A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. By rotating the prism, the minimum angle of deviation is measured to be `40^@`. What is the refractive index of the prism ? If the prism is placed in water `(mu = 1.33)`, predict the new angle of minimum deviation of the parallel beam. The refracting angle of prism is `60^@`.

Explanation

`mu=sin((A+delta_(m))/2)/sin(A/2)=sin((60^(@)+40^(@))/2)/sin(60^(@)/2)`
`(sin50^(@))/(sin30^(@))=0.7660/0.50=1.532~1.53`
Let `delta'_(m)` be the new angle of minimum deviation of a parallel beam of light when the prism is placed in water `""_(a)mu_(w)=1.33,""_(a)mu_(g)=1.532`
`therefore ""_(w)mu_(g)=(""_(a)mu_(g))/(""_(a)mu_(w))=1.532/1.33=1.152`
`""_(w)mu_(g)=sin((A+delta'_(m))/2)/sin(A/2)`
or `1.152=sin((60+delta'_(m))/2)/sin(60/2)`
or `sin(30^(@)+(delta'_(m))/2)=1.152xx1/2=0.5759=sin35.17^(@)`
or `30^(@)+(delta'_(m))/2=35.17^(@) or (delta'_(m))/2=35.17^(@)-30-5.17^(@)`
Angle of minimum deviation, `delta'_(m)=2xx5.17^(@)=10.34^(@)=10^(@)20'`