NCERT Exemplar for Class 12 Physics Chapter 10

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Access the NCERT Exemplar Class 12 Physics Chapter 10 Wave Optics

Short answer questions

1. In Young’s experiment, how does the interference fringes’ angular separation alter as the slit distance is increased?

Ans. The angular separation will decrease when the two slits are separated further. Angular separation, =d ∴ ∝ 1d

2. Give a justification for why two separate sources of light cannot be regarded as coherent sources.

Ans. Coherence between two separate sources of light is impossible. Atoms produce light when they return to their ground state. There are billions of atoms in even the tiniest source of light, and they can’t possibly all create light waves at the same time.

3. From the air, a light wave enters the glass. How will this impact:

(a) the wave’s energy?

(b) the wave’s frequency?

Ans. (a) A portion of the light wave that travels from the air to the glass is reflected back into the atmosphere. As a result, the wave’s energy would be decreased in the glass.

(b) The frequency of a light wave that enters glass from the air would not change.

4. How does the wavefront look when light is emitted from a point source?

Ans. The wavefront would exhibit a spherical shape when light diverges from a point source.

5. A plane surface of the glass with a refractive index receives unpolarized light at an angle i. Write the relationship between angle I and the refractive index if the reflected light becomes completely polarised.

Ans. =tan ip.

6. A glass slab has a refractive index of 3. At what angle of incidence should it be struck to make the reflected and refracted rays parallel to each other?

Ans. = tan ip under the given condition

∴ ip=tan-1 (3)=3 radian

7. Differentiate between a wavefront and a ray.

Ans. The path of light is determined by a ray, whereas a wavefront is an area in which all points in a light wave oscillate at the same time.

8. In Young’s double slit experiment, how would the angular spacing of interference fringes alter if the slits and screen were twice as far apart?

Ans. Slit-screen separation is unrelated to angular separation, = d.

Hence, nothing will change.

9. Describe the conditions for coherence between two light sources.

Ans. The following prerequisites must be met:

1. a) They must continually emit waves with the same wavelength.
2. b) There must be zero or a constant phase difference between the waves.
3. What distinguishes a plain polarized light from an unpolarized light?

Ans. When light is polarized, the electric field vector’s directions are limited to a single plane. However, when light is unpolarized, the electric field vector’s directions can be in any plane that is perpendicular to the direction of the propagation.

10. Which of the given waves can be polarized?

(i) heat waves

(ii) sound waves

Ans. Due to their transverse nature, heat waves are susceptible to polarization.

1. Define the word “coherent sources,” which are necessary for Young’s double slit experiment to produce an interference pattern.

Ans. Coherent sources are two monochromatic sources that generate light waves with a fixed phase difference.

11. On the boundary of two transparent media, a beam of unpolarized light is incident at Brewster’s angle, which is equal to iB. What degree of polarization occurs in the light that is reflected?

Ans. The polarization of the reflected light occurs at an incidence angle of = iB

Short 3-mark question

1. How will the fringe width change in the young’s double slit experiment under the following conditions

(a) a smaller frequency of light is used

(b) there is a reduction in the distance between the slits.

Ans. (a) We can represent the fringe width as =Dd

In this case, the fringe width would increase if a sampler frequency of light is used.

 

(b) The distance between the slits and the fringe width is inversely proportional, ∝ 1d

In that case, the fringe width will increase if there is a reduction between the slits.

2. Assume there is an interaction of two sources of intensities, I and 4I. What will the intensity be at the locations of the phase differences in the following cases:

(a) 2

(b)

(c) Is white light capable of producing interference, and what is its nature?

Ans (a) We know that I=a²+b²+2ascos

Here, the amplitude of the two coherent waves is a and b, and their phase difference is .

I= I+4I+2141cos

I= 5I+41cos

When =2

I= 5I+41cos2

∴ 5I

Hence, there will be five times more intensity.

(b) When, =

I= 5I+41cos

I= 5I-4I

I=I

Therefore, the intensity will be the same.

(c) Interference would result from white light. However, since the central bright fringe of the white light interference pattern overlaps with all the other colours present at the location, white is determined to be its colour. It will appear that a few coloured rings will surround the white central bright fringe.

3. (a) A glass has a refractive index of 1.5. If the light’s speed in a vacuum is 3.0108, what will be the speed of light within the glass?

(b) Is the light’s speed in glass free of the colour of light? If not, which of the two colours, red and violet, travel slower in a glass prism?

Ans. (a) We are given that,

The glass refractive index is =1.5

The speed of light is c= 3.0108

The speed of light within a glass can be given by the following relation:

v=c

If we substitute the given values, we get

v=3.01081.5

∴ 2108 m/s

(b) The colour of the light is found to affect the speed of light in glass. It is known that the violet component of white light has a higher refractive index than the red component. As a result, in glass, violet light travels more slowly than red light. As a result, in a glass prism, violet light moves more slowly than red light.

4. (i) Describe the principle upon which an optical fibre functions.

(ii) What conditions must exist in order for this phenomenon to take place?

Ans. (i) The theory of total internal reflection underlies how optical fibres function.

(ii) These are some conditions that must exist for this phenomenon to take place

• The light should move from a denser to a rarer medium.
• A greater angle of incidence is required.

Critical angle can be given as ic=sin-1 1

5. Give two conditions that must be met in to obtain coherent sources. Demonstrate that the dark and bright screen fringes are similarly spaced in Young’s arrangement to create an interference pattern.

Ans. Coherent sources must satisfy the following two requirements:

(i) Two sources must produce monochromatic light.

(ii) Some equipment should be able to produce coherent light sources from a single source.

The dark and bright fringe’s width is determined by

=Dd

Therefore, it will be the same for both bright and dark fringe. Hence, it is found that they are spaced equally.

6. What will the Brewster angle for the air-to-glass transition be? (Hint: the refractive index of the glass is 1.5)

Ans. We are given that

The refractive index of the glass is =1.5

Brewster angle=

The Brewster angle and refractive index are known to be related in the following way:

Tan =

= tan-1 (1.5)

∴ = 56.31°

Hence, the Brewster angle in case of air to glass transition will be 56.31°

7. The Doppler formula for frequency shift is slightly different for sound waves.

Between the two examples provided and justify your choice. Would you, in the event of light, anticipate that the formulas will be exactly the same for the two scenarios moving through a medium?

(i) The source is at rest, but the observer is moving, and

(ii) The source is moving, but the observer is stationary. But in each of these cases, the precise Doppler formulas for light waves in a vacuum are similar.

Ans. We know that a medium is the only way for sound waves to travel.

(i) Given conditions that are not equal in a scientific sense since an observer’s relative motion in the two examples provided, a medium is different in each case. The Doppler equations for the two instances described cannot be identical.

(ii) Now, sound can travel through vacuum spaces as light waves can. Since the speed of light is unaffected by motion, the two situations are determined to be similar.

Both the motion of the source and that of the observer. While, as light moves across a medium, Since the speed of light would now be dependent on the two examples above, they are not equivalent to the medium’s wavelength.

8. Provide answers to the given questions.

(a) A low-flying aircraft may occasionally cause us to experience seeing shakiness on our TV screen. Give a potential reason.

(b) The linear superposition of wave displacement is fundamental for understanding intensity distributions in diffraction and requires knowledge of patterns of interference

and intensity distribution in diffraction. What is the reason for this principle?

Ans. (a) We know that low-flying aircraft send weak radar signals, which can interfere with the antenna’s TV signals. The TV, as a result of this, could become distorted. Consequently, when an aeroplane passes overhead at a low altitude, we, on our TV screen, occasionally notice a tiny trembling of the image.

(b) Intensity distributions and interference patterns are important for our comprehension; it is crucial to understand the linear superposition of the wave displacement principle. This is due to superposition results from a differential equation’s linear nature, which is known to control wave motion. Let y1 and y2 be the answers are second-order wave solutions. In that case, any linear combination of y1 and y2 may also be the solution to the wave equation.

9. How is the angular width of the central bright fringe maximum altered in a single slit experiment when

(a) the slit width is increased?

(b) the screen’s distance from the slit is widened?

(c) the utilization of light is with a shorter wavelength?

Ans. (a) The fringe width for single-slit diffraction is =2Dd

As the slit’s width ‘d’ increases, lowers.

(b) As “D” increases, the width of the central bright fringe also increases since both quantities are proportional to one another.

(c) The width of the central bright maximum would diminish when light with a reduced wavelength was employed.

A 500 nm parallel beam of light strikes a small slit, and a screen 1-metre distant displays the diffraction pattern that results. It is noted that the initial minimum is located 2.5 mm from the screen’s centre. Determine the slit’s width.

Ans. We are given,

=500 nm =50010-9 m, D=1

xn=2.5 mm= 2.510-3 m, n=1

xmdDn

d=nDxn

d=1(50010-9)12.510-3=2.510-4 m

Answer the following:

(i) What relationship exists between the interference pattern in a double slit experiment and the diffraction from each slit?

(ii) When a little circular barrier is placed in the path of the light from a distant source, a small spot can be observed at the centre of the obstacle’s shadow. Explain why.

Ans. (i) In a double-slit experiment, interference pattern and diffraction from each slit are related in the following ways:

• There is no uniform intensity for all bright fringes for diffraction, but for interference, it is.
• The intensity of minima for interference is usually not zero, but for diffraction, it is.

(ii) As a result of the circular obstacle’s edge diffracting waves from the distant source, a brilliant spot is created in the centre of the shadow cast by the obstacle.

1. Young’s double slit experiment uses light with a wavelength of 400 nm, interference fringes with a width of 600 nm, and a halved slit distance. Find the ratio of the screen’s distance to the interference plane if you want the observed fringe width on the screen to be similar in the two scenarios.

Ans. Let’s take the distance between the sources and the screen when light having a wavelength of 400 nm is used to be D1

=Dd

X=D140010-9d ………….(1)

To obtain a similar fringe width

D260010-9d=X= ………….(2)

From equations (1) and (2)

D1D2=60010-940010-9

∴ D1D2=1.5

Hence, 3:2 will be the ratio of the distance that is found in the given arrangement.

Long answer questions

1. The angular width of a fringe on a screen situated 1m distant is found to be 0.2 in a double-slit experiment. The light used has a 600nm wavelength. What will the fringe’s angular width be if the whole experimental setup is submerged in water? Consider the water’s refractive index to be 43.

Ans. We are given the following information.

The distance that exists between the slits and the screen is D=1m

The light’s wavelength that is used is 1=600 nm

The fringe’s angular width in the air is 1=0.2°

The fringe’s angular width in the water, 2

The refractive index of the water is =43

The angular width and the refractive index is related as

=12

2=340.2= 0.15°

Hence, the fringe’s angular width in water would be reduced to 0.15°

1. When a parallel 500 nm laser beam strikes a small slit, the diffraction pattern that results is visible on a screen one meter distant. It is noted that the first minimum is situated 2.5 nm from the screen’s centre. Determine the slit’s width.

Ans. We are given the following information:

The light beam’s wavelength is =500 nm= 50010-9 m,

The distance that exists between the slits and the screen is given as D=1m

For the first minimum, n=1

The distance among the slits=d

The distance between the centre of the screen and the first minimum can be given as

x=2.5 mm= 2.510-3

The following equation shows its relation to the order of the minima

n=xdD

d=nDx

d= 150010-912.510-3= 210-4 m= 0.2 mm

Hence, 0.2 mm will be the width of the slits.

3. Differentiate between a light that is linearly polarized and unpolarized light. Explain, using a diagram, how scattering causes unpolarized light to become linearly polarized.

Ans. Unpolarized light is a type of light wave where the electric vector oscillates in all directions on a plane perpendicular to the path of propagation.

The linearly polarized light is referred to be linearly polarized light if the oscillations of the electric vectors are limited to one direction in a plane which is perpendicular to the direction of transmission.

It is brought on by the earth’s atmosphere’s molecules dispersing light.

The incident (unpolarized) wave’s electric field causes the molecules’ electrons to acquire components of motion in both of these directions. Since there is no transverse component to the acceleration of charges moving parallel to the double arrows, they do not emit energy in the direction of the observer.

Therefore, dots are used to depict the radiation that the molecules dispersed, and that was polarized perpendicular to the figure’s plane.

4. In Young’s double slit experiment, a monochromatic light that has a wavelength of 630 nm illuminates a slits pair, resulting in an interference pattern with two bright fringes spaced 8.1 mm apart. The interference pattern, in which the two successive bright fringes are separated by 7.2 mm, is created by a second monochromatic light source. Find the light’s wavelength from the second source. What happens to the interference fringes if a white light source is used in place of the monochromatic source?

Ans. The nth bright fringe’s position from the central bright can be represented as nDd

So, the separation existing between the two consecutive bright fringes is Dd

With 1= 630 nm, we have 1Dd= 8.1 mm ………….(i)

With 2, we have 2Dd= 7.2 mm …………..(ii)

When we divide (i) by (ii), we get 12=8.17.2

2=7.28.11=89630=560 nm

If the white light replaces a monochromatic, the following will happen:

• The central bright will continue to remain white.
• The least wavelength colour, violet, will produce its brightness near the central bright, with all the other colours generating their own maximas.

5. A Young’s double-slit experiment is used to obtain interference fringes using a beam of light with two wavelengths, 650 nm, and 520 nm. Answer the following:

(a) Calculate the distance between the third bright fringe on the screen and the central maximum having a wavelength of 650 nm.

(b) What is the shortest distance between the central maxima and the brilliant fringes caused by the overlap of the two wavelengths?

Ans. (a) The expression for fringe width is

x= Dd

Now, as per the condition given, x=3650Dd=1950Ddnm

Also, n2=(n-1)1

But, 1=650 nm

∴ n=5

The wavelength of the other light beam is 2=520 nm.

The distance that exists between the slits and the screen is given as= D.

The distance present between the two slits=d.

The distance of between the central maximum and the nth bright fringe present on the screen can be given as,

x=n1Dd

In case of third bright fringe, N=3

∴ x=3650Dd=1950Ddnm

(b) Let’s say that on the screen, the nth bright fringe due to wavelength 2 and the (n-1)thbright fringe owing to wavelength 1 coincide. The requirements for bright fringes might be related to

n2=(n-1)1

520n=650n-650

650=130n

∴ n=5

Therefore, the following equation can give us the shortest distance from the central maximum.

x= 2Dd

∴ x=5520Dd=260Ddnm

Please note that the values of both D and d are not given in this question.

6. Describe how corpuscular theory predicts that the speed of light is greater in a medium than in a vacuum, such as water. Is the experimental measurement of the speed of light in water consistent with the prediction? Which alternative theory of light, if any, is consistent with the experiment?

Ans. No, wave theory applies. We are aware of the fact that Newton’s corpuscular theory of light asserts that when light corpuscles collide with the interface of two media—from a rarer (air) to a denser (water) medium—the particles feel forces of attraction normal to the surface. As a result, the component of velocity along the normal increases while the component along the surface stays constant.

We can write the following equation

C sin i= v sin r ………….(1)

Here, i denotes the angle of incidence, r denotes the angle of reflection, c denotes the light’s velocity in the air, and v denotes the light’s velocity in the water.

The relationship between the respective refractive index of water and air is as follows:

=vc

Therefore, equation (1) can be written as

vc=sin isin r= ………….(2)

But, > 1

Therefore, equation (2) can be used to deduce that v > c. Since this prediction contradicts the experimental findings of c > v, it is not possible.

As a result, we discovered that the experimental findings and the wave model of light are consistent with each other.

7. Answer the following:

• Why is a sustained interference pattern only possible with coherent sources?
• When monochromatic light of wavelength is used in Young’s double slit experiment, the light intensity at a position on the screen where the path difference is is of K units. Find the light’s intensity at a location where the path difference is 3.

Ans. (a) In order to prevent the sites of maxima and minima from shifting over time, coherent sources are required. These form continuous interference patterns because there is a consistent phase difference between them.

(b) intensity, I=4I0 cos² 2

When the path difference is the phase difference will be

=2

Hence, K=4I0 cos²=4I0=K

When the path difference is 3, the phase difference will be

t=23=2 3=23

The intensity will be I’=4I0 cos² 3= 4I0122=I0

∴ I’=K4

8. Answer the following:

• What connection exists between the interference pattern that may happen in a double slit experiment and the diffraction from each slit?
• A single slit with an aperture of 2 10-4 m is utilized to analyse the diffraction occurring there using two wavelengths of a sodium light that are 590 nm, and 596 nm. The slit is 1.5 metres away from the screen. Determine the distance between the locations of the first maxima of the two cases’ produced diffraction patterns.

Ans. (a) The effect is typically referred to as interference when there are only a few sources, such as two interfering sources, but it appears that the word diffraction is more frequently used when there are many sources. We must keep in mind that in the double-slit experiment, the pattern that appears on the screen is actually a superposition of the interference patterns caused by single-slit diffraction from both slits and double slits. As a result, a larger diffraction peak is visible, and several fringes with smaller widths are caused by double-slit interference.

(b) The distance between the centre of the screen and the first secondary maximum is, x1=32Da

Therefore, the space between the screen and the first secondary maxima from the given wavelengths:

x=3D2a(2 – 1)

= 3152210-4(596-590)10-9

=4.5610-54=6.7510-5 m

NCERT Exemplar For Class 12 Science

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