DPP for NEET: Daily Practice Problems: Ray Optics- 2 (Solutions)

 Page 1


(1) (a) We know that  d = A (m – 1)
or  m = 1 + 
A
d
Here A = 6 °, d = 3° , therefore
m = 1 + 
3
6
 = 1.5
(2) (c) According to given problem
A = 30°, i
1
 = 60°  and  d = 30°  and as in a prism
d = (i
1
 + i
2
) – A,  30°  = (60 + i
2
) – 30 i.e.,  i
2
 = 0
So the emergent ray is perpendicular to the face from
which it emerges.
Now as  i
2
 = 0, r
2
 = 0
But as  r
1
 + r
2
 = A ,  r
1
 = A = 30°
So at first face
1× sin 60° = m sin 30°    i.e., m = 
3
(3) (a) As are know, m = 
c
v
 Þ v = 
c3
4
=
m
× 3 × 10
8
  = 2.25 × 10
8
 m/s
As, c = n l
0
  and v = n l
 Þ l / l
0
  = 
v1
c
=
m
i.e.,  l = l
0
/m = 
3
4
 × 6000Å = 4500Å
(4) (d) Herer + 90º +
 
r’ = 180º
Þ 
 
r’
 
= 90º – r
or, r’ = (90º – i)  as  Ð i = Ðr
Now, according to Snell’s law :
sin i = m sin r’ = m sin (90º – i)
or, tan i = m
or,  i = tan
–1
 m = tan
–1
 (1.5)
(5) (b) Here the requirement is that i > c
Þ sin i > sin c Þ sin i > 
2
1
m
m
From Snell's law m
1
 = 
sin
sin r
a
Also in DOBA
r + i = 90° Þ r = (90  – i)
Hence from equation (ii)
sin a = m
1
 sin(90 – i)
Þ cos i = 
1
sin a
m
sin i = 
2
1 cos i -
 = 
2
1
sin
1
æö a
-
ç÷
m
èø
...(iii)
From equation (i) and (ii)
Þ sin
2
 a < ( )
22
12
m -m
 sin a < 
22
12
m -m
a
max
 = sin
–1
 
22
12
m -m
a
O
r
i
B
A
(6) (b) From the information given, it is clear that the apparent
depth is 2.58 mm and the real depth is 4mm. Therefore,
the refractive index will be
m = 
R4
A 2.58
=
 = 1.55
(7) (d) The apparent shift of the bottom point upwards will be
x = x
1
 + x
2
   = t
1
 
1
1
1
æö
-
ç÷
m èø
 + t
2
 
2
1
1
æö
-
ç÷
m èø
   = 4 
1
1
(4 / 3)
æö
-
ç÷
èø
 + 2 
1
1
(3 / 2)
æö
-
ç÷
èø
   = 4 
3
1
4
æö
-
ç÷
èø
 + 2 
2
1
3
æö
-
ç÷
èø
 = 1.67 cm.
(8) (d) Since    v = 
C
n
The time taken are
t
2
 = 
20 (1.63)
C
,  t
1
 = 
20 (1.47)
C
Therefore , the difference is
t
2 
- t
1 
= 
8
20(1.63 1.47) 20 0.16
C
3 10
-´
=
´
= 1.07 × 10
–8
 sec.
(9) (b) As the beam just suffers TIR at interface of region III
and IV
Region I Region II Region III
q
n
0
0
n
2
0
n
6
0
n
8
q
0.2 m
0.6 m
0
1
n
nosin sin
2
qq = 
0
2
n
sin
6
q
 = 
0
n
sin90,
8
°
Page 2


(1) (a) We know that  d = A (m – 1)
or  m = 1 + 
A
d
Here A = 6 °, d = 3° , therefore
m = 1 + 
3
6
 = 1.5
(2) (c) According to given problem
A = 30°, i
1
 = 60°  and  d = 30°  and as in a prism
d = (i
1
 + i
2
) – A,  30°  = (60 + i
2
) – 30 i.e.,  i
2
 = 0
So the emergent ray is perpendicular to the face from
which it emerges.
Now as  i
2
 = 0, r
2
 = 0
But as  r
1
 + r
2
 = A ,  r
1
 = A = 30°
So at first face
1× sin 60° = m sin 30°    i.e., m = 
3
(3) (a) As are know, m = 
c
v
 Þ v = 
c3
4
=
m
× 3 × 10
8
  = 2.25 × 10
8
 m/s
As, c = n l
0
  and v = n l
 Þ l / l
0
  = 
v1
c
=
m
i.e.,  l = l
0
/m = 
3
4
 × 6000Å = 4500Å
(4) (d) Herer + 90º +
 
r’ = 180º
Þ 
 
r’
 
= 90º – r
or, r’ = (90º – i)  as  Ð i = Ðr
Now, according to Snell’s law :
sin i = m sin r’ = m sin (90º – i)
or, tan i = m
or,  i = tan
–1
 m = tan
–1
 (1.5)
(5) (b) Here the requirement is that i > c
Þ sin i > sin c Þ sin i > 
2
1
m
m
From Snell's law m
1
 = 
sin
sin r
a
Also in DOBA
r + i = 90° Þ r = (90  – i)
Hence from equation (ii)
sin a = m
1
 sin(90 – i)
Þ cos i = 
1
sin a
m
sin i = 
2
1 cos i -
 = 
2
1
sin
1
æö a
-
ç÷
m
èø
...(iii)
From equation (i) and (ii)
Þ sin
2
 a < ( )
22
12
m -m
 sin a < 
22
12
m -m
a
max
 = sin
–1
 
22
12
m -m
a
O
r
i
B
A
(6) (b) From the information given, it is clear that the apparent
depth is 2.58 mm and the real depth is 4mm. Therefore,
the refractive index will be
m = 
R4
A 2.58
=
 = 1.55
(7) (d) The apparent shift of the bottom point upwards will be
x = x
1
 + x
2
   = t
1
 
1
1
1
æö
-
ç÷
m èø
 + t
2
 
2
1
1
æö
-
ç÷
m èø
   = 4 
1
1
(4 / 3)
æö
-
ç÷
èø
 + 2 
1
1
(3 / 2)
æö
-
ç÷
èø
   = 4 
3
1
4
æö
-
ç÷
èø
 + 2 
2
1
3
æö
-
ç÷
èø
 = 1.67 cm.
(8) (d) Since    v = 
C
n
The time taken are
t
2
 = 
20 (1.63)
C
,  t
1
 = 
20 (1.47)
C
Therefore , the difference is
t
2 
- t
1 
= 
8
20(1.63 1.47) 20 0.16
C
3 10
-´
=
´
= 1.07 × 10
–8
 sec.
(9) (b) As the beam just suffers TIR at interface of region III
and IV
Region I Region II Region III
q
n
0
0
n
2
0
n
6
0
n
8
q
0.2 m
0.6 m
0
1
n
nosin sin
2
qq = 
0
2
n
sin
6
q
 = 
0
n
sin90,
8
°
DPP/ P 50
139
1
sin
8
q
 Þ q = 
1
1
sin
8
-
(10) (b) The ray of light returns back from the polished face
AC.
A
F
D
i
C B
E
30º
30º
60º
\ Ð ADE = 90º. From the figure it is clear that the angle
of refraction at face AB is 30º.  Hence from Snell’s law
m = 
sini
sinr
As  m = 
2
 and   r = 30º   \ 
2
  = 
sini
sin 30º
or  sin i = 
21
2 2
=
 = sin 45º
\  i = 45º
(11) (a) A = r
1
 + r
2
 = 60 .......(1)
In minimum deviation position
r
1
 = r
2
 .......(2)
From eqs. (1) and (2)
A = 2r
1
 = 60º
\ r
1
 = 30º
\ n = 
sin i sin 50º
sin r sin 30º
= = 1.532
m
A
sin
2
n
A
sin
2
+d
=  or 1.532 = 
m
60
sin
2
sin 30
+d
°
m
60 1.532
sin
22
+d
= = 0.766
\ d
m
 = 40º
(12) (b) d
1
 = d
2
(m
1 
– 1) A
1
 = (m
2
 – 1) A
2
(1.5 – 1) 6 = (1.6 – 1) A
2
A
2
 = 5º
(13) (a) The deviation produced by the crown prism is
d = (m – 1) A
and by the flint prism is
d ' = (m' – 1) A'
The prisms are placed with their angles inverted with
respect to each other . The deviations are also in
opposite directions. Thus, the net deviation is
D =  d – d ' = (m – 1) A – (m' – 1) A'     ...........(1)
If the net deviation  for the mean ray is zero,
(m – 1) A  = (m' – 1) A'
or,   A' = 
( 1)
( ' 1)
m-
m-
  A = 
1.517 1
5
1.620 1
-
´°
-
= 4.2°.
The angular dispersion produced by the crown prism
is
d
v
 – d
r
 = (m
v
 – m
r
) A
and that by the flint prism is
d '
v
 – d '
r
 = (m'
v
 – m'
r
) A'
The net angular dispersion is ,
d = (m
v
 – m
r
) A – (m'
v
 – m'
r
) A'
   = (1.523 –1.514) × 5° – (1.632–1.613) × 4.2°
    = – 0.0348°.
The angular dispersion has magnitude 0.0348°.
(14) (a) m
v
 = 1.5230, m
r
 = 1.5145, w = ?
Mean refractive index,
m = 
vr
1.5230 1.5145
22
m +m +
=
m = 1.5187
w = 
vr
1
m -m
m-
    = 
1.5230 1.5145 0.0085
1.5187 1 0.5187
-
=
-
 = 0.0163
(15) (c) Here, w = 0.021; m = 1.53;  w ' = 0.045;
m'  = 1.65;      A' = 4.2°
For no dispersion, w d + w ' d ' = 0
or w A (m – 1) + w ' A ' (m – 1) = 0
or A = –
( )
( )
'A' '1
1
w m-
w m-
       = – 
( )
( )
0.045 4.2 1.65 1
0.021 1.53 1
´ ´-
´-
 = – 11.04°
Net deviation,
d + d ' = A (m – 1)  + A' (m ' – 1)
   = – 11.04 (1.53 –1) + 4.2 (1.65 –1)
   = – 11.04 × 0.53 + 4.2 × 0.65
    = – 5.85 + 2.73 = 3.12°
(16) (d) At first face of the prism as i
1
 = 0,
sin 0 = 1.5 sin r
1
  i.e., ,  r
1
 = 0
And as for a prism
r
1
 + r
2
 = A so r
2
 = A  ......(1)
But at  second face, as the ray just fails to emerge
i.e., r
2
  = q
C
  ......(2)
So from Eqn,.(1) and (2)
A = r
2
 = q 
C
But as q
C
 = sin 
–1
 
1 éù
êú
m
ëû
 = sin 
–1
 
2
3
éù
êú
ëû
 = 42°
So A = 42°
(17) (a) Here, m
b
= 1.532 and  m
r
 = 1.514    A = 8° .
Angular dispersion
= (m
b
 – m
r
) A = (1.532 – 1.514) × 8
= 0.018 × 8 = 0.144° .
Page 3


(1) (a) We know that  d = A (m – 1)
or  m = 1 + 
A
d
Here A = 6 °, d = 3° , therefore
m = 1 + 
3
6
 = 1.5
(2) (c) According to given problem
A = 30°, i
1
 = 60°  and  d = 30°  and as in a prism
d = (i
1
 + i
2
) – A,  30°  = (60 + i
2
) – 30 i.e.,  i
2
 = 0
So the emergent ray is perpendicular to the face from
which it emerges.
Now as  i
2
 = 0, r
2
 = 0
But as  r
1
 + r
2
 = A ,  r
1
 = A = 30°
So at first face
1× sin 60° = m sin 30°    i.e., m = 
3
(3) (a) As are know, m = 
c
v
 Þ v = 
c3
4
=
m
× 3 × 10
8
  = 2.25 × 10
8
 m/s
As, c = n l
0
  and v = n l
 Þ l / l
0
  = 
v1
c
=
m
i.e.,  l = l
0
/m = 
3
4
 × 6000Å = 4500Å
(4) (d) Herer + 90º +
 
r’ = 180º
Þ 
 
r’
 
= 90º – r
or, r’ = (90º – i)  as  Ð i = Ðr
Now, according to Snell’s law :
sin i = m sin r’ = m sin (90º – i)
or, tan i = m
or,  i = tan
–1
 m = tan
–1
 (1.5)
(5) (b) Here the requirement is that i > c
Þ sin i > sin c Þ sin i > 
2
1
m
m
From Snell's law m
1
 = 
sin
sin r
a
Also in DOBA
r + i = 90° Þ r = (90  – i)
Hence from equation (ii)
sin a = m
1
 sin(90 – i)
Þ cos i = 
1
sin a
m
sin i = 
2
1 cos i -
 = 
2
1
sin
1
æö a
-
ç÷
m
èø
...(iii)
From equation (i) and (ii)
Þ sin
2
 a < ( )
22
12
m -m
 sin a < 
22
12
m -m
a
max
 = sin
–1
 
22
12
m -m
a
O
r
i
B
A
(6) (b) From the information given, it is clear that the apparent
depth is 2.58 mm and the real depth is 4mm. Therefore,
the refractive index will be
m = 
R4
A 2.58
=
 = 1.55
(7) (d) The apparent shift of the bottom point upwards will be
x = x
1
 + x
2
   = t
1
 
1
1
1
æö
-
ç÷
m èø
 + t
2
 
2
1
1
æö
-
ç÷
m èø
   = 4 
1
1
(4 / 3)
æö
-
ç÷
èø
 + 2 
1
1
(3 / 2)
æö
-
ç÷
èø
   = 4 
3
1
4
æö
-
ç÷
èø
 + 2 
2
1
3
æö
-
ç÷
èø
 = 1.67 cm.
(8) (d) Since    v = 
C
n
The time taken are
t
2
 = 
20 (1.63)
C
,  t
1
 = 
20 (1.47)
C
Therefore , the difference is
t
2 
- t
1 
= 
8
20(1.63 1.47) 20 0.16
C
3 10
-´
=
´
= 1.07 × 10
–8
 sec.
(9) (b) As the beam just suffers TIR at interface of region III
and IV
Region I Region II Region III
q
n
0
0
n
2
0
n
6
0
n
8
q
0.2 m
0.6 m
0
1
n
nosin sin
2
qq = 
0
2
n
sin
6
q
 = 
0
n
sin90,
8
°
DPP/ P 50
139
1
sin
8
q
 Þ q = 
1
1
sin
8
-
(10) (b) The ray of light returns back from the polished face
AC.
A
F
D
i
C B
E
30º
30º
60º
\ Ð ADE = 90º. From the figure it is clear that the angle
of refraction at face AB is 30º.  Hence from Snell’s law
m = 
sini
sinr
As  m = 
2
 and   r = 30º   \ 
2
  = 
sini
sin 30º
or  sin i = 
21
2 2
=
 = sin 45º
\  i = 45º
(11) (a) A = r
1
 + r
2
 = 60 .......(1)
In minimum deviation position
r
1
 = r
2
 .......(2)
From eqs. (1) and (2)
A = 2r
1
 = 60º
\ r
1
 = 30º
\ n = 
sin i sin 50º
sin r sin 30º
= = 1.532
m
A
sin
2
n
A
sin
2
+d
=  or 1.532 = 
m
60
sin
2
sin 30
+d
°
m
60 1.532
sin
22
+d
= = 0.766
\ d
m
 = 40º
(12) (b) d
1
 = d
2
(m
1 
– 1) A
1
 = (m
2
 – 1) A
2
(1.5 – 1) 6 = (1.6 – 1) A
2
A
2
 = 5º
(13) (a) The deviation produced by the crown prism is
d = (m – 1) A
and by the flint prism is
d ' = (m' – 1) A'
The prisms are placed with their angles inverted with
respect to each other . The deviations are also in
opposite directions. Thus, the net deviation is
D =  d – d ' = (m – 1) A – (m' – 1) A'     ...........(1)
If the net deviation  for the mean ray is zero,
(m – 1) A  = (m' – 1) A'
or,   A' = 
( 1)
( ' 1)
m-
m-
  A = 
1.517 1
5
1.620 1
-
´°
-
= 4.2°.
The angular dispersion produced by the crown prism
is
d
v
 – d
r
 = (m
v
 – m
r
) A
and that by the flint prism is
d '
v
 – d '
r
 = (m'
v
 – m'
r
) A'
The net angular dispersion is ,
d = (m
v
 – m
r
) A – (m'
v
 – m'
r
) A'
   = (1.523 –1.514) × 5° – (1.632–1.613) × 4.2°
    = – 0.0348°.
The angular dispersion has magnitude 0.0348°.
(14) (a) m
v
 = 1.5230, m
r
 = 1.5145, w = ?
Mean refractive index,
m = 
vr
1.5230 1.5145
22
m +m +
=
m = 1.5187
w = 
vr
1
m -m
m-
    = 
1.5230 1.5145 0.0085
1.5187 1 0.5187
-
=
-
 = 0.0163
(15) (c) Here, w = 0.021; m = 1.53;  w ' = 0.045;
m'  = 1.65;      A' = 4.2°
For no dispersion, w d + w ' d ' = 0
or w A (m – 1) + w ' A ' (m – 1) = 0
or A = –
( )
( )
'A' '1
1
w m-
w m-
       = – 
( )
( )
0.045 4.2 1.65 1
0.021 1.53 1
´ ´-
´-
 = – 11.04°
Net deviation,
d + d ' = A (m – 1)  + A' (m ' – 1)
   = – 11.04 (1.53 –1) + 4.2 (1.65 –1)
   = – 11.04 × 0.53 + 4.2 × 0.65
    = – 5.85 + 2.73 = 3.12°
(16) (d) At first face of the prism as i
1
 = 0,
sin 0 = 1.5 sin r
1
  i.e., ,  r
1
 = 0
And as for a prism
r
1
 + r
2
 = A so r
2
 = A  ......(1)
But at  second face, as the ray just fails to emerge
i.e., r
2
  = q
C
  ......(2)
So from Eqn,.(1) and (2)
A = r
2
 = q 
C
But as q
C
 = sin 
–1
 
1 éù
êú
m
ëû
 = sin 
–1
 
2
3
éù
êú
ëû
 = 42°
So A = 42°
(17) (a) Here, m
b
= 1.532 and  m
r
 = 1.514    A = 8° .
Angular dispersion
= (m
b
 – m
r
) A = (1.532 – 1.514) × 8
= 0.018 × 8 = 0.144° .
DPP/ P 50
140
(18) (c) V elocity of the ball when it reaches at the height of 12.8
m. above the surface is v = 
2 10 7.2 ´´
 = 12 m/s
Height of the ball from surface as seen by fish
h¢ = m
h
 Þ 
dh
dt
¢
= 
dh
dt
m
u = 12 m/s
u = 0
20
12.8 m
Þ v¢ = mv = 
4
12
3
´
 = 16 m/s.
(19) (a) Suppose, the angle of the crown prism needed is A and
that of the flint prism is A’. We have
w = 
vr
1
m -m
m-
 or,, m
v
 – m
r
 = (m – 1) w
The angular dispersion produced by the crown prism
is
(m
v
 – m
r
) A = (m –1) w A
Similarly , the angular dispersion produced by the flint
pris m is (m´ – 1) w ´ A´
For achromatic combination , the net dispersion should
be zero. Thus ,
(m – 1) w A = (m´ – 1) w´ A´
or , 
A ' ( 1)
A ( '1)'
m-w
=
m-w
 = 
0.517 0.03
0.621 0.05
´
´
 = 0.50   ......(1)
The deviation in the yellow ray produced by the crown
prism is d = (m – 1) A and by the flint prism is  d´ = (m´
– 1) A´. The net deviation produced by the combination
is
d – d´ = (m – 1) A – (m´ – 1) A´
or 1° = 0.517 A – 0.621 A´                             ......(2)
Solving (1) and (2 ), A = 4.8° and A´ = 2.4°. Thus, the
crown prism should have its refracting angle 4.8° and
that of the flint prism should be 2.4°.
(20) (a) For TIR at AC
q
q
C
B A
   q > C
sin sinC Þ q³
wg
1
sin
µ
Þ q³
w
g
µ
sin
µ
Þ q³
8
sin
9
Þ q³
(21) (c)
Prism
(µ 1)A (1.5 1)4 2 d = - = - °=°
Total Prism Mirror
\ d = d +d
= (µ – 1)A + (180 – 2i) = 2° + (180 – 2 × 2) = 178°
(22) (a) From the following figure
a
r
i
n
r + i = 90° Þ i = 90° – r
For ray not to emerge from curved surface i > C
Þ sin i > sin C Þ sin (90° – r) > sin C Þ cos r > sin C
Þ
2
1
1 sin r
n
->
1
sinC
n
ìü
\=
íý
îþ
Þ 
2
22
sin1
1
nn
a
->
 Þ  1 > ( )
2
2
1
1 sin
n
+a
Þ n
2
 > 1 + sin
2
 a Þ 
2 n >
{sin i ® 1}
Þ Least value = 
2
(23) (a) w depends only on nature of material.
(24) (a)
(25) (c);
26. (b),   27 (a)
The normal shift produced by a glass slab is,
12
1 1 (6)2
3
x t cm
æö
æö
D=- =- =
ç÷
ç÷
èø è mø
i.e., for the mirror, the object is placed at a distance
(32 – Dx) = 30 cm from it.
Applying mirror formula,
1 11
v uf
+=
1 11
30 10 v
+ =-
or, v = – 15 cm
(a) When x = 5 cm: The light falls on the slab on its return
journey as shown. But the slab will again shift it by a
distance.
6 cm
15 cm
Dx
I
Dx = 2 cm. Hence, the final real image is formed at a
distance (15 + 2) = 17 cm from the mirror.
Page 4


(1) (a) We know that  d = A (m – 1)
or  m = 1 + 
A
d
Here A = 6 °, d = 3° , therefore
m = 1 + 
3
6
 = 1.5
(2) (c) According to given problem
A = 30°, i
1
 = 60°  and  d = 30°  and as in a prism
d = (i
1
 + i
2
) – A,  30°  = (60 + i
2
) – 30 i.e.,  i
2
 = 0
So the emergent ray is perpendicular to the face from
which it emerges.
Now as  i
2
 = 0, r
2
 = 0
But as  r
1
 + r
2
 = A ,  r
1
 = A = 30°
So at first face
1× sin 60° = m sin 30°    i.e., m = 
3
(3) (a) As are know, m = 
c
v
 Þ v = 
c3
4
=
m
× 3 × 10
8
  = 2.25 × 10
8
 m/s
As, c = n l
0
  and v = n l
 Þ l / l
0
  = 
v1
c
=
m
i.e.,  l = l
0
/m = 
3
4
 × 6000Å = 4500Å
(4) (d) Herer + 90º +
 
r’ = 180º
Þ 
 
r’
 
= 90º – r
or, r’ = (90º – i)  as  Ð i = Ðr
Now, according to Snell’s law :
sin i = m sin r’ = m sin (90º – i)
or, tan i = m
or,  i = tan
–1
 m = tan
–1
 (1.5)
(5) (b) Here the requirement is that i > c
Þ sin i > sin c Þ sin i > 
2
1
m
m
From Snell's law m
1
 = 
sin
sin r
a
Also in DOBA
r + i = 90° Þ r = (90  – i)
Hence from equation (ii)
sin a = m
1
 sin(90 – i)
Þ cos i = 
1
sin a
m
sin i = 
2
1 cos i -
 = 
2
1
sin
1
æö a
-
ç÷
m
èø
...(iii)
From equation (i) and (ii)
Þ sin
2
 a < ( )
22
12
m -m
 sin a < 
22
12
m -m
a
max
 = sin
–1
 
22
12
m -m
a
O
r
i
B
A
(6) (b) From the information given, it is clear that the apparent
depth is 2.58 mm and the real depth is 4mm. Therefore,
the refractive index will be
m = 
R4
A 2.58
=
 = 1.55
(7) (d) The apparent shift of the bottom point upwards will be
x = x
1
 + x
2
   = t
1
 
1
1
1
æö
-
ç÷
m èø
 + t
2
 
2
1
1
æö
-
ç÷
m èø
   = 4 
1
1
(4 / 3)
æö
-
ç÷
èø
 + 2 
1
1
(3 / 2)
æö
-
ç÷
èø
   = 4 
3
1
4
æö
-
ç÷
èø
 + 2 
2
1
3
æö
-
ç÷
èø
 = 1.67 cm.
(8) (d) Since    v = 
C
n
The time taken are
t
2
 = 
20 (1.63)
C
,  t
1
 = 
20 (1.47)
C
Therefore , the difference is
t
2 
- t
1 
= 
8
20(1.63 1.47) 20 0.16
C
3 10
-´
=
´
= 1.07 × 10
–8
 sec.
(9) (b) As the beam just suffers TIR at interface of region III
and IV
Region I Region II Region III
q
n
0
0
n
2
0
n
6
0
n
8
q
0.2 m
0.6 m
0
1
n
nosin sin
2
qq = 
0
2
n
sin
6
q
 = 
0
n
sin90,
8
°
DPP/ P 50
139
1
sin
8
q
 Þ q = 
1
1
sin
8
-
(10) (b) The ray of light returns back from the polished face
AC.
A
F
D
i
C B
E
30º
30º
60º
\ Ð ADE = 90º. From the figure it is clear that the angle
of refraction at face AB is 30º.  Hence from Snell’s law
m = 
sini
sinr
As  m = 
2
 and   r = 30º   \ 
2
  = 
sini
sin 30º
or  sin i = 
21
2 2
=
 = sin 45º
\  i = 45º
(11) (a) A = r
1
 + r
2
 = 60 .......(1)
In minimum deviation position
r
1
 = r
2
 .......(2)
From eqs. (1) and (2)
A = 2r
1
 = 60º
\ r
1
 = 30º
\ n = 
sin i sin 50º
sin r sin 30º
= = 1.532
m
A
sin
2
n
A
sin
2
+d
=  or 1.532 = 
m
60
sin
2
sin 30
+d
°
m
60 1.532
sin
22
+d
= = 0.766
\ d
m
 = 40º
(12) (b) d
1
 = d
2
(m
1 
– 1) A
1
 = (m
2
 – 1) A
2
(1.5 – 1) 6 = (1.6 – 1) A
2
A
2
 = 5º
(13) (a) The deviation produced by the crown prism is
d = (m – 1) A
and by the flint prism is
d ' = (m' – 1) A'
The prisms are placed with their angles inverted with
respect to each other . The deviations are also in
opposite directions. Thus, the net deviation is
D =  d – d ' = (m – 1) A – (m' – 1) A'     ...........(1)
If the net deviation  for the mean ray is zero,
(m – 1) A  = (m' – 1) A'
or,   A' = 
( 1)
( ' 1)
m-
m-
  A = 
1.517 1
5
1.620 1
-
´°
-
= 4.2°.
The angular dispersion produced by the crown prism
is
d
v
 – d
r
 = (m
v
 – m
r
) A
and that by the flint prism is
d '
v
 – d '
r
 = (m'
v
 – m'
r
) A'
The net angular dispersion is ,
d = (m
v
 – m
r
) A – (m'
v
 – m'
r
) A'
   = (1.523 –1.514) × 5° – (1.632–1.613) × 4.2°
    = – 0.0348°.
The angular dispersion has magnitude 0.0348°.
(14) (a) m
v
 = 1.5230, m
r
 = 1.5145, w = ?
Mean refractive index,
m = 
vr
1.5230 1.5145
22
m +m +
=
m = 1.5187
w = 
vr
1
m -m
m-
    = 
1.5230 1.5145 0.0085
1.5187 1 0.5187
-
=
-
 = 0.0163
(15) (c) Here, w = 0.021; m = 1.53;  w ' = 0.045;
m'  = 1.65;      A' = 4.2°
For no dispersion, w d + w ' d ' = 0
or w A (m – 1) + w ' A ' (m – 1) = 0
or A = –
( )
( )
'A' '1
1
w m-
w m-
       = – 
( )
( )
0.045 4.2 1.65 1
0.021 1.53 1
´ ´-
´-
 = – 11.04°
Net deviation,
d + d ' = A (m – 1)  + A' (m ' – 1)
   = – 11.04 (1.53 –1) + 4.2 (1.65 –1)
   = – 11.04 × 0.53 + 4.2 × 0.65
    = – 5.85 + 2.73 = 3.12°
(16) (d) At first face of the prism as i
1
 = 0,
sin 0 = 1.5 sin r
1
  i.e., ,  r
1
 = 0
And as for a prism
r
1
 + r
2
 = A so r
2
 = A  ......(1)
But at  second face, as the ray just fails to emerge
i.e., r
2
  = q
C
  ......(2)
So from Eqn,.(1) and (2)
A = r
2
 = q 
C
But as q
C
 = sin 
–1
 
1 éù
êú
m
ëû
 = sin 
–1
 
2
3
éù
êú
ëû
 = 42°
So A = 42°
(17) (a) Here, m
b
= 1.532 and  m
r
 = 1.514    A = 8° .
Angular dispersion
= (m
b
 – m
r
) A = (1.532 – 1.514) × 8
= 0.018 × 8 = 0.144° .
DPP/ P 50
140
(18) (c) V elocity of the ball when it reaches at the height of 12.8
m. above the surface is v = 
2 10 7.2 ´´
 = 12 m/s
Height of the ball from surface as seen by fish
h¢ = m
h
 Þ 
dh
dt
¢
= 
dh
dt
m
u = 12 m/s
u = 0
20
12.8 m
Þ v¢ = mv = 
4
12
3
´
 = 16 m/s.
(19) (a) Suppose, the angle of the crown prism needed is A and
that of the flint prism is A’. We have
w = 
vr
1
m -m
m-
 or,, m
v
 – m
r
 = (m – 1) w
The angular dispersion produced by the crown prism
is
(m
v
 – m
r
) A = (m –1) w A
Similarly , the angular dispersion produced by the flint
pris m is (m´ – 1) w ´ A´
For achromatic combination , the net dispersion should
be zero. Thus ,
(m – 1) w A = (m´ – 1) w´ A´
or , 
A ' ( 1)
A ( '1)'
m-w
=
m-w
 = 
0.517 0.03
0.621 0.05
´
´
 = 0.50   ......(1)
The deviation in the yellow ray produced by the crown
prism is d = (m – 1) A and by the flint prism is  d´ = (m´
– 1) A´. The net deviation produced by the combination
is
d – d´ = (m – 1) A – (m´ – 1) A´
or 1° = 0.517 A – 0.621 A´                             ......(2)
Solving (1) and (2 ), A = 4.8° and A´ = 2.4°. Thus, the
crown prism should have its refracting angle 4.8° and
that of the flint prism should be 2.4°.
(20) (a) For TIR at AC
q
q
C
B A
   q > C
sin sinC Þ q³
wg
1
sin
µ
Þ q³
w
g
µ
sin
µ
Þ q³
8
sin
9
Þ q³
(21) (c)
Prism
(µ 1)A (1.5 1)4 2 d = - = - °=°
Total Prism Mirror
\ d = d +d
= (µ – 1)A + (180 – 2i) = 2° + (180 – 2 × 2) = 178°
(22) (a) From the following figure
a
r
i
n
r + i = 90° Þ i = 90° – r
For ray not to emerge from curved surface i > C
Þ sin i > sin C Þ sin (90° – r) > sin C Þ cos r > sin C
Þ
2
1
1 sin r
n
->
1
sinC
n
ìü
\=
íý
îþ
Þ 
2
22
sin1
1
nn
a
->
 Þ  1 > ( )
2
2
1
1 sin
n
+a
Þ n
2
 > 1 + sin
2
 a Þ 
2 n >
{sin i ® 1}
Þ Least value = 
2
(23) (a) w depends only on nature of material.
(24) (a)
(25) (c);
26. (b),   27 (a)
The normal shift produced by a glass slab is,
12
1 1 (6)2
3
x t cm
æö
æö
D=- =- =
ç÷
ç÷
èø è mø
i.e., for the mirror, the object is placed at a distance
(32 – Dx) = 30 cm from it.
Applying mirror formula,
1 11
v uf
+=
1 11
30 10 v
+ =-
or, v = – 15 cm
(a) When x = 5 cm: The light falls on the slab on its return
journey as shown. But the slab will again shift it by a
distance.
6 cm
15 cm
Dx
I
Dx = 2 cm. Hence, the final real image is formed at a
distance (15 + 2) = 17 cm from the mirror.
DPP/ P 50
141
(b) When x = 20 cm: This time also the final image is at a
distance of 17 cm from the mirror but it is virtual as
shown.
I
Dx
(28) (b) As rays of all colours emerge in the same direction (of
incidence of white light), hence there is no dispersion,
but only lateral displacement in a glass slab.
(29) (b) The velocity of light in a material medium depends upon
its colour (wavelength). If a ray of white light is incident
on a prism, then on emerging the different colours are
deviated through  different angles.
Also dispersive power 
( )
( ) 1
m -m
w=
m-
VR
Y
i.e. w depends upon only m
(30) (c) Apparent shift for different coloured letter is
1
1 dh
æö
=-
ç÷
m
èø
Þ l
R
 > l
V 
so m
R 
<  m
V
Hence d
R
 < d
V
 i. e. red coloured letter raised least.