# NCERT Solutions for Class 12 Physics Chapter 10

NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics Chapter end Exercises with solutions and Additional Exercises Solutions in PDF format free download updated for new academic session 2020-2021.

UP Board Solutions as well as Offline Apps are based on latest NCERT Books issued for 2020-21.

## NCERT Solutions for Class 12 Physics Chapter 10

Class: | 12 |

Subject: | Physics |

Chapter 10: | Wave Optics |

### Chapter 10 Wave Optics Solutions

- Study Online: Exercises Question Answers
- Study Online: Additional Exercises Answers
- Download Exercises in PDF form
- Download Additional Exercises in PDF
- Practice Questions for Competitive Exams
- NCERT Book Chapter 10
- NCERT Book Answers
- Revision Book Chapter 10
- Revision Book Answers
- Revision Notes 1
- Revision Notes 2
- Visit to 12th Physics Main Page

### Class 12 Physics Chapter 10 Solutions in English

NCERT Solutions for Class 12 Physics Chapter 10 in English Medium free to download in PDF form for new session 2020-2021. Join the discussion forum to discuss your queries about NIOS board as well as CBSE Board. Download NCERT Books 2020-21 for new session based on latest CBSE Syllabus.

#### Important Questions for practice

1. A ray of light passes through an equilateral prism in such a manner that the angle of incidence is equal to angle of emergence and each of these angles is equal to 3/4 of angle of prism. Find angle of deviation.

2. Critical angle for a certain wavelength of light in glass is 30°. Calculate the polarizing angle and the angle of refraction in glass corresponding to this.

3. An object of length 2.5cm is placed at a distance of 1.5f from a concave mirror where f is the focal length of the mirror. The length of object is perpendicular to principal axis. Find the size of image. Is the image erect or inverted?

4. A radio can tune into any station of frequency band 7.5 MHz to 10 MHz. Find the corresponding wave length range.

5. Describe an astronomical telescope and derive an expression for its magnifying power using a labelled ray diagram.

##### Questions from Board Papers

1. Show that maximum intensity in interference pattern is four times the intensity due to each slit if amplitude of light emerging from slits is same.

2. A person looking at a mesh of crossed wire is able to see the vertical wire more distinctly than the horizontal wire. Which defect he is suffering from? How can this defect be corrected?

3. How is a wave front different from a ray? Draw the geometrical shape of the wave fronts when.

(i) light diverges from a point source, (ii) light emerges out of convex lens when a point source is placed at

its focus.

4. Define diffraction. Deduce an expression for fringe width of the central maxima of the diffraction pattern, produced by single slit illuminated with monochromatic light source

5. State the condition under which the phenomenon of diffraction of light takes place. Derive an expression for the width of the central maximum due to diffraction of light at a single slit. Also draw the intensity pattern with angular position.

### Important Questions on 12th Physics Chapter 10

The refractive index of a violet component of white light is greater than the refractive index of a red component. Hence, the speed of violet light is less than the speed of red light in glass. Hence, violet light travels slower than red light in a glass prism.

Brewster angle = θ

Brewster angle is related to refractive index as:

tanθ = μ

⇒ θ = tan^(-1)(1.5)

= 56.31°

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

Newton’s corpuscular theory of light states that when light corpuscles strike the interface of two media from a rarer (air) to a denser (water) medium, the particles experience forces of attraction normal to the surface. Hence, the normal component of velocity increases while the component along the surface remains unchanged.

Hence, we can write the expression: c sini=v sinr … (i)

Where, i = Angle of incidence

r = Angle of reflection

c = Velocity of light in air

v = Velocity of light in water

We have the relation for relative refractive index of water with respect to air as: μ=v/c

Hence, equation (i) reduces to

v/c=sini/sinr =μ … (2)

But, μ>1

Hence, it can be inferred from equation (ii) that v > c. This is not possible since this prediction is opposite to the experimental results of c > v. The wave picture of light is consistent with the experimental results.

Sound waves can propagate only through a medium. The two given situations are not scientifically identical because the motion of an observer relative to a medium is different in the two situations. Hence, the Doppler formulas for the two situations cannot be the same.

In case of light waves, sound can travel in a vacuum. In a vacuum, the above two cases are identical because the speed of light is independent of the motion of the observer and the motion of the source. When light travels in a medium, the above two cases are not identical because the speed of light depends on the wavelength of the medium.

On the one hand, the wavelength of the light waves is too small in comparison to the size of the obstacle. Thus, the diffraction angle will be very small. Hence, the students are unable to see each other. On the other hand, the size of the wall is comparable to the wavelength of the sound waves. Thus, the bending of the waves takes place at a large angle. Hence, the students are able to hear each other.