A) \[ta{{n}^{-1}}\left[ 1.5 \right]\]
B) \[si{{n}^{-1}}\left[ 0.8 \right]\]
C) \[ta{{n}^{-1}}\left[ 1.6667 \right]\]
D) \[ta{{n}^{-1}}\left[ 0.6667 \right]\]
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 \[{{i}_{p}}={{\tan }^{-1}}(\mu )=ta{{n}^{-1}}\left( \frac{1}{0.6} \right)\] \[={{\tan }^{-1}}(1.6667)\]
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