COMEDK 2025 Morning Solved Paper

Concept:Total internal reflection occurs when light moves from a denser to a rarer medium and the angle of incidence exceeds the critical angle.Explanation:First, calculate the speed of light in the given medium: $v = \frac{3\ \text{cm}}{0.2\ \text{ns}} = \frac{0.03\ \text{m}}{2\times10^{-10}\ \text{s}} = 1.5\times10^8\ \text{m/s}$.The refractive index of the medium is $n = \frac{c}{v} = \frac{3\times10^8\ \text{m/s}}{1.5\times10^8\ \text{m/s}} = 2$.For a medium–air interface ($n_{\text{air}} \approx 1$), the critical angle $\theta_c$ satisfies $\sin\theta_c = \frac{n_{\text{air}}}{n_{\text{med}}} = \frac{1}{2}$; therefore $\theta_c = 30^\circ$.Now compare the angles of incidence: Ray A is $15^\circ$ (less than $30^\circ$), so no total internal reflection.Ray B is $42^\circ$ (greater than $30^\circ$), so total internal reflection does occur for this ray.Answer:Therefore, total internal reflection happens only for ray B. The correct option is C: Yes, B.