NCERT Solutions for Class 10 Science Chapter 9 Light – Reflection and Refraction

NCERT
SOLUTIONS
CLASS - 10th

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Class : 10th
Subject : Science
Chapter : 10
Chapter Name : Light – Re ection And Refraction

Q1 De ne the principal focus of a concave mirror.

Answer. The principal focus of a concave mirror is a point on its principal axis to which all the
light rays which are parallel and close to the axis, converge after re ection from the concave
mirror.

Page : 168 , Block Name : Questions

Q2 The radius of curvature of a spherical mirror is 20 cm. What is its focal length?

Answer. Focal length = 2 x Radius of curvature = 2 x 20 cm = 10 cm

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Q3 Name a mirror that can give an erect and enlarged image of an object.

Answer. Concave mirror.

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Q4 Why do we prefer a convex mirror as a rear-view mirror in vehicles?

Answer. We prefer a convex mirror as a rear-view mirror in vehicles because of two reasons -1.
A convex mirror always produces an erect image of the objects. 2 The image formed in a
convex mirror is highly diminished or much smaller than the object, due to which a convex
mirror gives a wide eld of view of the traf c behind. A convex mirror enables the driver to
view such larger area of the traf c behind him.

Page : 168 , Block Name : Questions

Q1 Find the focal length of a convex mirror whose radius of curvature is 32 cm.

Answer. R = +32cm and f =
R 32
= + = +16cm
2 2

Page : 171 , Block Name : Questions

Q2 A concave mirror produces three times magni ed (enlarged) real image of an object placed
at 10 cm in front of it. Where is the image located?

Answer.
′
h v
m = = − = −3 or v = 3u
h u

But u = −10cm
−v −v
Therefore, m = ⇒ −3 = −v
u −10
⇒ −3 =
−10
∴ v = −30cm

Page : 171 , Block Name : Questions

Q1 A ray of light travelling in air enters obliquely into water. Does the light ray bend towards
the normal or away from the normal? Why?

Answer. The light-ray bends towards the normal because the ray of light goes from a rarer
medium to a denser medium.

Page : 176 , Block Name : Questions

Q2 Light enters from air to glass having refractive index 1.50. What is the speed of light in the
glass? The speed of light in vacuum is 3 × 108 m s–1.

Answer. Refractive index of glass,n 8 = 1.50
8 −1
Speed of light in vacuum, c = 3 × 10 ms
8
c 3×10 m/s 8 −1
Speed of light in glass = = = 2 × 10 ms
ng 1.5

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Q3 Find out, from Table 10.3, the medium having highest optical density. Also nd the
medium with lowest optical density.

Answer.From table 10.3, diamond has highest refractive index (- 2.42), so it has highest optical
density. Air has lowest refractive index 1.0003), so it has lowest optical density.

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Q4 You are given kerosene, turpentine and water. In which of these does the light travel
fastest? Use the information given in Table 10.3.

Answer. For kerosene, n = 144 For turpentine, n 1.47 For water, n - 1.33 Because water has the
lowest refractive index, therefore light travels fastest in this optically rarer medium than
kerosene and turpentine oil.

Page : 176 , Block Name : Questions

Q5 The refractive index of diamond is 2.42. What is the meaning of this statement?

Answer. By saying that the refractive index of diamond is 2.42, we mean that the speed of light
in diamond is lower by a factor of 2.42 relative to that in vacuum.

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Q1 De ne 1 dioptre of power of a lens.

Answer. One dioptre is the power of a lens whose focal length is 1 metre.

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Q2 A convex lens forms a real and inverted image of a needle at a 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 the
object? Also, nd the power of the lens.

Answer. Question 2 A convex lens forms a real and inverted image of a needle at a 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 the object ? Also, nd the power of the lens. , Sol. Here, u — +50 cm Solution Here v
= +50cm Because the real image is of the same size as the object,
′
h v
Therefore, m = = = −1
h u

or u = −v = −50cm

1 1 1 1 1 2 1
Now, = − = − = + = +
f v u +50 −50 50 25

or f = 25cm = 0.25m

1 1
P = = + = +4D
f 0.25m

Page : 184 , Block Name : Questions

Q3 Find the power of a concave lens of focal length 2 m.

Answer. Because the focal length of a concave lens is negative, therefore f = -2 m
′
h v
Therefore, m = = = −1
h u

or u = −v = −50cm

1 1 1 1 1 2 1
Now, = − = − = + = +
f v u +50 −50 50 25

or f = 25cm = 0.25m

1 1
P = = + = +4D
f 0.25m

Page : 184 , Block Name : Questions

Q1 Which one of the following materials cannot be used to make a lens? (a) Water (b) Glass (c)
Plastic (d) Clay

Answer. (d) Clay

Page : 185 , Block Name : Exercise

Q2 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.

Page : 185 , Block Name : Exercise

Q3 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 in nity (d)

Between the optical centre of the lens and its principal focus.

Answer. (b) At twice the focal length.

Page : 185 , Block Name : Exercise

Q4 A spherical mirror and a thin spherical lens have each 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

Page : 185 , Block Name : Exercise

Q5 No matter how far you stand from a mirror, your image appears erect. The mirror is likely
to be (a) only plane. (b) only concave. (c) only convex. (d) either plane or convex.

Answer. (d) Either plane or convex.

Q6 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.

Q7 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. A concave mirror gives an erect image when the object is placed between the focus F
and the pole P of the concave mirror, i.e., between 0 and 1 5 cm from the mirror. The image
thus formed will be virtual, erect and larger than the object. A'

Q8 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) Concave mirrors are used as re ectors in headlights of cars. When a bulb is located
at the focus of the concave mirror, the light rays after re ection from the mirror travel over a
large distance as a parallel beam of high intensity.
(b) A convex mirror is used as a side/rear-view mirror of a vehicle because A convex mirror
always forms an erect, virtual and diminished image of an object placed anywhere in front it.
A convex mirror has a wider eld of view than a plane mirror of the same size.
(c) Large concave mirrors are used to concentrate sunlight to produce heat in solar furnaces.

Q9 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. A convex lens forms complete image of an object, even if its one half is covered with
black paper. It can be explained by considering following two cases. Case I When the upper
half of the lens is covered In this case, a ray of light coming from the object will be refracted by
the lower half of the lens. These rays meet at the other side of the lens to form the image of
the given object, as shown in the following gure.

Case II: When the lower half of the lens Is covered In this case, a ray of light coming from the

object is refracted by the upper half of the lens. These rays meet at the other side of the lens to
form the image of the given object, as shown in the given gure.

Q10 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 nd the position, size and the nature of the image formed.

Answer. Here : Object distance, u= -25 cm, Object height, h = 5 cm, Focal length, f = +10 cm
According to the lens formula,
1 1 1
= −
f ν u

We have ⇒
1 1 1 1 1 15 250
= − = − = or ν = = 16.66cm
ν f u 10 25 250 15

The positive value of v shows that the image is formed at the other side of the lens.
Image distance
For magni cation m = Object distance
=
v

u
=
16.66

−25
= −0.66

The negative sign shows that the image is inverted.
But magni cation
Image height ′
h
m = =
Object height 5

′
h = −0.66 × 5
Or
= −3.3cm

Q11 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 (as concave lens forms the image on

the same side of the lens) From the lens formula 1

f
=
1

ν
−
1

u
, we have
1 1 1 1 1
= − = −
u v f −10 −15

−3 + 2 1
= = −
30 30

Q12 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. Object distance, u = -30 cm The ne ative value of u indicates that the obéct is laced in
front of the lens.
1 1 1
From mirror formula, + = , we have
u v f

1 1 1 1 1 2 + 3 1
= − = − = = or v = 6cm
v f u u −10 30 6

Thus, image distance, v = + 6 cm Because v is +ve, so a virtual image is formed at a distance of
6 cm behind the mirror. Magni cation, m =
−v −6 1
= = ( i.e. < 1)
u −30 5

The positive value of m shows that image erect and its value, which is less than 1 , shows that
image is smaller than the object. Thus, image is virtual, erect and diminished.

Q13 The magni cation produced by a plane mirror is +1. What does this mean?

Answer. Since magni cation, m = Given, m = +1, so h' = h and v (i) m = 1 indicates
h −ν
=
h u

the size of image is same as that of object. (ii) positive sign of m indicates that an erect image
is formed. -u The opposite signs of v and u indicate that image is formed on the other side of
the mirror from where the object is placed i.e., image is formed behind the mirror and thus
image formed is virtual.

Q14 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.

Answer. Since object size, h = +5 cm, object distance, u -20 cm and radius of curvature, R = +30

cm
R 30
∴ Focal length, f = = + = 15cm
2 2
1 1 1 1 1 1
From mirror formula, = + , we have = −
f v u v f u

1 1 1 4 + 3 7 60
or = − = = or v = = 8.6cm
v +15 −20 60 60 7
′
v h
∵ Magnification, m = − =
u h
′ vh 8.6×5
∴ Image size , h = − = − = 2.15 ≅ 2.2cm
u −20

A virtual, erect image of height 2.2 cm is formed behind the mirror at a distance of 8.6 cm
from the mirror.

Q15 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 focussed
image can be obtained? Find the size and the nature of the image.

Answer. Here, object size, h = +7.0 cm, object distance, u = -27 cm and focal length, f -18 cm
Image distance, v = ? and image size, h' =?
For the mirror formula = − , 1

7
1

ν
1

u

We have 1

v
=
1

f
−
1

u
or
1

v
=
−18
1
−
1

−27
=
−3+2

54
= −
1

54
or v = −54cm

The screen should be placed at a distance of 54 cm on the object side of the mirror to obtain a
sharp image
′
h v
magnification, m = = −
h u
Now (−54) × (+7)
vh
′
image size, h = − = = −14cm.
u (−27)

The image is real, inverted and enlarged in size.

Q16 Find the focal length of a lens of power – 2.0 D. What type of lens is this?

Answer. Here, P = -2.0 D The type of lens is concave because the focal length is negative.
1 1
f = = = −0.5m
P −2.0D

Q17 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. Here P=+1.5 D
1 1 10
f = = = + m = +0.6667cm = +66.67cm
P +1.5D 15

Because the focal length is positive, the prescribed lens is converging.