NCERT Exemplar Class 12 Physics Solutions Chapter 10 Wave Optics 2021

Q: Are these solutions helpful in board exams?

Yes, these solutions of NCERT will be helpful in board exam preparation as one can understand the chapter and topics better.

Q: How these solutions can be used?

One can learn the how to solve questions for this chapter in board exam, can also cross check their answers while practicing.

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Yes, each and every question is solved as per the CBSE pattern in NCERT exemplar Class 12 Physics solutions chapter 10, which will help in understanding each step separately.

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Yes! The Careers360 NCERT Exemplar Solutions are step-by-step and based on the latest CBSE syllabus. They are great for revising key formulas, clearing doubts, and practicing high-level questions before your exams.

Q: What is the main focus of Chapter 10: Wave Optics in Class 12 Physics?

This chapter explains how light behaves like a wave. It covers important topics like interference (light adding up or cancelling), diffraction (bending around edges), and polarisation (filtering light direction). These help explain real-life effects like colors in soap bubbles or the working of polarised sunglasses.

NCERT Exemplar Class 12 Physics Solutions Chapter 10 Wave Optics

Vishal kumarUpdated on 12 Jul 2025, 10:21 AM IST

Have you ever seen how light bends when it passes through a slit or spreads out after passing an obstacle? This happens because of a property called wave nature of light. In Chapter 10: Wave Optics of Class 12 Physics you will learn about how light behaves like a wave. This Chapter covers important topics like interference, diffraction and polarization explaining how they work and where we see them in real life. These NCERT Exemplar Solutions are prepared by expert faculty as per the latest CBSE syllabus.

The NCERT Exemplar Solutions for Class 12 Physics Chapter 10 by Careers360 provide step by step answers to all examplar problems. These NCERT Exemplar Solutions for Class 12 include multiple choice questions(MCQs), short questions and long answer questions. Practicing these will help you understand wave optics better improve your problem solving skills and prepare well for exams like Board exam, JEE and NEET.

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This Story also Contains

NCERT Exemplar Class 12 Physics Solutions Chapter 10: MCQI
NCERT Exemplar Class 12 Physics Solutions Chapter 10: MCQII
NCERT Exemplar Class 12 Physics Solutions Chapter 10: Very Short Answer
NCERT Exemplar Class 12 Physics Solutions Chapter 10: Short Answer
NCERT Exemplar Class 12 Physics Solutions Chapter 10: Long Answer
Main Subtopics Covered in NCERT Exemplar Class 12 Physics Solutions Chapter 10 Wave Optics
NCERT Exemplar Class 12 Physics Solutions Chapter-Wise
NCERT Exemplar Class 12 Solutions

NCERT Exemplar Class 12 Physics Solutions Chapter 10 Wave Optics

NCERT Exemplar Class 12 Physics Solutions Chapter 10: MCQI

Question:1

Consider a light beam incident from air to a glass slab at Brewster’s angle, as shown in Fig. 10.1. A Polaroid is placed in the path of the emergent ray at point P and rotated about an axis passing through the centre and perpendicular to the plane of the polaroid.

A. For a particular orientation, there shall be darkness as observed through the polaroid.
B. The intensity of light, as seen through the polaroid, shall be independent of the rotation.
C. The intensity of light, as seen through the polaroid, shall go through a minimum but not zero for two orientations of the polaroid.
D. The intensity of light, as seen through the polaroid shall go through a minimum for four orientations of the polaroid.

Answer:

The answer is the option (c)
Explanation: Brewster’s law states that when a beam of unpolarized light is reflected in a transparent medium, the resultant reflected light is completely polarized at a certain angle of incidence. It is given by:
From the figure, it is clear that

θ p + θ r = 90^{\circ}

Also,

n = \tan θ p

(Brewster’s law)

(i) For $i < θ_{p} o r i > θ_{p}$
Both of the reflected and refracted light ray become partially polarized.

(ii) For glass $θ_{p} = 51^{\circ} f o r w a t e r θ_{p} = 53^{\circ}$

Question:2

Consider sunlight incident on a slit of width $10^{4}$ A. The image seen through the slit shall
A. be a fine sharp slit white in colour at the centre.
B. a bright slit white at the centre diffusing to zero intensities at the edges.
C. a bright slit white at the centre diffusing to regions of different colours.
D. only be a diffused slit white in colour.

Answer:

The answer is the option (a)
Diffraction is a phenomenon of bending of light rays around an obstacle or an aperture of a similar wavelength.

Question:3

Consider a ray of light incident from air onto a slab of glass (refractive index n) of width d, at an angle θ. The phase difference between the ray reflected by the top surface of the glass and the bottom surface is
A. $\frac{4 π d}{λ} {(1 - \frac{1}{n^{2}} \sin^{2} θ)}^{1 / 2} + π$
B. $\frac{4 π d}{λ} {(1 - \frac{1}{n^{2}} \sin^{2} θ)}^{1 / 2}$
C. $\frac{4 π d}{λ} {(1 - \frac{1}{n^{2}} \sin^{2} θ)}^{1 / 2} + \frac{π}{2}$
D. $\frac{4 π d}{λ} {(1 - \frac{1}{n^{2}} \sin^{2} θ)}^{1 / 2} + 2 π$

Answer:

The answer is the option (a)

Question:4

In a Young’s double-slit experiment, the source is white light. One of the holes is covered by a red filter and another by a blue filter. In this case
A. there shall be alternate interference patterns of red and blue.
B. there shall be an interference pattern for red distinct from that for blue.
C. there shall be no interference fringes.
D. there shall be an interference pattern for red mixing with one for blue.

Answer:

The answer is the option (c)
Here, in this case, due to the presence of red and blue filters. The waves of light will only be Red and Blue. In YDSE, the monochromatic light is used for the formation of fringe on the screen. Therefore, in this case, there will be no interference.

Question:5

Figure 10.2 shows a standard two-slit arrangement with slits S₁, S₂. P₁, P₂ are the two minima points on either side of P (Fig. 10.2). At P₂ on the screen, there is a hole and behind P₂ is a second 2- slit arrangement with slits S₃, S₄ and a second screen behind them.

A. There would be no interference pattern on the second screen but it would be lighted.
B. The second screen would be totally dark.
C. There would be a single bright point on the second screen.
D. There would be a regular two slit pattern on the second screen

Answer:

The answer is the option (d)
Wavefront is the plane where every point on given wave front is the source of disturbance which are known as secondary wavelets.

The wavefront emitted by a narrow source is divided in two parts by reflection, refraction or diffraction. The coherent sources so obtained are imaginary.

NCERT Exemplar Class 12 Physics Solutions Chapter 10: MCQII

Question:6

Two source S₁ and S₂ of intensity I₁ and I₂ are placed in front of a screen [Fig. 10.3 (a)]. The pattern of intensity distribution seen in the central portion is given by Fig. 10.3 (b). In this case which of the following statements are true.

A. S₁ and S₂ have the same intensities.
B. S₁ and S₂ have a constant phase difference.
C. S₁ and S₂ have the same phase.
D. S₁ and S₂ have the same wavelength.

Answer:

The correct answers are the options (a, b, c)

Key concept:

For getting the sustained interference the initial phase difference between the interfering waves must remain constant, i.e., sources should be coherent.
for two coherent sources, the resultant intensity is given by
$I = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \cos ϕ$

Resultant intensity at the point of observation will be maximum.
$I_{m a x} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} I_{m a x} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$

Resultant intensity at the point of observation will be minimum.

$I_{m i n} = I_{1} + I_{2} - 2 \sqrt{I_{1} I_{2}} I_{m i n} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$

Question:7

Consider sunlight incident on a pinhole of $10^{3}$ A. The image of the pinhole seen on a screen shall be
A. a sharp white ring.
B. different from a geometrical image.
C. a diffused central spot, white in colour.
D. diffused coloured region around a sharp central white spot.

Answer:

The correct answers are the options (b,d)
Diffraction of light can only be observed if the size of the aperture or obstacle is less than the wavelength of the light wave.
The given width of the pinhole is 103 angstrom. The wavelength of sunlight is 4000 angstrom to 8000 A. Therefore, the light is diffracted from the hole. And due to this, the image formed on the screen will be geometrically different.

Question:8

Consider the diffraction pattern for a small pinhole. As the size of the hole is increased
A. the size decreases.
B. the intensity increases.
C. the size increases.
D. the intensity decreases.

Answer:

The correct answers are the options (a, b)
Key concept: The 'shadow' of the hole of dimeter d is spread out over an angle

Δ θ = 1.22 \frac{λ}{D} \Rightarrow Δ θ \propto \frac{1}{D}

Question:9

For light diverging from a point source
A. the wavefront is spherical.
B. the intensity decreases in proportion to the distance squared.
C. the wavefront is parabolic.
D. the intensity at the wavefront does not depend on the distance

Answer:

The correct answers are the options (a, b)

Due to the point source light propagates in all direction symmetrically and hence, wavefront will be spherical.

As the intensity of the source will be

$I \propto \frac{1}{r^{2}}$

where r is the radius of the wavefront at any time.
Hence the intensity decreases in proportion to the distance squared.

NCERT Exemplar Class 12 Physics Solutions Chapter 10: Very Short Answer

Question:10

Is Huygens’s principle valid for longitudinal sound waves?

Answer:

The principle of Huygen’s is valid for longitudinal sound waves

Question:11

Consider a point at the focal point of a convergent lens. Another convergent lens of the short focal length is placed on the other side. What is the nature of the wavefronts emerging from the final image?

Answer:

Here the orientation of the rays is perpendicular to L1. It forms an image at I1, i.e. the focal length. Again, the image is converged and goes through L2 after which the final image is formed at I. The nature of the wavefronts emerging from the final image is Spherical.

Question:12

What is the shape of the wavefront on earth for sunlight?

Answer:

As the sun is at an exceptionally large distance from earth. Assuming it as a point source of light at infinity, as seen from earth, we can conclude that the radius of the wavefront from the sun to the earth is infinite. This would mean that the rays are perpendicular to earth and the wave front is almost a plane.

Question:13

Why is the diffraction of sound waves more evident in daily experience than that of a light wave?

Answer:

The wavelength of sound waves is 15 m to 15 mm for 20 Hz to 20,000 Hz respectively. Therefore, sound waves diffraction take place when the comparable size of the obstacle is confronted. While in the case of the light wave, the wavelength of visible light is 0.4 to 0.7 micron. So, the obstacles of this size are not easily present around us. Therefore, diffraction of light is not so evident in day to day life.

Question:14

The human eye has an approximate angular resolution of ? = 5.8×10^–4 rad and a typical photo printer prints a minimum of 300 dpi (dots per inch, 1 inch = 2.54 cm). At what minimal distance z should a printed page be held so that one does not see the individual dots.

Answer:

It is given, angular resolution of human eye

ϕ = 5.8 \times 10^{- 4}

rad and printer print 300 dots per inch.
The linear distance between the two dots is

l = \frac{2.54}{300} c m = 0.84 \times 10^{- 2} c m

At a distance of z cm, this subtends an angle,

ϕ = \frac{l}{z} z = \frac{l}{ϕ} = \frac{0.84 \times 10^{- 2} c m}{5.8 \times 10^{- 4}} = 14.5 c m

If a printed page be held at a distance of 14.5 cm, then one does not be able to see the individual dots.

Question:15

A polaroid (I) is placed in front of a monochromatic source. Another polaroid (II) is placed in front of this polaroid (I) and rotated till no light passes. A third polaroid (III) is now placed in between (I) and (II). In this case, will light emerge from (II)? Explain.

Answer:

Monochromatic source of light is kept behind polaroid (I). Then some other polaroid (II) is placed in front of polaroid (I). So, the axes of both polaroid are parallel to each other; the light passes through (II) unaffected.

Now polaroid (II) is rotated till no light passes. In this situation the pass axis of polaroid (II) is perpendicular to polaroid (I), then (I) and (II) are set in a crossed position. No light passes through a polaroid- (II)

Now third polaroid (III) is now placed in between (I) and (II). Only in the special cases when the pass axis of (III) is parallel to (I) or (II) there shall be no light emerging. In all other cases, there shall be light emerging because the pass axis of (II) is no longer perpendicular to the pass axis of (III).

Now, the polaroid (II) is rotated such that (I) and (II) are in a crossed position. Therefore, no light will pass through (II).

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NCERT Exemplar Class 12 Physics Solutions Chapter 10: Short Answer

Question:16

Can reflection result in plane-polarized light if the light is incident on the interface from the side with higher refractive index?

Answer:

If Brewster’s angle is equal to the incident angle, then the transmitted light is slightly polarized, while the reflected light is plane-polarized.

Polarisation by reflection occurs when the angle of incidence is Brewster's angle.
i.e.

\tan i_{R} =^{1} μ_{2} = \frac{μ_{2}}{μ_{1}} w h e r e μ_{2} < μ_{1}

When the light rays travel in such a medium, the critical angle is

\sin i_{e} = \frac{μ_{2}}{μ_{1}} w h e r e μ_{2} < μ_{1} A s | \tan i_{B} | > | \sin i_{e} | f o r l a r g e a n g l e s i_{B} > i_{e}

Thus the polarization by reflection occurs definitely.

Question:17

For the same objective, find the ratio of the least separation between two points to be distinguished by a microscope for the light of $5000 Å$ and electrons accelerated through 100V used as the illuminating substance.

Answer:

5000 Å = 5000 \times 10^{- 10} m \frac{1}{d} = \frac{2 \sin β}{1.22 λ} d m i n = \frac{1.22 λ}{2 \sin β} d_{m i n} = \frac{1.22 \times 5000 \times 10^{- 10}}{2 s i n β}

When we use 100V light,

λ_{d} = \frac{1.27}{\sqrt{V}} n m = \frac{1.27}{\sqrt{100}} n m

d^{'} m i n = \frac{1.22 λ_{d}}{2 \sin β}

d^{'} m i n = \frac{1.22 \times 1.27 \times 10^{- 10}}{2 \sin β}

The required ratio

= \frac{d m i n}{d^{'} m i n} = \frac{1.22}{5000} = 0.244 \times 10^{- 3}

Question:18

Consider a two-slit interference arrangement (Fig. 10.4) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of D in terms of such that the first minima on the screen fall at a distance D from the centre O.

Answer:

\begin{aligned} S_{1} P = \sqrt{D^{2}} + (D - x)^{2} \\ S_{2} P = \sqrt{D^{2}} + (D - x)^{2} \\ T_{2} P = D + x \\ T_{2} P = D - x \\ {[D^{2} + (D + x)^{2}]}^{1 / 2} - {[D^{2} + (D - x)^{2}]}^{- 1 / 2} = \frac{λ}{2} \\ D = 0.404 λ \end{aligned}

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NCERT Exemplar Class 12 Physics Solutions Chapter 10: Long Answer

Question:19

Figure 10.5 shown a two-slit arrangement with a source which emits unpolarised light. P is a polarizer with axis whose direction is not given. If I0 is the intensity of the principal maxima when no polarizer is present, calculate in the present case, the intensity of the principal maxima as well as of the first minima.

Answer:

The amplitude of wave in normal/perpendicular polarization

A ⊥= A_{⊥}^{0} (\sin (k x - ω t) + \sin (k x - ω t + ϕ))

The amplitude of wave in parallel polarization

A ∥= A_{∥}^{0} (\sin (k x - ω t) + \sin (k x - ω t + ϕ))

The intensity of the wave at the first minima with a polarizer

{| A_{⊥}^{0} |}^{2} (1 - 1) + \frac{{| A_{⊥}^{0} |}^{2}}{2} = \frac{I_{0}}{8}

Question:20

A small transparent slab containing material of μ =1.5 is placed along $A S_{2}$ (Fig.10.6). What will be the distance from O of the principal maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab?

Answer:

For the principal maxima,(path difference is zero)

$\begin{aligned} Δ x = 2 d \sin θ + [(μ - 1) L] = 0 \\ \sin θ_{0} = - \frac{L (μ - 1)}{2 d} = - \frac{- L (0.5)}{2 d} [∴ L = d / 4] \\ or \Rightarrow \sin θ_{0} = \frac{- 1}{16} \end{aligned}$

$θ_{0}$ is the angular position corresponding to the principal maxima.

$\Rightarrow O P = D \tan θ_{0} \approx D \sin θ_{0} = \frac{- D}{16}$

For the first minima, the path difference is

$\begin{aligned} \pm \frac{λ}{2} \\ Δ x = 2 d \sin θ_{1} + 0.5 L = \pm \frac{λ}{2} \\ \sin θ_{1} = \frac{\pm λ / 2 - 0.5 L}{2 d} = \frac{\pm λ / 2 - d / 8}{2 d} \\ \Rightarrow \sin θ_{1} = \frac{\pm λ / 2 - λ / 8}{2 λ} = \pm \frac{1}{4} - \frac{1}{16} \end{aligned}$

Question:21

Four identical monochromatic sources A, B, C, D, as shown in the (Fig.10.7) produce waves of the same wavelength and are coherent. Two receiver R1 and R2 are at great but equal distances from B.

(i) Which of the two receivers picks up the larger signal?
(ii) Which of the two receivers picks up the larger signal when B is turned off?
(iii) Which of the two receivers picks up the larger signal when D is turned off?
(iv) Which of the two receivers can distinguish which of the sources B or D has been turned off?

Answer:

(i) Let us consider the disturbances at the receiver R₁, which is at a distance d from B.
Let the equation of wave at R₁, because of A be

y_{A} = a \cos ω t

. . . . . . (i)

The path difference of the signal from A with that from B is

λ / 2

and hence, the phase difference

Δ ϕ = \frac{2 π}{λ} \times (p a t h d i f f e r e n c e) = \frac{2 π}{λ} \times \frac{λ}{2}

Thus, the wave equation at R1, because of B is

y_{B} = a \cos (ω t - π) = - a \cos ω t

. . . . . . (i i)

The path difference of the signal from C with that from A is

λ

and hence the phase difference

Δ ϕ = \frac{2 π}{λ} \times (p a t h d i f f e r e n c e) = \frac{2 π}{λ} \times λ = 2 π

Thus, the wave equation at R1 because of C is

y_{C} = a \cos ω t = a \cos (ω t - 2 π) = a \cos ω t

. . . . . . (i i i)

The path difference between the signal from D with that of A is

Δ_{x_{R_{1}}} = \sqrt{d^{2} + {(\frac{λ}{2})}^{2}} - (d - \frac{λ}{2}) = d {(1 + \frac{λ^{2}}{4 d^{2}})}^{1 / 2} - d + \frac{λ}{2}

= d (1 + \frac{λ^{2}}{8 d^{2}}) - d . \frac{λ}{2} = \frac{λ}{2} (∵ d >> λ)

Therefore, phase difference is

π

y_{D} = a \cos (ω t - π) = - a \cos ω t

. . . . . . (i i i)

The resultant signal picked up at R_1, from all the four sources is the summation of all four waves,

y_{R_{1}} = y_{A} + y_{B} + y_{C} + y_{D}

y_{R_{1}} = a \cos ω t - a \cos ω t + a \cos ω t - a \cos ω t = 0

Thus, the signal picked up at R₁ is zero.
Now let us consider the resultant signal received at R_2. Let the equation of wave at R₂. Let the equation
of wave at R₂, because of B be

y_{B} = a_{1} \cos ω t

The path difference of the signal from from D with that from B is

\frac{λ}{2}

and
hence, the phase difference

Δ ϕ = \frac{2 π}{π} \times (path difference) = \frac{2 π}{λ} \times \frac{λ}{2} = π

Thus, the wave equation at

R_{2},

because of D is

y_{B} = a_{1} \cos (ω_{t} - π) = - a_{1} \cos ω t . . . . (i i)

The path difference between signal at A and that at B is

Δ x_{R_{2}} = \sqrt{{(d)}^{2} + {(\frac{λ}{2})}^{2}} - d = d {(1 + \frac{λ^{2}}{4 d^{2}})}^{\frac{1}{2}} - d ≃ \frac{λ^{2}}{8 d^{2}}

d >> λ,

therefore this path differences

Δ x_{R_{2}} \to 0

and phase difference

Δ ϕ = \frac{2 π}{λ} \times (path difference)

= \frac{2 π}{λ} \times 0 \to 0 (very small) = ϕ (say)

Hence,

y_{B} = a_{1} \cos (ω_{t} - ϕ)

Similarly,

y_{B} = a_{1} \cos (ω_{t} - ϕ)

The resultant signal picked up at

R_{2},

from all the four sources is the summation of all four waves,

y_{R_{2}} = y_{A} + y_{B} + y_{C} + y_{D}

y_{R_{2}} = a_{1} \cos ω t - a_{1} \cos ω t + a_{1} \cos (ω t - ϕ) + a_{1} \cos (ω t - ϕ)

= 2 a_{1} \cos (ω t - ϕ)

∴

Signal picked up by

R_{2}

y_{R_{2}} = 2 a_{1} \cos (ω t - ϕ)

∴

{[V_{R_{2}}]}^{2} = 4 a_{1}^{2} \cos^{2} (ω t - ϕ) \Rightarrow ⟨ I_{R_{2}} ⟩ = 2 a_{1}^{2}

Thus,

R_{2}

picks up the larger signal.
(ii) If B is switched off,

R_{1}

, picks up

y = a \cos ω t

∴

⟨ I_{R_{1}} ⟩ = a^{2} < \cos^{2} ω t >= \frac{a^{2}}{2}

R_{2}

picks up

y = a \cos ω t

⟨ I_{R_{2}} ⟩ = a^{2} < \cos^{2} ω t >= \frac{a^{2}}{2}

Thus ,

R_{1}

and

R_{2}

pick up the same signal
(iii) If D is switched off.

R_{1}

picks up

y = a \cos ω t

∴

⟨ I_{R_{1}} ⟩ = \frac{1}{2} a^{2}

R_{2}

picks up

y = 3 a \cos ω t

∴

⟨ I_{R_{2}} ⟩ = 9 a^{2} < \cos^{2} ω t >= \frac{9 a^{2}}{2}

Thus,

R_{2}

picks up larger signal compared to

R_{1}

.
(iv) Thus, a signal at

R_{1}

indicates B has been switched off and an enhanced signal at

R_{2}

indicates D has been switched off.

Question:22

The optical properties of a medium are governed by the relative permittivity (εr) and relative permeability (μr ). The refractive index is defined as $\sqrt{μ, ϵ} = n$ For ordinary material εr > 0 and μr > 0, and the positive sign is taken for the square root. In 1964, a Russian scientist V. Veselago postulated the existence of material with εr < 0 and μr < 0. Since then, such ‘metamaterials’ have been produced in the laboratories and their optical properties studied. For such materials $n = \sqrt{μ, ϵ}$
As light enters a medium of such refractive index, the phases travel away from the direction of propagation.

(i) According to the description above shows that if rays of light enter such a medium from the air (refractive index = 1) at an angle in 2^nd quadrant, them the refracted beam is in the 3^rd quadrant.
(ii) Prove that Snell’s law holds for such a medium.

Answer:

Again consider figure (i), let AB represent the incident wavefront and DE represent the refracted wavefront. All point on a wavefront must be in same phase and in turn, must have the same optical path length.

Thus $- \sqrt{ϵ_{r} μ_{r}} A E = B C - \sqrt{ϵ_{r} μ_{r}} C D$
Or $B C = \sqrt{ϵ_{r} μ_{r}} (C D - A E)$

$B C > 0, C D > A E$

As showing that the postulate is reasonable. If however, the light proceeded in the sense it does for ordinary material (viz. in the fourth quadrant, Fig.2)

Then $- \sqrt{ϵ_{r} μ_{r}} A E = B C - \sqrt{ϵ_{r} μ_{r}} C D$
Or $B C = \sqrt{ϵ_{r} μ_{r}} (C D - A E)$
if $B C > 0$ , then $C D > A E$
Which is obvious from Fig. (i). Hence, the postulate is reasonable.
However, if the light proceeds in the sense it does for ordinary material, (going from the second quadrant to 4th quadrant)as shown in Fig. (i). then proceeding as above,

$- \sqrt{ϵ_{r} μ_{r}} A E = B C - \sqrt{ϵ_{r} μ_{r}} C D$

Or $B C = \sqrt{ϵ_{r} μ_{r}} (C D - A E)$
As $A E > C D$ Therfore $B C < 0$
Which is not possible. Hence, the postulate is correct

(ii) From Fig (i),

B C = A C \sin θ, a n d C D - A E = A C \sin θ A s B C = - \sqrt{μ_{r} ϵ_{r}} (A E - C D) ∴ A C \sin θ_{i} = - \sqrt{ϵ_{r} μ_{r}} A C \sin θ o r \frac{\sin θ_{i}}{\sin θ_{r}} = \sqrt{ϵ_{r} μ_{r}} n

Which proves Snell's law

Question:23

To ensure almost 100 per cent transitivity, photographic lenses are often coated with a thin layer of dielectric material. The refractive index of this material is intermediated between that of (which makes the optical element of the lens). A typically used dielectric film is MgF₂ (n = 1.38). What should the thickness of the film be so that at the center of the visible spectrum (5500 Å) there is a maximum transmission

Answer:

IA is an incident ray at point A is such that the incident angle I is formed from air to the film surface.
AR₁ and AD are the reflected and refracted rays, respectively. D is the point on which the partial reflection of glass and film interface. CR₂ and AR₁ are parallel.

\begin{aligned} μ (A D + C D) - A B \\ A D = A C = \frac{d}{\cos r} \\ d \tan r = \frac{A C}{2} \\ d = 1000 A \end{aligned}

Main Subtopics Covered in NCERT Exemplar Class 12 Physics Solutions Chapter 10 Wave Optics

Introduction
Huygens Principle
Refraction and reflection of plane waves using Huygens Principle
Refraction of a plane wave
Refraction at a rarer medium
Reflection of a plane wave by a plane surface
The Doppler Effect
Coherent and Incoherent Addition of Waves
Interference of light waves and Young’s experiment
Diffraction
The single slit
Seeing the single-slit diffraction pattern
Resolving power of optical instruments
The validity of ray optics
Polarisation
Polarisation by scattering
Polarisation by reflection

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NCERT Exemplar Class 12 Physics Solutions Chapter-Wise

Chapter 1 Electric Charges and Fields

Chapter 2 Electrostatic Potential and Capacitance

Chapter 3 Current Electricity

Chapter 4 Moving Charges and Magnetism

Chapter 5 Magnetism and Matter

Chapter 6 Electromagnetic Induction

Chapter 7 Alternating Current

Chapter 8 Electromagnetic Waves

Chapter 9 Ray Optics and Optical Instruments

Chapter 10 Wave Optics

Chapter 11 Dual Nature of Radiation and Matter

Chapter 12 Atoms

Chapter 13 Nuclei

Chapter 14 Semiconductor Electronics

Chapter 15 Communication Systems

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NCERT Exemplar Class 12 Solutions

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NCERT solutions for Class 12 Physics Chapter 1 Electric Charges and Fields

NCERT solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance

NCERT solutions for Class 12 Physics Chapter 3 Current Electricity

NCERT solutions for Class 12 Physics Chapter 4 Moving charges and magnetism

NCERT solutions for Class 12 Physics Chapter 5 Magnetism and Matter

NCERT solutions for Class 12 Physics Chapter 6 Electromagnetic Induction

NCERT solutions for Class 12 Physics Chapter 7 Alternating Current

NCERT solutions for Class 12 Physics Chapter 8 Electromagnetic Waves

NCERT solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments

NCERT solutions for Class 12 Physics Chapter 10 Wave Optics

NCERT solutions for Class 12 Physics Chapter 11 Dual nature of radiation and matter

NCERT solutions for Class 12 Physics Chapter 12 Atoms

NCERT solutions for Class 12 Physics Chapter 13 Nuclei

NCERT Solutions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices And Simple Circuit

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Frequently Asked Questions (FAQs)

Q: What is the main focus of Chapter 10: Wave Optics in Class 12 Physics?

This chapter explains how light behaves like a wave. It covers important topics like interference (light adding up or cancelling), diffraction (bending around edges), and polarisation (filtering light direction). These help explain real-life effects like colors in soap bubbles or the working of polarised sunglasses.

Q: Are the solutions provided by Careers360 good for last-minute revision?

Yes! The Careers360 NCERT Exemplar Solutions are step-by-step and based on the latest CBSE syllabus. They are great for revising key formulas, clearing doubts, and practicing high-level questions before your exams.

Q: Are these solutions helpful in board exams?

Yes, these solutions of NCERT will be helpful in board exam preparation as one can understand the chapter and topics better.

Q: How these solutions can be used?

One can learn the how to solve questions for this chapter in board exam, can also cross check their answers while practicing.

Q: Are these questions solved as per CBSE?

Yes, each and every question is solved as per the CBSE pattern in NCERT exemplar Class 12 Physics solutions chapter 10, which will help in understanding each step separately.

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Questions related to CBSE Class 12th

On Question asked by student community

Have a question related to CBSE Class 12th ?

Can i improve my 12 result this year in PCM

Hello,

If you want to improve the Class 12 PCM results, you can appear in the improvement exam. This exam will help you to retake one or more subjects to achieve a better score. You should check the official website for details and the deadline of this exam.

I hope it will clear your query!!

Read Complete Answer

Rahul Mandal

8 Sep'25

how may marks should i get in my 12th board exams to get fee concesion in sastra university

Hello Aspirant,

SASTRA University commonly provides concessions and scholarships based on merit in class 12 board exams and JEE Main purposes with regard to board merit you need above 95% in PCM (or on aggregate) to get bigger concessions, usually if you scored 90% and above you may get partial concessions. I suppose the exact cut offs may change yearly on application rates too.

Read Complete Answer

gopi

4 Sep'25

Which course is best for computer science students after 12th standard

Hello,

After 12th, if you are interested in computer science, the best courses are:

B.Tech in Computer Science Engineering (CSE) – most popular choice.
BCA (Bachelor of Computer Applications) – good for software and IT jobs.
B.Sc. Computer Science / IT – good for higher studies and research.
B.Tech in Information Technology (IT) – focuses on IT and networking.

All these courses have good career scope. Choose based on your interest in coding, software, hardware, or IT field.

Hope it helps !

Read Complete Answer

Prachi Kumari

4 Sep'25

I am a CBSE class 12 private candidate will my maksheet of CBSE supplement exam will be delivered on main exam form address for supplementary exam form address ,my both addresses are different please tell

Hello Vanshika,

CBSE generally forwards the marksheet for the supplementary exam to the correspondence address as identified in the supplementary exam application form. It is not sent to the address indicated in the main exam form. Addresses that differ will use the supplementary exam address.

Read Complete Answer

gopi

3 Sep'25

12 arts previous years board paper 2025

To find Class 12 Arts board papers, go to the official website of your education board, then click on the Sample Papers, Previous Years Question Papers(PYQ) or Model Papers section, and select the Arts stream. You will find papers for the various academic year. You can then select the year of which you want to solve and do your practice. There are many other educational websites that post pyqs on their website you can also visit that.

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT Exemplar Class 12 Physics Solutions Chapter 10 Wave Optics

NCERT Exemplar Class 12 Physics Solutions Chapter 10: MCQI

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NCERT Exemplar Class 12 Physics Solutions Chapter 10: Very Short Answer

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NCERT Exemplar Class 12 Physics Solutions Chapter-Wise

NCERT Exemplar Class 12 Solutions

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