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( \sin \theta _C )

1 \over \tan ^ - 1( \sin \theta _C )

Geometrical Optics PYQ — JEE Main 2017 · 8 Apr · Shift 1 · Q51 with Solution

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Correct answer: D

Given condition: reflected and refracted rays are perpendicular
When the angle between the reflected ray and refracted ray is $90^\circ$ , the incidence angle is the Brewster angle for the given pair of media.

So, $i+r=90^\circ$ where $i=A$ and $r$ is the angle of refraction.

Apply Snell’s law
Light goes from denser medium to rarer medium. If refractive index of denser medium with respect to rarer medium is $n_{12}$ , then $n_{12}=\frac{\sin r}{\sin i}$

Using $r=90^\circ-i$ , $n_{12}=\frac{\sin(90^\circ-i)}{\sin i}=\frac{\cos i}{\sin i}=\cot i$

Hence, $\tan i=\frac{1}{n_{12}}$

So, $A=i=\tan^{-1}\left(\frac{1}{n_{12}}\right)$

Use the critical angle relation
For critical angle $\theta_C$ when light goes from denser to rarer medium, $\sin\theta_C=\frac{1}{n_{12}}$

Substitute into the above expression: $A=\tan^{-1}(\sin\theta_C)$

Therefore, $\boxed{A=\tan^{-1}(\sin\theta_C)}$