Question:
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 degree . What is the refractive index of thematerial of the prism? The refracting angle of the prism is 60 degree . 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:
Angle of minimum deviation (δ) = 40°
Angle of the prism (A) = 60°
Refractive index of water (µ)= 1.33
Refractive index of the material of the prism = µ1
The refractive index
µ1 = sin (A + δ)/2/ sin A/2
= sin (60° + 40°)/2/ sin 60°/2
= sin 50°/sin 30°
= 1.532
Since the prism is placed in water, let δ1 be the new angle of minimum deviation for the same prism.
The refractive index of glass with respect to water is
δg = µ1 / µ = (A + δ1 )/2/ sinA/2
sin (A + δ1 )/2 = µ1 x sinA/2/ µ
= 1.532 x sin 60°/2/ 1.33
sin (A + δ1 )/2 = sin-1 0.57 = 35.16°
60° + δ1 = 70.32°
δ1 = 70.32° - 60° = 10.32°
Hence, the new minimum angle of deviation is 10.32°.
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