11

4
Reflection of Light

Q.1. (A)
i. (C)
ii. (D)
iii. (A)

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iv. (B)
(B)
i.
Group ‘A’ Group ‘B’

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a. Plane mirror 3. 1
b. Convex mirror 1. Always positive

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ii. Image will be formed behind the mirror.
iii. Periscope: Plane mirror :: Street lights: Convex mirror

iv.
1 1
f
= +
v
1
u
lic
Q.2. (A)
i. a. Concave mirror is a focusing mirror. The parallel rays of light incident on this mirror converge at the
ub
principal focus.
b. In this way, the concave mirrors in solar devices focus the heat and radiations coming from sun
in the focal plane.
c. Hence, to use the maximum amount of heat and radiation coming from the sun, concave mirrors are
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used in solar devices.

ii. a. In convex mirrors, the image of an object is erect and smaller.
b. It remains erect but becomes smaller and smaller when the mirror is taken away from the object.
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c. As a result, we can see the images of the surroundings in the mirror.

d. By using convex mirror on the outside of cars, driver can get an erect, diminished and clear
view (image) of the vehicles coming from behind.
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Hence, the mirrors fitted on the outside of cars are convex.

(B)
i.
Ta

C : Centre of curvature of the mirror

I
P F : Principal focus of the mirror
C F P : Pole of the mirror
I : Point image of the object

ii. Solution:
Given: Object distance (u) =  20 cm, image distance (v) =  40 cm
To find: Focal length (f)
1 1 1
Formula: + =
v u f
1
 

Std. IX: Perfect Science and Technology

Calculation: From formula,
1 1 1
+ =
40 20 f
1 1 1
   =
40 20 f
1  2 1
 =
40 f
1 3
 =
f 40
40
 f= = 13.33 cm.
3

ns
Negative sign of f shows that mirror is concave.
Neglecting the negative sign f = 13.33 cm
Ans: Focal length of the concave mirror is 13.33 cm.

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iii.
Position of object Between focus and pole
Position of image Behind the mirror

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Nature of image Virtual and erect
Size of image Magnified

Q.3. i.
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Magnification due to spherical mirror is given by, the ratio of the height of the image (h2) to the height
of the object (h1).
B
2cm

ub
m

A A
C P
F
2cm

B
Object placed at the
centre of curvature
et

Nature of image: Real, inverted and of same size as that of object.

ii. Given: Object distance (u) = 10 cm, radius of curvature (R) = 40 cm
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To find: a. Position of image (v), b. Nature of image,

c. Magnification (M)
R 1 1 1
Formulae: a. f = b. + =
Ta

2 v u f

v
c. Magnification (M) =
u
Calculation: From formula (a),
40
f= =  20 cm
2
From formula (b),
1 1 1
+ =
v 10 20
1 1 1 2 1 1
 =  = =
v 10 20 20 20
 v = 20 cm
22
 

Chapter 11: Reflection of Light

Positive sign of v shows that image is virtual.
From formula (c),
v 20
M= = =2
u 10
Positive sign of M shows that image is erect.
Ans: a. The position of the image is 20 cm.
b. The image formed is virtual and erect.
c. The magnification of the image is 2.
iii. a. In floodlights, the source of light is kept just beyond the centre of curvature of concave
mirror to obtain light confined in small cone of illumination.
b. In shaving mirror, the source of light or object is kept between pole and focus to obtain an
erect, virtual and magnified image.

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c. In solar devices, the source of light is sun which is at a very large distance and sun rays
converge at a single point after reflection.
360
Q.4. i. a. Angle between two mirrors A =

io
n +1
b. As number of images, n = 3
360
 A= = 90

at
31
c.
Image 1 lic Image 3

M1
(Mirror) 90

Image 2
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Object
(Mirror)
M2
d. Plane mirror is used for the given activity.
e. If two such mirrors are kept parallel to each other, then we will see infinite images of the object
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kept between the mirrors.

ii. a. As the image is to be projected on the wall (screen), the student should use a concave mirror. It
is because a concave mirror forms real images.
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b. To get a magnified image on the wall, he should place the arrow between the focus and the centre of
curvature i.e., anywhere between focal length (f) and twice the focal length (2f).
c. Ray diagram:
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B F P
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C B

A
Object placed between focus
and centre of curvature
d. Yes, he can use this mirror to project a diminished image of the arrow on the same wall. For
this, he needs to place the arrow beyond the centre of curvature (C) of the concave mirror.
e. The mirror used is of converging type.