Explanation:
C: Given,
\(\begin{aligned}
\lambda_x=3500 \text { वा } \\
\lambda_y=7000 \text { 四 }
\end{aligned}\)
Critical angle, \(\theta_{\mathrm{C}} \propto\) wavelength \((\lambda)\)
\(\frac{\left(\theta_{\mathrm{c}}\right)_{\mathrm{x}}}{\left(\theta_{\mathrm{c}}\right)_{\mathrm{y}}}=\frac{\lambda_{\mathrm{x}}}{\lambda_{\mathrm{y}}}=\frac{3500 \AA}{7000 \AA}=\frac{1}{2}\)
Also, \(\mu=\frac{1}{\lambda}\)
Then, \(\frac{\mu_{\mathrm{x}}}{\mu_{\mathrm{y}}}=\frac{\lambda_{\mathrm{y}}}{\lambda_{\mathrm{x}}}=\frac{7000}{3500}=\frac{2}{1}\)
We know that,
\(\begin{aligned}
\theta_C=\sin ^{-1}\left(\frac{1}{{ }_y \mu_x}\right) \\
\theta_C=\sin ^{-1}\left(\frac{\mu_y}{\mu_x}\right) \\
\theta_C=\sin ^{-1}\left(\frac{1}{2}\right) \\
\theta_C=30^{\circ}
\end{aligned}\)