To find the critical angle for the medium, we need to determine the refractive index of the medium relative to air.$\\$Given:$\\$• Light travels a distance $x$ in time $t_1$ in air.$\\$• Light travels a distance $10x$ in time $t_2$ in the denser medium.$\\$The speed of light in air $(c)$ is given by:$\\ c = \frac{x}{t_1} \\$The speed of light in the denser medium $(v)$ is given by:$\\ v = \frac{10x}{t_2} \\$The refractive index of the medium $(n)$ is the ratio of the speed of light in air to the speed in the medium:$\\ n = \frac{c}{v} = \frac{\frac{x}{t_1}}{\frac{10x}{t_2}} \\$Simplify the expression for $n: \\ n = \frac{x}{t_1} \cdot \frac{t_2}{10x} = \frac{t_2}{10t_1} \\$The critical angle $(\theta_c)$ is given by:$\\ \sin(\theta_c) = \frac{1}{n} \\$Substitute the expression for $n: \\ \sin(\theta_c) = \frac{1}{\frac{t_2}{10t_1}} = \frac{10t_1}{t_2} \\$Therefore, the critical angle is:$\\ \theta_c = \sin^{-1} \left(\frac{10t_1}{t_2} \right) \\$This corresponds to Option 4.