A monochromatic ray of light has a wavelength of 480 nm in a...

The refractive index of a medium is inversely proportional to the wavelength of light in it, $n \propto \dfrac{1}{\lambda}$.

The refractive indices of medium $X and medium$Y$are$n_X = \dfrac{c}{f \lambda_X}$and$n_Y = \dfrac{c}{f \lambda_Y}$.<br><br>The critical angle$\theta_c$for total internal reflection at the interface when light travels from a denser medium$X$to a rarer medium$Y$is given by$\sin \theta_c = \dfrac{n_Y}{n_X}$.<br><br>Substituting the expressions for refractive indices, we get$\sin \theta_c = \dfrac{\lambda_X}{\lambda_Y}$.<br><br>Given$\lambda_X = 480 \text{ nm}$and$\lambda_Y = 600 \text{ nm}$, we have$\sin \theta_c = \dfrac{480}{600} = \dfrac{4}{5}$.<br><br>Therefore,$\theta_c = \sin^{-1}\left(\dfrac{4}{5}\right)$.<br><br>Answer:$\sin^{-1}\left(\dfrac{4}{5}\right)