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Chapter 1:Chemical Reaction And Equations

Chapter 2:Acids, Bases and Salts

Chapter 3:Metals and Non-metals

Chapter 4:Carbon and Its Compounds

Chapter 5: Periodic Classification of Elements

Chapter 6: Life Processes

Chapter 10: Light Reflection and Refraction

Chapter 11:Human Eye and Colourful World

Chapter 12:Electric

Chapter 13:Magnetic Effects of Electric Current

Chapter 14:Sources of Energy

Chapter 15:Our Environment

Chapter 16:Sustainable Management of Natural Resources

NCERT Solutions for class 10 Science Chapter 10: Light Reflection and Refraction

NCERT Solutions for class 10 Science Chapter 10 Exercise Question

question 1
Which one of the following materials cannot be used to make a lens?

(a) Water

(b) Glass

(c) Plastic

(d) Clay

Answer

(d) Clay

question 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 centre of curvature

(b) At the centre of curvature

(c) Beyond the centre of curvature

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

Answer

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

question 3
Where should an object be placed in front of a 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 centre of the lens and its principal focus.

Answer

(b) Between the optical centre of the lens and its principal focus

question 4
A spherical mirror and a thin spherical lens have a focal length of -15 cm. The mirror and the 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, but the lens is concave

Answer

(a) Both concave.

question 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 plane or convex

Answer

(d) either plane or convex

question 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

Answer

(c) A convex lens of focal length 5 cm

question 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 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.

Answer
Chapter 10: Light Reflection and Refraction exercise solution nCERT solution

As we know an erect image of an object, using a concave mirror form when we place object between focus and pole So range of the distance of the object must be between 0 to 15 cm from the pole of the mirror.

Nature of the image is virtual, erect, and larger than the object.

question 8
Name the type of mirror used in the following situations.

(a) Headlights of a car

(b) Side/rear-view mirror of a vehicle

(c) Solar furnace

Support your answer with reason.

Answer

(a) We use Concave Mirror for Headlights of a car to get powerful parallel beams of light.

(b) Convex mirrors are used rear-view (wing) mirrors in vehicles because it always give an erect image. Also, it have a wider field of view

(c) Concave Mirror: concave mirrors are used to concentrate sunlight to produce heat in solar furnaces..

question 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 answer experimentally. Explain your observations.

Answer
 exercise solution Chapter 10: Light Reflection and Refraction NCERT solution

Yes, from above figure we observe that the lens will produce a complete image even half of a convex lens is covered with a black paper. however the intensity of the image may be less.

question 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.

Answer

Height of the Object, ho = 5 cm

object Distance u = -25 cm

Focal length f = 10 cm

We know lens formula,

1 / f
=
1 / v
-
1 / u

1 / 10
=
1 / v
-
1 / (-25)

1 / 10
=
1 / v
+
1 / 25

1 / v
=
1 / 10
-
1 / 25

1 / v
=
25 -10 / 250

1 / v
=
15 / 250

v =
250 / 15

v = 16.66
Again we know
m =
v / u
=
hi / ho

16.7 / (-25)
=
hi / 5

16.7 x 5 = hi x (-25)
83.5 / -25
= hi
hi = -3.34cm
minus sign show image is inverted

So Image is inverted and formed at a distance of 16.7 cm behind the lens and measures 3.3 cm. The ray diagram is shown below.

Chapter 10: Light Reflection and Refraction exercise solution nCERT solution

question 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.

Answer

Focal length f = – 15 cm

Image distance, v= – 10 cm

We know lens formula,

1 / f
=
1 / v
-
1 / u

1 / (-15)
=
1 / (-10)
-
1 / u

1 / u
=
1 / (-10)
-
1 / (-15)

1 / u
=
1 / (-10)
+
1 / 15

1 / u
=
15 -10 / (-150)

1 / u
= -
5 / 150

u = -
150 / 5
u = -30cm

The negative value indicates that the object is placed in front of the lens

Chapter 10 science class 10 solution nCERT solution

question 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 the image.

Answer

Focal length of convex mirror (f) = +15 cm

Object distance (u) = – 10 cm

we know mirror formula,

1 / f
=
1 / v
+
1 / u

1 / 15
=
1 / v
+
1 / (-10)

1 / 15
=
1 / v
-
1 / 10

1 / v
=
1 / 15
+
1 / 10

1 / v
=
15 + 10 / 150

1 / v
=
25 / 150

v =
150 / 25
=6
Also magnification =
-v / u
=
-6 / -10
=0.6cm

The image distance is 6 cm .

So image formed is virtual and erect and diminished.

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

Answer

The magnification is 1 means that the height of the image is equal to the height of the object and positive sign show image is virtual and erect

question 14
An object 5 cm is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position, nature and size of the image.

Answer

Object distance u = – 20 cm

Object height h = 5 cm

R = 30 cm

Also radius = 2 × Focal length

f = 15 cm

We know mirror formula,

1 / f
=
1 / v
+
1 / u

1 / 15
=
1 / v
+
1 / (-20)

1 / 15
=
1 / v
-
1 / 20

1 / v
=
1 / 15
+
1 / 20

1 / v
=
20 + 15 / 300

1 / v
=
35 / 300

v =
300 / 35
=8.5cm
magnification =
-v / u
=
-8.5 / -20
= 0.42 cm
Similarly
magnification =
hi / ho

0.42 =
hi / 5

hi = 0.42 x 5 = 2.12cm

So, the image formed is erect, virtual, and smaller in size.

question 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 sharply focused image can be obtained? Find the size and nature of the image.

Answer

Object distance u = – 27 cm

Object height h = 7 cm

Focal length f = – 18 cm

We know mirror formula

1 / f
=
1 / v
+
1 / u

1 / (-18)
=
1 / v
+
1 / (-27)

1 / -18
=
1 / v
-
1 / 27

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

1 / v
=
27 - 18 / -486

1 / v
=
9 / -486

v =
-486 / 9
= - 54cm
magnification =
-v / u
=
-(-54) / -27
= -2 cm
Similarly
magnification =
hi / ho

-2 =
hi / 7

hi = - 2 x 7 = -14cm

The image formed is inverted and larger than object.

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

Answer
Power of lens P =
1 / f

Here P = -2D

-2 =
1 / f
f = -
1 / 2
= -0.5 m

A concave lens has a negative focal length. So it is a concave lens.

question 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?

Answer
Power of lens P =
1 / f

P = 1.5D

f =
1 / P
=
1 / 1.5
= 0.66 m

. it is a convex lens because a convex lens has a positive focal length.