A) \[{{\sin }^{-1}}(\tan r)\]
B) \[{{\sin }^{-1}}(\tan r)\]
C) \[{{\tan }^{-1}}(\sin i)\]
D) \[{{\cot }^{-1}}(\tan i)\]
Correct Answer: B
Solution :
We know that, \[\frac{I}{\mu }=\frac{\sin i}{\sin r}\] But \[r+r={{90}^{o}}\] \[r={{90}^{o}}-r\] \[\Rightarrow \] \[\sin r=\sin \,({{90}^{o}}-r=\cos i\] \[\frac{1}{\mu }=\frac{\sin i}{\cos i}\Rightarrow \frac{1}{\mu }=\tan i\Rightarrow \sin {{i}_{C}}=\tan i=\tan r\] The critical angle for the pair of media \[{{i}_{C}}={{\sin }^{-1}}(\tan r)\]
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