forecast
• IFTMBC – Booking confirmation

EDIFACT Segment

• It is the subset of messages.
• A segment is a three-character alphanumeric code.
• These segments are listed in segment tables.
• Segments may contain one, or several related user data elements.

EDIFACT Elements

• The elements are the piece of actual data.
• These data elements may be either simple or composite.

EDI Separators
EDIFACT has the following punctuation marks that are used as standard separators.

12th Computer Applications Guide Electronic Data Interchange – EDI Additional Important Questions and Answers

Part A

Choose The Correct Answers

Question 1.
……………………. is the exchange of business documents between one trade partner and another electronically.
(a) EDI
(b) UDI
(c) FDI
(d) DDI
Answer:
(a) EDI

Question 2.
First EDI standards were released by ………..
a) EDI
b) EFT
c) EDIA
d) TDCC
Answer:
d) TDCC

Question 3.
……………………. is a paperless trade.
(a) EDI
(b) XML
(c) EDIF
(d) EFT
Answer:
(a) EDI

Question 4.
………… establishes a direct connection between various business stakeholders
and partners individually.
a) Direct EDI
b) EDI via VAN
c) Web EDI
d) Mobile EDI
Answer:
a) Direct EDI

Question 5.
Electronic data interchange architecture specifies ……………. different layers.
a) two
b) three
c) four
d) five
Answer:
c) four

Question 6.
TDCC was formed in the year …………………….
(a) 1964
(b) 1966
(c) 1968
(d) 1970
Answer:
(c) 1968

Question 7.
In ……………… UN created the EDIFACT to assist with the global reach of technology in E-Commerce.
a)1985
b)1978
c)1974
d)1975
Answer:
a)1985

Question 8.
Expand EDIA
(a) Electronic Data Interchange Authority
(b) Electronic Data Information Association
(c) Electronic Data Interchange Association
(d) Electronic Device Interface Amplifier
Answer:
(c) Electronic Data Interchange Association

Question 9.
Which of the following is for the exchange of EDI-based documents through the Internet?
a) FTP/VPN
b) SFTP
c) FTPS
d) All of the above
Answer:
d) All of the above

Question 10.
EDIA has become …………………….. committee.
(a) ANSIXI2
(b) ANSIXI3
(c) ANSIXI4
(d) ANSIX15
Answer:
(a) ANSIXI2

Fill In The Blanks:

1. ……….. was developed to solve the problems inherent in paper-based transaction processing.
Answer:
EDT

2. ………….. is also called as Point-to-Point EDI.
Answer:
Direct EDT

3. Interchange is also called…………..
Answer:
Envelope

4. EDT is ……………… Trade.
Answer:
Paperless

5. EFT is …………….. Payment
Answer:
Paperless

6. ………… is “the computer-to-computer interchange of strictly formatted messages.
Answer:
EDI

7. …………….. EDI is easy and convenient for small and medium organizations.
Answer:
Web-based

8. The …………. is the most critical part of the entire EDI.
Answer:
standard

Abbreviations

1. EDI – Electronic Data Interchange
2. EFT – Electronic Transfer
3. TDCC – Transportation Data Coordinating Committee
4. EDIA – Electronic Data Interchange Association
5. ANSI – American National Standards Institute
6. VAN – Value Added Network
7. ANSI ASC – American National Standards Institute Accredited Standard Committee
8. GTDI – Guideline for Trade Data Interchange
9. UN/ECE/ – United -Nations Economic Commission for Europe
10. UN/EDIFACT -United Nations / Electronic Data Interchange for Administration, Commerce, and Transport
11. UNSM -United Nations Standard Message

Assertion And Reason
Question 1.
Assertion (A): According to the National Institute of Standards and Technology, EDI is the computer-to-computer interchange of strictly formatted messages that represent documents other than monetary instruments.
Reason(R): The Electronic Data Interchange (EDI) is the exchange of business documents between one trade partner and another electronically.
a) Both (A) and (R) are correct and (R) ¡s the correct explanation of (A)
b) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)
c) (A) is true and (R) is false
d) (A) is false and (R) is true
Answer:
b) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)

Question 2.
Assertion(A): EFT is “Paperless Trade”
Reason(R): The Electronic Data Interchange (EDI) is the exchange of business documents between one trade partner and another electronically.
a) Both (A) and (R) are correct and (R) is the correct explanation of (A)
b) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)
c) (A) is true and (R) is false
d) (A) is false and (R) is true
Answer:
d) (A) is false and (R) is true

Question 3,
Assertion (A): United Nations / Electronic Data Interchange for Administration, Commerce, and Transport (UN / EDIFACT) is an international EDI – a standard developed under the supervision of the United Nations.
Reason(R): In 1985, the UN / EDIFACT syntax rules were approved as ISO: IS09735 standard by the International Organization for Standardization.
a) Both (A) and (R) are correct and (R) is the correct explanation of (A)
b) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)
c) (A) is true and (R) is false
d) (A) is false and (R) is true
Answer:
c) (A) is true and (R) is false

Question 4.
Assertion (A): The segment table lists the message tags.
Reason(R): It contains the tags, tag names, requirements designator, and repatriation field.
a) Both (A) and (R) are correct and (R) is the correct explanation of (A)
b) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)
c) (A) is true and (R) is false
d) (A) is false and (R) is true
Answer:
a) Both (A) and (R) are correct and (R) is the correct explanation of (A)

Question 5.
Assertion (A): The top level of EDIFACT structure is Interchange.
Reason(R): Interchange is also called an envelope. An interchange may contain multiple messages. It starts with UNB and ends with UNZ
a) Both (A) and (R) are correct and (R) is the correct explanation of (A)
b) Both (A) and (R) are correct, but (R) is not the correct explanation of (A)
c) (A) is true and (R) is false
d) (A) is false and (R) is true
Answer:
a) Both (A) and (R) are correct and (R) is the correct explanation of (A)

Short Answer Questions

Question 1.
Who is the father of EDI?
Answer:
Ed Guilbert is called the father of EDI

Question 2.
What is Paperless trade?
Answer:
The exchange of business documents between one trade partner and another electronically is called Paperless trade.

Question 3.
What is Paperless Payment?
Answer:
Transfer of money from one bank account to another, via computer-based systems, is known as Paperless payment

Question 4.
What is another name of Direct EDI?
Answer:
Another name of Direct EDI is Point-to-Point EDI.

Question 5.
How many alphabets require for EDI messages?
Answer:
Every EDI message requires six uppercase English Alphabets

Match The Following:

1. EDI – Booking confirmation
2. EFT – Paperless Trade
3. EDIFACT – Envelope
4. Interchange – Delivery forecast
5. CEFIC – Directories
6. EDIFURN – Chemical industry
7. EDIGAS – Technical error
8. CONTRL – Multiple credit advice
9. APERAK – Furniture industry
10. CREMUL – Arrival of Message
11. DELFOR – Gas business
12. IFTMBC – Paperless Payment

Answers
1. Paperless Trade
2. Paperless Payment
3. Directories
4. Envelope
5. Chemical industry
6. Furniture industry
7. Gas business
8. Arrival of Message
9. Technical error
10. Multiple credit advice
11. Delivery forecast
12. Booking confirmation

Find The Odd One On The Following

1. (a) Deliver/ Notes
(b) Invoices
(c) Advance Ship Notice
(d) EDIFACT
Answer:
(d) EDIFACT

2. (a) EDIFACT
(b) XML
(c) CDMA
(d) ANSI ASCX12
Answer:
(c) CDMA

3. (a) Direct EDI
(b) InDirectEDI
(c) Web EDI
(d) Mobile EDI
Answer:
(b) InDirectEDI

4. (a) FTP/VPN
(b) HTTP
(c) SFTPP
(d) FTPS
Answer:
(b) HTTP

5. (a) Dial-Up Line
(b) I way
(c) point to point
(d) Internet
Answer:
(c) point to point

6. (a) Email
(b) MIME
(c) HTTP
(d) ANSI X12
Answer:
(d) ANSI X12

7. (a) Transport Layer
(b) Semantic Layer
(c) Application Layer
(d) physical Layer
Answer:
(c) Application Layer

8. (a) Standards
(b) Catalogs
(c) TDCC
(d) guidelines
Answer:
(c) TDCC

9. (a) CREMUL
(b) DELFOR
(c) APERAK
(d) IFTMBC
Answer:
(c) APERAK

10. (a) Segment Terminator
(b) : – component data
(c) ? – data element separator
(d). – decimal point
Answer:
(c) ? – data element separator

Important Years To Remember:

Part B

Short Answers

Question 1.
What is VAN?
Answer:
A value-added network is a company, that is based on its own network, offering EDI services to other businesses. A value-added network acts as an intermediary between trading partners. The principal operations of value-added networks are the allocation of access rights and providing high data security.

Question 2.
What are the types of EDI?
Answer:

1. Direct EDI
2. EDI via VAN
3. EDI via FTP/VPN, SFTP, FTPS
4. Web EDI
5. Mobile EDI

Question 3.
Write a short note on the Segment Table?
Answer:
Segment Tables:
The segment table lists the message tags. It contains the tags, tag names, requirements designator, and repetitation field. The requirement designator may be mandatory (M) or conditional (C). The (M) denotes that the segment must appear atleast once. The (C) denotes that the segment may be used if needed.

Question 4.
Mention some International accepted EDI Standards.
Answer:

• EDIFACT
• XML
• ANSI
• ASC XI2,

Part C

Brief Answers

Question 1.
Write a short note on EDIFACT Structure.
Answer:

• EDIFACT is a hierarchical structure where the top level is referred to as an interchange, and lower levels contain multiple messages.
• The messages consist of segments, which in turn consist of composites.
• The final iteration is a data element.

Question 2.
What is EDI interchange?
Answer:

• The top-level of the EDIFACT structure is Interchange.
• An interchange may contain multiple messages.
• It starts with UNB and ends with UNZ

Question 3.
What is the EDI segment?
Answer:

• A segment is a three-character alphanumeric code.
• These segments are listed in segment tables.
• Segments may contain one, or several related user data elements.

Question 4.
Write a note on EDI Interchange?
Answer:
EDI Interchange:
Interchange is also called an envelope. The top-level of the EDIFACT structure is Interchange. An interchange may contain multiple messages. It starts with UNB and ends with UNZ.

Part D

Detailed Answers

Question 1.
Explain EDI standards?
Answer:
EDI Standards:

• The standard is the most critical part of the entire EDI. Since EDI is the data transmission and information exchange in the form of an agreed message format, it is important to develop a unified EDI standard.
• The EDI standard is mainly divided into the following aspects: basic standards, code-standards, message standards, document standards, management standards, application standards, communication standards, and security standards.
• The first industry-specific EDI standard was the TDCC published by the Transportation Data Coordinating Committee in 1975.
• Then other industries started developing unique standards based on their individual needs. E.g. WINS in the warehousing industry.
• Since the application of EDI has become more mature, the target of trading operations is often not limited to a single industry.
• In 1979, the American National Standards Institute Accredited Standard Committee (ANSI ASC) developed a wider range of EDI standards called ANSI XI2.
• On the other hand, the European region has also developed an integrated EDI standard. Known as GTDI (Guideline for Trade Data Interchange).
• ANSI X12 and GTDI have become the two regional EDI standards in North America and Europe respectively.
• After the development of the two major regional EDI standards and a few years after the trial, the two standards began to integrate and conduct research and development of common EDI standards.
• Subsequently, the United Nations Economic Commission for Europe (UN/ECE/WP.4) hosted the task of the development of international EDI standards. In 1986, UN/EDIFACT is officially proposed. The most widely used EDI message standards are the United Nations EDIFACT and the ANSI X12.

Question 2.
Draw the structure of the UN/EDIFACT message.

Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Physics Guide Pdf Chapter 6 Optics Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 12th Physics Solutions

Chapter 6 Optics

12th Physics Guide Optics Text Book Back Questions and Answers

Part – I:

Textbook Evaluation:

I. Multiple Choice Questions:

Question 1.
The speed of light in an isotropic medium depends on,
a) its intensity
b) its wavelength
c) the nature of propagation
d) the motion of the source w.r.t medium
Answer:
b) its wavelength

Question 2.
A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that its end closer to the pole is 20 cm away from the mirror. The length of the image is,
a) 2.5 cm
b) 5 cm
c) 10 cm
d) 15 cm
Answer:
b) 5 cm
Solution:

Question 3.
An object is placed in front of a convex mirror of focal length of f and the maximum and minimum distance of an object from the mirror such that the image formed is real and magnified.
a) 2f and c
b) c and ∞
c) f and O
d) None of these
Answer:
d) None of these

Question 4.
For light incident from the air onto a slab of refractive index 2. The maximum possible angle of refraction is,
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer:
(a) 30°
Hint:
From Snell’s law, µ = \(\frac { sin i }{ sin r }\)
Now consider an angle of incident is 90°
\(\frac { sin 90° }{ 2 }\)
sin r = sin^-1 (0.5)
r = 30°

Question 5.
If the velocity and wavelength of light in air is V_a and λ_a, and that in water is V_w. and λ_w, then the refractive index of water is,
a) \(\frac{V_{\mathrm{W}}}{V_{\mathrm{a}}}\)
b) \(\frac{V_{\mathrm{a}}}{V_{\mathrm{w}}}\)
c) \(\frac{\lambda_{\mathrm{W}}}{\lambda_{\mathrm{a}}}\)
d) \(\frac{V_{\mathrm{a}} \lambda_{\mathrm{a}}}{V_{\mathrm{w}} \lambda_{\mathrm{W}}}\)
Answer:
b) \(\frac{V_{\mathrm{a}}}{V_{\mathrm{w}}}\)
Solution:
µ = \(\frac{V_{\mathrm{a}}}{V_{\mathrm{w}}}\)

Question 6.
Stars twinkle due to,
a) reflection
b) total internal reflection
c) refraction
d) polarisation
Answer:
c) refraction

Question 7.
When a biconvex lens of glass having a refractive index 1.47 is dipped in a liquid, it acts as a plane sheet of glass. This implies that the liquid must have a refractive index,
a) less than one
b) less than that of glass
c) greater than that of glass
d) equal to that of glass
Answer:
d) equal to that of glass
Solution:

Question 8.
The radius of curvature of curved surface at a thin planoconvex lens is 10 cm and the refractive index is 1.5. If the plane surface is silvered, then the focal length will be
a) 5 cm
b) 10 cm
c) 15 cm
d) 20 cm
Answer:
b) 10 cm
Solution:

Question 9.
An air bubble in a glass slab of refractive index 1.5 (near-normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness of the slab is,
(a) 8 cm
(b) 10 cm
(c) 12 cm
(d) 16 cm
Answer:
(c) 12 cm
Hint:
Let d₁ = 5 cm and d₂ = 3 cm ; n = 1.5
Actual width is the sum of real depth from 2 sides
Thickness of slab = d₁n + d₂ n
= (5 x 1.5) +(3 x 1.5)= 12 cm

Question 10.
A ray of light travelling in a transparent medium of refractive index n falls, on a surface separating the medium from air at an angle of incidents of 45°. The ray can undergo total internal reflection for the following n,
a) n = 1.25
b) n = 1.33
c) n = 1.4
d) n = 1.5
Answer:
d) n = 1.5
Solution:
For total internal reflection i > i_c
sin i > sin i_c
sin45° > \(\frac{1}{\mu}\)
µ > √2
µ > 1.414
µ = 1.50

Question 11.
A plane glass is placed over a various coloured letters (violet, green, yellow, red) The letter which appears to be raised more is,
a) red
b) yellow
c) green
d) violet
Answer:
d) violet

Question 12.
Two-point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm approximately. The maximum distance at which these dots can be resolved by the eye is, [take wavelength of light, λ = 500 nm]
a) 1 m
b) 5 m
c) 3 m
d) 6m
Solution:

Question 13.
In Young’s double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to,
(a) 2D
(b) \(\frac { D }{ 2 }\)
(c) √2D
(d) \(\frac { D }{ √2 }\)
Answer:
(a) 2D
Hint:
Young’s double -slite experiment is
β = \(\frac { λD }{ d}\) ; β’ = \(\frac { λD’ }{ d’}\) ; d’ = 2d
Same fringe space, β = β’
⇒ \(\frac { λD }{ d}\) = \(\frac { λD’ }{ d’}\) ; D’ = 2D

Question 14.
Two coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum possible intensities in the resulting beam are
a) 5I and I
b) 5I and 3I
c) 9I and I
d) 9I and 3I
Answer:
c) 9I and I
Solution:
I_max = I + 4I + 2\(\sqrt{I \times 4 I}\)
= 5I + 4I = 9I
I_min = I × 4I – 2\(\sqrt{I \times 4 I}\)
= 5I – 4I = I

Question 15.
When light is incident on a soap film of thickness 5 × 10^-5 cm, the wavelength of light reflected maximum in the visible region is 5320 Å. Refractive index of the film will be,
a) 1.22
b) 1.33
c) 1.51
d) 1.83.
Answer:
b) 1.33
Solution:
2µt cos r = (2n + 1)\(\frac{\lambda}{2}\)
For normal incidence
cos r = 1
for n = 0,
λ₁ = 4µt = 26600 Å
n = 1,
λ₂ = \(\frac{4 \mu t}{3}\) = 8866 Å
n=2,
λ₃ = \(\frac{4 \mu t}{5}\) = 5320 Å
2µt = \(\frac{5 \lambda}{2}\)
µ = \(\frac{5 \lambda}{4 t}\)

α = 1.33

Question 16.
First diffraction minimum due to a single slit of width 1.0 × 10_-5 cm is at 30°. Then wavelength of light used is,
a) 400 Å
b) 500 Å
c) 600 Å
d) 700 Å
Answer:
b) 500 Å
Solution:
d sin θ = n λ
d sin 30° = λ
1 × 10^-7 × \(\frac{1}{2}\) = λ
0.5 × 10^-7 = λ
λ = 500 Å

Question 17.
A ray of light strikes a glass plate at an angle 60°. If the reflected and refracted rays are perpendicular to each other, the refractive index of the glass is,
(a) √3
(b) \(\frac { 3 }{ 2 }\)
(c) \(\sqrt { \frac { 3 }{ 2 } } \)
(d) 2
Answer:
(a) √3
Hint.
Angle of refraction r = 60° ; Angle of incident i = 30°
sin i =n x sin r
n = \(\frac {sin 30°}{ sin 60°} \) = √3

Question 18.
One of the of Young’s double slits is covered with a glass plate as shown in figure. The position of central maximum will, Glass slide Screen will,

a) get shifted downwards
b) get shifted upwards
c) will remain the same
d) data insufficient to conclude
Answer:
b) get shifted upwards

Question 19.
Light transmitted by Nicol prism is,
a) partially polarised
b) unpolarised
c) plane polarised
d) elliptically polarised
Answer:
c) plane polarised

Question 20.
The transverse nature of light is shown in,
a) interference
b) diffraction
c) scattering
d) polarisation
Answer:
d) polarisation

II. Short Answer Questions:

Question 1.
State the laws of reflection.
Answer:
(a) The incident ray, reflected ray and normal to the reflecting surface all are coplanar (ie. lie in the same plane).
(b) The angle of incidence i is equal to the angle of reflection r. i = r

Question 2.
What is angle of deviation due to reflection?
Answer:
The angle between the incident and deviated light ray is called angle of deviation.

Question 3.
Give the characteristics of image formed by a plane mirror.
Answer:

1. The image formed by a plane mirror is virtual, erect, and laterally inverted.
2. The size of the image is equal to the size of the object.
3. The image distance far behind the mirror is equal to the object distance in front of it.
4.  If an object is placed between two plane mirrors inclined at an angle 0, then the number of images n formed is as, n = \(\left( \frac { 360 }{ \theta } -1 \right) \)

Question 4.
Derive the relation between f and R for a spherical mirror.
Answer:
1. C be the center of curvature of the mirror
2. F be the principal focus.
3. Line CM is normal to the mirror at M.
4. i be the angle of incidence
∠MCP = i and ∠MFP = 2 i

Question 5.
What are the Cartesian sign conventions for a spherical mirror?
Answer:

1. The Incident light is taken from left to right (i.e. object on the left of the mirror).
2. All the distances are measured from the pole of the mirror (pole is taken as origin).
3. The distances measured to the right of pole along the principal axis are taken as positive.
4. The distances measured to the left of pole along the principal axis are taken as negative.
5. Heights measured in the upward perpendicular direction to the principal axis are taken as positive.
6. Heights measured in the downward perpendicular direction to the principal axis, are taken as negative.

Question 6.
What is optical path? Obtain the equation for optical path of a medium of thickness d and refractive index n.
Answer:
1. Optical path of a medium is defined as the distance d’ light travels in vacuum at the same time it travels a distance ‘d’ in the medium.
2. n = refractive index.
3. d = thickness
ν = speed of light in the medium

As ‘n’ is always greater than 1, the optical path d’ of the medium is always greater than 1.

Question 7.
State the laws of refraction.
Answer:
Law of refraction is called Snell’s law.
Snell’s law states that,
(a) The incident ray, refracted ray and normal to the refracting surface are all coplanar (i.e. lie in the same plane).
(b) The ratio of angle of the incident i in the first medium to the angle of reflection r in the second medium is equal to the ratio of the refractive index of the second medium n₂ to that of the refractive index of the first medium n₁.
\(\frac { sin i }{ sin r }\) = \(\frac { { n }_{ 2 } }{ { n }_{ 1 } } \).

Question 8.
What is angle of deviation due to refraction?
Answer:

1. When light travels from rarer to denser it deviates towards normal.
   d = i – r
2. When light travels from denser to rarer it deviates away from normal.
   d = r – i

Question 9.
What is the principle of reversibility?
Answer:
The principle of reversibility states that light will follow exactly the same path if its direction of travel is reversed.

Question 10.
What is relative refractive index?
Answer:
\(\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}\) is called relative refractive index. It is the ratio of refractive index of the second medium with respect to the first medium.

Question 11.
Obtain the equation for apparent depth.
Answer:
Light from object O at the bottom passes from denser (water) to rarer medium (air).
n₁ and n₂ are refractive of denser and rarer medium.
By snells law
n₁ sin i = n₂ sin r
sin i ≈ tan i, as i and r are small
n₁ tan i ≈ n₂ tan i

1. 
   Apparent depth

Question 12.
Why do stars twinkle?
Answer:

1. Light from the stars near the horizon reaches the earth obliquely through the atmosphere.
2. It’s path changes due to refraction.
3. Frequent atmosphere disturbances changes the path of light causes twinkling of stars.

Question 13.
What is critical angle and total internal reflection?
Answer:
The angle of incidence in the denser medium for which the refracted ray graces the boundary is called critical angle i_c.
The entire light is reflected back into the denser medium itself. This phenomenon is called total internal reflection.

Question 14.
Obtain the equation for critical angle.
Answer:
The angle of incidence in the denser medium for which the refracted ray grace the boundary is called critical angle.
Snell’s law in Product form
n₁ sin i_c = n₂ sin 90°
n₁ sin i_c = n₂
sin i_c = \(\frac{n_{2}}{n_{1}}\)
here n₁ > n₂,
If the rarer medium is air
n₁ = n and n₂ = 1
sin i_c = \(\frac{1}{n}\)
i_c = sin^-1 \(\frac{1}{n}\)

critical angle and total internal reflection

Question 15.
Explain the reason for glittering of diamond.
Answer:
The diamond appears dazzling because the total internal reflection of light happens inside the diamond. The refractive index of only diamond is about 2.417. It is much larger than that for ordinary glass which is about only 1.5. The critical angle of a diamond is about 24.4°. It is much less than that of glass. A skilled diamond cutter makes use of this larger range of angle of incidence (24.4° to 90° inside the diamond), to ensure that light entering the diamond is total internally reflected from the many cut faces before getting out. This gives a sparkling effect for diamonds.

Question 16.
What are mirage and looming?
Answer:

1. It is an optical illusion observed in deserts or over hot extended surfaces like a coal – tarred road, due to which a traveller sees a shimmering Pond of water some distances ahead of him and in which the surrounding objects like trees etc appear inverted.
2. In cold regions like glaciers and frozen lakes, the reverse effect of mirage will happen hence an inverted image is formed little above the surface. This phenomenon is called looming.

Question 17.
Write a short note on the prisms making use of total internal reflection.
Answer:

1. Prisms can be designed to reflect light by 90° or 180°.
2. It can also used to invert images without changing their size.
   
   Prisms making use of total internal reflection

Question 18.
What is Snell’s window?
Answer:
When light entering the water from outside is seen from inside the water, the view is restricted to a particular angle equal to the critical angle ic. The restricted illuminated circular area is called Snell’s window.

Question 19.
Write a note on optical fibre.
Answer:

1. Principle – Total internal reflection.
2. Inner part of optical fibre – core, outer part – cladding or sleeving.
3. ‘µ’ of core must be higher than cladding.
4. Light travels inside by means of total internal reflection with no appreciable loss in intensity of light.
5. Even while bending the wire, total internal reflection ensures every reflection.

Question 20.
Explain the working of an endoscope.
Answer:
An endoscope is an instrument used by doctors which has a bundle of optical fibres that are used to see inside a patient’s body. Endoscopes work on the phenomenon of total internal reflection. The optical fibres are inserted in to the body through mouth, nose or a special hole made in the body. Even operations could be carried out with the endoscope cable which has the necessary instruments attached at their ends.

Question 21.
What are primary focus and secondary focus of convex lens?
Answer:

1. Primary focus: It is defined as a point where an object should be placed to give parallel emergent rays to the principal axis.
2. Secondary focus: It is defined as a point where all the parallel rays travelling close to the principal axis converge to form an image on the principal axis.

Question 22.
What are the sign conventions followed for lenses?
Answer:

1. Sign of focal length is not decided on the direction of measurement of the focal length from the pole of the lens as they have two focal lengths (i.e Primary & Secondary focus)
2. Focal length of thin lenses is taken as positive for a converging lens and negative for a diverging lens.

Question 23.
Arrive at lens equation from lens maker’s formula.
Answer:

1. If the refractive index of the lens is n₂ and it is placed in air, then n₂ = n and n₁ = 1.
2. \(\frac{1}{f}=(n-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\) This equation is called lens maker formula. This can be altered as
3. \(\frac{1}{\mathrm{v}}-\frac{1}{u}=\frac{1}{f}\) This equation is known as lens equation.

Question 24.
Obtain the equation for lateral magnification for thin lens.
Answer:

Lateral Magnificient in thin lens

1. OO’ be an object
2. II’ be the inverted real image.
3. Lateral Magnificient is defined as the ratio of height of the image to that of object.
4. m = \(\frac{I I^{\prime}}{O O^{\prime}}\)
∆ POO’ and ∆ PII’

m is negative for real image, Positive for virtual image.
Concave lens, m is always positive and less than one.
m for combining lenses,
\(\mathrm{m}=\frac{h_{2}}{h_{1}}=\frac{f}{f+u}=\frac{f-\mathrm{v}}{f}\)

Question 25.
What is the power of a lens?
Power is reciprocal of focal length
P = \(\frac{1}{f}\)
Unit: diopter(D)

Question 26.
Derive the equation for effective focal length for lenses in contact.
Answer:

Lenses in Contact

1. Two lenses (1) and (2) having focal length are placed coaxially in contact.
2. Lens equation for first equation be
\(\frac{1}{v^{\prime}}-\frac{1}{u}=\frac{1}{f_{1}}\)

3. Lens equation for second lens be
\(\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{v}^{\prime}}=\frac{1}{f_{2}}\)

4. Adding the above equation
\(\frac{1}{\mathrm{v}}-\frac{1}{u}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\)

5. If the combinations acts as single lens then,
\(\frac{1}{\mathrm{v}}-\frac{1}{u}=\frac{1}{f}\)

6. On comparing,
\(\frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\)

∴ For any number of lens in contact
\(\frac{1}{f_{1}}+\frac{1}{f_{2}}+\frac{1}{f_{3}}+\)………..

Question 27.
What is the angle of minimum deviation?
Answer:
The minimum value of angle of deviation is called the angle of minimum deviation D.

Question 28.
What is dispersion?
Answer:
Dispersion is splitting of white light into its constituent colours. This band of colours is called spectrum.

Question 29.
How are rainbows formed?
Answer:
Rainbow is formed by dispersion of sunlight into its constituent colours by raindrops which disperse sunlight by refraction and deviate the colours by total internal reflection.

Question 30.
What is Rayleigh’s scattering?
Answer:
Scattering of light by a particle of size less than wavelength of light is called Rayleigh scattering which is inversely proportional to the fourth power of wavelength.
\(I \alpha \frac{1}{\lambda^{4}}\)

Question 31.
Why does sky appear blue?
Answer:

1. Blue colour of sky is due to scattering of sunlight by molecules of the atmosphere (eg. N₂, O₂)
2. ‘λ’_b is smaller hence it scattered more strongly based on Rayleigh Scattering law.
3. Hence sky appears blue.

Question 32.
What is the reason for reddish appearance of sky during sunset and sunrise?
Answer:
During sunrise or sunset, the sun is near the horizon. Sunlight has to travel a greater distance. So shorter waves of blue region are scattered away by the atmosphere. Red waves of longer wavelength are least scattered and reach the observer. So the sun appears red.

Question 33.
Why do clouds appear white?
Answer:

1. A cloud is composed of large-sized partic1e which do not obey Rayleigh scattering law
2. Therefore all colours are scattered nearly equal.
3. Hence clouds appear white.

Question 34.
What are the salient features of the corpuscular theory of light?
Answer:

• According to this theory, light is emitted as tiny, massless (negligibly small mass) and perfectly elastic particles called corpuscles.
• As the corpuscles are very small, the source of light does not suffer appreciable loss of mass even if it emits light for a long time.
• On account of high speed, they are unaffected by the force of gravity and their path is a straight line in a medium of uniform refractive index.
• The energy of light is the kinetic energy of these corpuscles. When these corpuscles impinge on the retina of the eye, the vision is produced.
• The different size of the corpuscles is the reason for different colours of light.
• When the corpuscles approach a surface between two media, they are either attracted or repelled.
• The reflection of light is due to the repulsion of the corpuscles by the medium and refraction of light is due to the attraction of the corpuscles by the medium.

Question 35.
What is the wave theory of light?
Answer:

1. Light travel as longitudinal waves in an invisible ether medium.
2. This theory explains reflection, refraction, interference and diffraction.
3. Existence of ether medium was disproved later.
4. It could not explain the propagation of light in vacuum, polarisation phenomenon etc.
5. It was proposed by Huygen.

Question 36.
What is electromagnetic wave theory of light?
Answer:

1. It was proposed by maxwell.
2. It is a transverse wave which does not require any medium.
3. All phenomenon was successfully explained.
4. It could not explain photoelectric effect and crompton effect.

Question 37.
Write a short note on the quantum theory of light.
Answer:
Light interacts with matter as photons to eject the electrons
A photon is a discrete packet of energy.
E = hυ
h = plancks constant
υ = frequency of radiation

Question 38.
What is a wavefront?
Answer:
A wavefront is the locus of points which are in the same state or phase of vibration.

Question 39.
What is Huygen’s principle?
Answer:
According to Huygens principle, each point of the wavefront is the source of secondary wavelets emanating from these points spreading out in all directions with the speed of the wave. These are called as secondary wavelets.

Question 40.
What is the interference of light?
Answer:
The phenomenon of addition or superposition of two light waves which produces increase intensity at some points and decrease in intensity at some other points is called interference of light.

Question 41.
What is a phase of a wave?
Answer:
Phase is the angular position of a vibration.

Question 42.
Obtain the relation between phase difference and path difference.
Answer:
Consider two waves of wavelength λ. A phase difference of 2π corresponds to a path difference of λ. If x is the path difference between two waves and φ is the phase difference then,
\(\delta=\frac{\lambda}{2 \pi} \phi\)

\(\delta\) – path difference, φ – Phase difference

Question 43.
What are coherent sources?
Answer:
Two light sources are said to be coherent if they produce waves which have same phase or constant phase difference, same frequency or wavelength (monochromatic), same waveform and preferably same amplitude.

Question 44.
What is intensity division?
Answer:
(i) Intensity or amplitude division:
1. If we allow light to pass through a partially silvered mirror (beam splitter), both reflection and refraction take place simultaneously.
2. As the two light beams are obtained from the same light source, the two divided light beams will be coherent beams.
3. They will be either in-phase or at constant phase difference.
4. Instruments like Michelson’s interferometer, Fabray- Perrot etalon work on this principle.

Question 45.
How does wavefront division provide coherent sources?
Answer:

1. A point source produces spherical wavefronts.
2. All the points are in the same phase.
3. If two points are chosen on the wavefront by using a double slit, It will act as coherent sources.

Question 46.
How do sources and images behave as coherent sources?
Answer:
Source and image will have waves in phase.
Fresnel biprism uses two virtual sources as two coherent sources.
Llyod’s mirror uses sources and its virtual image as two coherent sources.

Question 47.
What is the bandwidth of the interference pattern?
Answer:
The bandwidth (β) is defined as the distance between any two consecutive bright or dark fringes.

Question 48.
What is diffraction?
Answer:
Bending of light around sharp edges is called diffraction.

Question 49.
Differentiate between Fresnel and Fraunhofer diffraction.
Answer:

Fresnel diffraction	Fraunhofer diffraction
1. Spherical or cylindrical wavefront undergoes diffraction	Plane wavefront undergoes diffraction
2. Lightwave is from a source at finite distance	Lightwave is from a source at infinity
3. For Laboratory conditions, convex lenses need not be used	In laboratory conditions, convex lenses are to be used
4. Difficult to observe and analyse	Easy to observe and analyse
5.

Question 50.
Discuss the special cases on the first minimum in Fraunhofer diffraction.
Answer:
1. Divide the slit AB into two halts AC and CB.
2. Two different points on the slit which are separated by the same width is corresponding points.
3. The path difference of light waves from corresponding points meet at p and interfere destructive to make first minimum.
δ = \(\begin{array}{l}
\text { a } \\
\hline 2
\end{array}\) sin θ
> The condition for p to first minimum
\(\frac{a}{2} \sin \theta=\frac{\lambda}{2}\)
a sin θ = λ (first minimum)
a sin θ = 2 λ (second minimum)
a sin θ = 3 λ (third minimum)
\(\frac{\mathrm{a} \sin \theta}{2 \mathrm{n}}=\frac{\lambda}{2}\) (nth minimum)

Question 51.
What is Fresnel’s distance? Obtain the equation for Fresnel’s distance.
Answer:
1. Fresnel’s distance is the distance up to which the ray optics is valid in terms of rectilinear propagation of light.
2. Fresnel’s distance is the distance up to which ray optics is obeyed and beyond which ray optics is not obeyed but, wave optics becomes significant.
3. From the diffraction equation for first minimum,
sin θ = \(\frac{\lambda}{a}\);
θ = \(\frac{\lambda}{a}\)
4. From the definition of Fresnel’s distance,
sin 2θ = \(\frac{a}{z}\) ;
2θ = \(\frac{a}{z}\)

5. Equating the above two equation gives,
\(\frac{\lambda}{a}=\frac{a}{2 z}\)
6. After rearranging, we get Fresnel’s distance z as,
\(z=\frac{a^{2}}{2 \lambda}\)

Question 52.
Mention the differences between interference and diffraction.
Answer:

Question 53.
What is a diffraction grating?
Answer:
A diffraction grating is an optical component with a periodic structure that splits and diffracts light into several beams travelling in different directions.

Question 54.
What are resolution and resolving power?
Answer:

1. Resolution is the quality of the image which is decided by diffraction effect and Rayleigh criterion.
2. It is measured by the smallest distance which could be seen clearly without blur due to diffraction.
3. The ability of an optical instrument to seperate or distinguish small or closely adjacent objects through the image formation is said to be resolving power.

Question 55.
What is Rayleigh’s criterion?
Answer:
The images of two-point objects are just resolved when the central maximum of the diffraction pattern of one falls over the first minimum of the diffraction pattern of the other.

Question 56.
What is polarisation?
Answer:
Polarisation is restricting electric or magnetic field vibrations to one plane.

Question 57.
Differentiate between polarised and unpolarised light.
Answer:

Question 58.
Discuss polarisation by selective absorption.
Answer:

1. Selective absorption is the property of a material which transmits waves whose electric fields vibrate in a plane parallel to a certain direction of orientation and absorbs all other waves.
2. The polaroids or polarisers are thin commercial sheets that make use of the property of selective absorption to produce an intense beam of plane polarised light.
3. Selective absorption is also called dichroism.

Question 59.
What are polariser and analyser?
Answer:
Polariser: The Polaroid which plane polarises the unpolarised light passing through it is called a polariser.
Analyser: The polaroid which is used to examine whether a beam of light is polarised or not is called an analyser.

Question 60.
What are plane polarised, unpolarised and partially polarised light?
Answer:

1. In-plane ploraised light the intensity varies from maximum to zero for every rotation of 90° of the analyser.
2. If the intensity of light varies between maximum and minimum for every rotation of 90° of the analyser, the light is said to be partially polarised light.
3. A Transverse wave which has vibrations in all directions in a plane perpendicular to the direction of propagation is said to be unpolarised light.

Question 61.
State and obtain Malus’ law.
Answer:
When a beam of plane polarised light of intensity I₀ is incident on an analyser, the light transmitted of intensity I from the analyser varies directly as the square of the cosine of the angle θ between the transmission axis of polariser and analyser. This is known as Malus’ law.
I = I₀ cos²θ
Proof:
1. Let us consider the plane of polariser and analyser are inclined to each other at an angle θ.
2. Let I₀ be the intensity and ‘a’ be the amplitude of the electric vector transmitted by the polariser.
3. The amplitude ‘a’ of the incident light has two rectangular components, (acos θ) and (a sin θ) which are the parallel and perpendicular components to the axis of transmission of the analyser.
4. Only the component (a cosθ) will be transmitted by the analyser. The intensity of light transmitted from the analyser is proportional to the square of the component of the amplitude transmitted by the analyser.
5. I ∝ (a cos θ)² I = k(a cos θ)²
Where k is constant of proportionality

Malu’s Law

I = ka² cos² θ
I = I₀ cos² θ
Where, I₀ = ka² is the maximum intensity of light transmitted form the analyser.

Special cases:
Case (i):
When θ = 0°, cos 0 = 1, I = I₀

Case (ii):
When θ = 90°, cos 90° = 0, I = 0

Question 62.
List the uses of polaroids.
Answer:

1. Polaroids are used in goggles and cameras to avoid glare of light.
2. Polaroids are useful in three-dimensional motion pictures i.e., in holography.
3. Polaroids are used to improve contrast in old oil paintings.
4. Polaroids are used in optical stress analysis.
5. Polaroids are used as window glasses to control the intensity of incoming light.

Question 63.
State Brewster’s law.
Answer:
Tangent of Polarising angle is numerically equal to the refractive index of the medium.
tan i_p = µ

Question 64.
What is the angle of polarisation and obtain the equation for the angle of polarisation
Answer:
1. The angle of incidence at which the reflected beam is plane polarised light is called polarising angle (ip)

Polarisation by reflection

2. When ordinary light is incident on the surface of a transparent medium, the reflected light is partially plane polarised.
3. The extent of polarisation depends on the angle of incidence.
4. For a particular angle of incidence, the reflected light is found to be plane polarised.
5. The angle of incidence at which a beam of unpolarised light falling on a transparent surface is reflected as a beam of plane polarised light is called polarising angle or Brewster’s angle. It is denoted by i
6. At polarising angle, the reflected transmitted rays are perpendicular to other.
i_p + 90° + r_p = 180°
r_p = 90° – i_p
From Snell’s law,
\(\frac{\sin i_{p}}{\sin r_{p}}\) = n
7. Where n is the refractive index of the medium with respect to air.
8. Substitute the value of r from Equation, we get,
\(\frac{\sin i_{p}}{\sin \left(90^{\circ}-i_{p}\right)}\)}=\(\frac{\sin i_{p}}{\sin r_{p}}\) = n
tan i_p = n
This relation is known as Brewster’s law.

Question 65.
Discuss pile of plates.
Answer:
1. The phenomenon of polarisation by reflection is used in the construction of pile of plates.
2. It consists of a number of glass plates placed one over the other in a tube
3. The plates are inclined at an angle of 33.7°(90° – 56.3°) to the axis of the tube. A beam of unpolarised light is allowed to fall on the pile of plates along the axis of the tube. So, the angle of incidence of light will be at 56.3° which is the polarising angle for glass.

Pile of Plates

4. The vibrations perpendicular to the plane of incidence are reflected at each surface and those parallel to it are transmitted.
5. The larger the number of surfaces, the greater is the intensity of the reflected plane polarised light.
6. The pile of plates is used as a polarizer and also as an analyser.

Question 66.
What is double refraction?
Answer:
When a ray of unpolarised light is incident on a calcite crystal, two refracted rays are produced. Hence, two images of a single object are formed. This phenomenon is called double refraction.

Question 67.
Mention the types of optically active crystals with example.
Answer:
Uniaxial crystals :
Crystals with only one optic axis is called uniaxial crystals Ex: calcite, quartz, tourmaline.
Biaxial crystals:
Crystals with two optic axis is called biaxial crystal. Ex: Mica, topaz, selenite.

Question 68.
Discuss Nicol prism.
Answer:
Principle:
Double Refraction

Construction:
1. One of the most common forms of the Nicol prism is made by taking a calcite crystal which is a double refracting crystal with its length three times its breadth.
2. It is cut into two halves along the diagonal so that their face angles are 72° and 108°.

Nicol Prism
3. The two halves are joined together by a layer of Canada balsam, a transparent cement.
4. A ray of unpolarized light from monochromatic source such as a sodium vapour lamp is incident on the face AC of the Nicol prism. Double refraction takes place and the ray is split into ordinary and extraordinary rays.
5. They travel with different velocities.
6. The refractive index of the crystal for the ordinary ray (monochromatic sodium light) is 1.658 and for extraordinary ray is 1.486. The refractive index of Canada balsam is 1.523.
7. Canada balsam does not polarise light.
8. The ordinary ray is total internally reflected at the layer of Canada balsam and is prevented from emerging from the other face.
9. The extraordinary ray alone is transmitted through the crystal which is plane polarised.

Uses of Nicol prism:

• It produces plane polarised light and functions as a polariser
• It can also be used to analyse the plane polarised light (i.e) used as an analyser.

Drawbacks of Nicol prism:

• Its cost is very high due to the scarcity of large and flawless calcite crystals
• Due to extraordinary ray passing obliquely through it, the emergent ray is always displaced a little to one side.
• The effective field of view is quite limited
• Light emerging out of it is not uniformly plane polarised.

Question 69.
How is polarisation of light obtained by scattering of light?
Answer:
The light from a clear blue portion of the sky shows a rise and fall of intensity when viewed through a polaroid which is rotated. This is because of sunlight, which has changed its I direction (having been scattered) on encountering the molecules of the earth’s atmosphere. The electric field of light interact with the electrons present in the air molecules.

Under the influence of the electric field of the incident wave the electrons in the molecules acquire components of motion in both these directions. We have an observer looking at 90° to the direction of the sun. Clearly, charges accelerating parallel do not radiate energy towards this observer since their acceleration has no transverse component. The radiation scattered by the molecule is therefore polarized perpendicular to the plane.

Polarisation by Scattering

Question 70.
Discuss simple microscope and obtain the equations for magnification for near point focusing and normal focusing.
Answer:
(i) A simple microscope is a single magnifying (converging) lens of small focal length.
(ii) The idea is to get an erect, magnified and virtual image of the object. For this the object is placed between F and P on one side of the lens and viewed from other side of the lens.
(1) Near point focusing:
The image is formed at near point, i.e. 25 cm for normal eye. This distance is also called as least distance D of distinct vision. In this position, the eye feels comfortable but there is little strain on the eye.
(2) Normal focusing:
The image is formed at infinity. In this position, the eye is most relaxed to view the image.

Magnification in near point focusing:
(i) Object distance u is less than f. The image distance is the near point D. The magnification m is given by the relation,
m = \(\frac{v}{u}\)
(ii) The magnification can further be written as,
m = 1 – \(\frac{v}{f}\)(using lens equation)
Substituting for ν with sign convention,
ν = -D
m = \(1+\frac{D}{f}\)
This is the magnification for near point focusing.

Magnification in normal focusing (angular magnification):
(i) The magnification for image is formed at infinity.
(ii) The angular magnification is defined as the ratio angle θ_i, subtended by the image with an aided eye to the angle θ₀ subtended by the object with unaided eye.
m = \(\frac{\theta_{i}}{\theta_{0}}\)

Normal focusing

For unaided eye
tan θ₀ ≈ θ₀ = \(\frac{h}{D}\)
For aided eye
tan θ₀ ≈ θ₀ = \(\frac{h}{f}\)

The angular magnification is,
m = \(\frac{\theta_{i}}{\theta_{0}}=\frac{h / f}{h / D}\)

m = \(\frac{D}{f}\)
This is the magnification for normal focusing.

Question 71.
What are the near point and normal focusing?
Answer:
Near Point focusing:
If the image is formed at 25 cm which is the distance of distinct vision for normal eye, then the focusing is called near point focusing.
Normal focusing:
If the image is formed at infinity then the focusing is called normal focusing.

Question 72.
Why is oil-immersed objective preferred in a microscope?
Answer:
By placing an immersion oil with a refractive index equal to that of the glass slide in the space filled with air, more light is directed through the objective and a clearer image is observed.

Question 73.
What are the advantages and disadvantages of using a reflecting telescope?
Answer:
Advantages:

• The main advantage is reflector telescope can escape from chromatic aberration because wavelength does not effect reflection.
• The primary mirror is very stable because it is located at the back of the telescope and can be support in the back.
• More cost-effective than refractor of similar size.
• Easier to make a high quality mirror than lens because mirror need to only concern with one side of the curvature.

Disadvantages:

• Optical misalignment can occur quite easily.
• Require frequent cleaning because the inside is expose to the atmosphere.
• Secondary mirror can cause diffraction of original incoming light rays causing the “Christmas star effect” where a bright object have spikes.

Question 74.
What is the use of an erecting lens in a terrestrial telescope?
Answer:
It makes the final image to be erect.

Question 75.
What is the use of collimator?
Answer:
The collimator is an arrangement to produce a parallel beam of light.

Question 76.
What are the uses of spectrometer?
Answer:

1. To study the spectra of different sources of light.
2. To measure the refractive index of materials.

Question 77.
What is myopia? What is its remedy?
Answer:
A human eye is shortsighted or myopic if it can see near objects clearly but is unable to see the far objects.
Remedy:
By using a concave lens of the required focal length.

Question 78.
What is hypermetropia? What is its remedy?
Answer:
A human eye can see distant objects clearly but cannot see near objects clearly is said to be suffering by long-sightedness or hypermetropia.
Remedy: By using concave of required focal length.

Question 79.
What is presbyopia?
Answer:
This defect is similar to hypermetropia i.e., a person having this defect cannot see nearby objects distinctly, but can see distant objects without any difficulty. This defect occurs in elderly persons (aged persons).

Question 80.
What is astigmatism?
Answer:
It is a defect of vision in which, a person cannot simultaneously see both horizontal and vertical lines with the same clarity.

III. Long Answer Questions:

Question 1.
Derive the mirror equation and the equation for lateral magnification.
Answer:
1. AB is an object considered on the principal axis of a concave mirror.
2. Consider three paraxial rays from point B on the object.
3. The three reflected rays intersect at the point ‘B’.
4. A perpendicular drawn as A’ B’ to the principal axis is the real, inverted image of the object AB.
5. ∆ BPA and ∆ B’PA are similar

Mirror equation

Question 2.
Describe Fizeau’s method to determine the speed of light.
Answer:

speed of light by Fizeau’s method

Apparatus:

• The light from a source S was first allowed to fall on a partially silvered glass plate G kept at an angle of 45°
• It is then allowed to fall on a rotating tooth wheel with N teeth and N cuts of equal widths.
• Light passing through one cut will be reflected by a mirror M kept at a long distance d (about 8 km)
• If the wheel is not rotating, it gets reflected by a partially silvered glass plate.

Working:

• Angular speed of rotation of wheel was increased from zero to GO Until it blocked by an adjacent tooth.
• It is ensured by the disappearance of light while looking partially silvered glass plate.

Expression:

Question 3.
Obtain the equation for radius of illumination (or) Snell’s window.
Answer:

1. When light entering the water from outside is seen from inside the water, the view is restricted to a particular angle equal to the critical angle i_c.
2. This restricted illuminated circular area is called Snell’s window.
3. The angle of view for water animals is restricted to twice the critical angle 2i_c. The critical angle for water is 48.6°. The angle of view is 97.2°.
4. The radius R of the circular area depends on the depth d from which it is seen and the refractive indices of the media.
5. Light is seen from a point A at a depth d.

Snell’s law
n₁ sin i_c = n₂ sin 90°
n₁ sin i_c = n₂ ∵ sin 90° = 1
sin i_c = \(=\frac{n_{2}}{n_{1}}\)
6. From the Right angle triangle ∆ ABC,
sin i_c = \(\frac{C B}{A B}=\frac{R}{\sqrt{d^{2}+R^{2}}}\)
Equating the above two equations
\(\frac{R}{\sqrt{d^{2}+R^{2}}}=\left(\frac{n_{2}}{n_{1}}\right)\)

Radius of snell’s window

Squaring on both sides,
\(\frac{R^{2}}{R^{2}+d^{2}}=\left(\frac{n_{2}}{n_{1}}\right)^{2}\)
Taking reciprocal,
\(\frac{R^{2}+d^{2}}{R^{2}}=\left(\frac{n_{1}}{n_{2}}\right)^{2}\)
On further simplifying,

Question 4.
Derive the equation for acceptance angle and numerical aperture of optical fiber.
Answer:
1. To ensure the critical angle incident in the core-cladding boundary inside the optical fibre, the light should be incident at a certain angle at the end of the optical fiber while entering into it. This angle is called an acceptance angle.

Acceptance angle

Acceptance Cone

2. By Snell’s law
n₃ sin i_a = n₁ sin r_a
To have the internal reflection inside optical fibre,
n₁ sin i₁ = n₂ sin 90°
n₁ sin i_c = n₂ sin 90° = 1
∴ sin i_c = \(\frac{n_{2}}{n_{1}}\)
3. From the right angle triangle AABC,
i_c = 90° – r_a
Now, equation becomes
sin (90° – r_a) = \(\frac{n_{2}}{n_{1}}\)

4. Using trigonometry,
cos r_a = \(\frac{n_{2}}{n_{1}}\)
sin r_a = \(\sqrt{1-\cos ^{2} r_{a}}\)
Substituting for cos r_a

5. Light can have any angle of incidence from o to i_a with the normal at the end of the optical fibre forming a conical shape called acceptance cone. The term (n₃ sin i_a) is called
numerical aperture NA of the optical fibre
NA = n₃ sin i_a = \(\sqrt{n_{1}^{2}-n_{2}^{2}}\)
6. If outer medium is air, then n₃ = 1. The numeric aperture NA becomes,
NA = sin i_a = \(\sqrt{n_{1}^{2}-n_{2}^{2}}\)

Question 5.
Obtain the equation for lateral displacement of light passing through a glass slab.
Answer:
1. When a ray of light passes through a glass slab it refracts at two refracting surfaces.
2. When the light ray enters the slab it travels from rarer medium (air) to denser medium (glass), results in deviation of ray towards the normal. When the light ray leaves the slab it travels from denser medium to rarer medium resulting in deviation of ray away from the normal.

Refraction in glass slab

3. After the two refractions, the emerging ray has the same direction as that of the incident ray on the slab with a lateral displacement or shift L.
4. Consider a glass slab of thickness and refractive index n is kept in air medium.
5. In the right angle triangle ∆ BCE,
sin (i – r) = \(\frac{L}{B C}\);
BC = \(\frac{L}{\sin (i-r)}\) ……………(1)
6. In the right angle triangle ∆ BCF, ……………….(2)
cos(r) = \(\frac{t}{B C}\);

BC = \(\frac{t}{\cos (r)}\)
Equating equations (1) & (2)
\(\frac{L}{\sin (i-r)}=\frac{t}{\cos (r)}\)

7. After rearranging,
L = \(t\left[\frac{\sin (i-r)}{\cos (r)}\right]\)

Question 6.
Derive the equation for refraction at a single spherical surface
Answer:
1. Let us consider two transparent media having refractive indices n₁ and n₂ are separated by a spherical surface.
2. Let C be the centre of curvature of the spherical surface. Let a point object O be in the medium n₁.
3. The line OC cuts the spherical surface at the pole P of the surface.

Refraction at single Spherical Surface

4. As the rays considered are paraxial rays, the perpendicular dropped for the point of incidence to the principal axis is very close to the pole or passes through the pole itself.
5. Light from O falls on the refracting surface at N. The normal drawn at the point of incidence passes through the centre of curvature C.
6. As n₂ > n₁, light in the denser medium deviates towards the normal and meets the principal axis at I where the image is formed,
7. By Snell’s law,
n₁ sin i = n₂ sin r
As the angles are small,
n₁ i = n₂ r
Let the angles,
∠ NOP = α, ∠ NCP = β ∠ NIP = γ
tan α = \(\frac{P N}{P O}\);

tan β = \(\frac{P N}{P C}\);

tan γ = \(\frac{P N}{P I}\)
8. As these angles are small, tan of angle could be approximated to the angle itself.
α = \(\frac{P N}{P O}\);

β = \(\frac{P N}{P C}\);

γ = \(\frac{P N}{P I}\)
for the triangle, ΔONC,
i = α + β
for the triangle, ΔINC,
β = r + γ (or) r = β – γ
9. on sub i & r, we get
n₁ ( α + β) = n₂ (β – γ)
Rearranging
n₁α + n₂γ = (n₂ – n₁)β

Question 7.
Obtain the lens maker’s formula and mention its significance.
Answer:

1. Let us consider a thin lens made up of a medium of refractive index n₂ is placed in a medium of refractive index n₁. Let R₁ and R₂ be the radii of curvature of two spherical surfaces (1) and (2) respectively.
2. The general equation for the refraction at a spherical surface is given from Equation
\(\frac{n_{2}}{v}-\frac{n_{1}}{u}=\frac{\left(n_{2}-n_{1}\right)}{R_{1}}\)
For the refracting surface (1) , the light goes from n₁ to n₂

3. \(\frac{n_{2}}{v}-\frac{n_{1}}{u}=\frac{\left(n_{2}-n_{1}\right)}{R_{1}}\)
For the refracting surface (2), the light goes from medium n₂ to n₁.

4. By adding
\(\frac{n_{1}}{\mathrm{v}}-\frac{n_{1}}{u}=\left(n_{2}-n_{1}\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)

Further simplifying and rearranging,

5. If the object is at infinity, the image is formed at the focus of the lens.
Thus, for u = ∞, υ = f.
Then the equation becomes,
\(\frac{1}{\mathrm{v}}-\frac{1}{u}=\left[\frac{n_{2}}{n_{1}}-1\right]\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)

6. If the refractive index of the lensis n₂ and it is placed in air, then n₂ = n and n₁ = 1. So the equation becomes,

\(\frac{1}{f}=(n-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
The above equation is called the lens maker’s formula,

7. \(\frac{1}{\mathrm{v}}-\frac{1}{u}=\frac{1}{f}\)
This is called lens equation.

Question 8.
Derive the equation for a thin lens and obtain its magnification.
Answer:
1. Let us consider an object OO’ of height h₁ placed on the principal axis with its height perpendicular to the principal axis.
2. The ray OP passing through the pole of the lens goes undefeated.
3. The inverted real image II’ formed has a height h₂
4. The lateral or transverse magnification m is defined as the ratio of the height of the image to that of the object.
m = \(\frac{I I^{\prime}}{O O^{\prime}}\)

5. From the two similar triangle ∆POO’ and PII’, we can write,
\(\frac{I I^{\prime}}{O O^{\prime}}=\frac{P I}{P O}\)

Applying sign convention,
\(\frac{-h_{2}}{h_{1}}=\frac{\mathrm{v}}{-\mathrm{u}}\)

6. Substituting this in equation for magnification,
m = \(\frac{-h_{2}}{h_{1}}=\frac{\mathrm{v}}{-u}\)

After rearranging
m = \(\frac{h_{2}}{h_{1}}=\frac{\mathrm{v}}{\mathrm{u}}\)

7. The magnification is negative for real image and positive for the virtual image.
8. Magnification by combining the lens equation with the formula for magnification as,
\(m=\frac{h_{2}}{h_{1}}=\frac{f}{f+u}\) (or)
\(m=\frac{h_{2}}{h_{1}}=\frac{f-\mathrm{v}}{f}\)

Question 9.
Derive the equation for effective focal length for lenses in out of contact.
Answer:
Focal length of lens in out of contact:
1. Let O be a point object on the principal axis of a lens. OA is the incident ray on the lens at a point A at a height h above the optical centre. The ray is deviated through an angle δ and forms the image I on the principle axis.
2. The incident and refracted rays subtend the angles. ∠ AOP = α and ∠ AIP = β with the principle axis respectively.

3. In the triangle ∆ OAI, the angle of deviation δ can be δ = α + β
If the height h is small as compared to PO and PI the angles α, β and δ are also small then,

4. For a parallel ray that falls on the arrangement, the two lenses produce deviations δ₁ and δ₂ respectively and The net deviation δ is
δ = δ₁ + δ₂

δ₁ = \(\frac{h_{1}}{f_{1}}\)

δ₂ = \(\frac{h_{2}}{f_{2}}\)

δ = \(\frac{h_{1}}{f}\) …………(1)

\(\frac{h_{1}}{f}=\frac{h_{1}}{f_{1}}+\frac{h_{2}}{f_{2}}\)

Lens in out of contact

From the geometry,
h₂ – h₁ = P₂G – P₁G = CG
h₂ – h₁ = BG tan δ₁ ≈ BG δ₁
h₂ – h₁ = \(\frac{d h_{1}}{f_{1}}\)
h₂ = h₁ + \(\frac{d h_{1}}{f_{1}}\) ……………..(2)
\(\frac{h_{1}}{f}=\frac{h_{1}}{f_{1}}+\frac{h_{1}}{f_{2}}+\frac{h_{1} d}{f_{1} f_{2}}\)
On further simplification,

\(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}+\frac{d}{f_{1} f_{2}}\)
The above equation could be used to find the equivalent focal length.
To find the position of the equivalent lens, we can further write from the geometry.
PP₂ = EG = \(\frac{\mathrm{GC}}{\tan \delta}\)
PP₂ = EG = \(\frac{h_{1}-h_{2}}{\tan \delta}=\frac{h_{1}-h_{2}}{\delta}\)
From equation (1) and (2)
h₂ – h₁ = d\(\frac{\mathrm{h}_{1}}{\mathrm{f}_{1}}\) and
δ = \(\frac{\mathrm{h}_{1}}{\mathrm{f}}\)
PP₂ = [d \(\frac{f}{f_{1}}\)]
The position of the equivalent single lens from the second lens. Its position from the first lens.
PP₁ = d – [d \(\frac{f}{f_{1}}\)];
PP₁ = d[1 – \(\frac{f}{f_{1}}\)]

Question 10.
Derive the equation for angle of deviation produced by a prism and thus obtain the equation for the refractive index of material prism.
Answer:
1. The angle between the direction of the incident ray PQ and the emergent ray RS is called the angle of deviation d.
2. The deviation d₁ at the surface AB is,
angle ∠RQM = d₁ = i₁ – r₁
3. The deviation d₂ at the surface AC is, angle ∠QRM =
d₂ = i₂ – r₂

4. Total angle of deviation d produced is
d = d₁ + d₂
Substituting for d₁ and d₂,
d = (i₁ – r₁) + (i₂ – r₂)
After rearranging,
5. d = (i₁ + i₂) – (r₁ + r₂)
In the quadrilateral AQNR, two of the angles (at the vertices Q and R) are right angles. Therefore, the sum of the other angles of the quadrilateral is 180°.
6. ∠A + ∠QNR = 180°
From the triangle ∆ QNR,
r₁ + r₂ + ∠QNR = 180°
Comparing these two equations and we get,
r₁ + r₂ = A
substituting this in equation (6.98) for angle of deviation,
d = i₁ + i₂ – A

7. A graph plotted between the angle of incidence and angle of deviation.
8. The minimum value of angle of deviation is called the angle of minimum deviation D. At minimum deviation,
(a) the angle of incidence is equal to the angle of emergence, i₁ = i₂
(b) the angle of refraction at face one and face two are equal, r₁ = r₂

(c) the incident ray and emergent ray are symmetrical with respect to the prism.
(d) the refracted ray inside the prism is parallel to its base of the prism.
The case of the angle of minimum deviation is shown in figure
9. Refractive index of the material of the prism
At minimum deviation, i₁ = i₂ = i and r₁ = r₂ = r
D = i₁ + i₂ – A = 2i – A (or)
i = \(\frac{(\mathrm{A}+\mathrm{D})}{2}\)
r₁ + r₂ = A ⇒ 2r = A (or)
r = \(\frac{\mathrm{A}}{2}\)
n = \(\frac{\sin i}{\sin r}\)

n = \(\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)

Question 11.
What is dispersion? Obtain the equation for dispersive power of a medium.
Answer:
1. Dispersion is splitting of white light into its constituent colours. This band of colours of light is called its spectrum.
2. Consider a beam of white light passes through a prism; it gets dispersed into its constituent colours.
3. Let δ_V, δ_R are the angles of deviation for violet and red light. Let n_v and n_r are the refractive indices for the violet and red light respectively.

4. The refractive index of the material of a prism.
n = \(\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)
Here A is the angle of the prism and D is the angle of minimum deviation.
n = \(\frac{\sin \left(\frac{A+\delta}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)
For small angles of A and δ_m,

δ = (n – 1) A
Whn white light enters the prism, the deviation is different for different colours. Thus, the refractive index is also different for different colours.
For violet light, δ =(n_v – 1) A
For Red light, δ_R =(n_R – 1) A
5. As, angle of deviation for violet colour δ_v is greater than the angle of deviation for red colour, δ_v the refractive index for violet colour n_v is greater than the refractive index for red colour n_R.
Subtracting δ_v from δ_R we get,
δ_v – δ_R = (n_v – n_R) A
6. The term (δ_v – δ_R) is the angular separation between the two extreme colours (violet and red) in the spectrum is called angular dispersion. Clearly, the angular dispersion produced by a prism depends upon.
(i) Angle of the prism
(ii) Nature of the material of the prism.
7. It is defined as the ratio of the angular dispersion for the extreme colours to the deviation for any mean colour.
Dispersive power (ω),
ω = \(\frac{\text { angular dispersion }}{\text { mean deviation }}=\frac{\delta_{\mathrm{v}}-\delta_{\mathrm{R}}}{\delta}\)
Substituting for (δ_v – δ_R) and (δ),
ω = \(\frac{\left(n_{\mathrm{v}}-n_{R}\right)}{(n-1)}\)

Question 12.
Prove laws of reflection using Huygen’s principle.
Answer:

Laws of Reflection

1. Let us consider a parallel beam of light, incident on a reflecting plane surface such as a plane mirror XY.
2. The incident wavefront is A’B’ and the reflected wavefront is AB in the same medium.
3. These wavefronts are perpendicular to the incident rays L, M and reflected rays L ‘, M’ respectively.
4. By the time point A of the incident wavefront touches the reflecting surface, the point B is yet to travel a distance BB’ to touch the reflecting surface B’.
5. When point B falls on the reflecting surface at B’, the point A would have reached A’. This is applicable to all the points on the wavefront.
6. Thus, the reflected wavefront A’B’ emanates as a plane wavefront. The two normals N and N’ are considered at the points where the rays L and M fall on the reflecting surface. As reflection happens in the same medium, the speed of light is same before and after the reflection.
7. Hence, the time taken for the ray to travel from B to B’ is the same as the time taken for the ray to travel from A to A’. Thus, the distance BB’ is equal to the distance AA’ (AA’ = BB’)
(i) The incident rays, the reflected rays the normal are in the same plane.
(ii) Angle of incidence,
∠ i = ∠ NAL = 90° – ∠ NAB = ∠ BAB’
Angle of reflection,
∠ r =∠ N’B’M’ = 90° – ∠ N’B’A’=∠ A’B’A
8. For the two right angle triangles, ∆ ABB’ and ∆ B’A’A, the two triangles are congruent. As per the property of congruency; the two angles, ∠ BAB’ and ∠ A’B’A must also be equal.
i = r
hence the laws of reflection is proved.

Question 13.
Prove laws of refraction using Huygen’s principle.
Answer:

Law of refraction

1. Let us consider a parallel beam of light is incident on a refracting plane surface XY such as a glass surface.
2. The incident wavefront AB is in rarer medium (1) and the refracted wavefront ” AB is in denser medium (2).
3. These wavefronts are perpendicular to the incident rays L, M and refracted rays L’, M’ respectively.
4. By the time the point A of the incident wavefront touches the refracting surface, the point B is yet to travel a distance BB’ to touch the refracting surface at B’.
5. When point B falls on the refracting surface at, B, the point A would have reached A’ in the other medium.
6. The time taken t for the ray to travel from B to B is the same as the time taken for the ray to travel from A to A’
t = \(\frac{B B^{\prime}}{\mathrm{v}_{1}}=\frac{A A^{\prime}}{\mathrm{v}_{2}}\) (or)
t = \(\frac{B B^{\prime}}{A A^{\prime}}=\frac{v_{1}}{v_{2}}\)

(i) The incident rays, the refracted rays and the normal are in the same plane.
(ii) Angle of incidence,
i = ∠ NAL = 90° – ∠NAB = ∠ BAB’
Angle of refraction,
r = ∠N’B’M’= 90° – ∠N’B’A’ = ∠ A’B’A
7. For the two right angle triangle ABB’ and B’A’A,
\(\frac{\sin i}{\sin r}=\frac{B B^{\prime} / A B^{\prime}}{A A^{\prime} / A B^{\prime}}=\frac{B B^{\prime}}{A A^{\prime}}=\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{c / \mathrm{v}_{2}}{c / \mathrm{v}_{1}}\)

8. Here, c is speed of light in vacuum. The ratio c/υ is the constant, called refractive index of the medium.
\(\frac{\sin i}{\sin r}=\frac{n_{2}}{n_{1}}\)
In product form,
n₁ sin i = n₂ sin r
Hence, the laws of refraction are proved.

Question 14.
Obtain the equation for resultant intensity due to interference of light.
Answer:
1. The phenomenon of addition or superposition of two light waves which produces increase in intensity at some points and a decrease in intensity at some other points is called interference of light.
2. Let us consider two light waves from the two sources S₁ and S₂ meeting at a point P as shown
3. The wave from S₁ at an instant t and P is, y₁ = a₁ sin ωt
The wave from S₂ an instant t at P is
y₂ = a₂ sin(ωt + Φ)

Superposition principle

4. The two waves have different amplitudes a₁ and a₂, same angular frequency ω’ and a phase difference of Φ between them. The resultant displacement will be given by.
y = y₁ + y₂ = a₁ sin ωt + a₁ sin₂ (ωt + Φ) y = A sin (ωt + Φ)
Where, A =\(\sqrt{a_{1}^{2}+a_{2}^{2}+2 a_{1} a_{2} \cos \phi}\) ………….(1)
θ = \(\tan ^{-1} \frac{a_{2} \sin \phi}{a_{1}+a_{2} \cos \phi}\) ………….(2)

5. The resultant amplitude is maximum.
A_max = \(\sqrt{\left(a_{1}+a_{2}\right)^{2}}\);
when Φ = 0, ±2π, ± 4π …………….(3)
6. The resultant amplitude is minimum.
A_min = \(\sqrt{\left(a_{1}-a_{2}\right)^{2}}\);
when Φ = 0, ±π, ± 3π ± 5π ………..(4)
7. The intensity of light is proportional to the square of amplitude.
8. I α A² ……………(5)
Now equation (1) becomes
I α I₁ + I₂ + 2\(\sqrt{I_{1} I_{2}}\) cos Φ ………….(6)
9. If the phase difference, Φ = 0, ± 2π, ± 4π., it corresponds to the condition for maximum intensity of light called as constructive interference.
10. The resultant maximum intensity is,
I_max α (a₁ + a₂)² ……………….(7)
11. If the phase difference, Φ = + π, ± 3π, ± 5π …., it corresponds to the condition for the minimum intensity of light called destructive interference.
12. The resultant minimum intensity is I_min α
(a₁ – a₂)² α I₁ + I₂ – 2\(\sqrt{\mathrm{I}_{1} \mathrm{I}_{2}}\) ………….(8)

As a special case, if a₁ = a₂ = a, then equation (1) becomes,
A = \(\sqrt{2 a^{2}+2 a^{2} \cos \phi}\)
= \(\sqrt{2 a^{2}(1+\cos \phi)}\)
= \(\sqrt{2 a^{2} 2 \cos ^{2}(\phi / 2)}\)

13. A = 2 a cos(Φ/2) ……….(9)

I α 4a² cos² (Φ/2) [∴ I α A²] …………(10)
I α 4 I₀ cos² (Φ/ 2) [ΦI₀ α a²] ……….(11)
I_Max = 4I₀ when, Φ = o, ± 2π, 4π ……………(12)
I_min = 0 when, Φ = ± π, ± 3π, ± 5π ……………..(13)
Conclusion:
The phase difference between the two waves decides the intensity of light meet at a point.

Question 15.
Explain Young’s double-slit experimental setup and obtain the equation for path difference.
Answer:

Young’s double slit experiment

1. Wavefronts from S₁ and S₂ spread out and overlapping takes place to the right side of double slit.
2. When a screen is placed at a distance of about 1 meter from the slits, alternate bright and dark fringes which are equally spaced appear on the screen. These are called interference fringes or bands.

Equation for path difference:
3. The Let d be the distance between the double slits S₁ and S₂ which act as coherent sources of wavelength λ.
4. A screen is placed parallel to the double slit at a distance D from it. The mid-point of S₁ and S₂ is C and the mid-point of the screen O is equidistant from S₁ and S₂. P is any point at a distance y from O.
5. The waves from S₁ and S₂ meet at P either in-phase or out-of-phase depending upon the path difference between the two waves.

Young’s double slit experiment setup.

δ = S₂P – S₁P
δ = S₂P – MP = S₂M

6. The angular position of the point P from C is θ. ∠ OCP = θ.
7. From the geometry, the angles ∠ OCP and ∠ S₂S₁M are equal.
∠ OCP = ∠ S₂S₁M = θ.
In right angle triangle ∆ S₁S₂M, the path difference,
S₂M = d sin θ
δ = d sin θ
If the angle θ is small, sin θ »tan θ θ
From the right angle triangle ∆ OCP,
tan θ = \(\frac{\mathrm{y}}{\mathrm{D}}\)

The path difference, δ = \(\frac{\mathrm{d} \mathrm{y}}{\mathrm{D}}\)

Question 16.
Obtain the equation for bandwidth in Young’s double slit experiment.
Answer:
Condition for bright fringe (or) maxima:
The condition for the constructive interference or the point P to be have a bright fringe is,
Path difference, δ = nλ
where, n = 0,1,2,…….. ∴ \(\frac{\mathrm{d} \mathrm{y}}{\mathrm{D}}\) = nl
y = n\(\frac{\lambda D}{d}\) (or)
y_n = n\(\frac{\lambda D}{d}\)

Condition for dark fringe (or) minima:
The condition for the destructive interference or the point P to be have a dark fringe is,
Path difference δ = (2n -1) \(\frac{\lambda}{2}\)
when, n = 1, 2, 3,………..
∴ \(\frac{d y}{D}\) =(2n-l) \(\frac{\lambda}{2}\)
y = \(\frac{(2 n-1)}{2} \frac{\lambda D}{d}\) (or)
y_n = \(\frac{(2 n-1)}{2} \frac{\lambda D}{d}\)

Equation for bandwidth:
1. The bandwidth (β) is defined as the distance between any two consecutive bright or dark fringes.
2. The distance between (n+l)th and nth consecutive brigh fringes from O is given by,
β = y_{(n + 1)} – y_n ((n + 1) \(\frac{\lambda D}{d}\)) – (n\(\frac{\lambda D}{d}\))
β = \(\frac{\lambda D}{d}\)

Question 17.
Obtain the equations for constructive and destructive interference for transmitted and reflected waves in thin films.
Answer:
Interference in thin films:
1. Let us consider a thin film of transparent material of refractive index p and thickness d. A parallel beam of light is incident on the film at an angle i.
2. The wave is divided into two parts at the upper surface, one is reflected and the other is refracted. The refracted part, which enters into the film, again gets divided at the lower surface into two parts; one is transmitted out of the film and the other is reflected back into the film.

Interference in thin films

For transmitted light:
1. The light transmitted may interfere to produce a resultant intensity. Let us consider the path difference between the two light waves transmitted from B and D. The two waves moved together and remained in phase up to B where splitting occurred.
2. The extra path travelled by the wave transmitted from D is the path inside the film, BC + CD. If we approximate the incidence to be nearly normal (i = 0), then the points B and D are very close to each other. The extra distance travelled by the wave is approximately twice thickness of the film, BC + CD = 2d. As this extra path is travelled in a medium of refractive index p, the optical path difference is,
δ = 2μd.
3. The condition for constructive interference in transmitted ray is,
2μd = nλ
Similarly, the condition for destructive interference in transmitted ray is
2μd = (2n-l) \(\frac{\lambda}{2}\)

For reflected light:
1. Wave while travelling in a rarer medium and getting reflected by a denser medium, undergoes a phase change of n or an additional path difference of \(\frac{\lambda}{2}\).
2. Let us consider the path difference between the light waves reflected by the upper surface at A and the other wave coming out at C after passing through the film.
3. The additional path travelled by wave coming out from C is the path inside the film, AB + BC. For nearly normal incidence this distance could be approximated as, AB + BC = 2d. As this extra path is travelled in the medium of refractive index p, the optical path difference is, δ = 2μd.
4. The condition for constructive interference for reflected ray is,
2μd + \(\frac{\lambda}{2}\) = nλ (or)
2μd = (2n – 1) \(\frac{\lambda}{2}\)

5. The additional path difference λ/2 is due to the phase change of n in rarer to denser reflection taking place at A. The condition for destructive interference for reflected ray is,
2μd + \(\frac{\lambda}{2}\) = (2n + l)\(\frac{\lambda}{2}\) (or)
2μd = nλ

Question 18.
Discuss diffraction at single slit and obtain the condition for nth minimum.
Answer:
1. Let a parallel beam of light fall normally on a single slit AB of width. The diffracted beam falls on a screen kept at a distance. The center of the slit is C. A straight line through C perpendicular to the plane of slit meets the center of the screen at O.
2. All the waves start parallel to each other from different points of the slit and interfere at point P and other points to give the resultant intensities. The point P is in the geometrically shadowed region, up to which the central maximum is spread due to diffraction.

Condition for P to be first minimum:
1. Let us divide the slit AB into two half’s AC and CB. Now the width of AC is (a/2). They are called corresponding points.
2. The path difference of light waves from different corresponding points meeting at point P and interfere destructively to make it first minimum.
3. The path difference δ between waves from these corresponding points is,
δ = \(\frac{a}{2}\) sin θ
The Condition for P to be first minimum,
\(\frac{a}{2}\) sin θ = \(\frac{\lambda}{2}\)
a sin θ = λ (first minimum)

Condition for P to be n^th order minimum:
Dividing the slit into 2n number of (even number of) equal parts the condition for n^th order minimum is
\(\frac{a}{2 n}\) sin θ = \(\frac{\lambda}{2}\)
a sin θ = nλ (n^th minimum)

Diffraction at single slit

Corresponding points

Question 19.
Discuss the diffraction at a grating and obtain the condition for m^th maximum.
Answer:
1. Grating has multiple slits with equal widths of size comparable to the wavelength of diffracting light.
2. Grating is a plane sheet of transparent material on which opaque rulings are made with a fine diamond pointer.
3. The rulings act as obstacles having a definite width b and the transparent space between the rulings act as slit of width a.
4. The combined width of a ruling and a slit is called grating element (e = a + b). Points on successive slits separated by a distance equal to the grating element are called corresponding points.

Diffraction grating experiment

5. Let a plane wavefront of monochromatic light with wave length λ be incident normally on the grating.
6. As the slits size is comparable to that of wavelength, the incident light diffracts at the grating.
7. A diffraction pattern is obtained on the screen when the diffracted waves are focused on a screen using a convex lens.
8. The path difference δ between the diffracted waves from one pair of corresponding points is,
δ = (a + b) sin θ
This path difference is the same for any pair of corresponding points. The point P will be bright, when
δ = mλ where m = 0,1,2,3
Combining the above two equations, we get,
(a + b) sin θ = mλ
Here, m is called an order of diffraction.
Condition for zero order maximum, m = 0
For (a + b) sinθ = 0, the position, θ = 0. sin θ = 0 and m = 0. This is called zero order diffraction or central maximum.
Condition for first-order maximum, m = 1
If (a + b) sin θ₁ = λ, the diffracted light meet at an angle θ₁ to the incident direction and the first-order maximum is obtained.

Condition for second-order maximum, m = 2
Similarly, (a + b) sin θ₂ = 2λ forms the second-order maximum at the angular position θ₂.
Condition for higher-order maximum
On either side of central maxima, different higher orders of diffraction maxima are formed at different angular positions.
If we take,
N = \(\frac{1}{a+b}\)
\(\frac{1}{\mathrm{~N}}\) sin θ = mλ (or) sin θ = Nmλ

Question 20.
Discuss the experiment to determine the wavelength of monochromatic light using diffraction grating.
Answer:
1. The wavelength of a spectral line can be very accurately determined with the help of a diffraction grating and a spectrometer.
2. Initially all the preliminary adjustments of the spectrometer are made.
3. The slit of collimator is illuminated by a monochromatic light, whose wavelength is to be determined.
4. The telescope is brought in line with a collimator to view the image of the slit.
5. The given plane transmission grating is then mounted on the prism table with its plane perpendicular to the incident beam of light coming from the collimator
6. The telescope is turned to one side until the first-order diffraction image of the slit coincides with the vertical cross wire of the eyepiece.
7. The reading of the position of the telescope is noted.
8. Similarly the first-order diffraction image on the other side is made to coincide with the vertical cross wire and corresponding reading is noted. The difference between the two positions gives 2θ.
9. Half of its value gives θ, the diffraction angle for first-order maximum. The wavelength of light
λ = \(\frac{\sin \theta}{\mathrm{Nm}}\)
10. Here, N is the number of rulings per metre in the grating and m is the order of the diffraction image.

Determination of wavelength using grating and spectrometer

Question 21.
Discuss the experiment to determine the wavelength of different colours using diffraction grating.
Answer:
1. When white light is used, the diffraction pattern consists of a white central maximum and on both sides, continuous coloured diffraction patterns are formed.
2. The central maximum is white as all the colours meet here constructively with no path difference. As θ increases, the path difference, (a+b)sin θ, passes through the condition for maxima of diffraction of different orders for all colours from violet to red.
3. It produces a spectrum of diffraction pattern from violet to red on either side of central maximum
4. By measuring the angle at which these colours appear for various orders of diffraction, the wavelength of different colours could be calculated using the formula,
\(\lambda=\frac{\sin \theta}{\mathrm{Nm}}\)
Here, N is the number of rulings per metre in the grating and m is the order of the diffraction image.

Diffraction with white light

Question 22.
Obtain the equation for resolving power of optical instrument.
Answer:
1. For a single rectangular slit, the half-angle θ subtended by the spread of central maximum is given by the relation,
a sin θ = λ
2. Similar to a rectangular slit, when a circular aperture or opening forms an image of a point object, the image formed will not be a point but a diffraction pattern of concentric circles that become fainter while moving away from the centre. These are known as Airy’s discs. The circle of central maximum has the half angular spread given by the equation,
a sin θ = 1.22 λ
3. The numerical value 1.22 comes for central maximum formed by circular apertures.

Airy’s disc

For small angles, sin θ ≈ θ
a θ =1.22 λ
Rewriting further,
θ = 1.22 λ/a  and
\(\frac{\mathrm{r}_{0}}{\mathrm{f}}=\frac{1.22 \lambda}{\mathrm{a}}\)
r₀ = \(\frac{1.22 \lambda f}{\mathrm{a}}\)

4. This equation is called spacial resolution.
5. When two-point sources close to each another form image on the screen, the diffraction pattern of one point source can overlap with another and produce a blurred image. To obtain a good image of the two sources, the two-point sources must be resolved i.e., the point sources must be imaged in such a way that their images are sufficiently far apart that their diffraction patterns do not overlap.
6. According to Rayleigh’s criterion, for two-point objects to be just resolved, the minimum distance between their diffraction images must be in such a way that the central maximum of one coincides with the first minimum of the other and vice versa. Such an image is said to be just resolved image of the object. The Rayleigh’s criterion is said to be limit of resolution.

7. According to Rayleigh’s criterion the two-point sources are said to be just resolved when the distance between the two maxima is at least r₀. The angular resolution has a unit in radian (rad) and it is given by the equation,
θ = \(\frac{1.22 \lambda}{\mathrm{a}}\)
8. The ability of an optical instrument to separate or distinguish small or closely adjacent objects through the image formation is said to be resolving power of the instrument.

Question 23.
Discuss about simple microscope and obtain the equations for magnification for near point focusing and normal focusing.
Answer:
1. A simple microscope is a single magnifying (converging) lens of small focal length The idea is to get an erect, magnified and virtual image of the object.
2. (1) Near point focusing – The image is formed at near point, i.e. 25 cm for normal eye. This distance is also called as least distance D of distinct vision. In this position, the eye feels comfortable but there is little strain on the eye.
3. (2) Normal focusing – The image is formed at infinity. In this position the eye is most relaxed to view the image.
> Object distance u is less than f. The image distance is the near point D. The magnification m is given by the relation,
m = \(\frac{\mathrm{v}}{\mathrm{u}}\)
4. With the help of lens equation,
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) the magnification can further be written as,
m = 1 – \(\frac{v}{f}\)
5. Substituting for v with sign convention,
v = -D
6. m = 1 + \(\frac{\mathrm{D}}{f}\)
This is the magnification for near point focusing.

Near Point focusing

7. The angular magnification is defined as the ratio of angle θ_i subtended by the image with aided eye to the angle θ₀ subtended by the object with unaided eye.
m = \(\frac{\theta_{i}}{\theta_{0}}\)

Normal focusing

8. For unaided eye shown in Figure (a),
tan θ₀ ≈ θ₀ = \(\frac{\mathrm{h}}{\mathrm{D}}\)
9. For aided eye shown in Figure (b),
tan θ_i ≈ θ_i = \(\frac{\mathrm{h}}{f}\)
10. The angular magnification is,
m = \(\frac{\theta_{i}}{\theta_{0}}=\frac{\mathrm{h} / \mathrm{f}}{\mathrm{h} / \mathrm{D}}\)
m = \(\frac{\mathrm{D}}{f}\)
This is the magnification for normal focusing.

Question 24.
Explain about compound microscope and obtain the equation for magnification.
Answer:
1. The lens near the object, called the objective, forms a real, inverted, magnified image of the object. This serves as the object for the second lens which is the eyepiece.
2. Eyepiece serves as a simple microscope that produces finally an enlarged and virtual image.
3. The first inverted image formed by the objective is to be adjusted close to, but within the focal plane of the eyepiece so that the final image is formed nearly at infinity or at the near point.
4. The final image is inverted with respect to the original object. We can obtain the magnification for a compound microscope.

Compound microscope
5. From the ray diagram,
m₀ = \(\frac{\mathrm{h}^{1}}{\mathrm{~h}}\)
tan β = \(\frac{\mathrm{h}}{\mathrm{f}_{0}}=\frac{\mathrm{h}^{1}}{\mathrm{~L}}\) then
\(\frac{\mathrm{h}^{\prime}}{\mathrm{h}}=\frac{\mathrm{L}}{\mathrm{f}_{\mathrm{o}}}\)
m₀ = \(\frac{\mathrm{L}}{f_{0}}\)
Jo
6. Here, the distance L is between the first focal point of the eyepiece to the second focal point of the objective. This is called the tube length L of the microscope as f₀ and f_e are comparatively smaller than L.
7. If the final image is formed at P (near point focusing), the magnification me of the eyepiece is, D
m_e = 1 + \(\frac{\mathrm{D}}{f_{\mathrm{e}}}\)
8. The total magnification m in near point focusing is,
m = m₀ m_e = \(\left(\frac{L}{f_{0}}\right)\left(1+\frac{D}{f_{e}}\right)\)

9. If the final image is formed at infinity (normal focusing), the magnification m_e of the eyepiece is,
m_e = \(\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\)
10. The total magnification m in normal focusing is,
m = m₀ m_e = \(\left(\frac{\mathrm{L}}{f_{0}}\right)\left(\frac{\mathrm{D}}{f_{\mathrm{e}}}\right)\)

Question 25.
Obtain the equation for resolving power of microscope.
Answer:
1. A microscope is used to see the details of the object under observation. The ability of microscope depends not only in magnifying the object but also in resolving two points on the object separated by a small distance d_min. Smaller the value of d_min better will be the resolving power of the microscope.

r₀ = \(\frac{1.22 \lambda v}{a}\)
2. then the magnification m is,
m = \(\frac{\mathrm{r}_{0}}{\mathrm{~d}_{\min }}\)
d_min = \(\frac{r_{0}}{m}=\frac{1.22 \lambda v}{a m}=\frac{1.22 \lambda v}{a(v / u)}=\frac{1.22 \lambda u}{a}\)
[∴ m = v/u]
d_min = \(\frac{1.22 \lambda f}{a}\) [∴ u ≈f]
On the other side,
2 tan β ≈ 2 sin β = \(\frac{a}{f}\) ∴ [a = f 2sin β]
d_min = \(\frac{1.22 \lambda}{2 \sin \beta}\)
3. To further reduce the value of d_min the optical path of the light is increased by immersing the objective of the microscope into a bath containing oil of refractive index n.
d_min = \(\frac{1.22 \lambda}{2 n \sin \beta}\)
4. Such an objective is called oil immersed objective. The term n sin β is called numerical aperture NA.
d _min = \(\frac{1.22 \lambda}{2(N A)}\)

Question 26.
Discuss about astronomical telescope.
Answer:
1. An astronomical telescope is used to get the magnification of distant astronomical objects like stars, planets, moon etc. the image formed by the astronomical telescope will be inverted.
2. It has an objective of long focal length and a much larger aperture than the eyepiece.
3. Light from a distant object enters the objective and a real image is formed in the tube at its second focal point.
4. The eyepiece magnifies this image producing a final inverted image.

Astronomical Telescope

5. The magnification m is the ratio of the angle β subtended at the eye by the final image to the angle ∞ which the object subtends at the lens or the eye.
m = \(\frac{\beta}{\alpha}\)
From the diagram,
m = \(\frac{h / f_{e}}{h / f_{0}}\)
m = \(\frac{f_{0}}{f_{e}}\)
The length of the telescope is approximately, L = f₀ +f_e

Question 27.
Mention different parts of the spectrometer and explain the preliminary adjustments.
Answer:
The spectrometer is an optical instrument used to study the spectra of different sources of light and to measure the refractive indices of materials. It consists of basically three parts. They are
(i) collimator,
(ii) prism table and
(iii) Telescope.

(i) Collimator:
1. The collimator is an arrangement to produce a parallel beam of light. It consists of a long cylindrical tube with a convex lens at the inner end and a vertical slit at the outer end of the tube.
2. The distance between the slit and the lens can be adjusted such that the slit is at the focus of the lens. The slit is kept facing the source of light.
3. The width of the slit can be adjusted. The collimator is rigidly fixed to the base of the instrument.

(ii) Prism table:
1. The prism table is used for mounting the prism, grating etc. It consists of two circular metal discs provided with three levelling screws.
2. It can be rotated about a vertical axis passing through its centre and its position can be read with verniers V₁ and V₂. The prism table can be raised or lowered and can be fixed at any desired height.

(iii) Telescope:

1. The telescope is an astronomical type. It consists of an eyepiece provided with cross wires at one end and an objective lens at its other end.
2. The distance between the objective lens and the eyepiece can be adjusted so that the telescope forms a clear image at the cross wires when a parallel beam from the collimator is incident on it.
3. The telescope is attached to an arm which is capable of rotation about the same vertical axis as the prism table. A circular scale graduated in a half degree is attached to it.
4. Both the telescope and prism table is provided with radial screws for fixing them in a desired position and tangential screws for fine adjustments.

Adjustments of the spectrometer:
(a) Adjustment of the eyepiece:
The telescope is turned towards an illuminated surface and the eyepiece is moved to and fro until the cross wires are clearly seen.
(b) Adjustment of the telescope:
The telescope is adjusted to receive parallel rays by turning it towards a distant object and adjusting the distance between the objective lens and the eyepiece to get a clear image on the cross wire.
(c) Adjustment of the collimator:
The telescope is brought along the axial line with the collimator. The slit of the collimator is illuminated by a source of light. The distance between the slit and the lens of the collimator is adjusted until a clear image of the slit is seen at the cross wire of the telescope. Since the telescope is already adjusted for parallel rays, a well- defined image of the slit can be formed, only when the light rays emerging from the collimator are parallel.
(d) Levelling the prism table:
The prism table is adjusted or levelled to be in a horizontal position by means of levelling screws and a spirit level.

Question 28.
Explain the experimental determination of material of the prism using a spectrometer.
Answer:

Angle of prism

1. The prism is placed on the prism table with its refracting edge facing the collimator.
2. The slit is illuminated by a sodium light (monochromatic light). The parallel rays coming from the collimator fall on the two faces AB and AC.
3. The telescope is rotated to the position T₁ until the image of the slit formed by the reflection at the face AB is made to coincide with the vertical cross wire of the telescope.
4. The readings of the verniers are noted. The telescope is then rotated to the position T₂ where the image of the slit formed by the reflection at the face AC coincides with the vertical cross wire. The readings are again noted.
5. The difference between these two readings gives the angle rotated by the telescope, which is twice the angle of the prism. Half of this value gives the angle of the prism A.

(ii) Angle of minimum deviation (D):
1. The prism is placed on the prism table so that the light from the collimator falls on a refracting face, and the refracted image is observed through the telescope. The prism table is now rotated so that the angle of deviation decreases.
2. A stage comes when the image stops for a moment and if we rotate the prism table further in the same direction, the image is seen to recede and the angle of deviation increases.
3. The vertical cross wire of the telescope is made to coincide with the image of the slit where it turns back. This gives the minimum deviation position.

Angle of Minimum deviation:

Angle of minimum deviation.

1. The readings of the verniers are noted. Now, the prism is removed, and the telescope is turned to receive the direct ray and the vertical cross wire is made to coincide with the image. The readings of the verniers are noted.
2. The difference between the two readings gives the angle of minimum deviation D. The refractive index of the material of the prism n is calculated using the formula,
n = \(\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)

IV. Conceptual Questions:

Question 1.
Why are dish antennas curved?
Answer:
In order to receive parallel signal rays coming from same direction to fall on a single point; its focus where the receiver is placed.

Question 2.
What type of lens is formed by a bubble inside water?
Answer:
Air bubble has spherical surface and is surrounded by medium (water) of higher refractive index. When light passes from water to air, it gets diverged. So it behaves as a concave lens.

Question 3.
Is it possible for two lenses to produce zero power? Explain.
Answer:

1. Yes, possible
2. If one lens is converging lens with some focal length so that its power is P₁ = P.
3. The other lens is a diverging lens with the same focal length with a negative sign, so that its power is P₂ = -P.
4. The combination of these two lenses is given by
   P_total = P₁ + P₂ = P + (-P)
   P = 0

Question 4.
Why does the sky look blue and clouds look white?
Answer:

1. Blue colour of the sky is due to scattering of sunlight by air molecules. According to Rayleigh’s law, I α \(\frac{1}{\lambda^{4}}\) So blue light (λ_min) is scattered much more.
2. Clouds have large particles like dust and water droplets which scatter light of all colours almost equally. Hence clouds generally appear white.

Question 5.
Why is yellow light preferred to during fog?
Answer:
Yellow light has longer wavelength than green, blue or violet components of white light. According to Rayleigh scattering law yellow colour is least scattered and produces sufficient illumination.

Question 6.
Two independent monochromatic sources cannot act as coherent sources. Why?
Answer:

1. Two independent monochromatic sources emit waves of the same wavelength. But the waves are not in phase. So they are not coherent.
2. This is because atoms cannot emit light waves in same phase.

Question 7.
Does diffraction take place at Young’s double-slit?
Answer:
Yes, light waves arriving at each slit are diffracted and then diffracted waves interfere to produce interference pattern.

Question 8.
Is there any difference between coloured light obtained from prism and colours of soap bubble?
Answer:

1. Coloured light obtained from prism is due to the phenomenon of dispersion.
2. Colours of soap bubble are due to interference of light waves from the upper and lower surface of the soap film.

Question 9.
A small disc is placed in the path of the light from distance source. Will the center of the shadow be bright or dark?
Answer:
Waves diffracted from the edge of circular obstacle interfere constructively at the center of the shadow resulting in the formation of a bright spot.

Question 10.
When a wave undergoes reflection at a denser medium, what happens to its phase?
Answer:
Its phase changes at 180 degrees.

V. Numerical problems:

Question 1.
An object is placed at a certain distance from a convex lens of focal length 20 cm. Find the distance of the object if the image obtained is magnified 4 times.
Answer:
Given:
f = 20 cm
m = -4 = \(\frac{v}{u}\)
v = -4u

u = -15 cm

Question 2.
A compound microscope has a magnification of 30. The focal length of eyepiece is 5 cm. Assuming the final image to be at least distance of distinct vision, find the magnification produced by the objective.
Answer:
m_e = \(\frac{v}{u}=\frac{\mathrm{D}}{\mathrm{u}_{e}}=1+\frac{\mathrm{D}}{f_{\mathrm{e}}}\)
= 1 + \(\frac{25}{5}\)
= 6
for compound microscope,
m = m₀ × m_e
30 = m₀ × 6
m₀ = \(\frac{30}{6}\) = 5

Question 3.
An object is placed in front of a concave mirror of focal length 20 cm. The image formed is three times the size of the object. Calculate two possible distances of the object from the mirror.
Answer:
First case when the image is virtual, m = 3
m = \(-\frac{v}{u}\)
3 = \(-\frac{v}{u}\)
v = -3u
\(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)
\(-\frac{1}{20}=\frac{1}{u}-\frac{1}{3 u}\)
\(\frac{3 \mathrm{u}}{4}\) = – 20
3u = -40
u = –\(\frac{40}{3}\) cm
Second case when the image is real
m = -3
m = \(-\frac{\mathrm{v}}{\mathrm{u}}\)
-3 = \(-\frac{v}{u}\)
V = 3u
\(\frac{1}{f}=\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}\)
\(-\frac{1}{20}=\frac{1}{u}+\frac{1}{3 u}\)
\(\frac{3 \mathrm{u}}{4}\) = -20
u = – \(\frac{80}{3}\) cm

Question 4.
A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. (Consider the bulb to be a point source.)
Answer:

The light rays from the small bulb S which are incident at an angle i > i_c are totally internally reflected and cannot emerge out of water surface. The light from the bulb ‘S’ comes out through a circles path of radius r is given by

Area of the patch = πr²
= π h² tan² i_c
= 3.14 × 0.80² × \(\frac{9}{7}\)
= 2.58 m²
≈ 2.6 m²

Question 5.
A thin converging glass lens made of glass with refractive index 1.5 has a power of + 5.0 D. When this lens is immersed in a liquid of refractive index n, it acts as a divergent lens of focal length 100 cm. What must be the value of n?
Answer:

Question 6.
If the distance D between an object and screen is greater than 4 times the focal length of a convex lens, then there are two positions of the lens for which images are formed on the screen. This method is called conjugate foci method. If d is the distance between the two positions of the lens, obtain the equation for focal length of the convex lens.
Answer:

seperation between object screen is
D = u + v
m₁ = \(\frac{\mathrm{V}}{\mathrm{u}}\);
m₂ = \(\frac{u}{v}\)
m₁m₂ = 1
size of image 1 = m₁ times the size of object
S_I1 = m₁S₀
S_I2 = m₂S₀
S₀ = \(\sqrt{\mathrm{S}_{\mathrm{I}_{1}} \mathrm{~S}_{\mathrm{I}_{2}}}\)
here distance between its two positions of lens where sharp images are obtained
d = v – u
focal length of lens f = \(\frac{\mathrm{uv}}{\mathrm{u}+\mathrm{v}}\)
= \(\frac{(u+v)^{2}-(u-v)^{2}}{4(u+v)}\)
f = \(\frac{\mathrm{D}^{2}-\mathrm{d}^{2}}{4 \mathrm{D}}\)
when D > 4f

Question 7.
A beam of light of wavelength 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. What is the distance between the first dark fringe on either side of the central bright fringe?
Answer:
β = \(\frac{\lambda D}{d}\)
d = 1mm
λ = 600 × 10^-9 m
D = 2m
= \(\frac{6 \times 10^{2} \times 10^{-9} \times 2}{1 \times 10^{-3}}\)
= 1.2 mm
Distance between two dark fringes which are on either side of the central bright fringe = 2 × 1.2 mm = 2.4 mm

Question 8.
In Young’s double-slit experiment, the slits are 2 mm apart and are illuminated with a mixture of two-wavelength λ₀ = 750 nm and λ = 900 nm. What is the minimum distance from the common central bright fringe on a screen 2 m from the slits where bright fringe from one interference pattern coincides with a bright fringe from the other?
Answer:
Let n₁ bright fringe of λ₁ coincides with (n+1) bright fringe of λ₂.
\(\frac{\mathrm{n} \lambda_{1} \mathrm{D}}{\mathrm{d}}=\frac{(\mathrm{n}+1) \lambda_{2} \mathrm{D}}{\mathrm{d}}\)
n × 900 × 10^-9 = (n + 1) × 750 ×10^-9
minimum distance from central maxima = \(\frac{\mathrm{n} \lambda_{1} \mathrm{D}}{\mathrm{d}}\)
= \(\frac{5 \times 900 \times 10^{-9} \times 2}{2 \times 10^{-3}}\)
= 4.5 mm

Question 9.
In Young’s double-slit experiment, 62 fringes are seen in the visible region for sodium light of wavelength 5893 Å. If violet light of wavelength 4359 Å is used in place of sodium light, then what is the number of fringes seen? Answer:
n₁λ₁ = n₂ λ₂
62 × 5893 A° = n₂ × 4358 A°