A) \[{{\sin }^{-1}}\left( \frac{1}{\sqrt{\mu }} \right)\]
B) \[{{90}^{o}}\]
C) \[{{\sin }^{-1}}\left( \frac{1}{{{\mu }^{2}}} \right)\]
D) \[{{\sin }^{-1}}\left( \frac{1}{\mu } \right)\]
Correct Answer: C
Solution :
\[_{a}{{\mu }_{g}}=\frac{1}{\sin \theta }\]\[\Rightarrow \mu =\frac{1}{\sin \theta }\] .....(i) Now from Snell's law \[\mu =\frac{\sin i}{\sin r}\]\[=\frac{\sin \theta }{\sin r}\] \[\Rightarrow \sin r=\frac{\sin \theta }{\mu }\] .....(ii) From equation (i) and (ii) \[\sin r=\frac{1}{{{\mu }^{2}}}\]\[\Rightarrow r={{\sin }^{-1}}\left( \frac{1}{{{\mu }^{2}}} \right)\]
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