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.

Answer

`10^(@)12'`

Explanation

As `mu = (sin (A + delta_(m))//2)/(sin A//2)`
`:. "sin" (60^(@) + delta_(m))/(2) = mu xx "sin"(60^(@))/(2)`
`=(1.53)/(1.33) xx (1)/(2) = 0.5751 = sin 35^(@)6'`
`:. (60 + delta_(m))/(2) = 35^(@)6'`,
`delta_(m) = 10^(@)12'`