A ray of light undergoes minimum deviation in a glass prism of apex angle 60^ . The angle of minimum deviation in air is found to be 60^ . The prism is then immersed in an unknown transparent liquid, and the critical angle for light travelling from the prism to the liquid is measured to be 45^ . The refractive index of the liquid is :

Step-by-step solution

First, calculate the refractive index of the prism ( _p ) using the minimum deviation condition in air: _p = ( A+ _m 2 ) ( A 2 ) Substitute A = 60^ and _m = 60^ : _p = ( 60^ +60^ 2 ) ( 60^ 2 ) = 60^ 30^ = 3 2 1 2 = 3 Next, when the prism is immersed in the liquid, the critical angle _c for the prism-liquid interface is given by: _c = _l _p Given _c = 45^ , we have: 45^ = _l 3 1 2 = _l 3 _l = 3 2 Answer: 3 2