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# NCERT Solutions Class 12 Physics Chapter 10 Wave Optics

Last updated date: 08th Sep 2024
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## NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics - FREE PDF Download

Class 12 Physics Chapter 10 Wave Optics NCERT solutions by Vedantu introduces the concept of wave optics, which explains various optical phenomena using the wave theory of light. The chapter begins with Huygens' Principle, explaining how light propagates. This foundation is crucial for exploring interference. The chapter also covers diffraction, leading to characteristic patterns of maxima and minima. These principles collectively enhance our understanding of light's behaviour, extending beyond the simpler ray optics model to explain more complex phenomena and their practical applications in technology and scientific research. With Vedantu's Class 12 Physics NCERT Solutions, you'll find step-by-step explanations of all the exercises in your textbook, ensuring you understand the concepts thoroughly.

Table of Content
1. NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics - FREE PDF Download
2. Glance on Physics Chapter 10 Wave Optics Class 12 Solutions
3. Access NCERT Solutions for Class 12 Physics Chapter 10 – Wave Optics
4. Overview of Deleted Syllabus for CBSE Class 12 Physics Wave Optics
5. Other Study Material for CBSE Class 12 Physics Chapter 10
6. Chapter-Specific NCERT Solutions for Class 12 Physics
FAQs

## Glance on Physics Chapter 10 Wave Optics Class 12 Solutions

• Chapter 10 of Class 12 Physics Wave Optics NCERT Solutions introduces the concept of a wavefront as the locus of points having the same phase. Also describes the phenomenon of interference and Young's Double Slit Experiment.

• Explains diffraction as the bending of light waves around obstacles and the spreading out of light waves when they pass through small apertures.

• Single-slit diffraction pattern is discussed, highlighting the formation of central and secondary maxima and minima.

• Introduces the concept of polarization, where the vibrations of light waves are restricted to a single plane.

• Explains the concept of resolving power, which is the ability of an optical instrument to distinguish between two closely spaced objects.

• There are 6 fully solved questions in the exercise of class 12th Physics Chapter 10 Wave Optics.

Competitive Exams after 12th Science
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## Access NCERT Solutions for Class 12 Physics Chapter 10 – Wave Optics

1. Monochromatic light of wavelength $589nm$is incident from air on a water surface. What are the wavelength, frequency and speed of

(a) Reflected light?

Ans: Wavelength of incident monochromatic light is given as,

$\lambda =589nm=589\times {{10}^{9}}m$

Speed of light in air is $c=3\times {{10}^{8}}m$

Refractive index of water is $\mu =1.33$

In this case, the ray will reflect in the same medium as that of the incident ray. Hence, the wavelength, speed, and frequency of the reflected ray will be the equal to that of the incident ray.

Frequency of light can be given by the relation,

$\nu =\frac{c}{\lambda }$

Where,

$\nu =$Frequency of light

$c=$Speed of light

$\lambda =$Wavelength of light

$\Rightarrow \nu =\frac{3\times {{10}^{8}}}{589\times {{10}^{-9}}}$

$\Rightarrow \nu =5.09\times {{10}^{14}}Hz$

Hence, the speed, frequency, and wavelength of the reflected light are

$3\times {{10}^{8}}m/s$ , $5.09\times {{10}^{14}}Hz$ and $589nm$respectively.

(b) Refracted light? Refractive index of water is $1.33$.

Ans: The frequency of light is independent of the property of the medium in which it is travelling.

Hence, the frequency of the refracted ray in water will be equal to the frequency of the incident light or reflected light in air.

Frequency of the refracted light ray, $\nu =5.09\times {{10}^{14}}Hz$

Speed of light in water is related to the refractive index of water as given in the formula below:

$v=\frac{c}{\mu }$

$\Rightarrow v=\frac{3\times {{10}^{8}}}{1.33}$

$\Rightarrow v=2.26\times {{10}^{8}}m/s$

The formula below gives the relation of wavelength of light in water and the speed and frequency of light,

$\lambda =\frac{v}{\nu }$

$\Rightarrow \lambda =\frac{2.26\times {{10}^{8}}}{5.09\times {{10}^{14}}}$

$\Rightarrow \lambda =444.007\times {{10}^{-9}}m$

$\Rightarrow \lambda =444.01nm$

2. What is the shape of the wavefront in each of the following cases:

(a) Light diverging from a point source.

Ans: When a light diverges from a point source, the shape of the wavefront in this case is spherical. The wavefront originating from a point source is shown in the given figure.

(b) Light emerging out of a convex lens when a point source is placed at its focus.

Ans: The shape of the wavefront when a light emerges out of a convex lens when a point source is placed at its focus is a parallel grid. This can be represented as shown in the given figure.

(c) The portion of the wavefront of light from a distant star intercepted by the Earth.

Ans: In this case the portion of the wavefront of light from a distant star intercepted by the Earth is a plane.

3.(a) The refractive index of glass is $1.5$. What is the speed of light in glass? Speed of light in vacuum is $3.0\times {{10}^{8}}m/s$.

Ans: Refractive index of glass is given as,

$\mu =1.5$

Speed of light, $c=3.0\times {{10}^{8}}m/s$

Speed of light in glass is given by the formula,

$v=\frac{c}{\mu }$

$\Rightarrow v=\frac{3\times {{10}^{8}}}{1.5}=2\times {{10}^{8}}m/s$

Hence, the speed of light in glass is $2\times {{10}^{8}}m/s$.

(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: The speed of light in glass depends on the colour of light.

The refractive index of a violet component of white light is more than the refractive index of a red component. Hence, the speed of violet light is less than the speed of red light in glass as speed and refractive index are inversely related to each other.

Hence, violet light travels slower as compared to red light in a glass prism.

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

Ans: Distance between the slits is given as, $d=0.28mm=0.28\times {{10}^{-3}}m$

Distance between the slits and the screen, $D=1.4m$

Distance between the central fringe and the fourth $\left( n=4 \right)$fringe, $u=1.2cm=1.2\times {{10}^{-2}}m$

In case of a constructive interference, the relation for the distance between the two fringes can be given as: $u=n\lambda \frac{D}{d}$

where,

$n=$ Order of fringes $=4$

$\lambda =$ Wavelength of light used

$\lambda =\frac{ud}{nD}$

$\Rightarrow \lambda =\frac{1.2\times {{10}^{-2}}\times 0.28\times {{10}^{-3}}}{4\times 1.4}$

$\Rightarrow \lambda =6\times {{10}^{-7}}$

$\Rightarrow \lambda =600nm$

Hence, the wavelength of the light is $600nm$.

5. In Young's double-slit experiment using monochromatic light of wavelength $\lambda$. The intensity of light at a point on the screen where path difference is $\lambda$ , is $K$ units. What is the intensity of light at a point where path difference is $\frac{\lambda }{3}$?

Ans: The intensity of the two light waves be ${{I}_{1}}$ and ${{I}_{2}}$. Their resultant intensities can be evaluated as: ${I}'={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos \phi$

Where,

$\phi =$The phase difference between two waves for monochromatic light waves,

Since ${{I}_{1}}={{I}_{2}}$

So ${I}'=2{{I}_{1}}+2{{I}_{1}}\cos \phi$

The formula for phase difference can be given as:

$Phase\,difference=\frac{2\pi }{\lambda }\times \,Path\,difference$

Since, path difference is $\lambda$,

Phase difference is $\phi =2\pi$

${I}'=2{{I}_{1}}+2{{I}_{1}}=4{{I}_{1}}$

Given,

${{I}_{1}}=\frac{{{K}'}}{4}$ …… (1)

When path difference $=\frac{\lambda }{3}$

phase difference, $\phi =\frac{2\pi }{3}$

Hence, resultant intensity,

${{I}_{R}}={{I}_{1}}+{{I}_{1}}+2\sqrt{{{I}_{1}}{{I}_{1}}}\cos \frac{2\pi }{3}$

$\Rightarrow {{I}_{R}}=2{{I}_{1}}+2{{I}_{2}}\left( -\frac{1}{2} \right)={{I}_{1}}$

Using equation (1), we can state that

${{I}_{R}}={{I}_{1}}=\frac{K}{4}$

Hence, for monochromatic light waves, the intensity of light at a point where the path difference is $\frac{\lambda }{3}$ is $\frac{K}{4}$units.

6. A beam of light consisting of two wavelengths, $650nm$ and $520nm$, 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 $650nm$.

Ans: Given that,

Wavelength of the first light beam, ${{\lambda }_{1}}=650nm$

Wavelength of second light beam, ${{\lambda }_{2}}=520nm$

Distance of the slits from the screen $=D$

Distance between the two slits $=d$

Distance of the ${{n}^{th}}$ bright fringe on the screen from the central maximum is given by the formula below,

$x=n{{\lambda }_{1}}\left( \frac{D}{d} \right)$

For the third bright fringe, $n=3$

$x=3\times 650\times \frac{D}{d}=1950\left( \frac{D}{d} \right)nm$, which is nothing but the distance of the third bright fringe on the screen from the central maximum.

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

Ans: In this case, let the ${{n}^{th}}$bright fringe due to wavelength ${{\lambda }_{2}}$and ${{\left( n-1 \right)}^{th}}$ bright fringe due to wavelength ${{\lambda }_{1}}$ coincide on the screen. Equate the conditions for bright fringes as follows:

$n{{\lambda }_{2}}=\left( n-1 \right){{\lambda }_{1}}$

$\Rightarrow 520n=650n-650$

$\Rightarrow 650=130n$

$\Rightarrow n=5$

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

$x=n{{\lambda }_{2}}\frac{D}{d}$

$\Rightarrow x=5\times 520\frac{D}{d}=2600\frac{D}{d}nm$

Note: The value of $d$ and $D$ are not given in the question, hence the exact answer cannot be found.

## Overview of Deleted Syllabus for CBSE Class 12 Physics Wave Optics

 Chapter Dropped Topics Wave Optics 10.3.4 Doppler Effect Example 10.1 10.5 Interference of Light Waves and Young’s Experiment (retain the final expressions for dark and bright fringes but delete the derivation; delete expression for fringe width) 10.6 Diffraction (retain only qualitative treatment) 10.6.3 Resolving Power of Optical Instruments 10.6.4 Validity of Ray Optics 10.7.1 Polarisation by Scattering 10.7.2 Polarisation by Reflection Exercises 10.7–10.21

## Conclusion

NCERT Class 12 Physics Chapter 10 Exercise Solutions on Wave Optics provided by Vedantu explores an in-depth exploration of the wave nature of light and its implications for various optical phenomena. By mastering these concepts, students gain a solid foundation for advanced studies in optics and related fields. By studying concepts such as Huygens' Principle, interference, diffraction, and polarization, students gain a deeper insight into how light behaves in different contexts. These principles are not only fundamental to theoretical physics but also have practical applications in technology, such as in designing optical instruments, improving communication systems, and conducting scientific research. From previous year's question papers, typically around 3–4 questions are asked from this chapter. These questions test students' understanding of theoretical concepts as well as their problem-solving skills.

## Other Study Material for CBSE Class 12 Physics Chapter 10

 S.No. Important Links for Chapter 10 Wave Optics 1 Class 12 Wave Optics Important Questions 2 Class 12 Wave Optics Revision Notes 3 Class 12 Wave Optics Important Formulas 4 Class 12 Wave Optics NCERT Exemplar Solution

## Chapter-Specific NCERT Solutions for Class 12 Physics

Given below are the chapter-wise NCERT Solutions for Class 12 Physics. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions Class 12 Physics Chapter 10 Wave Optics

1. What are Optical Waves?

In Physics, Wave Optics generally refers to the branch of the optic from which interference, polarization, diffraction is studied. Magnetic fields and electrical fields of high frequency are combined to be termed as Optical Waves. The wave ranges from the infrared to the ultraviolet light. It helps to analyze the propagation of the light full of photons.

2. What are the Two Types of Optics?

Classical optics are mainly classified into two main parts. The first one is Ray or Geometrical optic and another one is physical or waves optics. The theory in which light is considered to travel a straight line comes under geometrical optics. The physical or Wave Optics include the light considered as an Electromagnetic way. As a whole in Physics, light normally indicates electromagnetic radiation of any wavelength, whether it is visible or not.

3. Is Vedantu the Best Option for Online Teaching?

Yes, of course, it is one of the best choices. Here, the subjects are taught or revised by the expert and they clear all the doubts easily. One can connect from anywhere at any time from all over the country. The live online classes arranged by Vedantu are so much useful because the teaching methods, animated videos, pictures attract the student's attention and also helps them to understand any difficult chapters easily. Apart from your respective curricular subject, one can also enrol for various competitive exams for preparatory courses like IIT, JEE, Neet, UPSC, WBCS, JEN-Paugh, CAT, GET, etc. These courses aspirants can get their desirable rank according to their merit.

4. How to solve the problems based on wave optics quickly in Chapter 10 of NCERT Solutions for Class 12 Physics?

When it comes to board examinations, NCERT is the most significant study resource you may use. On Vedantu, you can find NCERT Solutions for Class 12 Physics in PDF format. These solutions are available online, and you can download the Solutions PDF to use it offline. The best thing is that you can get the NCERT Solutions for Class 12 PDF for free. They are also available on the Vedantu app.

5. How should I use the NCERT Solutions for Class 12 Physics Chapter 10 PDF from Vedantu?

The NCERT Maths Solutions for Vedantu Class 10 are available in PDF format. Here's how to get them:

1. Visit NCERT Solutions for Class 12 Physics Chapter 10.

2. Scroll down to the exercise you want solutions for.

The PDF solutions will appear on your screen. To save the solutions for offline use, use the 'Download PDF' option. You can access them anytime you want to.

6.  Which website offers the best NCERT Solutions for Class 12 Physics?

The explanations for all of the subjects can be found in Vedantu's NCERT Solutions for Class 12 Physics. After extensive research and effort, subject matter experts prepare the answers. The PDF answers will assist you in gaining a thorough grasp of the chapter. The solutions have been created by experts and are presented in a very clear and precise manner so that you can comprehend the concepts fast.

7. How can Vedantu help in understanding wave optics for Class 12 Physics?

The reasoning and calculation are broken down into steps to assist you to comprehend each stage of the process. It will help students comprehend the chapter better and understand the idea of Wave Optics as well as all of its concepts. Students may find important questions, revision notes and NCERT solutions for Class 12 Physics Chapter Wave Optics. All these materials are prepared by the best Physics teachers in India and thus they are 100% reliable.

8. What exercises are covered in Wave Optics Class 12 Physics at Vedantu?

Exercises in Wave Optics Class 12 Physics at Vedantu covers the following topics:

• Introduction

• Diffraction

• Refraction And Reflection of Plane Waves Using Huygens Principle

• Refraction At a rarer medium

• Reflection of a Plane Wave By a Plane Surface

• Refraction of a Plane Wave

• Resolving Power of Optical Instruments

• Interference of Light Waves And Young’s Experiment

• The Single Slit, Seeing The Single Slit Diffraction Pattern

• The Doppler Effect, And Coherent And Incoherent Addition of Waves

• The Validity of Ray Optics

• Polarisation, Polarisation By Scattering and Polarisation By Reflection

9. What is important in wave optics NCERT solutions?

Wave optics NCERT solutions are essential for understanding the wave nature of light, which explains various phenomena that cannot be described by ray optics alone. Key topics include interference, diffraction, polarization, and the resolving power of optical instruments. These concepts are crucial for advanced studies in physics and have practical applications in designing optical devices, improving communication systems, and conducting scientific research.

10. What are the concepts of wave optics class 12 exercise solutions?

The main concepts of wave optics class 12 exercise solutions include:

• Wavefronts and Huygens' Principle: Describing how light propagates as waves.

• Interference: The phenomenon where two coherent light waves superpose to form a pattern of alternating bright and dark fringes.

• Diffraction: The bending and spreading of light waves when they encounter obstacles or apertures.

• Polarization: The orientation of light waves such that the vibrations occur in a single plane.

• Resolving Power: The ability of optical instruments to distinguish between two closely spaced objects.

11. What is the summary of wave optics class 12 NCERT solutions?

Wave optics class 12 NCERT solutions explore the wave nature of light and its implications for various optical phenomena. Key topics include Huygens' Principle, interference patterns as demonstrated by Young's Double Slit Experiment, diffraction effects from single slits, and the concept of polarization. The chapter also covers the resolving power of optical instruments and practical applications of wave optics in technology and research.

12. How are wave optics used in everyday life?

Wave optics have numerous everyday applications, including:

• Anti-reflective Coatings: Used in eyeglasses and camera lenses to reduce glare.

• Holography: Used in data storage and security features on credit cards and currency.

• Fibre Optics: Used in telecommunications to transmit data over long distances with minimal loss.

• Polarization Filters: Used in sunglasses and photographic equipment to reduce glare.

13. What is the difference between waves and wave optics in class 12 wave optics NCERT solutions?

• Waves: Refers to the general concept of wave phenomena, including sound waves, water waves, and electromagnetic waves.

• Wave Optics: Specifically deals with the wave nature of light, explaining optical phenomena such as interference, diffraction, and polarization that cannot be explained by ray optics alone.

14. What are the branches of waves and optics according to class 12 physics wave optics NCERT solutions?

• Branches of Waves:

• Mechanical Waves: Such as sound waves and water waves, which require a medium to propagate.

• Electromagnetic Waves: Such as light waves, radio waves, and X-rays, can propagate through a vacuum.

• Branches of Optics:

• Geometrical Optics: Also known as ray optics, deals with the propagation of light in terms of rays, focusing on reflection and refraction.

• Physical Optics: Also known as wave optics, deals with the wave nature of light, including interference, diffraction, and polarization.

• Quantum Optics: Studies the interaction of light with matter at the quantum level, including phenomena such as photon emission and absorption.