A monochromatic light beam travels through two different...

Let $\lambda_A and$\lambda_B$be the wavelengths of light in medium A and medium B respectively.<br><br>The number of waves in a given distance is the distance divided by the wavelength.<br><br>For medium A:<br>$n_A = \dfrac{y}{\lambda_A} \Rightarrow \lambda_A = \dfrac{y}{n_A}$<br><br>For medium B:<br>$n_B = \dfrac{3y}{\lambda_B} \Rightarrow \lambda_B = \dfrac{3y}{n_B}$<br><br>The refractive index of a medium is inversely proportional to the wavelength of light in it. Thus, the relative refractive index of medium B with respect to medium A is:<br>$\dfrac{\mu_B}{\mu_A} = \dfrac{\lambda_A}{\lambda_B}$<br><br>Substituting the values of$\lambda_A$and$\lambda_B$:<br>$\dfrac{\mu_B}{\mu_A} = \dfrac{\dfrac{y}{n_A}}{\dfrac{3y}{n_B}} = \dfrac{n_B}{3n_A}$<br><br>The critical angle$\theta_c$for total internal reflection at the interface of a denser medium A and a rarer medium B is given by:<br>$\sin \theta_c = \dfrac{\mu_B}{\mu_A}$<br><br>$\sin \theta_c = \dfrac{n_B}{3n_A}$<br><br>$\theta_c = \sin^{-1}\left(\dfrac{n_B}{3n_A}\right)$<br><br>Answer:$\sin^{-1}\left(\dfrac{n_B}{3n_A}\right)