The angle of minimum deviation produced by a `60^(@)` prism is `40^(@)`. Calculate the refractive index of the material of the prism.

Explanation

`A=60^(@),delta=40^(@),mu=?`
The refractive index of the materail of the prism,
`mu=(sin[(A+delta)/2])/("sin"A/2)=(sin[(60^(@)+40^(@))/2])/("sin"[60^(@)/2])=(sin50^(@))/(sin30^(@))=(0.7660)/(1/2)=(0.7660)/1xx2`
= 1.532