Question

# The reflected and refracted rays are observed to be perpendicular to each other, when ray of light is incident at an angle of $$60^{\circ}$$ on a transparent block. The refractive index of that block is:

A
32
B
12
C
23
D
3

Solution

## The correct option is B $$\sqrt{3}$$So, by Snell's Law$$\dfrac{\mu_{1}}{\mu_{2}} = \dfrac{\sin \theta_{2}}{\sin \theta_{1}}$$$$=\dfrac{\sin 30^{\circ}}{\sin60^{\circ}}$$$$=\dfrac{1/2}{\sqrt{3}/2}$$                       $$\left ( \sin 60^{\circ} = \sqrt{3}/2;\; \sin 30^{\circ} = 1/2 \right )$$$$=\dfrac{1}{\sqrt{3}}$$So, $$\dfrac{\mu_{2}}{\mu_{1}} = \sqrt{3}$$or $$\mu_{2} = \sqrt{3} \left ( \mu_{1} \right )$$ $$=\sqrt{3} \left ( 1 \right )$$               $$\left (\because \mu_{1} = 1 \right )$$$$= \sqrt{3}$$Physics

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