Page 1
26. Reflection of Light
Introductory Exercise 26.2
1. Total deviation produced
d q = °  + °  180 2 180 2 i
d q = °  + 360 2( ) i
From figure
q = °  90 i
Þ d = °  +  360 2 90 [ ] i i
= ° 180
Hence rays 1 and 2 are parallel (anti
parallel).
2. v
0
2 = m/s for plane mirror v
i
= 2 m/s.
Velocity of approach = + v v
i 0
= 4 m/s.
3. In figure, AB is mirror, G is ground, CD is
pole and M is the man. The minimum height
to see the image of top of pole is = EN
Introductory Exercise 26.1
1. Since c =
1
0 0
m e
where c is the speed of light
in vacuum hence unit of
1
0 0
m e
is m/s.
2. Hence
B x t
y
= ´ + ´

2 10 500 10
7 11
T 1.5 sin [ ]
Comparing this equation with the standard
wave eqution B B kx t
y
= +
0
sin [ ] w
k =

500
1
m Þ k =
2p
l
Þ l
p
=
2
k
m
Þ l
p
=
2
500
m =
p
250
metre
w = ´ 1.5 10
11
rad/s
Þ 2 10
11
pn = ´ 1.5
Þ n = ´
1.5
2
10
11
p
Hz
Speed of the wave v
k
= =
´ w 1.5 10
500
11
= ´ 3 10
8
m/s
Let E
0
be the amplitude of electric field.
Then E cB
0 0
8 7
3 10 2 10 = = ´ ´ ´

= 60 V/m
Since wave is propagating along xaxis and
B along yaxis, hence E must be along zaxis
Þ E = 60 V/m sin [ ] 500 10
11
x t + ´ 1.5
90°
90°
90°
i
q
q
i
180°–2i
N
1
1
2
180°–2q
N
2
Page 2
26. Reflection of Light
Introductory Exercise 26.2
1. Total deviation produced
d q = °  + °  180 2 180 2 i
d q = °  + 360 2( ) i
From figure
q = °  90 i
Þ d = °  +  360 2 90 [ ] i i
= ° 180
Hence rays 1 and 2 are parallel (anti
parallel).
2. v
0
2 = m/s for plane mirror v
i
= 2 m/s.
Velocity of approach = + v v
i 0
= 4 m/s.
3. In figure, AB is mirror, G is ground, CD is
pole and M is the man. The minimum height
to see the image of top of pole is = EN
Introductory Exercise 26.1
1. Since c =
1
0 0
m e
where c is the speed of light
in vacuum hence unit of
1
0 0
m e
is m/s.
2. Hence
B x t
y
= ´ + ´

2 10 500 10
7 11
T 1.5 sin [ ]
Comparing this equation with the standard
wave eqution B B kx t
y
= +
0
sin [ ] w
k =

500
1
m Þ k =
2p
l
Þ l
p
=
2
k
m
Þ l
p
=
2
500
m =
p
250
metre
w = ´ 1.5 10
11
rad/s
Þ 2 10
11
pn = ´ 1.5
Þ n = ´
1.5
2
10
11
p
Hz
Speed of the wave v
k
= =
´ w 1.5 10
500
11
= ´ 3 10
8
m/s
Let E
0
be the amplitude of electric field.
Then E cB
0 0
8 7
3 10 2 10 = = ´ ´ ´

= 60 V/m
Since wave is propagating along xaxis and
B along yaxis, hence E must be along zaxis
Þ E = 60 V/m sin [ ] 500 10
11
x t + ´ 1.5
90°
90°
90°
i
q
q
i
180°–2i
N
1
1
2
180°–2q
N
2
= + = + EK KN KN 6
Now in DNKB,
NK
KB
= f tan Þ NK KB = tan f
= 4 tan f
In DBC C ¢ we get,
tan f =
¢
¢
= =
BC
CC
2
2
1
f = ° 45
So, NK = ´ ° = 4 45 4 tan m
Hence in minimum height
= + = 6 4 10 m m m
In DAC C ¢
tan q = =
4
2
2
In DL LA ¢ we get,
LL
LA
¢
= tan q
Þ
LL¢
=
4
2
Þ LL¢ = 8 m
Maximum height = + ¢ = + = CA LL 8 8 16 m
Introductory Exercise 26.3
1. Here f =  10 cm (concave mirror)
(a) u =  25 cm
Using mirror formula,
1 1 1
v u f
+ =
Þ
1 1 1
v f u
=  =  +
1
10
1
25
Þ
1 5 2
50 v
=
 +
Þ v =  = 
50
3
167 . cm
Hence image is real, inverted and less
height of the object.
(b) Since u =  10 cm,
Hence object is situated on focus of the
image formed at ¥.
(c) u  5, f =  10
1 1 1 1
10
1
5 v f u
=  =  +
Þ
1 1 2
10 v
=
 +
Þ v = 10 cm
Hence, image is virtual, erect and two time
of the object.
2. Here u =  3 m, f = 
1
2
m,
we have,
(a)
1 1 1
v f u
= 
Þ
1
2
1
3 v
=  +
Þ v =  0.6 m
As ball moves towards focus the image
moves towards ¥ and image is real as the
distance decreases by focal length image
become virtual which moves from + ¥ to
zero.
(b) The image of the ball coincide with ball,
when u R =  =  1 m
2
B
A
C
K
N
L'
L
8 m
q
q
f
f
f
C' C
D
2 m
pole
= 4 m
M
N
1
2 m
6 m
2 m
E
Page 3
26. Reflection of Light
Introductory Exercise 26.2
1. Total deviation produced
d q = °  + °  180 2 180 2 i
d q = °  + 360 2( ) i
From figure
q = °  90 i
Þ d = °  +  360 2 90 [ ] i i
= ° 180
Hence rays 1 and 2 are parallel (anti
parallel).
2. v
0
2 = m/s for plane mirror v
i
= 2 m/s.
Velocity of approach = + v v
i 0
= 4 m/s.
3. In figure, AB is mirror, G is ground, CD is
pole and M is the man. The minimum height
to see the image of top of pole is = EN
Introductory Exercise 26.1
1. Since c =
1
0 0
m e
where c is the speed of light
in vacuum hence unit of
1
0 0
m e
is m/s.
2. Hence
B x t
y
= ´ + ´

2 10 500 10
7 11
T 1.5 sin [ ]
Comparing this equation with the standard
wave eqution B B kx t
y
= +
0
sin [ ] w
k =

500
1
m Þ k =
2p
l
Þ l
p
=
2
k
m
Þ l
p
=
2
500
m =
p
250
metre
w = ´ 1.5 10
11
rad/s
Þ 2 10
11
pn = ´ 1.5
Þ n = ´
1.5
2
10
11
p
Hz
Speed of the wave v
k
= =
´ w 1.5 10
500
11
= ´ 3 10
8
m/s
Let E
0
be the amplitude of electric field.
Then E cB
0 0
8 7
3 10 2 10 = = ´ ´ ´

= 60 V/m
Since wave is propagating along xaxis and
B along yaxis, hence E must be along zaxis
Þ E = 60 V/m sin [ ] 500 10
11
x t + ´ 1.5
90°
90°
90°
i
q
q
i
180°–2i
N
1
1
2
180°–2q
N
2
= + = + EK KN KN 6
Now in DNKB,
NK
KB
= f tan Þ NK KB = tan f
= 4 tan f
In DBC C ¢ we get,
tan f =
¢
¢
= =
BC
CC
2
2
1
f = ° 45
So, NK = ´ ° = 4 45 4 tan m
Hence in minimum height
= + = 6 4 10 m m m
In DAC C ¢
tan q = =
4
2
2
In DL LA ¢ we get,
LL
LA
¢
= tan q
Þ
LL¢
=
4
2
Þ LL¢ = 8 m
Maximum height = + ¢ = + = CA LL 8 8 16 m
Introductory Exercise 26.3
1. Here f =  10 cm (concave mirror)
(a) u =  25 cm
Using mirror formula,
1 1 1
v u f
+ =
Þ
1 1 1
v f u
=  =  +
1
10
1
25
Þ
1 5 2
50 v
=
 +
Þ v =  = 
50
3
167 . cm
Hence image is real, inverted and less
height of the object.
(b) Since u =  10 cm,
Hence object is situated on focus of the
image formed at ¥.
(c) u  5, f =  10
1 1 1 1
10
1
5 v f u
=  =  +
Þ
1 1 2
10 v
=
 +
Þ v = 10 cm
Hence, image is virtual, erect and two time
of the object.
2. Here u =  3 m, f = 
1
2
m,
we have,
(a)
1 1 1
v f u
= 
Þ
1
2
1
3 v
=  +
Þ v =  0.6 m
As ball moves towards focus the image
moves towards ¥ and image is real as the
distance decreases by focal length image
become virtual which moves from + ¥ to
zero.
(b) The image of the ball coincide with ball,
when u R =  =  1 m
2
B
A
C
K
N
L'
L
8 m
q
q
f
f
f
C' C
D
2 m
pole
= 4 m
M
N
1
2 m
6 m
2 m
E
Using h ut gt = +
1
2
2
Þ t
h
g
= =
´ 2 2 2
9.8
= 0.639 s
Similarly again images match at t = 0.78 s.
3. Since image is magnified, hence the mirror
is concave.
Here, m
v
u
=

Þ

=
v
u
5
Þ v u =  5 …(i)
Let distance between mirror and object is x.
Since image is formed at a distance 5 m
from mirror
v x =  + ( ) 5 …(ii)
From Eqs.(i) and (ii), we get
 + =  ( ) 5 5 x x
Þ 4 5 x =
Þ x = 1.25
Hence mirror is placed at 1.25 m on right
side of the object by mirror formula
1 1 1
v u f
+ = ,
we have
1 1 1
f
=  
6.25 1.25
Þ f =
 6.25
6
,
Hence R f = 2 Þ R =  = 
6.25
2.08
3
m
Thus mirror is concave mirror of radius of
curvature 2.08 m.
4. Since the incident rays and reflected rays
are parallel to each other therefore mirror is
plane mirror.
5. Let us solve the first case :
By applying the geometry we can prove that,
PA v ¢ = =
40
3
cm
Further, in triangles ABP and PA B ¢ ¢ we
have,
AB A B
40 40 3
=
¢ ¢
( / )
\ A B
AB
¢ ¢ = =
3
2
3
cm
Similary, we can solve other parts also.
6. Simply apply :
1 1 1
v u f
= =
and m
I
o
v
u
= =

for lateral magnification. If
magnitfication is positive, image will be
virtual. If magnification is negative, image
will be real.
3
40 cm
20 cm
a
a
a
2q
q
q
q
q
A' F C
M
B
A
20 cm
Page 4
26. Reflection of Light
Introductory Exercise 26.2
1. Total deviation produced
d q = °  + °  180 2 180 2 i
d q = °  + 360 2( ) i
From figure
q = °  90 i
Þ d = °  +  360 2 90 [ ] i i
= ° 180
Hence rays 1 and 2 are parallel (anti
parallel).
2. v
0
2 = m/s for plane mirror v
i
= 2 m/s.
Velocity of approach = + v v
i 0
= 4 m/s.
3. In figure, AB is mirror, G is ground, CD is
pole and M is the man. The minimum height
to see the image of top of pole is = EN
Introductory Exercise 26.1
1. Since c =
1
0 0
m e
where c is the speed of light
in vacuum hence unit of
1
0 0
m e
is m/s.
2. Hence
B x t
y
= ´ + ´

2 10 500 10
7 11
T 1.5 sin [ ]
Comparing this equation with the standard
wave eqution B B kx t
y
= +
0
sin [ ] w
k =

500
1
m Þ k =
2p
l
Þ l
p
=
2
k
m
Þ l
p
=
2
500
m =
p
250
metre
w = ´ 1.5 10
11
rad/s
Þ 2 10
11
pn = ´ 1.5
Þ n = ´
1.5
2
10
11
p
Hz
Speed of the wave v
k
= =
´ w 1.5 10
500
11
= ´ 3 10
8
m/s
Let E
0
be the amplitude of electric field.
Then E cB
0 0
8 7
3 10 2 10 = = ´ ´ ´

= 60 V/m
Since wave is propagating along xaxis and
B along yaxis, hence E must be along zaxis
Þ E = 60 V/m sin [ ] 500 10
11
x t + ´ 1.5
90°
90°
90°
i
q
q
i
180°–2i
N
1
1
2
180°–2q
N
2
= + = + EK KN KN 6
Now in DNKB,
NK
KB
= f tan Þ NK KB = tan f
= 4 tan f
In DBC C ¢ we get,
tan f =
¢
¢
= =
BC
CC
2
2
1
f = ° 45
So, NK = ´ ° = 4 45 4 tan m
Hence in minimum height
= + = 6 4 10 m m m
In DAC C ¢
tan q = =
4
2
2
In DL LA ¢ we get,
LL
LA
¢
= tan q
Þ
LL¢
=
4
2
Þ LL¢ = 8 m
Maximum height = + ¢ = + = CA LL 8 8 16 m
Introductory Exercise 26.3
1. Here f =  10 cm (concave mirror)
(a) u =  25 cm
Using mirror formula,
1 1 1
v u f
+ =
Þ
1 1 1
v f u
=  =  +
1
10
1
25
Þ
1 5 2
50 v
=
 +
Þ v =  = 
50
3
167 . cm
Hence image is real, inverted and less
height of the object.
(b) Since u =  10 cm,
Hence object is situated on focus of the
image formed at ¥.
(c) u  5, f =  10
1 1 1 1
10
1
5 v f u
=  =  +
Þ
1 1 2
10 v
=
 +
Þ v = 10 cm
Hence, image is virtual, erect and two time
of the object.
2. Here u =  3 m, f = 
1
2
m,
we have,
(a)
1 1 1
v f u
= 
Þ
1
2
1
3 v
=  +
Þ v =  0.6 m
As ball moves towards focus the image
moves towards ¥ and image is real as the
distance decreases by focal length image
become virtual which moves from + ¥ to
zero.
(b) The image of the ball coincide with ball,
when u R =  =  1 m
2
B
A
C
K
N
L'
L
8 m
q
q
f
f
f
C' C
D
2 m
pole
= 4 m
M
N
1
2 m
6 m
2 m
E
Using h ut gt = +
1
2
2
Þ t
h
g
= =
´ 2 2 2
9.8
= 0.639 s
Similarly again images match at t = 0.78 s.
3. Since image is magnified, hence the mirror
is concave.
Here, m
v
u
=

Þ

=
v
u
5
Þ v u =  5 …(i)
Let distance between mirror and object is x.
Since image is formed at a distance 5 m
from mirror
v x =  + ( ) 5 …(ii)
From Eqs.(i) and (ii), we get
 + =  ( ) 5 5 x x
Þ 4 5 x =
Þ x = 1.25
Hence mirror is placed at 1.25 m on right
side of the object by mirror formula
1 1 1
v u f
+ = ,
we have
1 1 1
f
=  
6.25 1.25
Þ f =
 6.25
6
,
Hence R f = 2 Þ R =  = 
6.25
2.08
3
m
Thus mirror is concave mirror of radius of
curvature 2.08 m.
4. Since the incident rays and reflected rays
are parallel to each other therefore mirror is
plane mirror.
5. Let us solve the first case :
By applying the geometry we can prove that,
PA v ¢ = =
40
3
cm
Further, in triangles ABP and PA B ¢ ¢ we
have,
AB A B
40 40 3
=
¢ ¢
( / )
\ A B
AB
¢ ¢ = =
3
2
3
cm
Similary, we can solve other parts also.
6. Simply apply :
1 1 1
v u f
= =
and m
I
o
v
u
= =

for lateral magnification. If
magnitfication is positive, image will be
virtual. If magnification is negative, image
will be real.
3
40 cm
20 cm
a
a
a
2q
q
q
q
q
A' F C
M
B
A
20 cm
AIEEE Corner
¢ Subjective Questions (Level 1)
1. Here v = 39.2 cm, hence v =  39.2 cm
and magnification m = 1
Þ h h
i o
= = 4.85
Hence image is formed at 39.2 cm behind
the mirror and height of image is = 4.85 cm.
2. From figure, angle of incident = ° 15
Let reflected ray makes an angle q with the
horizontal, then
q + ° + ° = ° 15 15 90 Þ q = ° 60
3.
Since mirror are parallel to each other ¥
image are formed the distance of five closet
to object are 20 cm, 60 cm, 80 cm, 100 cm and
140 cm.
4. The distance of the object from images are
2 4 6 b b b , , ..... etc.
Hence the images distance are 2 nb, where
n = 1 2 , ,K . Ans.
5. Suppose mirror is rotated at angle q about
its axis perpendicular to both the incident
ray and normal as shown in figure
In figure (b) I remain unchanged N and R
shift to N¢ and R¢.
From figure (a) angle of rotation = i,
From figure (b) it is i  2q
Thus, reflected ray has been rotated by
angle 2q.
6. I is incident ray Ð = ° = Ð i r 30
From D PA A ¢ , we get
4
i i
I
N
R
y
x
(a)
IV
R'
y
x
(b)
I
q
q
i–q
i–q
i–2q
O''
b
O'''
4b
O' O''' b b
1
B D
A C
4b
2 b
O''
30 cm
30 cm
O'''
50 cm
1o cm
1o cm
O' O'''
70 cm
40 cm
B D
A C
50 cm
150°
q
150°
90°
N
Reflacted ray
Horizontal
15°
15°
Incident ray
Mirror
30°30°
20 cm
B
B' x
R
I
A
1.6m
P
A'
Page 5
26. Reflection of Light
Introductory Exercise 26.2
1. Total deviation produced
d q = °  + °  180 2 180 2 i
d q = °  + 360 2( ) i
From figure
q = °  90 i
Þ d = °  +  360 2 90 [ ] i i
= ° 180
Hence rays 1 and 2 are parallel (anti
parallel).
2. v
0
2 = m/s for plane mirror v
i
= 2 m/s.
Velocity of approach = + v v
i 0
= 4 m/s.
3. In figure, AB is mirror, G is ground, CD is
pole and M is the man. The minimum height
to see the image of top of pole is = EN
Introductory Exercise 26.1
1. Since c =
1
0 0
m e
where c is the speed of light
in vacuum hence unit of
1
0 0
m e
is m/s.
2. Hence
B x t
y
= ´ + ´

2 10 500 10
7 11
T 1.5 sin [ ]
Comparing this equation with the standard
wave eqution B B kx t
y
= +
0
sin [ ] w
k =

500
1
m Þ k =
2p
l
Þ l
p
=
2
k
m
Þ l
p
=
2
500
m =
p
250
metre
w = ´ 1.5 10
11
rad/s
Þ 2 10
11
pn = ´ 1.5
Þ n = ´
1.5
2
10
11
p
Hz
Speed of the wave v
k
= =
´ w 1.5 10
500
11
= ´ 3 10
8
m/s
Let E
0
be the amplitude of electric field.
Then E cB
0 0
8 7
3 10 2 10 = = ´ ´ ´

= 60 V/m
Since wave is propagating along xaxis and
B along yaxis, hence E must be along zaxis
Þ E = 60 V/m sin [ ] 500 10
11
x t + ´ 1.5
90°
90°
90°
i
q
q
i
180°–2i
N
1
1
2
180°–2q
N
2
= + = + EK KN KN 6
Now in DNKB,
NK
KB
= f tan Þ NK KB = tan f
= 4 tan f
In DBC C ¢ we get,
tan f =
¢
¢
= =
BC
CC
2
2
1
f = ° 45
So, NK = ´ ° = 4 45 4 tan m
Hence in minimum height
= + = 6 4 10 m m m
In DAC C ¢
tan q = =
4
2
2
In DL LA ¢ we get,
LL
LA
¢
= tan q
Þ
LL¢
=
4
2
Þ LL¢ = 8 m
Maximum height = + ¢ = + = CA LL 8 8 16 m
Introductory Exercise 26.3
1. Here f =  10 cm (concave mirror)
(a) u =  25 cm
Using mirror formula,
1 1 1
v u f
+ =
Þ
1 1 1
v f u
=  =  +
1
10
1
25
Þ
1 5 2
50 v
=
 +
Þ v =  = 
50
3
167 . cm
Hence image is real, inverted and less
height of the object.
(b) Since u =  10 cm,
Hence object is situated on focus of the
image formed at ¥.
(c) u  5, f =  10
1 1 1 1
10
1
5 v f u
=  =  +
Þ
1 1 2
10 v
=
 +
Þ v = 10 cm
Hence, image is virtual, erect and two time
of the object.
2. Here u =  3 m, f = 
1
2
m,
we have,
(a)
1 1 1
v f u
= 
Þ
1
2
1
3 v
=  +
Þ v =  0.6 m
As ball moves towards focus the image
moves towards ¥ and image is real as the
distance decreases by focal length image
become virtual which moves from + ¥ to
zero.
(b) The image of the ball coincide with ball,
when u R =  =  1 m
2
B
A
C
K
N
L'
L
8 m
q
q
f
f
f
C' C
D
2 m
pole
= 4 m
M
N
1
2 m
6 m
2 m
E
Using h ut gt = +
1
2
2
Þ t
h
g
= =
´ 2 2 2
9.8
= 0.639 s
Similarly again images match at t = 0.78 s.
3. Since image is magnified, hence the mirror
is concave.
Here, m
v
u
=

Þ

=
v
u
5
Þ v u =  5 …(i)
Let distance between mirror and object is x.
Since image is formed at a distance 5 m
from mirror
v x =  + ( ) 5 …(ii)
From Eqs.(i) and (ii), we get
 + =  ( ) 5 5 x x
Þ 4 5 x =
Þ x = 1.25
Hence mirror is placed at 1.25 m on right
side of the object by mirror formula
1 1 1
v u f
+ = ,
we have
1 1 1
f
=  
6.25 1.25
Þ f =
 6.25
6
,
Hence R f = 2 Þ R =  = 
6.25
2.08
3
m
Thus mirror is concave mirror of radius of
curvature 2.08 m.
4. Since the incident rays and reflected rays
are parallel to each other therefore mirror is
plane mirror.
5. Let us solve the first case :
By applying the geometry we can prove that,
PA v ¢ = =
40
3
cm
Further, in triangles ABP and PA B ¢ ¢ we
have,
AB A B
40 40 3
=
¢ ¢
( / )
\ A B
AB
¢ ¢ = =
3
2
3
cm
Similary, we can solve other parts also.
6. Simply apply :
1 1 1
v u f
= =
and m
I
o
v
u
= =

for lateral magnification. If
magnitfication is positive, image will be
virtual. If magnification is negative, image
will be real.
3
40 cm
20 cm
a
a
a
2q
q
q
q
q
A' F C
M
B
A
20 cm
AIEEE Corner
¢ Subjective Questions (Level 1)
1. Here v = 39.2 cm, hence v =  39.2 cm
and magnification m = 1
Þ h h
i o
= = 4.85
Hence image is formed at 39.2 cm behind
the mirror and height of image is = 4.85 cm.
2. From figure, angle of incident = ° 15
Let reflected ray makes an angle q with the
horizontal, then
q + ° + ° = ° 15 15 90 Þ q = ° 60
3.
Since mirror are parallel to each other ¥
image are formed the distance of five closet
to object are 20 cm, 60 cm, 80 cm, 100 cm and
140 cm.
4. The distance of the object from images are
2 4 6 b b b , , ..... etc.
Hence the images distance are 2 nb, where
n = 1 2 , ,K . Ans.
5. Suppose mirror is rotated at angle q about
its axis perpendicular to both the incident
ray and normal as shown in figure
In figure (b) I remain unchanged N and R
shift to N¢ and R¢.
From figure (a) angle of rotation = i,
From figure (b) it is i  2q
Thus, reflected ray has been rotated by
angle 2q.
6. I is incident ray Ð = ° = Ð i r 30
From D PA A ¢ , we get
4
i i
I
N
R
y
x
(a)
IV
R'
y
x
(b)
I
q
q
i–q
i–q
i–2q
O''
b
O'''
4b
O' O''' b b
1
B D
A C
4b
2 b
O''
30 cm
30 cm
O'''
50 cm
1o cm
1o cm
O' O'''
70 cm
40 cm
B D
A C
50 cm
150°
q
150°
90°
N
Reflacted ray
Horizontal
15°
15°
Incident ray
Mirror
30°30°
20 cm
B
B' x
R
I
A
1.6m
P
A'
x
20
30 = ° tan Þ x = ° 20 30 tan
No. of reflection = =
´ °
AB
x
160
20 30
cm
cm tan
= » 8 3 14
Hence the reflected ray reach other end
after 14 reflections.
7. The deviation produced by mirror M
1
is
= °  180 2 a
and the deviation produced by mirror M
2
is
=  180 2
Hence total deviation
=  +  f 180 2 180 2 a
=  + f 360 2 ( ) a
In D ABC we get,
90 90 180  + +  f = a q
Þ a q + f =
Hence deviation produces =  180 2q.
8. Here f
R
=  =  = 
2
22
2
11 cm
Object height h
0
6 = mm
u =  16.5 cm
(a) The ray diagram is shown in figure
Using mirror formula,
1 1 1
v u f
+ =
Þ
1 1 1
v f u
= 
Þ
1 1
11
1 11
11 v
=  + =
 +
´ 16.5
165
16.5
Þ v = 
´
= 
16.5
5.5
11
33 cm
Hence the image is formed at 33 cm from the
pole (vertex) of mirror on the object side the
image is real, inverted and magnified. The
absolute magnification
  m
v
u
=
½
½
½
½
= =
33
2
16.5
Hence size of image is h h
i
= ´ 2
0
= ´ = 2 6 12 mm.
9. Here u =  12 cm, f
R
= +
2
= + 10 cm
Using mirror formula
1 1 1
v u f
+ =
we get
1 1 1 1
10
1
12 v f u
=  = +
=
+ 6 5
60
Þ v =
60
11
cm = 5.46 cm
The image is formed on right side of the
vertex at a distance
60
11
cm. the image is
virtual and erect the absolute magnification
is given by   m
v
u
=
½
½
½
½
Þ  
( )
m =
´ 
½
½
½
½
½
½
=
60
11 12
5
11
Q m < 1
Hence image is demagnified.
Height of image h m h
i
= ´  
0
5
A
A'
B'
u = 16.5 cm
f
B
a
90°–a
180°–2a
A
Z'
I
1
M
1
R
2
180°–2q
C
R
1
90°–f q
a
f
f
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