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)  \[{{\sin }^{-1}}\left( \tan \,i \right)\]

B)  \[ta{{n}^{-1}}\left( sin\,i \right)\]

C)  \[si{{n}^{-1}}\left( \cot \,i \right)\]

D)  \[{{\cos }^{-1}}\left( \tan \,i \right)\]

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. (ii)\[\frac{\sin ({{90}^{o}}-i)}{\sin i}=\frac{{{\mu }_{d}}}{{{\mu }_{r}}}\] 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) \[\therefore \]\[\cot i=\sin C\Rightarrow C={{\sin }^{-1}}(coti)\]