A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be `40^(@)`. What is the refractive index of the material of the prism? The refracting angle of the prism is `606^(@)`. If the prism is placed in water (refractive idex 1.33), predict the new angle of minimum deviation of a parallel beam of light.

Explanation

`mu=sin((A+delta_(m))/(2))/(sin(A//2))`
`A=606^(@), delta_(m)=40^(@)`
`mu=(sin 50^(@))/(sin 30^(@))`
`=2xx0.766=1.532~~1.53.`
Now, `w_(mug)=(alpha_(mug))/(alpha_(muw))=(1.53)/(1.33)=1.15`
`1.15=(sin((60^(@)+delta_(m))/(2)))/(sin 30^(@))`
`sin((60^(@)+delta_(m))/(2))=1.15xxsin 30^(@)`
`=0.575`
`(60^(@)+delta_(m))/(2)=35^(@)6'`
`60^(@)+delta_(m)=70^(@)12' or delta_(m)=10^(@)12'`