Question

A prism is made of glass of unknown refractive index. A parallel beam of light is incident on the 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.

Answer 1

Angle of minimum deviation, δ_m = 40°

Angle of the prism, A = 60°

Refractive index of water, µ = 1.33

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 sin  "A"/2`

`sin  (("A" + δ_"m"^"'"))/2 = 1.532/1.33 xx sin  (60°)/2 = 0.5759`

` (("A" + δ_"m"^"'"))/2 = sin^-1 0.5759` = 35.16°

60° + `δ_"m"^"'"` = 70.32°

∴ `δ_"m"^"'"` = 70.32° − 60° = 10.32°

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