## In This post you will get answer of Q.

A disc is placed on a surface of pond which has refractive index 5/3. A source of light is placed 4 m below the surface of liquid. The minimum radius of disc needed so that light is not coming out is

Solution:

## $theta $ is the critical angle.

$therefore , , , , theta = sin ^{-1} (1/ mu ) = sin ^{-1} (3/5)$

or $ , , , sin theta = 3/5 $

$ therefore , , , tan theta = 3/4 = r/4 , , or , , r=3 m $