With respect to air, the critical angle in a medium for red light of wave length $\lambda_1$ is $\theta$.…
With respect to air, the critical angle in a medium for red light of wave length $\lambda_1$ is $\theta$. Other facts remaining same, critical angle for yellow light of wave length $\lambda_2$ will be
$\theta$
more than $\theta$
less than $\theta$
$\frac{\theta \lambda_1}{\lambda_2}$
Solution
Critical angle $=\sin ^{-1}\left(\frac{1}{\mathrm{n}}\right)=\sin ^{-1}\left(\frac{\mathrm{v}}{\mathrm{c}}\right)$
Critical angle $=\sin ^{-1}\left(\frac{\mathrm{f} \lambda}{\mathrm{c}}\right) \quad\left(\because \mathrm{v}=\frac{\mathrm{c}}{\mathrm{n}}\right)$
for red light, $\theta=\sin ^{-1}\left(\frac{f \lambda_1}{c}\right)(\because v=f \lambda)$
for yellow light $\theta^{\prime}=\sin ^{-1}\left(\frac{\mathrm{f} \lambda_2}{\mathrm{c}}\right)$
Since $\lambda_1>\lambda_2$
then $\theta^{\prime} < \theta$