A prism of refractive index 1.53 is place in water of refractive index `(4)/(3)`. If the angle of prism is `60^(@)`, calculate the angle of minimum deviation in water.

Explanation

Here `.^(a)mu_(g) = 1.33, .^(a)mu_(w) = 1.53, A=60^(@), delta_(m) = ? .^(w)mu_(g)= (.^(a)mu_(g))/(.^(a)mu_(w)) = (1.53)/(1.33) = 1.15 because .^(w)mu_(g) = (sin(A + delta_(m))/(2))/(sin(A)/(2))`
`therefore (sin(A+delta_(m)))/(2) = .^(w)mu_(g)xx "sin"(A)/(2) = 1.15"sin"(60)/(2) = 0.575 implies (A+delta_(m))/(2) = sin^(-1)(0.575) = 35.1^(@)`
`therefore delta_(m) = 35.1 xx 2-60 = 10.2 ^(@)`