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) =(sin(60^@ + 40^@)//2)/(sin 60^@//2) :. mu = (sin 50^@)/(sin 30^@) = (0.766)/(1//2) = 1.532`
when prism is placed in water, `.^w mu_g = (sin(A + delta'_m)//2)/(sin A//2)`
`sin (A + delta'_m)//2 = 1^w mu_g xx "sin" (A)/(2) = (.^a mu_g)/(.^a mu_w) sin 60^2//2 = (1.532)/(1.33) xx sin 30^@ = (1.532)/(1.33) = (1.532)/(1.33) xx (1)/(2) = 0.5759 = sin 35^@ 10'`
`:. (A + delta'_m)/(2)= 35^@ 10'`
`A + delta'_m = 70^@ 20'`
`delta'_m = 70^@ 20' - A = 70^@ 20' - 60^@ = 10^@ 20'`.