JEE Main 2017 · 8 Apr · Shift 1 · Physics Q22

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PhysicsSingle correct+4 / -1Question 22 of 90

Let the refractive index of a denser medium with respect to a rarer medium be $n_{12}$ and its critical angle be $\theta_C$ . 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 90°. Angle A is given by :

\dfrac1{\cos^{-1}(\sin\theta_C)}

\dfrac1{\tan^{-1}(\sin\theta_C)}

\cos^{-1}(\sin\theta_C)

\tan^{-1}(\sin\theta_C)

Answer and worked solution

Official answer: D

Reflected and refracted rays at 90° is the Brewster condition. Going from denser to rarer, $\tan A=\dfrac1{n_{12}}=\sin\theta_C$ , so $A=\tan^{-1}(\sin\theta_C)$ .

Topics:Polarisation and Brewster's law Total internal reflection

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JEE Main 2017 · 8 Apr · Shift 1, Physics question 22