A) \[{{\tan }^{-1}}\left( \frac{1}{{{\mu }^{2}}-1} \right)\]
B) \[{{\tan }^{-1}}({{\mu }^{2}}-1)\]
C) \[{{\tan }^{-1}}\sqrt{{{\mu }^{2}}-1}\]
D) \[{{\tan }^{-1}}\left( \frac{1}{\sqrt{{{\mu }^{2}}-1}} \right)\]
Correct Answer: D
Solution :
There is no deviation at P because ray is incident perpendicular to surface AB. Now at Q, By Snell?s Law \[{{\mu }_{1}}\sin {{i}_{1}}\,={{\mu }_{2}}\sin {{i}_{2}}\] \[\mu \,\times \sin A=1\times \sin \,{{90}^{0}}\] \[\sin A=\frac{1}{\mu },\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\tan A=\frac{1}{\sqrt{{{\mu }^{2}}-1}}\] \[A={{\tan }^{-1}}\left[ \frac{1}{\sqrt{{{\mu }^{2}}-1}} \right]\]You need to login to perform this action.
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