Q. The $x - y$ plane is the boundary between two transparent media. A medium I has a refractive index $\mu_{1} = \sqrt{2}$ and medium II has a refractive index $\mu_{2} = \sqrt{3}$. A ray of light in medium I, given by vector, $\overrightarrow{A} = \sqrt{3}\widehat{i} - \widehat{k}$ is incident on the plane of separation. The unit vector in the direction of the refracted ray in medium II is
A) $\frac{1}{\sqrt{2}}(\widehat{i} + \widehat{k})$
B) $\frac{1}{\sqrt{2}}(\widehat{i} + \widehat{j})$
C) $\frac{1}{\sqrt{2}}(\widehat{k} - \widehat{i})$
D) $\frac{1}{\sqrt{2}}(\widehat{i} - \widehat{k})$
Explanation
(d) From figure here $\overrightarrow{A} = \sqrt{3}\widehat{i} - \widehat{k}$
so tan $i_{1} = \frac{\sqrt{3}}{1} = \sqrt{3} = tan60{^\circ}$
$i_{1} = 60{^\circ}$
Using snell’s law $\mu_{1}$ sin $i_{1} = \mu_{2}$ sin $r$
or sin$\ r = \frac{\mu_{1}}{\mu_{2}}$ sin $i_{1} = \frac{\sqrt{2}}{\sqrt{3}}$ sin 60${^\circ} = \frac{1}{\sqrt{2}} = \ $sin45${^\circ}$
or $r =$45${^\circ}$
The unit vector in the direction of the refracted ray will be
$\widehat{n} = 1$ sin45${^\circ}\widehat{i} - 1$cos$45{^\circ}\widehat{k} = \frac{1}{\sqrt{2}}(\widehat{i} - \widehat{k})$
[image:4_image97.emf]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.