A) \[{{\sin }^{-1}}(\tan r')\]
B) \[{{\sin }^{-1}}(\tan r)\]
C) \[{{\tan }^{-1}}(\sin i)\]
D) \[{{\cot }^{-1}}(\tan i)\]
Correct Answer: D
Solution :
We know that, \[\frac{I}{\mu }=\frac{\sin i}{\sin r'}\] But \[r+r'={{90}^{\circ }}\]\[r'={{90}^{\circ }}-r\] \[\Rightarrow \] \[\sin r'=\sin ({{90}^{\circ }}-r)\] \[\sin r'\cos r\Rightarrow \sin r'\cos i\] \[\frac{1}{\mu }=\frac{\sin i}{\cos i}\Rightarrow \frac{1}{\mu }=\tan i\Rightarrow \sin {{i}_{C}}=\tan i=\tan r\] The critical angle for the pair of media \[{{i}_{C}}={{\sin }^{-1}}(\tan r)\]
You need to login to perform this action.
You will be redirected in
3 sec
