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Step 1: Calculate the Refractive Indices The refractive index $n_1$ of medium $M_1$ is given by $n_1 = \frac{c}{v_1}$, where $c = 3 \times 10^8 \, \text{m/s}$ is the speed of light in a vacuum and $v_1 = 1.5 \times 10^8 \, \text{m/s}$ is the speed of light in medium $M_1$. $n_1 = \frac{3 \times 10^8}{1.5 \times 10^8} = 2$ Similarly, the refractive index $n_2$ of medium $M_2$ is given by $n_2 = \frac{c}{v_2}$, where $v_2 = 2 \times 10^8 \, \text{m/s}$. $n_2 = \frac{3 \times 10^8}{2 \times 10^8} = 1.5$ Step 2: Use the Condition for Total Internal Reflection Total internal reflection occurs when the angle of incidence $\theta$ is greater than the critical angle $\theta_c$. The critical angle is given by $\theta_c = \sin^{-1}\left(\frac{n_2}{n_1}\right)$. $\theta_c = \sin^{-1}\left(\frac{1.5}{2}\right) = \sin^{-1}\left(\frac{3}{4}\right)$ Step 3: Determine the Angle of Incidence For total internal reflection, $\theta > \theta_c$. Therefore, $\theta > \sin^{-1}\left(\frac{3}{4}\right)$. Hence, the correct answer is option A.

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