JEE MAIN - Physics (2022 - 29th July Evening Shift - No. 18)
Light enters from air into a given medium at an angle of $$45^{\circ}$$ with interface of the air-medium surface. After refraction, the light ray is deviated through an angle of
$$15^{\circ}$$ from its original direction. The refractive index of the medium is:
1.732
1.333
1.414
2.732
Explanation
Let, refractive index of medium = $$\mu$$
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$$\therefore$$ $$r + 15^\circ = 45^\circ $$
$$ \Rightarrow r = 30^\circ $$
Using Snell's law,
$$1\,.\,\sin 45^\circ = \sin 30^\circ \times \mu $$
$$ \Rightarrow {1 \over {\sqrt 2 }} = {1 \over 2} \times \mu $$
$$ \Rightarrow \mu = {2 \over {\sqrt 2 }} = \sqrt 2 = 1.414$$
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