Refraction questions

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

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=  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.

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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]

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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]

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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