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JEE MAIN - Physics (2017 - 8th April Morning Slot - No. 2)

Let the refractive index of a denser medium with respect to a rarer medium be n12 and its critical angle be θC . At an angle of incidence A when light is travelling from denser medium to rarer medium, a part of the light is reflected and the rest is refracted and the angle between reflected and refracted rays is 90o. Angle A is given by :
$${1 \over {{{\cos }^{ - 1}}\left( {\sin {\theta _C}} \right)}}$$
$${1 \over {{{\tan }^{ - 1}}\left( {\sin {\theta _C}} \right)}}$$
$${\cos ^{ - 1}}\,\left( {\sin {\theta _C}} \right)$$
$${\tan ^{ - 1}}\,\left( {\sin {\theta _C}} \right)$$

Explanation

JEE Main 2017 (Online) 8th April Morning Slot Physics - Geometrical Optics Question 185 English Explanation

Refractive index,

n12 = $${{{n_D}} \over {{n_R}}}$$ = $${1 \over {\sin {\theta _c}}}$$

$$ \Rightarrow $$ $$\,\,\,$$ sin$$\theta $$c = $${{{n_R}} \over {{n_D}}}$$ . . . . . (1)

From Snell's law,

nD sinA = nR sin r

$${{{n_R}} \over {{n_D}}}$$ = $${{\sin A} \over {\sin \left( {{{90}^o} - A} \right)}}$$    [as    r = 90o $$-$$ A]

$$ \Rightarrow $$$$\,\,\,$$ $${{{n_R}} \over {{n_D}}} = \tan A$$

$$\therefore\,\,\,$$ From (1) we get,

tan A = sin $$\theta $$c

$$ \Rightarrow $$$$\,\,\,$$ A = tan$$-$$1 (sin $$\theta $$c)

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