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

Angle of minimum deviation, δ_m = 40° and angle of the prism, A = 60°

Refractive index of water, μ = 1.33 and refractive index of the material of the prism =μ’

The angle of deviation is related to refractive index (μ’) as:

μ’ = [sin (A+δ_m)/2]/sin A/2

= [sin (60°+40°)/2]/sin 60°/2

= [sin (50°)]/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 δ’_m be the new angle of minimum deviation for the same prism. The refractive index of glass with respect to water is given by the relation:

μ_g^w  =  μ’/μ = [sin (A+δ’_m)/2]/sin A/2

=>  [sin (A+δ’_m)/2] =    μ’/μ x  (sin A/2)

=>  [sin (A+δ’_m)/2] = 1.532/1.33 x sin 60°/2= 0.5759

=> (A+δ’_m)/2 = sin⁻¹0.5759=35.16°

60° +δ’_m = 70.32°

Therefore ,

δ’_m = 70.32° –  60°

= 10.32°

Hence, the new minimum angle of deviation is 10.32°.