Question

A ray of light passes from vacuum into a medium of refractive index u., the angle of incidence is found to be twice the angle of refraction. Then the angle of incidence is

• A. $ 2{{\cos }^{-1}}\left( \frac{\mu }{2} \right) $
• B. $ {{\sin }^{-1}}\left( \mu \right) $
• C. $ {{\sin }^{-1}}\left( \frac{\mu }{2} \right) $
• D. $ {{\cos }^{-1}}\left( \frac{\mu }{2} \right) $

Answer 1

Answer: (A)

The refractive index $ (\mu ) $ of a material is the factor by which the phase velocity of electromagnetic radiation is slowed down in that material.
From Snells law $ \mu =\frac{\sin i}{\sin r} $ Given, $ i=2r $ $ \therefore $ $ \mu =\frac{\sin 2r}{\sin r} $ Using $ \sin 2\theta =2\sin \theta \cos \theta $ $ \therefore $ $ \mu =\frac{2\sin r\cos r}{\sin r}=2\cos r $ $ \Rightarrow $ $ r={{\cos }^{-1}}\left( \frac{\mu }{2} \right) $ Hence, angle of incidence is $ i=2{{\cos }^{-1}}\left( \frac{\mu }{2} \right) $