Show that the angle of deviation produced by a thin prism is reduced to one fourth (w.r.t. air) when it is immersed in water. Given `.^a mu_g = 3//2` and `.^a mu_g = 4//3`.

Explanation

For deviation in air, `mu = (mu_(g))/(mu_(a)) = (3//2)/(1)`
and for deviation in water, `mu = (mu_(g))/(mu_(w)) = (3//2)/(4//3) = (9)/(8)`
Use `delta = (mu - 1)A` to arrive at the result.