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

Here, `delta_(m) = 40^(@), mu = ?, A = 60^(@)`
As `mu = (sin (A + delta_(m))//2)/(sin A//2)`
`:. mu = (sin(60^(@) + 40^(@))//2)/(sin 60^(@)//2) = (sin 50^(@))/(sin 30^(@)) = (0.7660)/(0.5000)`
`:. mu = 1.532`
When prism is immersed in water, we have to take refractive index of glass w.r.t. water :
`.^(w)mu_(g) = (.^(a)mu_(g))/(.^(a)mu_(w)) = (1.532)/(1.33) = 1.149`
if `delta'_(m)` is angle of minimum deviation in
water, then `.^(w)mu_(g) = (sin(A + delta'_(m))//2)/(sin A//2)`
`sin(A + delta'_(m))//2 = .^(w)mu_(g) xx sin A//2`
`= 1.149 xx sin 30^(@) = (1.149)/(2) = 0.5745`
`(A + delta'_(m))/(2) = sin^(-1)(0.5745) = 35^(@)4'`
` A + delta'_(m) = 70^(@)8'`
`delta'_(m) = 70^(@)8' - A = 70^(@)8' - 60^(@) = 10^(@)8'`