# NCERT Solutions Class 12 Physics Chapter 10

## NCERT Solutions Class 12 Physics Chapter 10- Wave Optics

The NCERT Solutions Class 12 Physics Chapter 10 Wave Optics provided by Extramarks is an important resource. Students appearing for the Class 12 examinations are advised to refer to these notes for effective preparation. Every minute detail provided is based on the latest guidelines issued by the CBSE board.

Students can rely on the chapter 10 Physics class 12 NCERT solutions for their competitive entrance exam preparation. With the help of the notes prepared by subject matter experts, students can get a firm grip on theoretical knowledge and enhance their problem-solving abilities.

The Wave Optics ch 10 Physics class 12 carries maximum marks and is regarded as the most important chapter from the entire CBSE syllabus. In the chapter, students will learn different laws of refraction and reflection. In addition, students will gain insight into Huygens, Young, and Fresnel’s experiments to understand the behaviour of light and wave theory. The Extramarks NCERT Solutions Class 12 Physics Chapter 10 also covers topics like Dopplers effect and polarization in detail. Besides derivations and theory, the calculation of numerical problems from various topics is explained step-by-step.

The class 12 Physics chapter 10 NCERT solutions aim to provide students with an opportunity to clear their doubts and enable the all-round development of young learners. In addition, Extramarks provides primary and secondary class study material like NCERT Solutions Class 1, NCERT Solutions class 2, and NCERT Solutions Class 3 for school students.

### Key Topics Covered In NCERT Solutions Class 12 Physics Chapter 10

The NCERT Solutions Class 12 Physics Chapter 10 provides many MCQs, exemplary problems, in-text questions, and exercises that help students gain knowledge about the topics. These notes are prepared by experts having years of experience by analyzing the CBSE marking scheme. In the NCERT Solutions Class 12 Physics Chapter 10, all minute detail and apt knowledge about the following topics are included:

 10.1 10.2 10.3 10.4 10.5 10.6 10.7 Introduction Huygens Principle Refraction And Reflection Of Plane Waves Using Huygens Principle Coherent And Incoherent Addition Of Waves Interference Of Light Waves And Young’s Experiment Diffraction Polarisation

10.1 Introduction

Ch 10 Physics class 12 is based on the wave theory of light. The chapter has a detailed explanation of refraction and reflection of waves with the help of the Huygens Principle, coherent and incoherent addition of waves, interference of light waves, and validity of ray optics.

Students will learn the following topics in the NCERT Solutions Class 12 Physics Chapter 10:

• Huygen’s principle
• Derivation of laws of reflection and refraction
• Interference based on the superposition principle
• Diffraction based on the principle of HuygensFresnel
• Polarization

10.2 Huygens Principle

Before studying the Huygens Principle, there is a simple introduction of wavefront and its types based on the shape of the source of light, i.e. Spherical and Cylindrical wavefront. Students are introduced to Huygens’s Principle and its assumptions in this section. The NCERT Solutions Class 12 Physics Chapter 10 includes many numerical questions depending on the relation between intensity and amplitude.

10.3 Refraction And Reflection Of Plane Waves Using Huygens Principle

Under this section in the Class 12 Physics chapter 10 NCERT solutions, students will learn about refraction of a plane wave, refraction at a rarer medium, reflection by a plane surface, and the Doppler Effect. These derivations are often asked in exams.

The concepts included in this sector are explained in detail in the NCERT Solutions Class 12 Physics Chapter 10. Students are advised to focus on this section to understand the next concepts easily.

10.4 Coherent And Incoherent Addition Of Waves

Based on the above topic, students learn the theory of the interference pattern produced by

the superposition of two waves. Students also learn the difference between Coherent And Incoherent waves. To understand this section, students should recall the superposition principle explained in the NCERT Solutions Class 11. The step-by-step and detailed explanation of the derivation will help students perceive the concept better to attempt all related questions in the exam.

Students are advised to use the chapter 9 Physics class 12 NCERT solutions to significantly enhance their understanding and learning process.

10.5 Interference Of Light Waves And Young’s Experiment

This section of NCERT Solutions Class 12 Physics Chapter 10 includes brief information about the Interference of Light. They also learn about the different types of interference, such as constructive and destructive interference. The resultant topic intensity due to two identical waves gives the relation between intensity for two coherent sources. Under this section, students have to solve various sums based on average intensity and the ratio of maximum and minimum intensities. The Young’s Double Slit Experiment includes a discussion of various topics like path difference, bright fringe, dark fringe, and Shifting of Fringe Pattern in YDSE.

Here are some of the formulas included in this section:

• Width of one fringe: =Dd
• Fringe width  1d where d is the separation between the sources
• Position of nth bright fringe from central maxima: xn =nDd = n
• Position of nth dark fringe from central maxima: xn =(2n-1)Dd = (2n-1)2

10.6 Diffraction

This section in NCERT Solutions Class 12 Physics Chapter 10 helps students understand the phenomenon of diffraction and its types, i.e. Fresnel and Fraunhofer Diffraction. It further includes the Single Slit, seeing the Single Slit Diffraction pattern, Resolving Power of optical instruments, and the validity of Ray Optics. In addition, students will learn more about the Objective lens of instruments like telescopes or microscopes. Sums based on angular half-width of Airy disc and the lateral width of the image will be asked in the exams.

10.7 Polarisation

Students will get a clear understanding of polarisation by Scattering and Reflection. Concepts like polarised light, polaroid, and Fresnel distance have been explained lucidly. Then, with the help of the NCERT Solutions Class 12 Physics Chapter 10, students will learn about Brewster’s law which gives the relationship between the angle of polarization and refractive index and Malus law.

### List of NCERT Solutions Class 12 Physics Chapter 10 Exercise & Answer Solutions

The NCERT Solutions Class 12 Physics Chapter 10 study materials are available on the Extramarks website. They include key points, important formulas, derivations, and chapter summaries. These notes ensure a thorough understanding of all essential topics included in the chapter. In addition, the chapter 10 Physics class 12 NCERT solutions provide a quick revision and evaluate learning to score better marks in exams. Students may click on the below links to get more information on ch 10 Physics class 12:

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### NCERT Exemplar Class 12 Physics

Physics is usually regarded as one of the most difficult subjects. However, students must gain high-level knowledge of fundamental concepts. To help students study Physics easily, Extramarks provides the NCERT solutions Class 12 Physics Chapter 10. Students can get apt and minute details of all topics included in this chapter. In addition, students can solve many MCQs, Numerical Problems, and CBSE extra questions enlisted in the notes. The NCERT Solutions help students prepare for competitive examinations held on state, national and international levels.

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### Key Features of NCERT Solutions Class 12 Physics Chapter 10

The key features of NCERT Solutions Class 12 Physics Chapter 10 include

• The NCERT Solutions are considered important reference material that enables quick revision.
• Subject Experts have extensively prepared these notes by analyzing several previous year questions papers and sample papers.
• The NCERT Solutions Class 12 Physics Chapter 10 includes important, repetitive, and those questions which are very likely to be asked in the exams.
• These notes can be accessed from anywhere on any device, namely, mobile, laptop, tablet, or PC.
• Students can clear their doubts and score well in exams with detailed explanations and step-by-step solutions included in the NCERT Solutions Class 12 Physics Chapter 10.

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Q.1 Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected, and (b) refracted light?
Refractive index of water is 1.33.

Ans.

Here, wavelength of incident monochromatic light,
λ = 589 nm
= 589 × 10−9 m
Speed of light in air, c = 3 × 108 ms-1
Refractive index of water, μ = 1.33
Frequency of light is given as:

$\begin{array}{l}\text{ν =}\frac{\text{c}}{\text{λ}}\\ \text{=}\frac{{\text{3 × 10}}^{\text{8}}{\text{ms}}^{\text{-1}}}{{\text{589 ×10}}^{\text{-9}}\text{m}}\\ {\text{= 5.09×10}}^{\text{14}}\text{ Hz}\end{array}$

(a) Since the light reflects back into the air after striking the water surface, therefore, the change of medium does not take place. Thus, the wavelength, speed, and frequency of the reflected light will remain the same as that of the incident light.

(b) Since the frequency of light is independent of the properties of the medium in which it is travelling, therefore, the frequency of the refracted ray in water will remain the same as that of the frequency of the incident or reflected light in air.
Hence, refracted frequency, ν = 5.09×1014 Hz

$\begin{array}{l}\text{Speed of light in water is given as:}\\ \text{v’ =}\frac{\text{c}}{\text{μ}}\\ \text{=}\frac{{\text{3×10}}^{\text{8}}}{\text{1.33}}\\ {\text{= 2.25×10}}^{\text{8}}{\text{ ms}}^{\text{-1}}\\ \text{Wavelength of light will be eqaul to}\frac{\text{v’}}{\mathrm{\nu }\text{‘}}\\ \mathrm{\lambda }=\frac{2.25×10\text{}{\mathrm{ms}}^{-1}}{\text{5}.0\text{9}×\text{1}{0}^{\text{14}}\text{Hz}}\\ =\text{444.01 nm}\end{array}$

Q.2 What is the shape of the wave front in each of the following cases:
(a) Light diverging from a point source.
(b) Light emerging out of a convex lens when a point source is placed at its focus.
(c) The portion of the wavefront of light from a distant star intercepted by the Earth.

Ans.

(a) The wavefront of a light diverging from a point source is diverging spherical wavefront as shown in the given figure.

(b) When we put a point source at the focus of the convex lens, the rays of light emerging out of the convex lens become parallel. Thus, the wavefront of the light emerging out of the convex lens when a point source is placed at its centre is a plane wavefront as shown in the given figure.

(c) Since the star is at infinite distance from Earth, therefore, the portion of the wavefront of light from a distant star intercepted by Earth is a plane wavefront.

Q.3 (a) The refractive index of glass is 1.5. What is the speed of light in glass? (Speed of light in vacuum is 3.0 × 108 ms−1)
(b) 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?

Ans.

(a) Given, refractive index of glass, μ = 1.5
Speed of light, c = 3 × 108 ms-1

$\begin{array}{l}\text{Refractive index is given as:}\\ \text{μ =}\frac{\text{c}}{\text{v}}\\ \therefore \mathrm{S}\text{peed of light in medium, v=}\frac{\text{c}}{\text{μ}}\\ \text{=}\frac{{\text{3×10}}^{\text{8}}{\text{ms}}^{-1}}{\text{1.5}}{\text{= 2×10}}^{\text{8}}{\text{ ms}}^{\text{-1}}\end{array}$

(b) No, the speed of light in glass is not independent of the colour of light. Out of red and violet lights, violet light travels slower than the red light because, the former has a shorter wavelength.

Q.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.

Ans.

Here, separation between the slits, d = 0.28 mm
= 0.28 × 10−3 m
Distance between the screen and the slits, D = 1.4 m
Distance between the central bright fringe and the fourth bright fringe (n = 4), x = 1.2 cm
= 1.2 × 10−2 m
Let wavelength of light used in the experiment = λ
For constructive interference, the distance between the two fringes is given as:

$\begin{array}{l}\text{x = nλ}\frac{\text{D}}{\text{d}}\text{ }\\ \therefore \text{λ =}\frac{\text{xd}}{\text{nD}}\\ \text{=}\frac{{\text{1.2 ×10}}^{\text{-2}}{\text{m × 0.28 m × 10}}^{\text{-3}}\mathrm{m}}{\text{4 × 1.4 m}}\\ {\text{= 6×10}}^{\text{-7}}\text{ m }\end{array}$

Therefore, wavelength of the light = 600 nm

Q.5 In Young’s double-slit experiment using 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 path difference is λ/3?

Ans.

$\begin{array}{l}{\text{Let the intensity of the two light waves be I}}_{\text{1}}{\text{ and I}}_{\text{2}}\\ \text{respectively.}\\ \text{Let phase difference between the two waves =}\mathrm{\varphi }\\ \text{Their resultant intensity is given as:}\\ {\text{I’ = I}}_{\text{1}}{\text{+ I}}_{\text{2}}\text{+ 2}\sqrt{{\text{I}}_{\text{1}}{\text{I}}_{\text{2}}}\text{cos}\mathrm{\varphi }\text{}\\ {\text{In case of monochromatic light waves, I}}_{\text{1}}{\text{= I}}_{\text{2}}\\ \therefore {\text{I’ = I}}_{\text{1}}{\text{+ I}}_{\text{1}}\text{+ 2}\sqrt{{\text{I}}_{\text{1}}{\text{I}}_{\text{1}}}\text{cos}\mathrm{\varphi }\\ \text{Phase difference =}\frac{\text{2π}}{\text{λ}}\text{× Path difference}\\ \text{As, path difference = λ}\\ \text{Phase difference, }\mathrm{\varphi }\text{= 2}\mathrm{\pi }\\ \therefore {\text{I’ = 2I}}_{\text{1}}{\text{+ 2I}}_{\text{1}}{\text{= 4I}}_{\text{1}}\text{\hspace{0.17em} \hspace{0.17em}}\to \left(\text{1}\right)\\ \text{Given, I’ = K}\\ \text{When path difference =}\frac{\text{λ}}{\text{3}}\\ \text{Phase difference, }\mathrm{\varphi }\text{=}\frac{\text{2}\mathrm{\pi }}{\text{3}}\\ \therefore {\text{Resultant intensity, I’}}_{\text{R}}{\text{=I}}_{\text{1}}{\text{+I}}_{\text{1}}\text{+2}\sqrt{{\text{I}}_{\text{1}}{\text{I}}_{\text{1}}}\text{cos}\frac{\text{2}\mathrm{\pi }}{\text{3}}\\ {\text{= 2I}}_{\text{1}}{\text{+ 2I}}_{\text{1}}\left(\text{–}\frac{\text{1}}{\text{2}}\right){\text{= I}}_{\text{1}}\\ \text{From equation}\left(\text{1}\right)\text{, we obtain:}\\ {\text{I}}_{\text{R}}{\text{= I}}_{\text{1}}\text{=}\frac{\text{K}}{\text{4}}\\ \therefore \text{The intensity of light at a point where the path difference is }\frac{\text{λ}}{\text{3}}\text{is}\frac{\text{K}}{\text{4}}\text{ units.}\end{array}$

Q.6 A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a 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 the wavelengths coincide?

Ans.

Wavelength of the beam of light,
λ1 = 650 nm
Wavelength of another beam of light,
λ2 = 520 nm
Let distance of the slits from the screen = D
Let separation between the slits = d

(a) Distance of the nth bright fringe on the screen from the central maximum is given as

$\begin{array}{l}\text{x = 3 × 650}\frac{\text{D}}{\text{d}}\\ \text{= 1950 }\left(\frac{\text{D}}{\text{d}}\right)\text{ nm}\\ \left(\text{b}\right){\text{Let the n}}^{\text{th}}{\text{bright fringe due to wavelength λ}}_{\text{1}}\text{and}{\left(\text{n-1}\right)}^{\text{th}}{\text{bright fringe due to wavelength λ}}_{\text{2}}\\ \text{coincide on the screen. We can equate the conditions for bright fringes as:}\\ {\text{nλ}}_{\text{2}}\text{=}\left(\text{n-1}\right){\text{λ}}_{\text{1}}\\ \text{520n = 650n – 650}\\ \text{650= 130n}\\ \therefore \text{n =}\frac{\text{650}}{\text{130}}\text{= 5}\\ \therefore {\text{Least distance from the central maximum, x = nλ}}_{\text{2}}\frac{\text{D}}{\text{d}}\text{ }\\ \text{= 5 × 520}\frac{\text{D}}{\text{d}}\\ \text{= 2600}\frac{\text{D}}{\text{d}}\text{ nm}\end{array}$

Q.7 In a double-slit experiment the angular width of a fringe is found to be 0.2° 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 refractive index of water to be 4/3.

Ans.

Here, distance between the screen and the slits,
D = 1 m
Wavelength of light used,
λ1= 600 nm
Angular width of the fringe in air,
θa = 0.2o
Let angular width of the fringe in water = θw

$Refractive index of water, μ = 4 3 Relation between refractive index and angular width is given as: μ = θ 1 θ 2 ∴ θ 2 = 3 4 θ 1 = 3 4 × 0.2 = 0.15 ∴Angular width of the fringe in water = 0 .15 o$

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

Ans.

$\begin{array}{l}\text{Given,\hspace{0.17em}refractive index of glass},\text{\hspace{0.17em}}\mathrm{\mu }=1.5\\ \text{Let\hspace{0.17em}Brewster angle}=\text{}\mathrm{\theta }\\ \text{Relation\hspace{0.17em}connecting\hspace{0.17em}Brewster angle and refractive index is\hspace{0.17em}given\hspace{0.17em}as:}\\ \mathrm{tan\theta }=\mathrm{\mu }\\ \therefore \mathrm{\theta }={\mathrm{tan}}^{-1}\left(1.5\right)\\ =56{.31}^{\mathrm{o}}\\ \therefore \text{Brewster angle for air to glass transition = 56}.\text{31}\mathrm{°}\end{array}$

Q.9 Light of wavelength 5000 Å 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?

Ans.

$\begin{array}{l}{\text{Wavelength of incident light, λ = 5000 Å = 5000 × 10}}^{\text{-10}}\text{m}\\ {\text{Speed of light, c = 3 × 10}}^{\text{8}}{\text{ms}}^{\text{-1}}\\ \text{Frequency of incident light is given as}\\ \mathrm{\nu }\text{=}\frac{\text{c}}{\text{λ}}\\ \mathrm{\nu }\text{=}\frac{{\text{3×10}}^{\text{8}}{\text{\hspace{0.17em}ms}}^{\text{-1}}}{{\text{5000×10}}^{\text{-10}}\text{\hspace{0.17em}m}}{\text{= 6×10}}^{\text{14}}\text{\hspace{0.17em}Hz}\\ \text{Wavelength and frequency of incident light is equal to the wavelength and frequency of reflected ray.}\\ \text{Therefore, wavelength of reflected light = 5000 Å}\\ {\text{Frequency of reflected light = 6×10}}^{\text{14}}\text{\hspace{0.17em}Hz}\\ \text{When reflected ray is normal to incident ray,}\\ \text{Sum\hspace{0.17em}of\hspace{0.17em}angle\hspace{0.17em}of\hspace{0.17em}incidence\hspace{0.17em}and\hspace{0.17em}angle\hspace{0.17em}of\hspace{0.17em}reflection=90°}\\ \therefore \angle \text{i +}\angle \text{r = 90°\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{(i)}\\ \text{According\hspace{0.17em}to\hspace{0.17em}the\hspace{0.17em}law\hspace{0.17em}of\hspace{0.17em}reflection:}\\ \angle \text{i =}\angle \text{r\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{(ii)}\\ \text{Substituting\hspace{0.17em}equation\hspace{0.17em}(ii)\hspace{0.17em}in\hspace{0.17em}equation\hspace{0.17em}(i),\hspace{0.17em}we\hspace{0.17em}obtain:}\\ \angle \text{i +}\angle \text{i = 90°}\\ \text{2}\angle \text{i = 90°}\\ \therefore \angle \text{i =}\frac{\text{90°}}{\text{2}}\text{= 45°}\\ \text{Therefore, the angle of incidence = 45}\mathrm{°}\end{array}$

Q.10 Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and wavelength 400 nm.

Ans.

Given, aperture, a = 4 mm
Wavelength of light, λ = 400 nm = 400 ×10-9 m
Since ray optics is a good approximation for Fresnel’s distance (ZF)
Fresnel’s distance (ZF) is given as:

$\begin{array}{l}{\text{Z}}_{\text{F}}\text{=}\frac{{\text{a}}^{\text{2}}}{\text{λ}}\\ \therefore {\text{Z}}_{\text{F}}\text{=}\frac{{\left({\text{4 × 10}}^{\text{-3}}\text{m}\right)}^{\text{2}}}{{\text{400 × 10}}^{\text{-9}}\text{m}}\\ \text{= 40 m}\\ \therefore \text{Distance for which the ray optics is good approximation = 40 m}\end{array}$

Q.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.

Ans.

Wavelength of Hα line emitted by hydrogen, λ = 6563 Å = 6563 × 10−10 m
Red-shift of star, (λ’- λ) =15 Å
=15×10−10 m
Speed of light, c=3×108 ms-1
Let the speed of the star receding away from the Earth = v
The relation connecting red shift and velocity is given as:

$\begin{array}{l}\text{λ’ – λ =}\frac{\text{v}}{\text{c}}\text{λ}\\ \text{v =}\frac{\text{c}}{\text{λ}}\left(\text{λ’-λ}\right)\\ \text{=}\frac{{\text{3 × 10}}^{\text{8}}{\text{ms}}^{-1}{\text{× 15 ×10}}^{\text{-10}}\mathrm{m}}{{\text{6563 × 10}}^{\text{-10}}\text{}\mathrm{m}}\\ {\text{= 6.87 × 10}}^{\text{-5}}{\text{ms}}^{\text{-1}}\end{array}$

Therefore, speed with which the star is receding away from the Earth = 6.87 × 105 ms-1

Q.12 Explain how Corpuscular theory predicts the speed of light in a medium, say, water, to be greater than the speed of light in 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 experiment?

Ans.

According to Newton’s corpuscular theory of light, when corpuscles of light strike the interface of two media from a rarer (air) to a denser (water) medium, a normal attractive force acts on them. As a result, the component of velocity along the normal increases while the component of velocity along the surface remains unchanged.

Let i and r be the angle of incidence and angle of refraction respectively as shown in the figure.
Let vr and vd be the velocity of light in the rarer medium and denser medium respectively.
Let velocity of light in air=c
Let velocity of light in water=v
From the given figure:
Component of vr along the interface XY= vrsini
Component of vd along the interface XY= vdsinr
Since,
Component of vr along the interface XY= Component of vd along the interface XY
Therefore, we have:

$\begin{array}{l}{\text{v}}_{\text{r}}{\text{sini = v}}_{\text{d}}\text{sinr}\\ \therefore \frac{\text{sini}}{\text{sinr}}\text{=}\frac{{\text{v}}_{\text{d}}}{{\text{v}}_{\text{r}}}\text{= μ}\\ \text{The relation for refractive index of water w.r.t. air is given as:}\\ \text{μ =}\frac{\text{v}}{\text{c}}\\ \text{In this case, μ>1}\\ \therefore {\text{v}}_{\text{d}}{\text{> v}}_{\text{r}}\end{array}$

From equation (i), it can be concluded that vd > vr. This prediction is not consistent with the experimental results. The wave picture of light is consistent with the experimental results of c > v.

Q.13 You have learnt 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.

Ans.

Consider that a point object P is placed in front of a plane mirror M1M2 as shown in the given figure.

Draw a circle from the centre of the object P such that it touches the plane mirror at point O. As per Huygens principle, ST is the spherical wavefront of the incident light. In the absence of the plane mirror, a similar wavefront S’T’ would appear from behind point O whose distance is equal to the object distance from the centre of the mirror. S’T’ can be treated as the virtually reflected image of ST. Therefore, we conclude that a point object placed in front of the plane mirror produces a virtual image whose distance from the mirror is equal to the distance of the object from the mirror.

Q.14 Let us list some of the factors, which could possibly influence the speed of wave propagation:
(i) nature of the source.
(ii) direction of propagation.
(iii) motion of the source and/or observer.
(iv) wave length.
(v) intensity of the wave.

On which of these factors, if any, does
(a) the speed of light in vacuum,
(b) the speed of light in a medium (say, glass or water), depend?

Ans.

(a) The speed of light in vacuum i.e., 3 × 108 ms-1 (approximately) is an absolute constant. It is not affected by any of the factors listed above.

(b) Among the tabulated factors, the speed of light in a medium depends only on the wavelength of light in that medium.

Q.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 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 case of light travelling in a medium?

Ans.

Sound waves cannot propagate in vacuum. Sound waves require a material medium for their propagation. In case of the given two situations, the motion of the observer w.r.t. the medium is different. That is why, the Doppler’s formula for given two situations cannot be scientifically identical to each other in case of sound waves.

However, light waves can travel in vacuum as well. The velocity of light waves is constant in the vacuum. Since the velocity of light in the vacuum is independent of the speed of observer and speed of the source, the Doppler’s formulae for light waves in vacuum are exactly similar in case of the given two situations.

However, when light propagates in a medium, the given two situations are not exactly identical. This is due to the fact that the speed of light in a medium depends on the optical density of the medium.

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

Ans.

$\begin{array}{l}\text{Given, }\mathrm{w}\text{avelength of light},\text{}\mathrm{\lambda }=\text{6}000\text{nm = 6}00×\text{1}{0}^{-\text{9}}\text{m }\\ \text{Angular width of fringe},\text{ }\mathrm{\theta }=0{.1}^{\mathrm{o}}\\ =0.1×\frac{\mathrm{\pi }}{180}=\frac{3.14}{1800}\mathrm{rad}\\ \text{Let spacing between the two slits}=\mathrm{d}\\ \text{The relation connecting the }\mathrm{a}\text{ngular width of a fringe}\\ \text{to slit spacing is given as}:\\ \mathrm{\theta }=\frac{\mathrm{\lambda }}{\mathrm{d}}\\ \therefore \mathrm{d}=\frac{\mathrm{\lambda }}{\mathrm{\theta }}\\ \mathrm{d}=\frac{\text{6}00×\text{1}{0}^{-\text{9}}\mathrm{m}}{\frac{3.14}{1800}}=3.44×\text{1}{0}^{-4}\text{ m}\\ \therefore \text{Spacing between the slits =}3.44×\text{1}{0}^{-4}\text{ m}\end{array}$

(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?

Ans.

(a) In a single slit diffraction experiment, if the width of the slit is doubled, the size of the central diffraction band reduces to half and its intensity becomes four times the original value.

(b) If the width of each slit is of the order of the wavelength of the light used, then the diffraction of light from each slit will modulate the interference pattern in the double slit experiment.

(c) When a tiny circular obstacle is placed in the path of light from a distant source, the light waves diffract from the edges of the circular obstacle to produce the constructive interference at the centre of the shadow of the obstacle. which is responsible for the bright spot in the centre of the shadow.

(d) The bending of waves from an obstacle by a large angle can take place only if the size of the obstacle is comparable to the wavelength of waves.

The wavelength of light waves is very small as compared to the size of the wall,i.e. wavelength of light is about 6 x 10-7 m so the bending of the light waves by a large angle is not possible in this case. Hence, the students separated by the partition wall are unable to see each other.

On the other hand, the wavelength of the sound waves is comparable to the size of the wall i.e. wavelength of sound waves with frequency 1 Khz is around 0.34 m, thus allowing bending of sound waves by large angles. That is why the students separated by the partition wall are able to hear each other.

(e) Since the wavelength of the light used is very small as compared to the aperture of the optical instruments, therefore, the diffraction effects are not significant in this case. That is why; the ray optics assumption is used to understand the location and other properties of the images in optical instruments.

Q.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?

Ans.

Distance between the towers, d = 40 km
Height of the line joining the hills, d = 50 m.
Here, Fresnel’s distance, ZF = 20 km
= 2 × 104 m
Aperture, a = d
= 50 m
The relation for Fresnel’s distance is given as:

$\begin{array}{l}{\text{Z}}_{\text{F}}\text{=}\frac{{\text{a}}^{\text{2}}}{\text{λ}}\text{ }\\ \therefore \text{λ =}\frac{{\text{a}}^{\text{2}}}{{\text{Z}}_{\text{F}}}\\ \text{=}\frac{{\left(\text{50 m}\right)}^{\text{2}}}{{\text{2 × 10}}^{\text{4}}\text{}\mathrm{m}}\\ {\text{= 1250 × 10}}^{\text{-4}}\text{ m}\\ \text{= 0.125 m}\end{array}$

Therefore, wavelength of radio waves =12.5 cm

Q.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.

Ans.

Wavelength of the beam of light, λ = 500 nm
= 500×10−9 m
Distance of the screen from the slit,
D = 1 m
In case of first minima,
n = 1
Separation between slits = d
Distance of the first minima from the centre of screen,
x = 2.5 mm
= 2.5 × 10−3 m
Let width of the slit = a
The condition for minima is given as:

$a x D = nλ ∴ a = nλD x = 1×5 ×10 -7 m ×1 m 2 .5×10 -3 m = 2×10 -4 m= 0.2 mm ∴Width of slit, a = 0.2 mm$

(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 of this principle?

Ans.

(a) When a low flying aircraft passes overhead the reflected radiowaves may interfere with the normal TV signal coming from the antenna and result in shaking of the picture on our TV screen.

(b) Since principle of superposition follows from the linear character of the differential equation of wave motion, therefore, the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns. For example: if x1 and x2 are the solutions of a wave equation, then the linear combination of x1 and x2 will also be the solution of that wave equation.

Q.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.

Ans.

$\begin{array}{l}\text{Let}\text{\hspace{0.17em}}\text{a}\text{\hspace{0.17em}}\text{slit}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{width}\text{\hspace{0.17em}}\text{a}\text{\hspace{0.17em}}\text{be}\text{\hspace{0.17em}}\text{divided}\text{\hspace{0.17em}}\text{into}\text{\hspace{0.17em}}\text{n}\text{\hspace{0.17em}}\text{smaller}\text{\hspace{0.17em}}\text{slits}:\text{\hspace{0.17em}}\\ \therefore \text{Width}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{each}\text{\hspace{0.17em}}\text{small}\text{\hspace{0.17em}}\text{slit},\text{\hspace{0.17em}}a‘=\frac{a}{n}\to \left(i\right)\\ \text{The}\text{\hspace{0.17em}}\text{relation}\text{\hspace{0.17em}}\text{for}\text{\hspace{0.17em}}\text{angle}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{diffraction}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{given}\text{\hspace{0.17em}}\text{as}:\\ \theta =\frac{n\lambda }{a}\to \left(ii\right)\\ \text{From}\text{\hspace{0.17em}}\text{equation}\text{\hspace{0.17em}}\text{(i)}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{equation}\text{\hspace{0.17em}}\text{(ii),}\text{\hspace{0.17em}}\text{we}\text{\hspace{0.17em}}\text{obtain:}\\ \theta =\frac{n\lambda }{a‘n}\\ =\frac{\lambda }{a‘}\end{array}$

Thus, the intensity given by each of these smaller slits is equal to zero in the direction of θ . Therefore, the entire single slit will give zero intensity in the direction of θ .