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

`mu_(g)=1.53`
`mu_(w)=1.33`
`A=60^(@)`
`delta_(m)=?`
`""_(w)mu_(g)=(mu_(g))/(mu_(w))=(1.53)/(1.33)=(153)/(133)`
`""_(w)mu_(g)=(sin((A+delta_(m))/(2)))/(sin""(A)/(2))`
`(153)/(133)=(sin((60^(@)+delta_(m))/(2)))/(sin 30^(@))`
`(153)/(133)xx(1)/(2)=sin((60^(@)+delta_(m))/(2))`
`0.57=sin((60^(@)+delta_(m))/(2))`
`sin 34^(@)45.=sin((60^(@)+delta_(m))/(2))`
`34^(@)45.=30^(@)+(delta_(m))/(2)`
`4^(@)45.=(delta_(m))/(2)`
`9^(@)30.=delta_(m)`