A circular beam of light of diameter `d = 2 cm` falls on a plane refractive of glass. The angle of incidence is `60^(@)` and refractive index of glass is `mu = 3//2`. The diameter of the refracted beam is

Options

A. 4.00cm
B. 3.0cm
C. 3.26cm
D. 2.52cm

Answer

c.

Explanation

Let `d^(')` be the diameter of refracted beam.
Then,
`d=PQcos60^(@)`
and `d'=PQ cosr`
i.e. `(d^('))/(d)=(cosr)/(cos60^(@))=2cosr `
or `d'=2d cosr`
`sin r = (sini)/(mu)=(sqrt(3)//2)/(3//2)=(1)/(sqrt(3))`
`:. cosr =sqrt(1-sin^(2)r)=sqrt((2)/(3))`
`:. d=(2)(2)sqrt((2)/(3))`
`=4sqrt((2)/(3))cm~~3.26cm`