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°. What is the refractive index of material of 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

Answer on Question #52317-Physics-Optics

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 $40{}^{\circ}$. What is the refractive index of material of prism? The refracting angle of the prism is $60{}^{\circ}$. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.

Solution

Angle of minimum deviation is $\delta_{m} = 40{}^{\circ}$.

Angle of the prism is $A = 60{}^{\circ}$.

Refractive index of water is $\mu = 1.33$.

Refractive index of the material of the prism is $\mu'$.

The angle of deviation is related to refractive index as:

$$\mu' = \frac{\sin \frac{(A + \delta_m)}{2}}{\sin \frac{A}{2}} = \frac{\sin \frac{(60{}^{\circ} + 40{}^{\circ})}{2}}{\sin \frac{60{}^{\circ}}{2}} = \frac{\sin 50{}^{\circ}}{\sin 30{}^{\circ}} = 1.532.$$

Hence, the refractive index of the material of the prism is 1.532.

Since the prism is placed in water, let $\delta_{m}^{\prime}$ 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:

$$\mu_g^w = \frac{\mu'}{\mu} = \frac{\sin \frac{(A + \delta_m^{\prime})}{2}}{\sin \frac{A}{2}}.$$ $$\sin \frac{(A + \delta_m^{\prime})}{2} = \frac{\mu'}{\mu} \sin \frac{A}{2}.$$ $$\sin \frac{(A + \delta_m^{\prime})}{2} = \frac{1.532}{1.33} \sin \frac{60{}^{\circ}}{2} = 0.5759.$$ $$\frac{(A + \delta_m^{\prime})}{2} = \sin^{-1} 0.5759 = 35.16{}^{\circ}.$$ $$60{}^{\circ} + \delta_m^{\prime} = 70.32{}^{\circ}.$$ $$\delta_m^{\prime} = 10.32{}^{\circ}.$$

Hence, the new minimum angle of deviation is $10.32{}^{\circ}$.

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