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) = 1.53 , A= 60^(@) , delta _(m ) = ?`
`.^(w )_(g ) = (.^(a)mu_(g ))/( .^(a)mu_(w)) = (1.53)/( 1.33) = 1.15 :. , .^(w ) mu _( g) = (sin .(A+ delta )/( 2 ))/( sin .(A)/(2))`
`:. , (sin (A+ delta ))/( 2 ) = .^(w)mu_(g ) xx sin (A)/(2) = 1.15 sin .(60^(@))/( 2)`
` = 0.575`
`(A+ delta )/( 2) = sin ^(-1) (0.575) = 35.1^(@)`
`:. , delta _( m) = 35.1 xx 2 - 60 -10-. 2^(@)`