## Refractive index of a rectangular glass slab is μ=√3. A light ray incident at an angle 60° is displaced laterally …..

Q. Refractive index of a rectangular glass slab is μ = √3. A light ray incident at an angle 60° is displaced laterally through 2.5cm. Distance travelled by light in the slab is
(a) 4cm
(b) 5cm
(c) 2.5√3
(d) 3cm
Ans: B

Sol:Angle of incidence = 60°

By applying Snell’s law ,

1×sin60 = √3×sinr

√3/2 = √3×sinr ,

sinr = 1/2

and r= 30°

By applying formula ,

Lateral Shift = t sin(i-r)/cosr

2.5 = tsin(60-30)/cos30

= t (tan30)

2.5 = t(1/√3)

t= 2.5 √3 cm

distance travelled by light in the slab

= t/cosr

= (2.5√3)/cos30 = 5 cm

## A monochromatic light passes through a glass slab (μ = 1.5) of thickness 9cm in time t1. If it takes a time t2 to travel…..

Q. A monochromatic light passes through a glass slab (μ = 1.5) of thickness 9 cm in time t1. If it takes a time t2 to travel the same distance through water (μ = 4/3). The value of (t1 − t2) is

(a) 5 × 10-11 sec

(b) 5 × 10-8 sec

(c) 2.5 × 10-10 sec

(d) 5 ×  10-10 sec

Ans: (a)

Sol: Distance d= 9 × 10−2 m , Speed in medium v= c/μ and

time t= distance/ speed

$\displaystyle t_1 = \frac{9 \times 10^{-2}}{c/1.5}$

$\displaystyle t_2 = \frac{9 \times 10^{-2}}{c/(4/3)}$

t1 – t2 = 5 x 10-11 sec

## A ray of light is incident at 50° on the middle of one of the two mirrors arranged at an angle of 60° between them….

Q. A ray of light is incident at 50° on the middle of one of the two mirrors arranged at an angle of 60° between them. The ray then touches the second mirror, get reflected back to the first mirror, making an angle of incidence of
(a) 50°
(b) 60°
(c) 70°
(d) 80°
Ans: C

## Two mirrors labeled L1 for left mirror and L2 for right mirror in the figure are parallel to each other and 3.0m apart…..

Q.Two mirrors labeled L1 for left mirror and L2 for right mirror in the figure are parallel to each other and 3.0 m apart. A person standing 1.0 m from the right mirror (L2) looks into this mirror and sees a series of images. The second nearest image in the right mirror is situated at a distance

(a) 2.0m from the person
(b) 4.0m from the person
(c) 6.0m from the person
(d) 8.0m from the person
Ans: C
Sol:

First image in L2 is 1m right fron L2 hence , 2m from man .

First image in L1 is 2m left from L1 , hence 5m from L2 .

Its image in L2 is 5m right from L2 , hence 6m from man .

## An object moves with 5 m/s towards right while the mirror moves 1 m/s towards the left as shown. Then the velocity of image

Q. An object moves with 5 m/s towards right while the mirror moves 1 m/s towards the left as shown. Then the velocity of image

(a) 7m/s towards left

(b) 7m/s towards right

(c) 5m/s towards right

(d) 5m/s towards left

Ans : (a)

Sol:

$\displaystyle \vec{V_{im}} =$ velocity of image w.r.t mirror

$\displaystyle \vec{V_{om}} =$ velocity of object w.r.t mirror

Hence ,

$\displaystyle \vec{V_{im}} = - \vec{V_{om}}$

$\displaystyle \vec{V_i} - \vec{V_m} = -(\vec{V_o} - \vec{V_m})$

$\displaystyle \vec{V_i} = 2 \vec{V_m} - \vec{V_o}$

$\displaystyle \vec{V_i} = 2 (-1) - 5 = -7$

VI= 7m/s towards left