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

Angle of minimum deviation, `(delta)_(m)=40^(@)`
Angle of the prism, A=`60^(@)`
Refractive index of water, `mu=1.33`
Refractive index of the material of the prism = `mu^(')`
The angle of deviation is related to refractive index `mu^(')` as:
`mu^(') = (sin(A+delta_(m))/(2))/(sinA/2) = (sin(60^(@)+40^(@))/(2))/(sin60^(@)/2)` = `(sin50^(@))/(sin 30^(@))` = 1.532
Hence, the refractive index of the material of the prism is 1.532
Since, the prism is placed in water, let `delta_(m)` be the new angle of minimum deviatin for the same prism.
The refractive index of glass with respect to water to water is given by the relation:
`mu_(g)^(w)=mu^(')/mu=(sin(A + delta_(m)^('))/(2))/(sinA/2)`
`sin(A+delta_(m)^('))/2 = mu^(')/musin A/2`
`sin(A+delta_(m)^('))/2 = 1.52/1.33 xx sin 60^(@)/2 = 0.5759`
`(A+delta_(m)^('))/2= sin^(-1) 0.5759 = i35.16^(@)`
`60^(@) + delta_(m)^(') = 70.32^(@)`.
`therefore delta_(m)^(') = 70.32^(@)-60^(@) = 10.32^(@)`.
Hence, the new minimum angle of deviation is `10.32^(@)`.