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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
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B
12
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C
23
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D
3
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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 $$\mu_{2} = \sqrt{3} \left ( \mu_{1} \right )$$

 $$=\sqrt{3} \left ( 1 \right )$$               $$\left (\because \mu_{1} = 1 \right )$$
$$= \sqrt{3}$$

54349_4992_ans.png

Physics

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