# Ch 9: Light – Reflection and Refraction (NCERT Solutions)

NCERT Solutions for Class 10 Science Ch 9: Light – Reflection and Refraction, is given below.

You will get solutions for;

• In-text questions as well as,
• Exercises questions.

So let’s dive into it!

## In-text Questions Set-1

1) Define the principle focus of a concave mirror.

Solution:
The principal focus of a concave mirror is the point on its principal axis where light rays that are parallel to the axis meet after reflecting off the mirror.

2) The radius of curvature of a spherical mirror is 20 cm. What is its focal length?

Solution:
The focal length (f) is calculated as half of the radius (R). So, if the radius is 20 cm, the focal length is:

f = R/2 = 20 cm / 2 = 10 cm.

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

Solution:
A concave mirror can give an erect and enlarged image of an object.

4) Why do we prefer a convex mirror as a rear-view mirror in vehicles?

Solution:
We use a convex mirror as a rear-view mirror in vehicles because it provides an upright, smaller image, giving the driver a much wider field of view.

## In-text Questions Set-2

1) Find the focal length of a convex mirror whose radius of curvature is 32 cm.

Solution:
The radius of curvature (R) is 32 cm.
The focal length (f) is half of the radius of curvature, so:
f = R/2 = 32 cm / 2 = 16 cm.

2) A concave mirror produces three times magnified real image of an object placed at 10 cm in front of it. Where is the image located?

Solution:
Distance of object from concave mirror (u) = -10 cm.
Magnification (m) = -3

m = -v/u

v = -mu = -(3) x (-10) = -30 cm.

## In-text Questions Set-3

1) A ray of light traveling in air enters obliquely into water. Does the light ray bend towards or away from the normal? Why?

Solution:
The light bends toward the normal when it enters water because water is optically denser than air.

2) Light enters from air to glass having refractive index 1.50. What is the speed of light in glass? The speed of light in vacuum is 3×108 m/s.

Solution:

Speed of light in vacuum (c) = 3 x 108 m/s.
Refractive index = c/v.
Speed of light in glass = 3 x 108 m/s/ 1.50
= 2 x 108 m/s

3) Find out, from Table (10.3), the medium having highest optical density. Also, find the medium with lowest optical density.

Solution:
According to the table, diamond has the highest optical density (2.42), while air has the lowest optical density (1.0003).

4) You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in table 10.3.

Solution:
The refractive index of water is the lowest of the three substances, which means that light travels fastest in water.

5) The refractive index of diamond is 2.42. What is the meaning of this statement?

Solution:
This means that light travels 2.42 times slower in diamond than in air.

## In-text Questions Set-4

1) Define 1 dioptre of power of lens.

Solution:
One diopter is the power of a lens with a focal length of 1 meter.

2) A convex lens forms a real and inverted image of a needle at distance of 50 cm. from it. Where is the needle placed in front of the convex lens if the image is equal to the size of objects? Also, find the power of lens.

Solution:
Image distance (v) = +50 cm, hi = ho

hi/ho = v/u

u = v × ho / hi
= 50 × ho / hi
= 50 cm.

Now,
u = -50 cm
v = + 50 cm.
f = ?

1/f = 1/v – 1/u
1/f = 1/50 + 1/50
f = + 25 cm = 0.25 m

Power of lens (P) = 1/f
= 1/ 0.25 = + 4D.

3) Find the power of a concave lens of focal length 2 m.

Solution:
Focal length of concave lens = – 2 m.

P = 1/f = 1/ (-2m)
= -0.5 D

## Exercise Questions

1) Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay

Solution:
(d) Clay

2) The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?
(a) Between the principal focus and the center of curvature
(b) At the center of curvature
(c) Beyond the center of curvature
(d) Between the pole of the mirror and its principal focus

Solution:
(d) Between the pole of the mirror and its principal focus

3) Where should an object be placed in front of convex lens to get a real image of the size of the object?
(a) At the principal focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical center of the lens and its principal focus.

Solution:
(b) At twice the focal length

4) A spherical mirror and thin spherical lens have each of focal length of -15 cm. the mirror and lens are likely to be
(a) Both concave
(b) Both convex
(c) The mirror is concave and the lens is convex
(d) The mirror is convex and lens is concave

Solution:
(a) Both concave

5) No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
(a) Plane
(b) Concave
(c) Convex
(d) Either concave or convex

Solution:
(d) Either concave or convex

6) 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 50 cm
(b) A concave lens of focal length 50 cm
(c) A convex lens of focal length 5 cm
(d) A concave lens of focal length 5 cm

Solution:

(c) A convex lens of focal length 5 cm

7) 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 mirror? What is the nature of image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.

Solution:

To get a virtual, erect image that’s larger than the object, place the object in front of a concave mirror, between its pole and principal focus, at a distance of less than 15 cm.

Here is the ray diagram:

8) Name the type of mirror used in the following situations:
(b) Side/rear-view mirror of a vehicle
(c) Solar furnace

Solution:

(a) Headlights of a car: A concave mirror is used to focus light into a parallel beam.
(b) Side/rear-view mirror of a vehicle: A convex mirror is used to create a virtual, upright, and smaller image, providing a wider field of view.
(c) Solar furnace: A concave mirror concentrates sunlight to produce heat.

9) 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.

Solution:

Yes, it will create a complete image of the object, as shown in the figure.

You can test this by covering the lower half of the lens with black paper and observing the image of a distant object, like a tree, on a screen. The image will still be complete, but it will be less bright.

10) An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.

Solution:

Height of object, ho = 5 cm
Object distance, u = -25 cm
Focal length, f = 10 cm

According to lens formula,
1/v – 1/u = 1/f

1/v = 1/f + 1/u = (1/10) – (1/25) = 15/250

V = 250/15 = 16.66 cm

Also for converging lens, hi/ho = v/u

hi = (v/u) × ho = [(250/15)/(-25)] × 5 = -10/3 = -3.3 cm

So, the image formed is inverted and it is obtained at 16.7 cm behind the lens, and its height is 3.3 cm.

The ray diagram is shown below:

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

Solution:

f= -15 cm, v= -10 cm

1/v -1/u = 1/f
1/u = 1/15 – 1/10 = -1/30

So, u = -30 cm.

Ray diagram as follows:

12) 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 image.

Solution:

f = +15 cm, u = -10 cm.

1/f = 1/v +1/u
1/v = 1/15 +1/10
1/v = 5/30

So, v = + 6 cm.

Now,
m = -v/u
= -(6)/(-10)
= 0.6

The image is formed 6 cm behind the mirror, it is a virtual and erect image. And the image is also diminished.

13) The magnification produced by a plane mirror is +1. What does this mean?

Solution:
m = hi/ho= v/u

For a plane mirror, the magnification is +1, which means the size of the image is the same as the size of the object.

14) An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.

Solution:

Radius of curvature (R) = 30 cm
f = R/2 = 30/2 = 15 cm

u = –20 cm, ho= 5 cm.

1/v + 1/u = 1/f
1/v = 1/15 + 1/20 = 7/60
v = 60/7 = 8.6 cm.

Image is virtual and erect and formed behind the mirror.

hi/ho = -v/u
hi/5 = -8.6/(-20)
hi = 2.15 cm.

Size of image is 2.15 cm.

15) An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focused image can be obtained? Find the size and the nature of the image.

Solution:

u = – 27 cm, f = – 18 cm. ho= 7.0 cm

1/v = 1/f- 1/u

1/v = -1/18 + 1/27 = -1/54

v = – 54 cm.

Screen must be placed at a distance of 54 cm from the mirror in front of it.

hi/ho = -v/u

hi/7 = +54/-27

hi = -2 x 7 = -14 cm.

Thus, the image is of 14 cm length and is inverted image.

16) Find the focal length of a lens of power -2.0 D. What type of lens is this?

Solution:
Power of lens (P) = -2.0 D
P = 1/f or f = 1/m

f = 1/-2.0 = -0.5 m.

The -ve sign of focal length means that the lens is concave lens.

17) A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Solution:
P = +1.5 D

f = 1/P = 1/+1.5 = 0.67 m.

As the power of lens is (+ve), the lens is a converging lens.