### JEE Main On 8 April 2017 Question 24

##### Question: Let the refractive index of a denser medium with respect to a rarer medium be $ n _{12} $ and its c 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{}^\circ $ . Angle A given by - [JEE Online 08-04-2017]

#### Options:

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

B) $ \frac{1}{{{\tan }^{-1}}(\sin {\theta_C})} $

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

D) $ \frac{1}{{{\cos }^{-1}}(\sin {\theta_C})} $

## Show Answer

#### Answer:

Correct Answer: A

#### Solution:

- $ \mu =\frac{{\mu_R}}{{\mu_D}}=\frac{\sin i _{c}}{\sin 90^{o}} $ $ \frac{{\mu_R}}{{\mu_D}}=\sin i _{i} $ $ \mu =\frac{{\mu_R}}{{\mu_D}}=\frac{sinA}{sinr} $ $ =\frac{\sin A}{\sin (90-A)}=\frac{sinA}{sinA} $ $ \frac{{\mu_R}}{{\mu_D}}=\tan A $ $ \tan A=\sin {\theta_C} $ $ A={{\tan }^{-1}}(sin{\theta_C}) $