CPP-4
Batches - PHONON
Class - XI
REFLECTION
1.
Sol.
Half the surface of a transparent sphere of refractive index 2 is silvered. A narrow, parallel beam of light is incident on
the unsilvered surface, symmetrically with respect to the silvered part. The light finally emerging from the sphere will
be a :
(A*) Parallel beam
(B) Converging beam
(C) Slightly divergent beam (D) Widely divergent beam
2 1
(2  1)
Refraction from surface 1 : v   = ( R )
1
 v1 = +2R
Reflection from silvered surface 2 : Since object is at pole so image also is at pole.
1
2
1 2
Refraction from surface 2 : v  2 R = ( R )
1
 v2 = 
Therefore the final emergent light will emergent as parallel beam.
2.
R
R
(D)
 1
 1
This question can be solved by symmetry as is shown in figure. If d be the distance of object from first surface then
refracted ray from first surface should become parallel.

(B) µR
(  1)
1
=
R
d
R
d=
Ans.
(  1)
Sol.
(C*)

1
(  1)

=
 (d )
(  R)
0+
3.
Surface 1 Surface 2
A transparent sphere of radius R and refractive index µ is kept in air. At what distance from the surface of the sphere
should a point object be placed so as to form a real image at the same distance from the sphere ?
(A) R/µ
Sol.
u=2
d
d
A ray of monochromatic light is incident on the plane surface of separation between two media x & y with angle of
incidence 'i' in the medium x and angle of refraction 'r' in the medium y. The graph shows the relation between sin r
and sin i :
sin r
(A) The speed of light in the medium y is (3)1/2 times then in medium x.
(B*) The speed of light in the medium y is (1/3)1/2 times then in medium x.
(C) The total internal reflection can take place when the incidence is in x.
30º
(D*) The total internal reflection can take place when the incidence is in y.
sin i
From graph
sin r
1
= tan 30° =
sin i
3

sin i = 3 sin r
From snell's law :
µx sin i = µy = sin r
Page 14
=

4.
c
c
3 sin r =
sin r
vx
vy
v y 3 = vx
A point source is placed at a depth h below the surface of water (refractive index = ). The medium above the surface
of water is air ( = 1). Find the area on the surface of water through which light comes in air from water.
h2
]
2  1
u × sin  = 1 × sin 90°
Ans. [
Sol.
sin  =
 tan  =
1

R
h
1
2  1
1
R = h tan  =

5.
Sol.
2  1
h 2
 Area = R2 =
 Area =
90°
( 2  1)
h 2
( 2  1)
vy =
1
3
Ans.
 vx
Given that, v velocity of light in quartz = 1.5 × 108 m/s and velocity of light in
glycerine = (9/4) × 108 m/s. Now a slab made of quartz is placed in glycerine as
shown. The shift of the object produced by slab is :
(A*) 6 cm
(B) 3.55 cm
(C) 9 cm
(D) 2 cm
  
Normal shift = hact  1  r 
 d 
here :
hact = 18 cm
3 108
hr = hglycerine =
hd = hquartz =
8
(9 / 4) 10
3  108
1.5  108
=
4
3
=2
 4/3
 4
 Normal shift = 18  1 
 = 18 1  
2 

 6
1
= 18   = 6 cm
 3
 Normal shift = 6 cm Ans.
Page 15
6.
Two refracting media are separated by a spherical interface as shown in the figure. P P' is he principal axis, µ1 and µ2
are the refractive indices of medium of incidence and medium of refraction respectively. Then :
(A*) If µ2 > µ1, then there cannot be a real image of real object
µ2
µ1
(B) If µ2 > µ1, then there cannot be a real image of virtual object
P
P'
(C*) If µ > µ , then there cannot be a virtual image of virtual object
1
Sol.
2
(D) If µ1 > µ2, then there cannot be a real image of real object
Object is in medium µ1 at a distance x from pole
2

  1
 1 = 2
v ( x )
( R )
1  2 1
2

=
R
x
v


v=
2 Rx
=
(1  2 ) x  (1 ) R
2 R
   R 
1  1  2   
 1  x 
Case-1 : Object is real  x negative  Let x = – d
   R
So (1A) Image is real  v positive  1  2   > 0
 1  d
1–
d>
2
R
>
1
d
1 R
(1   2 )
   R
(1B) Image is virtual  v negative  1  2   < 0
 1  d
2
R
<
1
d
Case-2 : Object is virtual  x negative  Let x = –d
1–
   R
So (2A) Image is real  v positive = 1  2   > 0
 1  d

 R
  2  1 <
 1
 d
d<
1 R
( 2  1 )
   R
(2B) Image is virtual  v negative  1  2   < 0
 1  d

 R
  2  1 >
 1
 d
1 R
( 2  1 )
Looking at four cases
If : µ2 > µ1  Real object can not have real image
If : µ2 < µ1  Virtual object can not have virtual image
d>
7.
A spherical surface of radius 30 cm separates two transparent media A and B with refractive Indices 4/3 and 3/2
respectively. The medium A is on the convex side of the surface. Where should a point object be placed in medium A
so that the paraxial rays become parallel after refraction at the surface ?
Ans. [240 cm away from the separating surface]
Page 16
µB µA
µB – µA
–
=
u

R
Sol.
O–
A
µ=4/3
3/ 2 – 4 / 3
4/3
=
(30)
u
µ = –240 cm.
8.
Sol.
9.
Sol.
B
µ=3/2
R = 30 cm
A ray incident at a point at an angle of incidence of 60º enters a glass sphere of  = 3 and it is reflected and refracted
at the farther surface of the sphere. The angle between reflected an refracted rays at this surface is :
(A) 50º
(B*) 90º
(C) 60º
(D) 40º
1 sin 60 = 3 sin r
1
60° 60°
sin r =
2
90 =
r = 30°
r
 = 180 – (60 + r) = 90°
A ray of light is incident on a parallel slab of thickness t and refractive index n. If the angle of incidence  is small, than
the displacement in the incident and emergent ray will be :
t ( n  1)
t
t n
(A*)
(B)
(C)
(D) None
n
n
n 1
sin  = n sin r
x = t sin r
t
x + y = t sin 

and
t
sin 
+ y = t sin 
n
d
y
d
= cos 
y
x
r
r
sin 
cos  

or
t
+ d = t
n
1

d = t   – 
n

=
10.
t ( n – 1)
n
A stationary swimmer S, inside a liquid of refractive index µ1, is at a distance d from a fixed point P inside the liquid. A
rectangular block of width t and refractive index µ2 (µ2 < µ1) is now placed between S and P.S will observe P to be at
a distance :

(A) d  t  1  1

 2 
Sol.
n

(B) d  t 1  2 
 1 

(C) d  t 1  2 
 1 
 µ2 
distance = (d – t) + t /  µ 
 1
µ1

(D*) d  t  1  1

 2 
µ2
P
S
 µ1

= d + t  µ –1
 2 
t
Page 17
CPP-5
Batches - PHONON
Class - XI
REFLECTION
1.
Sol.
2.
Sol.
A ray of light falls on a transparent sphere with centre at C as shown in figure.
The ray emerges from the sphere parallel to line AB. The refractive index of the
sphere is :
(A) 2
(B*) 3
(C) 3/2
(D) 1/2
i2 = r1
(isoceles triangle CDE)
E r

r2 = 60º
2

r1 + x2 = 60º
r
A
D
C
or
r1 = 30º
r1
60º
60º
sin 60º
3/2

µ=
=
= 3
sin 30º
1/ 2
A beam of diameter 'd' is incident on a glass hemisphere as shown. If the radius of curvature of the
hemisphere is very large in comparison to d, then the diameter of the beam at the base of the
hemisphere will be :
3
(A) d
(B) d
4
d
2
(C)
(D*) d
3
3
For objects at infinity image will be formed at :
3 / 2 1/ 2
=
V
R
or
V = 3R.
D
D'
=
2 R 3R

r
D
D' =
E
R
from similer triangles ABC and ADE
3.
B
B
C
3R
2R
2
D.
3
A
Rays incident on an interface would converge 10 cm below the interface if they
continued to move in straight lines without bending. But due to refraction, the
rays will bend and meet some where else. Find the distance of meeting point of
refracted rays below the interface, assuming the rays to be making small angles
with the normal to the interface.
Ans. [25 cm]
Page 18
x
Sol.
sin i =
2
10  x 2
i
x
sin r =
y  x2
y
sin i
5
= =
sin r
2
2
x y
or
Sol.
6.
Sol.
x
x 2  102
x
Find the apparent depth of the object seen by observer A ?
Ans. [
5.
5/2
y
5
=
2 10
y = 25 cm

Sol.
10
2
For small i and r
x << 10 and x << y
4.
i
r
2
68
cm]
3
Apparent depth =
d1
d2
+ µ
µ
air 1
1 2
=
15
25
+ (2.5) /(1.5)
1.5
=
50
68
+6=
3
3
A point source of light at the surface of a sphere causes a parallel beam of light to emerge from the opposite surface
of the sphere. The refractive index of the material of the sphere is :
(A) 1.5
(B) 5/3
(C*) 2
(D) 2.5
µsin  = 1 × sin 2
2
µ sin = 2sin cos
µ = 2 cos 
2
for is small
C
µ= 2
An object is placed 30 cm (from the reflecting surface) in front of a block of glass 10 cm thick having its farther side
silvered. The final image is formed at 23.2 cm behind the silvered face. The refractive index of glass is :
(A) 1.41
(B) 1.46
(C*) 100/ 132
(D) 1.61
BI1 = (10 + 20µ)
BI2 = (10 + 20µ)
AI2 = (20 + 20µ)
1
AI3 = (23.2) + 10 = 33.2 = (20 × 20µ) µ
µ=
7.
200
132
I1
O 20cm A
B
I3
I2
10cm
The x-y plane is the boundary between two transparent media. Media-1 with z > 0 has refractive index 2 and medium

– 2 with z < 0 has a refractive index 3 . A ray of light in medium-1 given by the vector A = 6 3iˆ  8 3 ˆj  10kˆ is
incident on the plane of separation. Find the unit vector in the direction of refracted ray in medium –2.
Page 19

Ans. [ r =
Sol.
3 ˆ 2 2 ˆ 1 ˆ
i
j
k (angle of incidence = 60º, r = 45º]
5
5 2
2
Unit vecto of in xy plane for incident ray and unit vector in xy plane for refracted will be same plane becuase they need
to be in same plane.
Here component of incident ray in xy plane is 6 3î  5 3 ˆj
Unit vector of incident ray in xy plane is
3
4
î  ˆj angle with x axis is 0 = 53º
5
5
Component of unit refractracted vection in xy plane is
Here refracted ray is
1
2
3
5 2
8.
Sol.
9.
cosî +
î
1
2
sin ĵ –
1
2
1
2
k̂
2 2
1 ˆ
–
k
5
2
A ray of light passes through four transparent media with refractive indices µ1,
µ2, µ3, and µ4 as shown in the figure. The surfaces of all media are parallel. If the
emergent ray CD is parallel to the incident ray AB, we must have :
(A) µ1 = µ2
(B) µ2 = µ3
(C) µ3 = µ4
(D*) µ4 = µ1
If CD is parallel to AB then µ1 = µy
(C)
5
2
(B*)
µ2
µ3
µ4
C
B
A
3
2
(D)
5
2
3h
3
2
h
2h
h
tan  =
= 1  = 45º
h
Sol.
h
1
h
1
tan =
= sin  =
2h
5
2
sin =
1
h
C
h
A

µ×
5
=1×
1
2
E
2h
h
B
1
D
h
2
Snell's law
µ × sin = 1 × sin 
Sol.
µ1
An observer can see through a pin hole the top of a thin rod of height h, placed as shown
in the figure. The beaker height is 3h and its radius is h. When the beaker is filled with a
liquid upto a height 2h, he can see the lower end of the rod. Then the refractive index of the
liquid is :
(A)
10.
D
µ=
5
Ans.
2
h
h
A solid, transparent sphere has a small, opaque dot at its centre. When observed from outside, the apparent position
of the dot will be :
(A) Closer to the eye than its actual position
(B) Farther away from the eye than its actual position
(C*) The same as its actual position
(D*) Independent of the refractive index of the sphere
u = –R
µ
v=?
1
O
Page 20
1 
1


=
( R )
v (  R)
1  1 
 =
v R
R
v = –R
Hence dot will appear at its original position.
Page 21
CPP-6
Batches - PHONON
Class - XI
REFLECTION
1.
Given that, velocity of light in quartz = 1.5 × 108 m/s and velocity of light in glycerine
= (9/4) × 108 m/s. Now a slab made of quartz is placed in glycerine as shown. What is
the shift produced by slab ?
(A*) 6 cm
(B) 3.55 cm
(C) 9 cm
(D) 2 cm

µrare 

Shift = t  1– µ
denser 

Sol.
µrare = µglycerine =
µdense = µquarter =
4
3 108
8 =
3
9 / 4  10
3  108
=2
1.5  108
 4/3
shift = 18 1 –
 = 6 cm
2 

2.
A rectangular glass slab ABCD, of refractive index n1, is immersed in water of refractive index n2 (n1 > n2). A ray of light
in incident at the surface AB of the slab as shown. The maximum value of the angle of incidence max, such that the ray
comes out only from the other surface CD is given by :
n

n 
(A*) sin 1 1 cos sin 1 2 
n1 
 n2

 1 1 
1 

(B) sin n1 cos  sin
n2 


Sol.
A
D
n1
max
B
n2
C
n 
(C) sin 1 1 
 n2 
n 
(D) sin 1 2 
 n1 
Condition will be satisfied it on lateral surface T.I.R takes place. Forthat 9 –   c

 90 – c

sin sin(90 – c)
 n2 
[c = sin–1  n  ]
 1
 n2 
sin  cot (c)  [sin  cos[sin–1  n  ]
 1
Snell's law
Page 22
n2 sin = n1 sin 
3.
 n2 
n2
sin = sin  cos [sin–1  n  ]
n1
 1

 n2 
n1
sin n cos [sin–1  n  ]
2
 1

 n1
 –1  n2
max = sin–1  n cos sin  n

 2
 1
 
 
  
A light beam is traveling from Region I to Region IV (Refer
figure). The refractive index in Regions I, II, III and IV are n0,
n0 n0
n
,
and 0 , respectively. The angle of incidence  for
2 6
8
which the beam just misses entering Region IV is Figure :
(A) sin
Sol.
–1
 3
 4
 
n0sin =
(B*) sin
–1
1
8
 
(C) sin
Region I
Region II

n
—0
2
Region III
Region IV
n
—0
6
n
—0
8
n0
0
–1
1
 4
 
0.2 m
(D) sin
0.6 m
–1
1
3
 
n0
1
n0
sin1 =
sin2  sin = sin2
6
6
2
n0 / 8
3
2  (c)3-4  sin2  n / 6  sin2 
4
0
4.
Sol.
1 3
1
  sin
6 4
8

sin

1
min = sin–1  
8
A hemispherical portion of the surface of a solid glass sphere (µ = 1.5) of radius r is silvered to make the inner side
reflecting. An object is placed on the axis of the hemisphere at a distance 3r from the centre of the sphere. The light
from the object is refracted at the unsilvered part, then reflected from the silvered part and again refracted at the
unsilvered part. Locate the final image formed.
Ans. [At the pole of reflecting surface of the sphere]
Refraction at sngle surface no. (1)
Surface
Surface
(1)
(2)
3
3 
µ=3/2
–1
2


1
µ=1
 
2 
– (–2r ) =
v1
r
r
r
O
3r
 v1 
Reflection from silvered concave surface (2)
1 1
2
r
+ = –  v2 = –
v2 
r
2
Refraction from surface (1) emerging out:
1
3/ 2
1 – 3 / 2 
–
=
 v3 = – 2r
v3 –(3r / 2)
(– r )
This means taot final image is formed at pole of refleting surface (2)
Page 23
5.
In ray of light (GH) is incident on the glass-water interface DC at an angle i.
It emerges in air along the water-air interface EF (see figure). If the refractive
index of water µw is 4/3, the refractive index of glass µg is :
(A)
3
4sin i
(B*)
air
µa = 1
F
Water
1
sin i
D
Glass
µ0 = ?
4sin i
4
(D)
3
3sin i
Refraction at CD:
µg × sini = µw × sin r
Refraction at F2F
µw × sin r = 1× sin90º

µg sin i = 1 × sin 90º
Sol.
i
A
G
B
1
sin i
µg =
6.
C
H
(C)
Sol.
E
µw = 4/3
A parallel beam of light is incident normally on the flat surface of a hemisphere
of radius 6 cm and refractive index 1.5, placed in air as shown in figure-(i).
Assume paraxial ray approximation :
(A*) The rays are focussed at 12 cm from the point P to the right, in the
situation as shown in figure-(i)
(B) The rays are focussed at 16 cm from the point P to the right, in the situation as shown in figure-(i)
(C) If the rays are incident at the curved surface (figure-(ii)) then these
are focused at distance 18 cm from point P to the right.
(D*) If the rays are incident at the curved surface (figure-(ii)) then these
are focused at distance 14 cm from point P to the right.
Figure (i)
Figure (ii)
µ=3/2
µ=3/2
P
P
(1)
Surface (1) :
R=6
R=6
(2)
3/ 2 1
3/ 2 –1
–
=
v1


3/ 2 1
3/ 2 –1
– =
v1

(6)
Surface (1):
v1  .
Surface (2):
v1 = + 18 cm
1 3/ 2 1 – 3/ 2
–
= (–6)  r2 = + 12
v2

Surface (2):
1
3/ 2
1 – 3/ 2
–
=
v2 (12)

option (A) and option (B) are correct.
7.
v2 = +8 cm
A point source of light is placed at a distance h below the surface of a large deep lake.
(a) Show that the fraction f of the light energy that escapes directly from the water surface is independent of h and is
given by f  1  1 n 2  1 where n is the index of refraction of water..
2 2n
(Note : Absorption within the water and reflection at the surface; except where it is total, have been neglected)
(b) Evaluate this ratio for n = 4/3.
Sol.
Ans. [(b) (4 –
7 ) / 8]
n + sin = 1 × sin 90º
1
sin =
n
90º
y
r
h
R
Page 24
1–
cos =
1
n2
Note:
Fore sphere
dA = (2R cos) (Rd)
dA = 2R 2 cosd
For area of out side sphere,  varies from (90 – ) to 90º
90

2
Aarea of side water =  2R cos d 
90 – 
90
= 2R2  cos d 
90 – 
2
= 2R [sin 90 – sin (90 – )]
= 2R2 [1 – cos]
1
Area of side water = 2R2 [1– 1 – n ]
2
Praction:
(b)
F=
Area of sphere outside water
Total area of sphere
F=
2R 2 1– 1 – 1  

 n2  

 
4R 2 
F=
1
2

1

1– 1 – 2
 n

n = 4/3  F =
 1
1
 = –
  2 2n
( n 2 – 1)
4– 7
8
45º
Air
8.
Sol.
A ray of light is incident from air to a glass rod at point A. The angle of incidence being 45º.
The minimum value of refractive index of the material of the rod so that T.I.R. takes place at
B is_____.
Ans. [ 3 / 2 ]
r = 90 –   = 90 – r  qc
sin (90 – r)  sin c
45º
Air
1
A
cos r 
n
r
1 – sin2r 
n
9.
nmin =
B
1×sin45º=n×sinr
1
sinr =
n 2
1
1
1
1–
³
n
2n 2 n 2
(3/ 2) ,
A
(3/ 2) Ans.
A man observes a coin placed at the bottom of a beaker which contains two
immiscible liquids of refractive indices 1.2 and 1.4 as shown in the figure. A
plane mirror is also placed on the surface of liquid. The distance of image
(from mirror) of coin in mirror as seen from medium A of refractive index 1.2 by
an observer just above the boundary of the two media is :
(A) 18 cm
(B*) 12 cm
(C) 9 cm
(D) None of these
Page 25
Sol.
1.2
A
3cm
1.4
B
7cm
coin
 4 
Image of coin due to AB interface is at depth 
  1.2 from interface AB.
 1.4 
 7
 1
For light moving from nod A to air image is at  1.2  3 
from mirror and nod A interface.
1.4
 1.2
Now final image will from observer will be at
 7
 1 
  1.2  3   × 1.2 + 3 = [6 + 3 ] + 3 = 12 cm
1.4

 1.2 

10.
Sol.
A container contains water upto a height of 20 cm and there is a point source at the centre of the bottom of the
container. A rubber ring of radius r floats centrally on the water. The ceiling of the room is 2.0 m above the water
surface. (a) Find the radius of the shadow of the ring formed on the ceiling if r = 15 cm. (b) Find the maximum value of
r for which he shadow of the ring is formed the ceiling. Refractive index of water = 4/3.
3
169
Ans. [(a)
m = 2.8 m; (b)
m = 22.6 cm]
60
5 7
2m
r'
µsin q = 1·sinr'
q = 37º
4 3
 = sinr'
3 5
sinr' =
tan 53º =
20cm
4
 r' = 53º
5
x
2
4
8
x
= x = m
3 2
3
Net r adius will be
15 8
 = 2.8 m
100 3
Shadow will form till critical incidence
 3
C = sin–1  
 4
i.e.,
tan  =
r
20

sin  =
3
4

tan  =

r=
3
7
3
7
× 20 cm or
3
5 7
m
Page 26
CPP-7
Class - XI
Batches -
PHONON
REFLECTION
1.
A
A small rod ABC is put in water making an angle 6º with vertical. If it is viewed paraxially
from above, it will look like bent shaped ABC'. The angle of bending CBC') will be in
4
degree …  nw   :
3

(A*) 2º
(B) 3º
(C) 4º
(D) 4.5º
=
Sol.
B
C'
C
v
v
,=
OC
OC '
6º
O
a OC ' 1
=
= µ
 OC
 4
 = µ =   (6º) = 8º
 3
6º
v
6º
C'
x/µ
x
C
 –  = 2º
2.
If the observer sees the bottom of vessel at 8 cm, find the refractive index
of the medium in which observer is present.
16
Ans. [ ]
15
µsel.=
Sol.
10
8
4 / 3 10
4 8
16
=
µ= 
=
µ
8
3 10 15
Q
3.
A man starting from point P crosses a 4 km wide lagoon and reaches
point Q in the shortest possible time by the path shown. If the
person swims at a speed of 3 km/hr and walks at a speed of 4 km/hr,
then his time of journey (in minutes) is ?
Ans. [250]
3
sin 1 =
5
Sol.
6km
3km
LAGOON
(Salt water lake)
4km
P
sin 2 sin 1
= v
v2
1

6
3
315
sin 2
=
3
4
4
5
Page 27

sin 2 =
4
5
distance travelled in water
= 5 km
time taken in water =
5
×60 hr
3
= 100 min
distance travelled in ground
6
6
=  =
3
/5
2
time =

4.
Sol.
10  60
= 150 min.

150 + 100 = 250 min.
The figure shows a parallel slab of refractive index n2 which is surrounded by media of refractive
indices n1 and n3. Light is incident on the slab at angle of incidence (0). The time taken by the ray
to cross the slab is 't1' if incidence is from 'n1' and it is 't2' if the incidence is from 'n3'. then assuming
that n2 > n1, n2 > n3 and n3 > n1. then value of t1/t2 :
(A) = 1
(B) > 1
(C*) < 1
(D) Cannot be decided
n
n
n
1
2
3
n1 sin  = n2 sin 1

n1
sin 1 = n sin
2
n3 sin  = n2 sin 2

n3
sin 2= n sin 
2
n1
n2
n3
But n3 > n1

sin 2 > sin 1
thus
t2 > t1
and
5.
t1
<1
t2
A pole of length 2.00 m stands half dipped in a swimming pool with water level 1 m higher than the bed (bottom). The
refractive index of water is 4/3 and sunlight is coming at an angle of 37º with the vertical. Find the length of the shadow
of the pole on the bed : Use sin–1 (0.45) = 26.8º, tan(26.8º) = 0.5.
Ans. [1.25 m]
Sol.
sin37º =
4
sin r
3
3 4
= sin r
5 3


37º
_4
=µ
3
r
9
sin r =
20
r = 26.8
3
+ 1 × tan26.8
4
= 0.75 + 0.5
= 1.25 m
Length of shadow=
Page 28
6.
Sol.
A ray of light is incident from air to a glass rod at point A. The angle of incidence being 45º. The
minimum value of refractive index of the material of the rod so that T.I.R. takes place at B is
_____ :
(A) 1.2
(B*) 1.5
(C) 1.7
For surface B
µ sin (90 – ) = 1 sin 90º
µ cos  = 1
1
cos  =
µ
For surface A
1
·1 = µ sin 
2
1
sin  =
2µ
From (1) and (2)
(D)
7.
A
...(1)
B
90–
...(2)
1
1
tan  =
µ=
45º
1
1
2µ
=
1
2
µ
tan  =
2.3
3
2
2
1
=
cos 
3
2
=
3
= 1.5 Ans.
2
3
is moved toward a stationary observer. A
2
point 'O' is observed by the observer with the help of paraxial rays through the slab. Both
'O' and observer lie in air. The velocity with which the image will move is :
In the figure shown a slab of refractive index
4
m/s towards left
3
(D*) Zero
(A) 2 m/s towards left
(B)
(C) 3 m/s towards left
Sol.
µ = 3/2
O
obs
x
Shift in the position of image is independent of the position of slab hence due to movement of slab object do not gain
any velocity.
8.
Sol.
A beam of light is incident on a spherical drop of water at an angle i. Find the angle between the incident ray & the
emergent ray after one reflection from internal surface. Is this possible by total internal reflection ?
Ans. [ = 2i – 4 sin–1  3 sin i  + , No]
4

For first deviation
S1 = i – r
For second deviation S2 = – 2r
For third deviation
S3 = i – r
Total deviation will be
i
i
Page 29
 = S1 + S2 + S3
= 2i +  – 4r
...(1)
and from Snell's law
1.sin i = µ sin r
sin i =
4
sin r
3
3

 = sin–1  sin i  ...(2)
4

Hence from (1) and (2)
3

 = 2i + – 4 sin–1  sin i 
4


9.
Sol.
A glass hemisphere of refractive index 4/3 and of radius 4 cm is placed on a
plane mirror. A point object is placed on axis of this sphere at a distance 'd'
from as shown. If the final image is formed at infinity, then find the value of
'd' in cm.
Ans. [3 cm]
Image of object O due to refraction at AB is at du from surface AB
Now due to ACB surface.
I'

1
1– 4/3
3
–
=
v (4  du)
–4
O
d
A
r
4cm
d –9
v = (12)(3  d )
d×µ
B
C
d –9
Now suppose d > 9 then image is formed below the mirror at distance v' = (12)(3  d ) from mirror
Now this image of mirror will act as object for refraction again at surface ACB (light goes from air to water)
So,
(d – 9)
4
4 / 3 –1
1
–
=
=
3v (12)(3  d )
4
12
For final image to be at infinity v = 
So,
–(d – 9)
1
=
 – d + 9 = 3+d
12(3  d ) 12
= 2d – C  d = 3 cm
10.
Sol.
An observer observers a fish moving upwards in a cylindrical container of cross section area 1 m2 filled with water up
to a height of 5 m. A hole is present at the bottom of the container having cross section area 1/1000 m2. Find out the
speed of the fish observed by observer when the bottom hole is just opened. (Given : The fish is moving with the
speed of 6 m/s towards the observer, µ of water = 4/3) Ans. [4.4975 m/s]
Velocity of liquid from the hole is 10 m/sec. So using equation of continuity
A1V2 = A2V2
v2
5m
x
cm
__
600 sec
v = 2 gh
Page 30
U2 =
y=
1
m/sec = 1 cm/sec
10
x
µ
1  dx 
dy
= µ  
dt
 dt 
Here x decreases so
dx
is – 601 cm/sec
dt
3
× (–601) = – 450.75 m/sec
4
This
dy
is velocity of image from surface so velocity of image will be 450.75 – 1
dt
= 449.75 cm/sec = 4.4975 m/sec
Page 31