NCERT Solutions for Class 10 Physics Chapter 10 : Light - Reflection and Refraction

Q: Question 1. Explain Reflection and Refraction of Light in short.

Answer. Reflection of Light refers to a phenomenon where the light waves bounce off the surface that does not has the capability of absorbing energy produced by radiation, after striking the surface. In reflection, the light entering the medium or surface returns in the same direction. This phenomenon usually occurs in mirrors. Also, the angle of incidence of the light is equal to the angle of reflection. Refraction of light refers to a process of the shift of light when it passes through a medium leading to the bending of light. In this process, the light entering the medium travels from one medium to another. This phenomenon usually occurs in Lenses.

Q: Question 2. What are Spherical mirrors?

Answer. Spherical mirrors refer to the mirrors that have curved surfaces that are painted on one of their sides. Spherical mirrors that have inward surfaces painted are convex mirrors while those that have outward surfaces painted are concave mirrors. The rays converge after falling on the concave mirrors and thus they are also called converging mirrors. Whereas, the rays diverge after falling on the convex mirror and thus they are also known as diverging mirrors.

Q: Question 3. Which type of mirror is used as rearview mirrors in vehicles?

Answer. Convex mirrors are used as rearview mirrors in vehicles. In convex mirrors, the light diverges after falling on the surface. Therefore, the objects appear smaller and closer to the vehicle. This provides essential information to the driver of the vehicle as he gets a better picture of the vehicles behind him. Also, these mirrors have a larger field of view.

Q: Question 4. What kind of image is formed by Concave mirrors?

Answer. Both real and virtual images are formed by the Concave mirrors. When we place the concave mirror very close to the object, it forms a virtual and magnified image. However, if we increase the distance between the object and the mirror, the size of the image reduces and it forms a real image.

Question 1

Define the principal focus of a concave mirror.

Hard

Solution

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Answer

Explanation :

The principal focal point of a spherical mirror is a point to which every one of the light beams which are equal will unite after reflection from the curved mirror.
After reflection from a curved mirror, the beams will join to the point of convergence of the mirror which is meant by point F in the picture.

Answer

Explanation :

The principal focal point of a spherical mirror is a point to which every one of the light beams which are equal will unite after reflection from the curved mirror.
After reflection from a curved mirror, the beams will join to the point of convergence of the mirror which is meant by point F in the picture.

Question 2

If the radius of curvature of a spherical mirror is 20 cm, then its focal length is _______ cm.

A

20

B

40

C

10

D

None of these

Medium

Solution

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Answer

Radius of curvature = 2 x focal length.
So, focal length =

\frac{1}{2} \times

radius of curvature =

\frac{1}{2} \times 20

cm = 10 cm

Answer

Radius of curvature = 2 x focal length.
So, focal length =

\frac{1}{2} \times

radius of curvature =

\frac{1}{2} \times 20

cm = 10 cm

Question 3

Name a mirror that can give an erect and enlarged image of an object.

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Solution

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Answer

Concave mirror forms an erect, virtual and enlarged image when the object is between focus and pole of mirror.

Answer

Concave mirror forms an erect, virtual and enlarged image when the object is between focus and pole of mirror.

Question 4

Which of the following lenses would you prefer to use while reading small letters found in a dictionary?

A

A convex lens of focal length of 50 cm

B

A concave lens of focal length of 50 cm

C

A convex lens of focal length of 5 cm

D

A concave lens of focal length of 5 cm

Medium

Solution

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Answer

$∙$ A convex lens forms a magnified image of the object on the other side of the lens when the object is placed between the radius of curvature and the focus of the lens.

∙

The magnification of the convex lens is larger for the smaller value of the focal length. Therefore, in order to read the small letters, the convex lens of smaller focal length will be used.

Correct option: C

Answer

$∙$ A convex lens forms a magnified image of the object on the other side of the lens when the object is placed between the radius of curvature and the focus of the lens.

∙

The magnification of the convex lens is larger for the smaller value of the focal length. Therefore, in order to read the small letters, the convex lens of smaller focal length will be used.

Correct option: C

Question 5

We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.

Hard

Solution

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Answer

Object distance should be less than the focal length for the formation of an erect image. Hence, the range of distance of object from the mirror should be less than 15 cm i.e. from 0 to 15 cm in the front of mirror from the pole. The nature of image so formed will be virtual and erect. The size of the image will be larger than object. Ray diagram is attached as image to the question.

Answer

Object distance should be less than the focal length for the formation of an erect image. Hence, the range of distance of object from the mirror should be less than 15 cm i.e. from 0 to 15 cm in the front of mirror from the pole. The nature of image so formed will be virtual and erect. The size of the image will be larger than object. Ray diagram is attached as image to the question.

Question 6

One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answers experimentally. Explain your observations.

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Solution

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Answer

A point on the object needs only two rays to form an image. Since the entire lens is not covered, at least two rays from each point does pass through the lens and hence, image is formed. However, since the number of rays involved in the image formation reduces as the lens is covered, the image formed is of lesser intensity and less brighter.

Experiment:

Apparatus needed: convex lens, candle, two black sheets of paper.

Procedure:

Mount a lens on a table. Light a candle with a match stick and place it on one side of the lens on the table. Turn of the lights of the room so that it gets dark. Take the sheet of paper and place it on the other side of the lens as the candle. Now focus the image formed on the sheet of paper by moving the sheet only. Observe the image. Mark the positions of candle, lens and screen. Now, turn off the light. Take the lens and cover half of it with the second sheet of black paper. Mount the lens on the table in the original position and repeat the above experiment.

Observations:

A focussed image is obtained in both the experiments. However, the image formed with half the lens covered is less bright as compared to when it is uncovered.

Answer

A point on the object needs only two rays to form an image. Since the entire lens is not covered, at least two rays from each point does pass through the lens and hence, image is formed. However, since the number of rays involved in the image formation reduces as the lens is covered, the image formed is of lesser intensity and less brighter.

Experiment:

Apparatus needed: convex lens, candle, two black sheets of paper.

Procedure:

Mount a lens on a table. Light a candle with a match stick and place it on one side of the lens on the table. Turn of the lights of the room so that it gets dark. Take the sheet of paper and place it on the other side of the lens as the candle. Now focus the image formed on the sheet of paper by moving the sheet only. Observe the image. Mark the positions of candle, lens and screen. Now, turn off the light. Take the lens and cover half of it with the second sheet of black paper. Mount the lens on the table in the original position and repeat the above experiment.

Observations:

A focussed image is obtained in both the experiments. However, the image formed with half the lens covered is less bright as compared to when it is uncovered.

Question 7

An object

5 c m

in length is held

25 c m

away from converging lens of focal length

10 c m

. Draw the ray diagram and find the position, size and nature of the image formed.

Medium

Solution

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Answer

Given,

Height of object

= 5 c m

Position of object,

u = - 25 c m

Focal length of the lens,

f = 10 c m

Position of image,

v = ?

We know that,

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

\frac{1}{v} + \frac{1}{2 5} = \frac{1}{1 0}

\frac{1}{v} = \frac{1}{1 0} - \frac{1}{2 5}

So,

\frac{1}{v} = \frac{( 5 - 2 )}{5 0}

That is,

\frac{1}{v} = \frac{3}{5 0}

So,

v = \frac{5 0}{3} = 16.66 c m

Thus, distance of image is

16.66 c m

on the opposite side of lens.

Now, magnification

= \frac{v}{u}

That is,

m = \frac{1 6 . 6 6}{- 2 5} = - 0.66

Also,

m = \frac{h e i g h t o f i m a g e}{h e i g h t o f o b j e c t}

or

- 0.66 = \frac{h e i g h t o f i m a g e}{5 c m}

Therefore, Height of image

= 3.3 c m

The negative sign of the height of image shows that an inverted image is formed.

Thus, position of image is at

16.66 c m

on opposite side of lens.

Size of image

= - 3.3 c m

at the opposite side of lens

Nature of image is real and inverted.

Answer

Given,

Height of object

= 5 c m

Position of object,

u = - 25 c m

Focal length of the lens,

f = 10 c m

Position of image,

v = ?

We know that,

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

\frac{1}{v} + \frac{1}{2 5} = \frac{1}{1 0}

\frac{1}{v} = \frac{1}{1 0} - \frac{1}{2 5}

So,

\frac{1}{v} = \frac{( 5 - 2 )}{5 0}

That is,

\frac{1}{v} = \frac{3}{5 0}

So,

v = \frac{5 0}{3} = 16.66 c m

Thus, distance of image is

16.66 c m

on the opposite side of lens.

Now, magnification

= \frac{v}{u}

That is,

m = \frac{1 6 . 6 6}{- 2 5} = - 0.66

Also,

m = \frac{h e i g h t o f i m a g e}{h e i g h t o f o b j e c t}

or

- 0.66 = \frac{h e i g h t o f i m a g e}{5 c m}

Therefore, Height of image

= 3.3 c m

The negative sign of the height of image shows that an inverted image is formed.

Thus, position of image is at

16.66 c m

on opposite side of lens.

Size of image

= - 3.3 c m

at the opposite side of lens

Nature of image is real and inverted.

Question 8

A concave lens of focal length 15cm forms an image 10 cm from the lens . How far is the object placed from the lens? Draw the ray diagram.

Medium

Solution

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Answer

Given:

f = - 15

v = - 10

From mirror formula,

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

\frac{1}{- 1 0} - \frac{1}{u} = \frac{1}{- 1 5}

\frac{1}{u} = \frac{1}{1 5} - \frac{1}{1 0}

u = - 30 c m

Answer

Given:

f = - 15

v = - 10

From mirror formula,

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

\frac{1}{- 1 0} - \frac{1}{u} = \frac{1}{- 1 5}

\frac{1}{u} = \frac{1}{1 5} - \frac{1}{1 0}

u = - 30 c m

Question 9

An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.

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Solution

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Answer

It is to be remembered that a convex mirror always forms the virtual, erect and diminished image.

Given: Object distance,

u = - 10 c m

Focal length,

f = 15 c m

To find: Image distance,

v

From mirror formula :

\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

\frac{1}{v} - \frac{1}{1 0} = \frac{1}{1 5}

\frac{1}{v} = \frac{1}{1 5} + \frac{1}{1 0} = \frac{2 5}{1 5 0}

v = 6 c m

Hence image will be formed 6 cm beyond the mirror, i.e virtual magnification

m = \frac{- v}{u} = \frac{- 6}{- 1 0} = 0.6

Hence it will be erect and diminished.

Answer

It is to be remembered that a convex mirror always forms the virtual, erect and diminished image.

Given: Object distance,

u = - 10 c m

Focal length,

f = 15 c m

To find: Image distance,

v

From mirror formula :

\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

\frac{1}{v} - \frac{1}{1 0} = \frac{1}{1 5}

\frac{1}{v} = \frac{1}{1 5} + \frac{1}{1 0} = \frac{2 5}{1 5 0}

v = 6 c m

Hence image will be formed 6 cm beyond the mirror, i.e virtual magnification

m = \frac{- v}{u} = \frac{- 6}{- 1 0} = 0.6

Hence it will be erect and diminished.

Question 10

An object

5.0 c m

in length is placed at a distance of

20 c m

in front of a convex mirror of radius of curvature

30 c m

. Find the position of the image, its nature, and its size.

Medium

Solution

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Answer

Given:

u = - 20 c m

f = R / 2 = 15 c m

Size of image,

h = 5 c m

Let size of image be

h^{'}

From mirror formula,

\frac{1}{u} + \frac{1}{v} = \frac{1}{f}

\frac{1}{- 2 0} + \frac{1}{v} = \frac{1}{1 5}

v = \frac{6 0}{7} = 8.57 c m

The image is formed

8.57 c m

behind the mirror. It is virtual and erect.

Magnification,

m = - v / u = 8.57 / 20 = 0.428

Also,

m = h^{'} / h = h^{'} / 5

From above,

h^{'} / 5 = 0.428

h^{'} = 2.14 c m

Answer

Given:

u = - 20 c m

f = R / 2 = 15 c m

Size of image,

h = 5 c m

Let size of image be

h^{'}

From mirror formula,

\frac{1}{u} + \frac{1}{v} = \frac{1}{f}

\frac{1}{- 2 0} + \frac{1}{v} = \frac{1}{1 5}

v = \frac{6 0}{7} = 8.57 c m

The image is formed

8.57 c m

behind the mirror. It is virtual and erect.

Magnification,

m = - v / u = 8.57 / 20 = 0.428

Also,

m = h^{'} / h = h^{'} / 5

From above,

h^{'} / 5 = 0.428

h^{'} = 2.14 c m

NCERT Solutions for Class 10 Physics Chapter 10 : Light - Reflection and Refraction

Access NCERT Solutions for Class 10 Physics Chapter 10 : Light - Reflection and Refraction

NCERT Solutions for Class 10 Physics Chapter 10 : Light - Reflection and Refraction

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