A circular beam of light (diameter d) falls on a plane surface of a liquid. The angle of incidence is `45^(@)` and refractive index of the liquid is `mu`. The diameter of the refracted beam is

Options

A. d
B. `(mu-1)d`
C. `(sqrt(2mu^(2)-1))/dd`
D. `(sqrt(mu^(2)-1))/mud`

Answer

C

Explanation

(c ) From the figure,AB`=d/(cosi)=d'/(cosr)`

`therefored'=(cosr)/(cosi)d=sqrt2cosr.d`
Further, `sinr=(sini)/mu=1/(sqrt2mu)`
Hence, `cosr=sqrt(1-1/(2mu^(2)))=(sqrt(2mu^(2)-1))/(sqrt2mu)`
Substituting in Eq.(i), we get
`d'=(sqrt(2mu^(2)-1))/mucdotd`