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Reflection Of Light 2025 - 26
Time : 60 Min Phy : Ray Optics Marks : 180
Hints and Solutions
01) Ans: A) The virtual image formed in a plane
mirror can be photographed.
Sol: As virtual image is seen on the photograph.
02) Ans: A) 3d
Sol: From the following ray diagram,
HI AB d and
d DS CD 2
As, AH 2AD
2d GH 2CD d 2
Similarly IJ = d,
GJ = GH + HI + IJ = d + d + d = 3d
03) Ans: C) 60 m
Sol: In this case,
h tan 45 60
h 60 m
04) Ans: C) A q and r; B q and r;
C q and s; D p and s
Sol: Magnification in the mirror,
v m u
m 2 v 2u
As v and u have same signs so the mirror is
concave and image formed is real.
1 u m v 2 2
Concave mirror and real image.
m 2 v 2u
As v and u have different signs but magnification is
2 so the mirror is concave and image formed is
virtual.
1 u m v 2 2
As v and u have different signs with magnification
so the mirror is convex and image formed is
virtual.
05) Ans: C)
f m f u
Sol: As we know,
v m u
Also,
f v u
u u 1 f v
u u 1 v f
v f
u f u
f m f u
06) Ans: D) 2v cos
Sol: From the adjoining figure, it is clear that
relative velocity between object and it's image
= 2v cos
07) Ans: C) 90 o
Sol: Incident ray and finally reflected ray are
parallel to each other. It implies that
o 180.
From, 360 2 180 360 2
o 90
08) Ans: D) 0.4 cm away from the mirror.
Sol: Mirror formula is given by
f cm. f v u f 20 ( 10) 3
If object moves towards the mirror by 0.1 cm, then
u = (10 (^) 0.1)=9.9 cm.
Thus, again from mirror formula,
v 20.4 cm 20 /3 v 9.
means image
shifts away from the mirror by 0.4 cm.
09) Ans: D) concave mirror
Sol: In case of concave mirror, virtual image formed
is larger in size.
10) Ans: D) 10 : 9
Sol: The illuminance on the screen without mirror
is (^1 )
L
I
r
The illuminance on the screen with mirror is given
by
(^2 2 2 )
L L 10 L
I
r (3r) 9 r
Thus,
2
1
I 10
I 9
11) Ans: D) 8 cm
Sol: Suppose, distanceu.
Now,
v 16 u
and v u 120
120 u 16 15u 120 u
u 8 c m
12) Ans: D) 7.5 cm
Sol: Given that, f 15 cm, m 2
(Positive as image is virtual.)
Since,
v m v 2u u
By using mirror formula,
u 7.5 cm 15 ( 2u) u
13) Ans: A) Concave mirror
Sol: Plane mirror as well as convex mirror always
forms erect images. Image formed by concave
mirror may be erect or inverted depending on
position of object.
14) Ans: B) 90 cm
Sol: By using,
f m (f u)
u 90 cm 4 ( 30) u
^
15) Ans: B) 2
Sol: The reflected ray deviate from its original path
by angle 2 , when a mirror is rotated by an angle
.
16) Ans: B) 20 cm/s
Sol: Let at any instant, plane mirror lies at a
distance x from object. Image will be formed behind
the mirror at the same distance x.
When the mirror shifts towards the object by
distance 'y' the image shifts x y (x y) 2y
Speed of image = 2 x speed of mirror
17) Ans: C) 1.78 cm
Sol: Using,
I f
O f u
I (25 / 2)
I 1.78 cm (7.5) 25 ( 40) 2
18) Ans: D)
o 50
Sol: Different angles as shown in the figure.
o o 40 90
o o o 90 40 50
19) Ans: C) f/
Sol: In this case, focal length f and u f
Putting these values in
f u v
1 1 1 f v f f v 2
20) Ans: A) convex.
Sol: Diminished, erect image is formed by using
convex mirror.
21) Ans: A) a real, inverted, same-sized image can
be formed using a convex mirror.
Sol: As we know, convex mirror always forms,
virtual, erect and smaller image.
22) Ans: A) 0.25 m, 1m, 0.5 m
Sol: From the given problem,
For surface P,
1
v f u 3 3
v m 2
For surface Q,
2
v f u 5 5
2
v m 4
1 2 v v 0.25m
Now, Magnification of
v 1 3 / 2 1 P u 3 2
Height of
P 2 1m 2
Now, by using mirror formula,
f v u 30 v ( 30)
v 15 cm , behind the mirror
36) Ans: B) 2 cm
Sol: Here,
I f
O f u
I f
6 f ( 4f )
I 2 cm.
37) Ans: A) 240 o
Sol: Deviation, (360 2 )
o (360 2 60) 240
38) Ans: D) f
Sol: The focal length of the mirror does not change.
39) Ans: B) 90 cm
Sol: From the following ray diagram,
Length of mirror
(10 170) 90 cm 2
40) Ans: C) convex mirror.
Sol: Given size is
It can't be plane and concave
mirror, as both conditions are not satisfied in plane
or concave mirror. Convex mirror can meet all the
requirements.
41) Ans: D) I-B, II-D, III-A, IV-E
42) Ans: A) 40 cm
Sol: Focal length,
R
f R 40 cm 2
43) Ans: B) 60 o
Sol: According to the problem, the angle between
mirror and reflected ray is shown in the following
figure.
44) Ans: D) 4 nR
Sol: As plane mirror rotates through an angle ,
then the reflected ray rotates through an angle 2 .
Therefore spot on the screen will make 2n
revolution per second.
45) Ans: B) 12 cm
Sol: Image is virtual, therefore m = + 3 and
R
f 18 cm 2
^.
^ From
f ( 18) m 3 f u ( 18) u
u 1 2 c m.
