A) \[ta{{n}^{-1}}[1.5]\]
B) \[{{\sin }^{-1}}[0.8]\]
C) \[{{\tan }^{-1}}[1.6667]\]
D) \[{{\tan }^{-1}}[0.6667]\]
Correct Answer: C
Solution :
Critical angle, \[C={{\sin }^{-1}}(0.6)\] \[\sin (C)=0.6\] \[\mu =\frac{1}{\sin C}=\frac{1}{0.6}\] Polarising angle \[{{i}_{p}}={{\tan }^{-1}}(\mu )={{\tan }^{-1}}\left( \frac{1}{0.6} \right)\] \[={{\tan }^{-1}}(1.6667)\]
You need to login to perform this action.
You will be redirected in
3 sec
