question_answer

A ray of light enters from a rarer to a denser medium. The angle of incidence is \[i\]. Then the reflected and refracted rays are mutually perpendicular to each other. The critical angle for the pair of media is

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) \[\therefore \]  \[\cot i=\sin C\,\,\,\Rightarrow \,\,\,C={{\sin }^{-1}}(\cot i)\]