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 \[\hat{n}=\frac{1}{\sqrt{2}}\left( \hat{i}-\hat{k} \right)\]You need to login to perform this action.
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