A prism of refractive index `1.53` is placed in water of refractive index `1.33`. If the angle of prism is `60^@`, calculate the angle of minimum deviation in water.

Explanation

Here, `.^(a)mu_(g) = 1.53`,
`.^(a)mu_(w) = 1.33, A = 60^@, delta_m = ?`
`.^(w)mu_(g) = (.^(a)mu_(g))/(.^(a)mu_(w)) = (1.53)/(1.33) = 1.15`
As `.^(w)mu_(g) = (sin (A + delta_m)//2)/(sin A//2)`
`:. (sin(A + delta_m))/(2) = .^(w)mu_(g) xx sin (A)/(2)`
=`1.15 sin (60^@)/(2) = 0.575`
`(A + delta_m)/(2) = sin^-1 (0.575) = 35.1^@`
`:. delta_m = 35.1 xx 2 - 60 = 10.2^@`.