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

Given that, `.^(a)mu_(g) = 1.53, .^(a)mu_(w) = (4)/(3), A = 60^(@)`
`delta_(m) = ?`
`.^(w)mu_(g) = (.^(a)mu_(g))/(.^(a)mu_(w)) = (1.53)/(1.33) = 1.15`
`.^(w)mu_(g) = (sin.((A+delta_(m)))/(2))/(sin.(A)/(2)) = sin.((A+delta_(m)))/(2) = ._(w)mu_(g) 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) + A = 35.1^(@) xx 2 = 70.2^(@)`
`delta_(m) = 70.2^(@) - A = 70.2^(@) - 60^(@) = 10.2^(@)`