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)/(mu)d`
D. `sqrt(mu^(2)-1)/(mu)d`

Explanation

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

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