question_answer

    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)  \[{{\cos }^{-1}}\left( \frac{\mu }{2} \right)\]

Correct Answer: A

Solution :

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