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 $i=2{\cos }^{-1}\left(\frac{\mu }{2}\right)$