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'`.