A) \[{{\sin }^{-1}}(\tan i)\]
B) \[{{\tan }^{-1}}(\sin i)\]
C) \[{{\sin }^{-1}}(\cot i)\]
D) \[{{\cos }^{-1}}(\tan i)\]
Correct Answer: C
Solution :
From law of reflection,\[\angle i=\angle r\] ... (i) and \[\frac{\sin r'}{\sin i}=\frac{{{\mu }_{d}}}{{{\mu }_{r}}}\] ... (ii) From the figure \[r=r'+{{90}^{o}}={{180}^{o}}\] \[\Rightarrow \] \[r+r'={{90}^{o}}\] or \[i=r'={{90}^{o}}\] \[r'=({{90}^{o}}-i)\] ... (iii) From Eq.(i) \[\frac{\sin \,({{90}^{o}}-i)}{\sin i}=\frac{{{\mu }_{d}}}{{{\mu }_{r}}}\] or \[\frac{\cos i}{\sin i}=\frac{{{\mu }_{d}}}{{{\mu }_{r}}}\,\,\,\,\Rightarrow \,\,\,\cot \,i=\frac{{{\mu }_{d}}}{{{\mu }_{r}}}\] But \[\frac{{{\mu }_{d}}}{{{\mu }_{r}}}=\sin C\] (where C is critical angle) \[\therefore \] \[\cot i=\sin C\,\,\,\Rightarrow \,\,\,C={{\sin }^{-1}}(\cot i)\]You need to login to perform this action.
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