KCET · Physics · Ray Optics
Wavelength of given light waves in air and in a medium are \(6000 Å\) and \(4000 Å\) respectively. The critical angle is
- A \(\tan ^{-1}\left(\frac{2}{3}\right)\)
- B \(\tan ^{-1}\left(\frac{3}{2}\right)\)
- C \(\sin ^{-1}\left(\frac{2}{3}\right)\)
- D \(\sin ^{-1}\left(\frac{3}{2}\right)\)
Answer & Solution
Correct Answer
(C) \(\sin ^{-1}\left(\frac{2}{3}\right)\)
Step-by-step Solution
Detailed explanation
\[
\begin{aligned}
\frac{1}{\sin C} &=\frac{\lambda_{1}}{\lambda_{2}} \\
\frac{1}{\sin C} &=\frac{6000}{3000} \\
\frac{1}{\sin C} &=\frac{6}{3}=\frac{3}{2} \\
\sin C &=\frac{2}{3} \\
C &=\sin ^{-1}\left(\frac{2}{3}\right)
\end{aligned}
\]
\begin{aligned}
\frac{1}{\sin C} &=\frac{\lambda_{1}}{\lambda_{2}} \\
\frac{1}{\sin C} &=\frac{6000}{3000} \\
\frac{1}{\sin C} &=\frac{6}{3}=\frac{3}{2} \\
\sin C &=\frac{2}{3} \\
C &=\sin ^{-1}\left(\frac{2}{3}\right)
\end{aligned}
\]
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