Solveeit Logo
Login

Question

Physics Question on Wave optics

Critical angle for certain medium is Sin−1Sin ^{-1} (0.6)(0.6). The polarizing angle of that medium is .......

A

Tan−1[1.6667]Tan ^{-1} [1.6667]

B

Tan−1[0.6667]Tan ^{-1} [0.6667]

C

Tan−1[1.5]Tan ^{-1} [1.5]

D

Sin−1[0.8]Sin ^{-1} [0.8]

Answer

Tan−1[1.6667]Tan ^{-1} [1.6667]

Explanation

Solution

Critical angle, C=sin⁡−1(0.6)C=\sin ^{-1}(0.6) sin⁡(C)=0.6\sin (C)= 0.6 μ=1sin⁡C=10.6\mu= \frac{1}{\sin C}=\frac{1}{0.6} Polarising angle ip=tan⁡−1(μ)=tan⁡−1(10.6)i_{p}=\tan ^{-1}(\mu)=\tan ^{-1}\left(\frac{1}{0.6}\right) =tan⁡−1(1.6667)=\tan ^{-1}(1.6667)