Light enters from air into a given medium at an angle of 45° with interface of the air-medium surface. After refraction, the light ray is deviated through an angle of 15° from its original direction.The refractive index of the medium is

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1.414

1.732

1.333

2.732

answer is C.

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Detailed Solution

$\begin{array}{l} i = 45^{\circ} \\ D = i - r \\ 15^{\circ} = 45 - r \Rightarrow r = 30^{\circ} \\ n_{1} \sin i = n_{2} \sin r \\ 1 \sin 45^{\circ} = μsin 30^{\circ} \\ μ = \sqrt{2} = 1 .414 \end{array}$

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