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.
Given, Angle of minimum deviation , δm=40∘. Angle of the prism, A=60∘. Refractive index of water, μw=1.33. Let the refractive index of the material of the prism be μg . Using the formula, μg=Sin(2A)Sin2(A+δm) we have, μg=Sin(260)Sin2(60+40) μg=Sin30Sin50=1.532 Thus, the refractive index of the material of the prism is 1.532. Since the prism is placed in water, let δmw 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: wμg=Sin(2A)Sin2(A+δmw) wμg=μwμg=Sin(2A)Sin2(A+δmw) wμg=1.331.532=Sin(260)Sin2(60+δmw) 1.331.532×Sin(260)=Sin2(60+δmw) 0.575=Sin2(60+δmw) 0.575=Sin2(60+δmw) 2(60+δmw)=Sin−1(0.575) 60+δmw=35.16×2 δmw=70.33−60 δmw=10.33∘ Thus, new angle of minimum deviation is 10.33∘.
