Question

A circular beam of light having a diameter 4 cm falls on a
plane glass slab at angle of incidence 60°. If refractive index
3
of the material of slab is u =3/2, then the diameter of the refracted beam is
A) 10√2/3 cm
B) 2 cm
C) 8√2/3 cm
D) 4√ 3/2 cm​

Answer 1

Given : A circular beam of light having a diameter 4 cm falls on a plane glass slab at angle of incidence is 60° . if the refractive index of the material of slab is 3/2.

To find : The diameter of the refracted beam.

solution : see ray diagram,

let d' is the diameter of reflected beam.

here, d = PQ cos60°

and d' = PQ cos r [ r is angle of reflection ]

now d'/d = (PQ cosr)/(PQ cos60°) = 2cosr

⇒d' = 2d cosr ........(1)

from Snell's law,

1 × sini = μ × sinr

⇒sinr = sini/μ = sin60°/(3/2) = 1/√3

⇒cosr = √(1 - sin²r) = √(1 - 1/√3²) = √(2/3)

putting, cos r = √(2/3) in equation (1) we get,

d' = 2d × √(2/3)

= 2 × 4 × √(2/3) [ given d = 4cm ]

= 8√(2/3) cm

Therefore diameter of the reflected beam is 8√(2/3) cm.