KCET · Physics · Wave Optics
Critical angle for certain medium is \(\sin ^{-1}(0.6)\). The polarizing angle of that medium is
- A \(\tan ^{-1}\) [1.5]
- B \(\sin ^{-1}[0.8]\)
- C \(\tan ^{-1}\) [1.6667]
- D \(\tan ^{-1}\) [0.6667]
Answer & Solution
Correct Answer
(C) \(\tan ^{-1}\) [1.6667]
Step-by-step Solution
Detailed explanation
Critical angle, \(C=\sin ^{-1}(0.6)\)
\[
\begin{aligned} \sin (C) &=0.6 \\ \mu &=\frac{1}{\sin C}=\frac{1}{0.6} \\ \text { Polarising angle } i_{p} &=\tan ^{-1}(\mu)=\tan ^{-1}\left(\frac{1}{0.6}\right) \\ &=\tan ^{-1}(1.6667) \end{aligned}
\]
\[
\begin{aligned} \sin (C) &=0.6 \\ \mu &=\frac{1}{\sin C}=\frac{1}{0.6} \\ \text { Polarising angle } i_{p} &=\tan ^{-1}(\mu)=\tan ^{-1}\left(\frac{1}{0.6}\right) \\ &=\tan ^{-1}(1.6667) \end{aligned}
\]
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