Question

Calculate the minimum deviation produced by a 60° glass prism if the refractive index of the glass is

    1.50. What is the angle of incidence at which minimum deviation occurs?

Answer 1

Answer

Minimum angle of reflection should be half of prism angle

$r=\frac{A}{2}=30°$

$n_1sini=n_2sinr$

So

$i=sin^{-1}(\frac{n_2sinr}{n_1}) \\i=sin^{-1}(\frac{(1.5)sin30°}{1})=48.6°$