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QJEEM17Ph2745: JEE Main 2017 Physics - Class 12 JEEM17Ph2745 Solution: Let the refractive index of a denser medium with respect to a…

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 :

Multiple Choice Options

  • A 1cos⁡−1(sin⁡θC){1 \over {{{\cos }^{ - 1}}\left( {\sin {\theta _C}} \right)}}
  • B 1tan⁡−1(sin⁡θC){1 \over {{{\tan }^{ - 1}}\left( {\sin {\theta _C}} \right)}}
  • C cos⁡−1 (sin⁡θC){\cos ^{ - 1}}\,\left( {\sin {\theta _C}} \right)
  • D tan⁡−1 (sin⁡θC){\tan ^{ - 1}}\,\left( {\sin {\theta _C}} \right)

Step-by-Step Verified Solution

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

Refractive index,

n12 = nDnR{{{n_D}} \over {{n_R}}} = 1sin⁡θc{1 \over {\sin {\theta _c}}}

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

From Snell's law,

nD sinA = nR sin r

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

⇒\Rightarrow    \,\,\, nRnD=tan⁡A{{{n_R}} \over {{n_D}}} = \tan A

∴   \therefore\,\,\, From (1) we get,

tan A = sin θ\theta c

⇒\Rightarrow    \,\,\, A = tan −- 1 (sin θ\theta c)

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