To solve this problem, we need to find the critical angle for the ray of light moving from medium A to medium B.$\\ \\$The critical angle $\theta_c$ is given by the formula:$\\ \sin(\theta_c) = \frac{n_B}{n_A} \\$where $n_A$ and $n_B$ are the refractive indices of media A and B, respectively.$\\ \\$The refractive index $n$ of a medium is related to the speed of light $v$ in that medium by:$\\ n = \frac{c}{v} \\$where $c$ is the speed of light in vacuum, approximately $3.0 \times 10^8$ m/s.$\\ \\$Let's calculate the refractive indices for media A and B:$\\ n_A = \frac{3.0 \times 10^8}{1.5 \times 10^8} = 2.0 \\ n_B = \frac{3.0 \times 10^8}{2.0 \times 10^8} = 1.5 \\ \\$Now, substitute these values into the critical angle formula:$\\ \sin(\theta_c) = \frac{n_B}{n_A} = \frac{1.5}{2.0} = 0.75 \\ \\$Thus, the critical angle $\theta_c$ is:$\\ \theta_c = \sin^{-1}(0.75) \\ \\$Therefore, the correct option is Option 2: $\sin^{-1}(0.750)$.