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

Given, angle of minimum deviation, `delta_(m)=40^(@)`
Angle of prism, `A=60^(@)`
Refractive index of material of prism is given by
`mu=(sin((A+delta_(m))/(2)))/(sin((A)/(2)))`
`mu=(sin50^(@))/(sin30^(@))=2xx0.766=1.532~~1.53`
Now, when the prism is placed in water, refractive index is gievn by
`""_(w)mu_(g)=(""_(a)mu_(g))/(""_(a)mu_(w))=(1.53)/(1.33)=1.15`
Let `delta._(m)` be the angle of minimum deviation now. Then,
`therefore" "1.15=(sin((60^(@)+delta._(m))/(2)))/(sin30^(@))`
`sin((60^(@)+delta._(m))/(2))=1.15xxsin30^(@)=0.575`
`(60^(@)+delta._(m))/(2)=35^(@)6.`
`60^(@)+delta._(m)=70^(@)12. or delta._(m)=10^(@)12.`