Question

A ray of light is incident at angle 60° on one face of a prism which has an apex angle of 30° . The ray emerging out of the prism make an angel of 30° with the incident ray . The refractive index of the material of prism is
(1)√2 (2)√3. (3)1.5 (4) 1.6

Answer 1

Answer on Question #74434, Physics / Optics

Question. A ray of light is incident at angle $60{}^{\circ}$ on one face of a prism which has an apex angle of $30{}^{\circ}$. The ray emerging out of the prism make an angel of $30{}^{\circ}$ with the incident ray. The refractive index of the material of prism is

(1) $\sqrt{2}$;

(2) $\sqrt{3}$;

(3) 1.5;

(4) 1.6.

Given. $\theta = 60{}^{\circ}; A = 30{}^{\circ}; \delta = 30{}^{\circ}$.

Find. $n - ?$

Solution.

For a prism

$$\delta = \theta + \gamma - A,$$

where

$$\sin \gamma = n \cdot \sin \left(A - \arcsin \left(\frac {\sin \theta}{n}\right)\right)$$

So,

$$3 0 {}^ {\circ} = 6 0 {}^ {\circ} + \gamma - 3 0 {}^ {\circ} \rightarrow \gamma = 0.$$ $$\sin 0 {}^ {\circ} = n \cdot \sin \left(A - \arcsin \left(\frac {\sin 6 0 {}^ {\circ}}{n}\right)\right) \rightarrow 0 = \sin \left(A - \arcsin \left(\frac {\sin 6 0 {}^ {\circ}}{n}\right)\right)\rightarrow$$ $$A - \arcsin \left(\frac {\sin 6 0 {}^ {\circ}}{n}\right) = 0 \rightarrow \arcsin \left(\frac {\sin 6 0 {}^ {\circ}}{n}\right) = 3 0 {}^ {\circ} \rightarrow$$ $$\frac {\sin 6 0 {}^ {\circ}}{n} = \sin 3 0 {}^ {\circ} \rightarrow \frac {\sqrt {3}}{2 n} = \frac {1}{2} \rightarrow n = \sqrt {3}$$

Answer. $n = \sqrt{3}$

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