A light ray incidents normally on one surface of an equilateral prism. The angle of deviation of the light ray is (refractive index of the material of the prism $=\sqrt{2}$ )

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Aicharya · Accepted Answer

Given, $i=0^{\circ}$ So, $ \begin{aligned} \frac{\sin i}{\sin r_1} & =\sqrt{2} \\ \frac{\sin 0^{\circ}}{\sin r_1} & =\sqrt{2} \\ \sin r_1 & =0 \text { or } r_1=0 \end{aligned} $ Now, $\frac{\sin r_2}{\sin e}=\frac{1}{\mu}$ $ \Rightarrow \quad \sin e=\sqrt{2} \sin r_2 $ But, $\quad r_1+r_2=60^{\circ}$ So, $ \begin{aligned} \sin e & =\sqrt{2} \sin 60^{\circ} \\ & =\sqrt{2} \times \sqrt{3} / 2>1 \end{aligned} $ So, light is incidenting at more than critical angle and totally internally reflected. $ \begin{aligned} \therefore \text { Deviation angle } & =(i+e)-\left(r_1+r_2\right) \\ & =0-60^{\circ}=60^{\circ}(\text { in magnitude }) \end{aligned} $

A light ray incidents normally on one surface of an equilateral prism. The angle of deviation of the light…

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