CPP-8
Batches - PHONON
Class - XI
PRISM
1.
Sol.
2.
Sol.
A ray of light is incident at angle i on a surface of a prism of small angle A & emerges normally from the opposite
surface. If the refractive index of the material of the prism is , the angle of incidence i is nearly equal to :
(A) A/
(B) A/(2 )
(C*) A
(D) A/2
i=i
r =A
A
1 × sin i = µ sin r
90°
sin i = µ sin A
i
A 90°
For small angle
µ

i = µA
Find the angle of deviation suffered by the light rays shown in figure for following two
condition the refractive index for the prism material is  = 3/2.
(i) When the prism is placed in air ( = 1)
(ii) When the prism is placed in water ( = 4/3)
3º
Ans. [(i) 1.5º; (ii)
]
8
i = 3° =
3

=
180 60
=r–i
For small angles = sin   
r = sin r
i = sin i
(i) Prism is in air
3

sin i = 1 sin r
2
 3

3
 =
r=i· =
40
2 60 2
 

180 3

=r–i=
=
=
= = 1.5°
40 60 120 120 2

 = 1.5°
(ii) Prism is in water
4
3

sin i = sin r
3
2
3 3
sin r =
sin i
2 4
9 
3
r= 
=
8 60 160
3 

180
3

=r–i=
=
=
=
160 60
480
480 8
3

 = ° Ans.
8
3º
90º
30°
i
3º
r i
Page 1
3.
A prism of refractive index 2 has refracting angle 60º. Answer the following questions
(i) In order that a ray suffers minimum deviation it should be incident at an angle :
(A*) 45º
(B) 90º
(C) 30º
(D) None
(ii) Angle of minimum deviation is :
(A) 45º
(B) 90º
(C*) 30º
(iii) Angle of maximum deviation is :
(A) 45º
(B) sin–1 ( 2 sin 15º)
(C*) 30º + sin –1( 2 sin 15º) (D) None
(D) None
60°
Sol.
i1
r1 r2
i2
r1 + r2 = 60°
(i) For minimum deviation

r1 – r2 =
60
= 30°
2
1 × sin i1 = 2 sin 30°

i1 = 45° Ans.
(ii) at minimum deviation : r1 = r2 and i1 = i2
min = (i1 + i2) – (r1 + r2)
= (45 + 45) – 60
= 90 – 60
min = 30° Ans.
(iii) For maximum deviation : emergent ray should become parallel to emergent surface, for that :
 1 
r2 = c = sin–1 
 = 45°
 2

r1 = 60° – 45° = 15°
1 × sin i1 = 2 sin (15°)
i1 = sin–1 ( 2 sin 15°)
and
i2 = 90°

max = i1 + i2 – A
= sin–1 ( 2 sin 15°) + 90° – 60°
max = 30° + sin–1 ( 2 sin 15°) Ans.
4.
Sol.
5.
At what values of the refractive index of a rectangular prism can a ray travel as shown in figure.
The section of the prism is an isosceles triangle & the ray is normally incident onto the face
AC.
Ans. [n > 2 ]
45° > c
A
sin 45° > sin c
45°
1
1
>
n
2
B
C
n> 2
A
B
C
The cross section of a glass prism has the form of an equilateral triangle. A ray is incident onto one of the faces
perpendicular to it. Find the angle  between the incident ray and the ray that leaves the prism. The refractive index of
glass is  = 1.5.
Ans. [ = 60º]
Page 2
60°
2
sin c = < 0.86
3
Sol.
90°
1.5
(2)
60° 60°
60° 30°
60°
60°
r
A prism having refractive index 2 and refracting angle 30º, has one of the refracting surfaces polished.
A beam of
(1)
light incident on the other refracting surface will retrace its path if the angle of incidence is :
(A) 0º
(B) 30º
(C*) 45º
(D) 60º
 > c
Angle between ray (1) and (2) is 60° as shown in figure.
6.
Sol.
1 sin  = 2 sin 3
30°
1
sin  =
30°
2
 = 45°
7.
Sol.
8.
A prism (n = 2) of apex angle 90º is placed in air (n = 1). What should be the angle of incidence so that light ray strikes
the second surface at an angle of incidence 60º.
Ans. [90º]
 × sin i = µ sin 30º

sin i = 2 ×


i=
i = 90º
1
2
i
60º
30º
30º
60º
Light is incident normally on face AB of a prism as shown in figure. A
liquid of refractive index µ is placed on face AC of the prism. The prism is
made of glass of refractive index 3/2. The limits of µ for which total internal
reflection takes place on face AC is :
(A) µ >
3
2
3 3
4
3
(D) µ <
2
(B*) µ <
(C) µ > 3
 3/ 2 
1

 =
µ

 sin C
Sol.
µ
60º

µ=
=
9.
Sol.
3
3 /2 ×
2
30º
3/2
3 3
3 3
 µC
4
4
The wavelength of light in vacuum is 6000 Å and in a medium it is 4000 Å. The refractive index of the medium is :
(A) 2.4
(B*) 1.5
(C) 1.2
(D) 0.67
µ=
6000
4000
= 1.5
10.
Ref. index of a prism (A = 60º) placed in air (n = 1) is n = 1·5.
(i) If light ray is incident on this prism at an angle of 60º. Find the angle of deviation. State whether this is a minimum
deviation.
(ii) A light ray emerges from the prism at the same angle as it is incident on it. Determine the angle by which the rays
is deflected from its initial direction as a result of its passage through the prism :
1
Given : sin–1
= 30º, sin–1 0.4 = 25º, sin–1 0.6 = 37º.
3
Ans. [(i) 37º, This deviation is not minimum; (ii) 38º = m = 2 sin–1(3/4) – 60º]
Page 3
sin 60º
= 1.5
sin r1
Sol.

sin r1=
=
3
2
1
3
 r1 = sin–1
60º
 1 


 3
e
60º
i2 = (60 – r1)
sin e = 1.5 + sin (60 – r1)
= 1.5 (sin 60º cosr1 – cos60º sin r1)
 sine =
2 6 –3
4 3
r1
r2
120º
e = sin–1 (0.4) = 25
= i + e – A
= 60 +25 – 60
i  e, hence this is not minimum deviation.
Page 4
CPP-9
Class - XI
Batches - PHONON
PRISM
1.
Sol.
2.
Sol.
A beam of white light is incident on hollow prism of glass. Then :
(A*) The light emerging from prism gives no dispersion
(B) The light emerging from prism gives spectrum but the bending of all colours is
away from base
(C) The light emerging from prism gives spectrum, all the colours bend towards base,
the violet the most and red the least
(D) The light emerging from prism gives spectrum, all the colours bend towards base,
the violet the least and red the most.
Surface and hollow prism behaves like slabs thus there is no disperism.
A triangular glass wedge is lowered into water ( = 4/3). The refractive
index of glass is g = 1.5. At what angle  will the beam of light normally
incident on AB reach AC entirely ?
8
Ans. [ > sin–1 ]
9
 > C
 4/3
8
> sin–1 
 = sin–1  
 3/ 2 
a
3.
Sol.
For a prism of apex angle 45º, it is found that the angle of emergence is 45º for grazing incidence. Calculate the
refractive index of the prism :
(A) (2)1/2
(B) (3)1/2
(C) 2
(D*) (5)1/2
r1 + r2 = A = 45º
45º
r1 = C
r2 = 45 – C
sin r2 = sin (45 – C)
µ
µ sin r2 = sin i2 =
1
2
1
sin r2 =
2µ
and sin C =
1
µ
1
2µ
= sin 45º cosC – cos 45º sin C
1
=
µ
1

1 
1 – 2  –
 µ  µ
2
4
1
1
1– 2  2 = 1 – 2
=
µ
µ
µ
µ
µ = (5)1/2
Page 5
4.
Sol.
The maximum refractive index of a material, of a prism of apex angle 90º, for which light will be transmitted is :
(A) 3
(B) 1.5
(C*) 2
(D) None of these
r1 = r2 = C
r1 + r2 = A = 90º
90º
So,
r1 = r2 = 45º = C
90º
90º
r r
1
2
1
1
sinC =
=
n
2
n=
5.
D
2
A prism of refractive index n1 and another prism of refractive index n2
are stuck together without a gap as shown in the figure. The angles of
the prisms are as shown, n1 and n2 depend on , the wavelength of light
C
70º
n2
n
20º
1
10.8 104
1.80 104
according to n1 = 1.20 +
and
n
=
1.45
+
where

2
60º
40º
2
2
A
B
is in nm.
(i) calculate the wavelength 0 for which rays incident at any on the interface BC pass through without bending at
the interference.
Ans. [0 = 600 nm, n = 1.5]
Sol.
n1 = 1.2 +
10.8 104
1.8  104
and
n
=
1.45
+
2
2
2
the incident ray will not deviate at BC if
n1 = n 2
9  10 4
= 0.25  l0 = 600 nm
 20
n = 1.5
6.
Sol.
A parallel beam of light is incident on the upper part of a prism of angle 1.8º & R.I.
3/2. The light coming out of the prism falls on a concave mirror of radius of
curvature 20 cm. The distance of the point (where the rays are focused after
reflection from the mirror) from the principal axis is :
(A) 9 cm
(B*) 1.5 7 mm
(C) 3.14 mm
(D) None of these
 = (µ – 1) A = 0.9º = 0.9 –

rod
180
y
 


= 0.9 
× 10 cm
  y= 0.9 ×
(10cm) 
180 
180
y = 1.57 mm
7.
Sol.
The refractive indices of the crown glass for blue and red lights are 1.51 & 1.49 respectively and those of the flint glass
are 1.77 & 1.73 respectively. An isosceles prism of angle 6º is made of crown glass. A beam of white light is incident at
a small angle on this prism. The other flint glass isosceles prism is combined with the crown glass prism such that there
is no deviation of the incident light. Determine the angle of the flint glass prism. Calculate the net dispersion of the
combined system.
Ans. [A = 4º,  = 0.04]
For croun glass
For flirit glass
flirit glass
µblue = 1.51
µred = 1.49
µblue= 1.77
µred = 1.73
60º
A
Croun glass
Page 6
µy =
1.51  1.49
= 1.5
2
µy =
µ ' y –1
A
=
Al
µy –1
1.77  1.73
= 1.75
2
= 1+
7
–1
A
4
=
Al 1.5 – 1
=
3
4
7
4
6
3/ 4
3 2
3
=
=
=
Al
1/ 2
4
2
62
=4
3
Net dispersion if system is
v – r = (µv – µr)A – (µ'v – µ'r)A'
= (0.02)6º – (0.04)4º
= 0.12º – 0.16
= –0.04º
A' =
8.
Sol.
9.
Sol.
An equilaterial prism is kept on a horizontal surface. A typical ray of light PQRS is shown
in the figure. For minimum deviation :
(A) The ray PQ must be horizontal
(B) The ray RS must be horizontal
(C*) The ray QR must be horizontal
(D) Any one of them can be horizontal
60º
For minimum
R
 = 30º
S
Q
So,
(1)1 = (1)2
Hence R is || to box.
P 60º
60º
R
Q
S
P
B
The faces of prism ABCD made of glass with a refractive Index n from dihedral angles
A = 90º,  B = 75º,  C = 135º &  D = 60º (The Abbe's prism). A beam of light falls on face
AB & after total internal reflection from face BC escapes through face AD. Find the range of
n and angle of incidence  of the beam onto face AB, if a beam that has passed through the
prism in this manner is perpendicular to the incident beam.
A
Ans. [r +  = 75º;  = 45º (geometry), c < 45º 2 > n > 2 , 45º <  < 90º (snell's law)]
EBFH EHF = 180º = 75º = 105º
EHF r + + 105º = 180º +r = 75º
CDGF 135º + 60º + 90º + r + 90º – = 360º – r = 15º
B
 = 45º
r = 30º
75º
C
Let C = Critical angle So > C
(90 – )
135º
C < 45º
sinC < sin45º
r 105º
C
D
E
1
1
<
n >
n
2
H
2
90º
Snell's law

1 × sin  = n × sin r = n + sin 30º =
n

sin =
 2sin  = n >
2
nmax = 2sin = 2sin 90º
n
2
r
90+r
90º
2
A
G
60º
D
Page 7
nmax = 2

sin  >
1
2
 > 45º
Range of   45º <  < 90º
Range of n 
10.
Sol.
2 <n<2
A given ray of light suffers minimum deviation in an equilateral prism P. Additional prisms Q and R of identical shape
and of same material as P are now added as shown in the figure. The ray will now suffer :
(A) Greater deviation
Q
(B) No deviation
P
(C*) Same deviation as before
R
(D) Total internal reflection
Two prism in pair will cancel effect of each other and third will give same deviation.
Page 8
CPP-10
Batches - PHONON
Class - XI
PRISM
1.
Sol.
A ray of light in air is incident on face AB of an irregular block made of
material with refractive index 2 , as shown in figure. The face CD
opposite to AB is a spherical surface of radius of curvature 0.4 m. From
this face the refracted ray enters a medium of refractive index 1.514 and
meets the axis PQ at point E. Determine the distance OE correct to two
decimal places.
Ans. [OE = 6.06 m]
applying Snell's law at AB  1 × sin 45º =
A
45º
P
O
E
n=1.514
n=1
60º n= 2
B
A
2 sin 

 = 30º
BMN =
90 +  = 120º
DBM + BMN = 60 + 120 = 180º Þ therefore MN || BD
Refraction curved surface CD.
C
45º
n=1
C
N
O
E
n=1.514
n= 2
60º
B
D
M
90º
P
Q
Q
D
1.514
2
1.514 – 2
–
=
(OE ) (–)
(40)
2.
Sol.

1.514
1.514 – 1.414
0.1
1
–O=
=
=
OE
40
40
400


OE = 1.514 × 400 = 605.6 cm
OE = 6.06 m
A glass prism with a refracting angle of 60º has a refractive index 1.52 for red and 1.6 for violet light. A parallel beam of
white light is incident on one face at an angle of incidence, which gives minimum deviation for red light. Find.
(a) The angle of incidence
(b) Angular width of the spectrum
(c) The length of the spectrum if it is focussed on a screen by lens of focal length 100 cm.
[Use : sin (49.7º) = 0.760; sin (31.6º) = 0.520; sin (28.4º) = 0.475; sin (56º) = 0.832;  = 22/7]
Ans. [(a) 49.7º; (b) 7.27º; (c) f  = 12.68 cm]
(a) For red.
A 60
=
= 30º
2
2
sin i = µsin r = 1.52 × sin30º = 0.76
i = 49.7º Ans.
(b) For red:
1 × sin(49.7º) = 1.52 sinr1  r1 = 30º
r2 = 60º – r1 = 60 – 30 = 30º
1.52 × sin3º = 1 × sine  e = 49.7º

red = i + e – A = 49.7 + 49.7 – 60 = 39.4º
For violet :
At min 
r=
 0.76 
1 × sin 49.7º = 1.6 sin r1  r1 = sin–1 

 1.6 


r1 = sin–1 (0.475) = 28.4º
r2 = 60 – r1 = 60 – 28.4 = 31.6º
Page 9
1 × sine = 1.6 × sin r2 = 1.6 sin 31.6 - 0.838
e = sin–1 (0.838) = 56.97º

violet = i + e – A = 49.7 + 56.97 – 60 = 46.67º
 Angular width of spectrum
 = violet – red
= 46.67 – 39.4
 = 7.27º
(c)
length = f 
 

= (100)  7.27 
 cm
180


= 12.68 cm
3.
In an experiment performed with a 60º prism where angle of minimum deviation for sodium light is 60º in air. The
following experiment was done. When sodium light enters at one face at grazing incidence from a certain liquid, it
emerges from the other face (in air) at 60º from the normal to edge of the prism. Are the observations correct ?
Ans. [No]
 A  
sin 

sin 60º
 2 
µ=
=
=

sin 30º
sin
2
Sol.
3
For energing at 60º,
sin60º =
3 sin i
60º
60º



4.
Sol.
1
sin i = Þ i = 30º
2
C = 30º  µsin 90º
µl =
50º i
3
3
(not possible)
2
The following figure represents a wavefront AB which passes
from air to another transparent medium and produces a new
wavefront CD after refraction. The refractive index of the
medium is (PQ is the boundary between air and the medium) :
cos 1
cos 4
(A)
(B)
cos 4
cos 1
sin 2
sin 1
(C*)
(D)
sin 3
sin 4
µ=
AD sin 1
BD
= AD sin 
AC
4
sin 1
= sin 
4
5.
In the figure two triangular prisms are shown each of refractive index 3 .
(a) Find the angle of incidence on the face AB for minimum deviation from the
prism ABC ?
(b) Find the angle through which the prism DCE should be rotated about the edge
passing through point C so that there should be minimum deviation from the
system ?
Ans. [(a) i = 60º, (b) 60º]
A
D
60º
60º
i
B
60º
60º
C
E
Page 10
Sol.
For minimum deviation
i=e
60º
180 –120
2
r1 = 30º = r2
sin i = 4 sin r1

30º
r1 = r2 =

Thus
e
r1 r2
i
1
3× 2

i = 60º
For minimum deviation but both lens together to radius the deviation to zero.

6.
Sol.
sin i =
60º
In the figure ABC is the cross section of a right angled prism and BCDE is the cross
section of a glass slab. The value of  so that light incident normally on the face AB does
not cross the face BC is (given sin–1 (3/5) = 37º) :
(A)  37º
(B*)  < 37º
(C)  53º
(D)  < 53º
Boundary conditions
C = 90 – 

sin C = cos 
C
6/5

= cos
3/ 2

cos = 4/5
or
sin = 3/5

= 37º
For T.I.R., < 37º
7.
The refractive index of a prism is µ. the maximum angle of the prism for which a ray incident on it will be transmitted
through other face without total internal reflection is _________.
1
]

A = r1 + r2
For A to be max r1 and r2 both should be more r1 is max if light is incident on AB at angle and r2 is max of light energy
at 90º.
So for surface AB
A
1 sin 90º = µ sin r1
Ans. [ 2 sin 1
Sol.
1
1
sin r1 = µ  r1 = sin–1 µ
µ
Similarly for surface AC
r1 r2
i
1
1
sinr2 = µ  r2 = sin–1 µ
A = r1 + r2

8.
B
C
1
A =  sin–1  
µ
Two mirrors, placed perpendicularly, form two sides of a vessel filled
with water. A light ray is incident on the water surface at an angle  and
emerges at an angle  after getting reflected from both the mirrors inside.
The relation between  and  is expressed as :
(A*)  = 
(B)  > 
(C)  < 
(D) All are possible, depending upon 
Page 11
Sol.
A
C
D
45º
45º
r
90–
90–
B
From ABC
90 –  + 45º + 90 –  =
 + =
From (1) and (2) we get
=
Hence
=
9.
Sol.

45º...(1)
F
E
90–
From DEF
45º +  + 90 + r = 
 + r = 45º...(2)
r

O is a point object kept on the principal axis of a concave mirror M
of radius of curvature 20 cm. P is a prism of angle  = 1.8º. Light
falling on the prism (at small angle of incidence) get refracted
through the prism and then fall on the mirror. Refractive index of
prism is 3/2. Find the distance between the images formed by the
concave mirror due to this light :
2

(A)
cm
(B)
cm
5
10

3
(C*)
cm
(D)
cm
20
20
O1
O2
=1.8º
10 cm
20 cm
Due to prism two images of object O, O1 and O2 an formed
Distance 6/n, O1, O2 is [(2)(µ–1)A]d
=

cm
10
Here if we take v = – 30 cm
f = – 10 cm
then by mirror fomula
1
1 1
 =
f
v v
v = – 15 cm
m= –
v
–15
0.1
=
=
u
–30
2
Hence net length between I1 and I2 of object O1 and O2 respectively is
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I1 I2 = m(O1 O2) = m ×
10.
Sol.

 
1
= ×
=
10 2 10 20
Light travelling in air falls at an incidence angle of 2º on one refracting surface of a prism of refractive index 1.5 and
angle of refraction 4º. The medium on the other side is water (n = 4/3). Find the deviation produced by the prism.
Ans. [1º]
1 = i – r1
4º
2 = e – r2
 = 1× 2 = i – r1 – r2 + e
2º
e r2
= i + e – r1 – r2
r1
r2
 = 2i – 4
1·sin2º =
2×
3
sin r1
2
 2
 =r
1
180 3
r1 =
r2 = A – r1 =
4
3 180
4
4 
8

–
= 
180º
3 180º
3 180º
4
3
sinr2 = sin e
3
2
9 8 
3
5
4

=ee=
So =
–
= 1º
8 3 180º
180º
180º 180º
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