question_answer

The x-y plane is the boundary between two transparent media. Medium-1 with \[z\ge 0\] has refractive index \[\sqrt{2}\] and medium-2 with \[z\le 0\]has a refractive index \[\sqrt{3}\] . A ray of light in medium given by vector \[\vec{A}=\sqrt{3}\,\,\hat{i}-\hat{k}\]is incident on the plane of separation. The unit vector in the direction of the refracted ray in medium-2 is

A) \[\frac{1}{\sqrt{2}}(\hat{k}-\hat{i})\]      

B)        \[\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})\]

C) \[\frac{1}{\sqrt{2}}(\hat{i}-\hat{k})\]      

D)        \[\frac{1}{\sqrt{2}}(\hat{i}+\hat{k})\]

Correct Answer: C

Solution :

[c] Given \[\vec{A}=\sqrt{3}\,\,\hat{i}-\hat{k}\] i.e.      \[\tan {{i}_{1}}=\frac{\sqrt{3}}{1}=\sqrt{3}\] or       \[{{i}_{1}}=60{}^\circ \] Snell?s law gives             \[{{\mu }_{1}}\sin {{i}_{1}}={{\mu }_{2}}\sin {{i}_{2}}\] or         \[\sin {{i}_{2}}=\frac{{{\mu }_{1}}}{{{\mu }_{2}}}\sin {{i}_{1}}=\frac{\sqrt{2}}{\sqrt{3}}\sin 60{}^\circ \] or         \[\sin {{i}_{2}}=\frac{1}{\sqrt{2}}\] or         \[{{i}_{2}}=45{}^\circ \] \[\therefore \]  The unit vector in the direction of the refracted ray will be \[\hat{i}=\frac{1}{\sqrt{2}}(\hat{i}-\hat{k})\] and \[\hat{n}=\frac{1}{\sqrt{2}}\left( \hat{i}-\hat{k} \right)\]