Wave Optics Chapter 10: NCERT Solutions for Class 12 Physics

Wave Optics Chapter 10 deals is an essential part of the Class 12th Physics for scoring good marks in upcoming board examinations. Students must make a habit of consistent practice and revision to perform well. Students struggling with chapter or having doubts must not worry too hard.

 

SimplyAcad is always there for you to support you throughout, here are the provided NCERT Solutions Class 12 Wave Optics Chapter 10, so the students use to solve the various questions provided in the prescribed Physics textbook.

Students can easily access the free material and take its advantage to clear all the existing confusions on the concepts and save time.

 

NCERT Solutions Class 12 Wave Optics Chapter 10

Question 1:

Monochromatic light having a wavelength of 589 nm from the air is incident on a water surface. Find the frequency, wavelength and speed of (i) reflected and (ii) refracted light. [1.33 is the refractive index of water]

Answer:

Monochromatic light incident having wavelength,

𝜆

= 589 nm = 589 x 10-9 m

Speed of light in air, c = 3 x 108 m s-1

Refractive index of water,

𝜇

= 1.33

(i) The ray will be reflected back to the same medium through which it passed.

Therefore, the wavelength, speed and frequency of the reflected ray will be the same as that of the incident ray.

The frequency of light can be found from the relation:

Hence, c = 3 x 108 m s-1, 5.09 × 1014 Hz, and 589 nm are the speed, frequency and wavelength of the reflected light.

(b) The frequency of light which is travelling never depends upon the property of the medium. Therefore, the frequency of the refracted ray in water will be equal to the frequency of the incident or reflected light in the air.

Refracted frequency, v = 5.09 x 1014 Hz

Following is the relation between the speed of light in water and the refractive index of the water:

Below is the relation for finding the wavelength of light in water:

Therefore, 444.007 × 10-9 m, 444.01nm, and 5.09 × 1014 Hz are the speed, frequency, and wavelength of the refracted light.

Question 2: What is the shape of the wavefront in each of the following cases: \begin{enumerate} \item Light diverging from a point source. \item Light emerging out of a convex lens when a point source is placed at its focus. \item The portion of the wavefront of light from a distant star intercepted by the Earth. \end{enumerate} \section*

{Solution} \begin{enumerate} \item \textbf{Light diverging from a point source:} \\ The wavefront of light diverging from a point source is spherical. This is because the point source emits light uniformly in all directions, creating spherical wavefronts that expand outward as the light travels. \\ \textbf{Shape of the wavefront:} Spherical \item \textbf{Light emerging out of a convex lens when a point source is placed at its focus:} \\ When a point source is placed at the focus of a convex lens, the lens converges the light into parallel rays. Consequently, the shape of the wavefront after passing through the lens becomes planar (or nearly planar) as the light emerges in parallel lines. \\ \textbf{Shape of the wavefront:} Planar \item \textbf{The portion of the wavefront of light from a distant star intercepted by the Earth:} \\ Light from a distant star is essentially parallel by the time it reaches Earth. Hence, the portion of the wavefront intercepted by Earth is nearly planar. This is because the curvature of the wavefront is negligible over such vast distances. \\ \textbf{Shape of the wavefront:} Planar

Question 3:

(i) The refractive index of glass is 1.5. What is the speed of light in glass? The speed of light in a vacuum is ( 3.0 x 108 m s-1 )

(ii) Is the speed of light in glass independent of the colour of light? If not, which of the two colours, red and violet, travels slower in a glass prism?

Answer:

(i) Refractive Index of glass,

𝜇

= 1.5

Speed of light, c = 3 × 108 ms-1

The relation for the speed of light in a glass is:

Hence, the speed of light in glass is 2 × 108 m s-1

(ii) The speed of light is dependent on the colour of the light. For a white light, the refractive index of the violet component is greater than the refractive index of the red component. So, the speed of violet light is less than the speed of red light in the glass. This will reduce the speed of violet light in a glass prism when compared with red light.

Question 4: In a Young’s double-slit experiment, the slits are separated by \(0.28\) mm and the screen is placed \(1.4\) m away. The distance between the central bright fringe and the fourth bright fringe is measured to be \(1.2\) cm. Determine the wavelength of light used in the experiment. \section*

{Solution} Given: \begin{itemize} \item Distance between the slits, \(d = 0.28 \text{ mm} = 0.28 \times 10^{-3} \text{ m}\) \item Distance between the slits and the screen, \(D = 1.4 \text{ m}\) \item Distance between the central fringe and the fourth fringe, \(u = 1.2 \text{ cm} = 1.2 \times 10^{-2} \text{ m}\) \end{itemize} In the case of constructive interference, the distance between the central fringe and the \(n\)-th fringe is given by: \[ u = n \frac{\lambda D}{d} \] Here, \(n = 4\). Rearranging the formula to solve for the wavelength \(\lambda\): \[ \lambda = \frac{u \cdot d}{n \cdot D} \] Substitute the given values: \[ \lambda = \frac{1.2 \times 10^{-2} \times 0.28 \times 10^{-3}}{4 \times 1.4} \] Calculate: \[ \lambda = \frac{1.2 \times 0.28 \times 10^{-5}}{5.6} \] \[ \lambda = \frac{0.336 \times 10^{-5}}{5.6} \] \[ \lambda = 0.06 \times 10^{-5} \text{ m} = 6 \times 10^{-7} \text{ m} = 600 \text{ nm} \] Therefore, the wavelength of light used in the experiment is \(600\) nm.

Question 5:

In Young’s double-slit experiment using the monochromatic light of wavelength

𝜆

, the intensity of light at a point on the screen where path difference is

𝜆

is K units. What is the intensity of light at a point where the path difference is

Answer:

Let

𝐼1

and

𝐼2

be the intensity of the two light waves. Their resultant intensities can be obtained as:

= Phase difference between the two waves

For monochromatic light waves:

𝐼1

=

𝐼2

Therefore

Phase difference =

When path difference=

𝜆3

Hence, the intensity of light at a point where the path difference is

Question 6: A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in Young’s double-slit experiment.

(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.

(b) What is the least distance from the central maximum where the bright fringes due to both wavelengths coincide?

 Answer:

Wavelength of the light beam,

𝜆1

= 650 nm

Wavelength of another light beam,

𝜆2

= 520 nm

Distance of the slits from the screen = D

Distance between the two slits = d

(i) Distance of the nth bright fringe on the screen from the central maximum is given by the relation,

x =

For the third bright fringe, n = 3

Therefore, x =

(b) Let, the

bright fringe due to wavelength

𝜆2

and

bright fringe due to wavelength

𝜆2

coincide on the screen. The value of n can be obtained by equating the conditions for bright fringes:

520n = 650n – 650

650 = 130n

Therefore, n = 5

Hence, the least distance from the central maximum can be obtained by the relation:

Note: The value of d and D are not given in the question.

Question 7:

In a double-slit experiment, 0.2° is found to be the angular width of a fringe on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take the refractive index of water to be

Answer:

Distance of the screen from the slits, D = 1 m

Wavelength of light used,

𝜆1

= 600 nm

Angular width of the fringe in air

𝜃1

= 0.2°

Angular width of the fringe in water=

𝜃2

Refractive index of water,

is the relation between the refractive index and the angular width

Therefore, 0.15° is the reduction in the angular width of the fringe in water.

Question 8: What is the Brewster angle for air-to-glass transition? (Refractive index of glass = 1.5.)

 Answer:

Refractive index of glass,

𝜇=1.5

Consider Brewster angle =

𝜃

Following is the relation between the Brewster angle and the refractive index:

Therefore, the Brewster angle for air-to-glass transition is 56.31°

Question 9: Light of wavelength 5000 Armstrong falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light? For what angle of incidence is the reflected ray normal to the incident ray?

Answer:

Wavelength of incident light,

= 5000 Armstrong = 5000 x 10-10 m

Speed of light, c =3 x 108 m

Following is the relation for the frequency of incident light:

v =

The wavelength and frequency of incident light are equal to the reflected ray. Therefore, 5000 Armstrong and

Hz is the wavelength and frequency of the reflected light. When the reflected ray is normal to the incident ray, the sum of the angle of incidence,

∠𝑖

and angle of reflection,

∠𝑟

is 90°.

From the laws of reflection, we know that the angle of incidence is always equal to the angle of reflection

Therefore, 45° is the angle of incidence.

Question 10:

Estimate the distance for which ray optics is a good approximation for an aperture of 4 mm and wavelength of 400 nm.

 Answer:

Fresnel’s distance (

) is the distance which is used in ray optics for a good approximation. Following is the relation,

Where,

Aperture width, a = 4 mm = 4 × 10-3 m

Wavelength of light,

𝜆

= 400 nm = 400 × 10-9 m

Therefore, 40 m is the distance for which the ray optics is a good approximation.

Question 11: The 6563 Å Hα line emitted by hydrogen in a star is found to be red-shifted by 15 Å. Estimate the speed with which the star is receding from the Earth.

Answer:

λ = 6563 Å

Δλ = 15 Å

Since the star is receding, the velocity (v) is negative.

Δλ = – vλ/c

v = – cΔλ/λ

= – (3 x 108) x (15 Å/ 6563 Å)

= – 6.86 x 105 m/s

Question 12: Explain how Corpuscular theory predicts the speed of light in a medium, say, water, to be greater than the speed of light in a vacuum. Is the prediction confirmed by experimental determination of the speed of light in water? If not, which alternative picture of light is consistent with the experiment?

Answer:

According to Newton’s Corpuscular theory, the velocity of light in the denser medium (water) is greater than the velocity of light in the rarer medium (vacuum). This was experimentally wrong.

At the angle of incidence (i) of the light of velocity v, the angle of refraction is r.

Due to the change in the medium, the change in the velocity of light in water is v.

Using Snell’s law,

c sin i = v sin r ——(1)

The relation between the velocities and the refractive index is

v/c = μ ———(2)

v/c = sin i/sin r = μ ——-(3)

But μ> 1, so v > c is not possible.

Huygens’ wave theory is consistent with the experiment.

Question 13: You have learned in the text how Huygens’ principle leads to the laws of reflection and refraction. Use the same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equal to the object distance from the mirror. \section*

{Solution} Consider a point object \(O\) placed in front of a plane mirror \(MO’\) at a distance \(r\) from it. To apply Huygens’ principle, follow these steps: 1. **Wavefront Construction:** – Treat the object \(O\) as the center of a spherical wavefront. – Draw a circle with center \(O\) and radius \(r\) which touches the plane mirror at point \(O’\). This circle represents the wavefront of the incident light. 2. **Virtual Wavefront Formation:** – According to Huygens’ principle, each point on a wavefront can be considered as a source of secondary wavelets. – If the mirror \(MO’\) were not present, a similar wavefront \(X’Y’\) would form behind the mirror at a distance \(r\) from the mirror, effectively creating a virtual wavefront as if the mirror were a real source of light. 3. **Image Formation:** – The wavefront \(X’Y’\) behind the mirror acts as if it were the reflected wavefront. – This virtual wavefront \(X’Y’\) appears to originate from a point \(O’\) at a distance \(r\) behind the mirror. – Thus, the virtual image formed is situated at a distance \(r\) from the mirror, which is the same distance as the object \(O\) from the mirror. Therefore, according to Huygens’ principle, a point object placed in front of a plane mirror produces a virtual image at a distance from the mirror equal to the object distance from the mirror.

Question 14: Let us list some of the factors which could possibly influence the speed of wave propagation:
(i) Nature of the source
(ii) The direction of propagation
(iii) The motion of the source and/or observer
(iv) Wavelength
(v) The intensity of the wave. On which of these factors, if any, does (a) the speed of light in a vacuum (b) the speed of light in a medium (say, glass or water) depend?

Answer:

(a) The speed of light in the vacuum does not depend on any of the factors listed.

(b) The speed of light in the medium depends on the wavelength of the light in that medium.

Question 15: For sound waves, the Doppler formula for frequency shift differs slightly between the two situations: (i) source at rest; observer moving, and (ii) source moving; observer at rest. The exact Doppler formulas for the case of light waves in a vacuum are, however, strictly identical for these situations. Explain why this should be so. Would you expect the formulas to be strictly identical for the two situations in the case of light travelling in a medium?

Answer:

The Doppler formula differs slightly between the two situations because sound waves can travel only through a medium. The motion of the observer relative to the medium is different in both cases. Hence, the doppler formula is different.

Light waves can propagate in a vacuum. In a vacuum, the speed of light does not depend on the motion of the observer and the source.

When light travels in the medium, the doppler formula for the two cases will be different.

Question 16: In a double-slit experiment using the light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1°. What is the spacing between the two slits?

Answer:

The wavelength of the light, λ = 600 nm

The angular width of the fringe formed, θ = 0.10 = 0.1 π/180

Spacing between the slits, d = λ/θ = (600 x 10-9 x 180)/(0.1 x 3.14)

= 108000 x 10-9/0.314

d =3.44 x 10-4 m

Question 17: Answer the following questions:
(a) In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?
(b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?
(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?
(d) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily?
(e) Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification?

Answer:

(a) In a single slit diffraction experiment, if the width of the slit is made double the original width, the size of the central diffraction band reduces to half, and the intensity of the band increases four times.

(b) If the width of each slit is comparable to the wavelength of light used, the interference pattern thus obtained in the double-slit experiment is modified by diffraction from each of the two slits.

(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. The bright spot is formed due to the constructive interference of the light waves that get diffracted from the edges of the circular obstacle.

(d) The obstacle bends the waves by a large angle if the wavelength of the wave is comparable with the size of the obstacle. The wavelength of the light wave is much smaller than the size of the wall. Therefore, the diffraction angle is also very small. As a result, the students will not be able to see each other. On the other hand, the size of the wall and the wavelength of the sound wave is comparable. Hence, the diffraction angle is large. Therefore, students can hear each other.

(e) The size of the aperture in the optical instruments is much larger than the wavelength of light. Therefore, the diffraction effect of light is negligible in these instruments. Thus, the assumption of light travelling in a straight line can be used in these instruments.

Question 18: Two towers on top of two hills are 40 km apart. The line joining them passes 50 m above a hill halfway between the towers. What is the longest wavelength of radio waves which can be sent between the towers without appreciable diffraction effects?

Answer:

Distance between the towers = 40 km

Height of the line joining the hills, d = 50 m

The radial spread of the radio waves must not exceed 50 m

Aperture a = d = 50 m

The hill is located halfway between the towers. Therefore, Fresnel’s distance is Zp = 20 km

Fresnel’s distance can be given by the equation, Zp = a2/λ

λ = a2/Zp

= (50)2/(20 x 103)

= 250/20 = 12.5 x 10-3 m = 12.5 cm

Question 19: A parallel beam of light of wavelength 500 nm falls on a narrow slit, and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit.

Answer:

The wavelength of the beam of light, λ = 500 nm

Distance between the slit and the screen, D= 1 m

Distance of the first minimum from the centre of the screen,  x = 2.5 mm = 2.5 x 10-3 m

First minima, n = 1

Consider the equation, nλ = xd/D

⇒ d = nλD/x = ( 1 x 500 x 10-9 x 1) /( 2.5 x 10-3 )

= 200 x 10-6 m

Question 20: Answer the following questions:
(a) When a low-flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Suggest a possible explanation.
(b) As you have learnt in the text, the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns. What is the justification for this principle?

Answer:

(a) The weak radar signals from the aircraft interfere with the TV signal received by the antenna.

(b)  This is because superposition follows from the linear character of a differential equation that governs wave motion.
​

Question 21: In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of nλ/a. Justify this by suitably dividing the slit to bring out the cancellation.

Answer:

Let “a”  be the width of a single slit. The single slit is further divided into n smaller slits of width a’.

a’ = a/n

Each of the smaller slits should produce zero intensity for the single slit to produce zero intensity.

For this to happen, the angle of diffraction, θ = λ/a’

⇒ θ  = λ/(a/n)

or, θ = nλ/a

Therefore, in deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of nλ/a.