To solve this problem, we need to understand the concept of total internal reflection and the critical angle.$\\$• Total internal reflection occurs when light travels from a medium with a higher refractive index to a medium with a lower refractive index, and the angle of incidence is greater than the critical angle.$\\$• The critical angle is defined as the angle of incidence for which the angle of refraction is $90^\circ. \\$Let's denote:$\\$• $n_1$ as the refractive index of the first medium (higher refractive index).$\\$• $n_2$ as the refractive index of the second medium (lower refractive index).$\\$According to Snell's Law:$\\ n_1 \sin(\theta_c) = n_2 \sin(90^\circ) \\$Since $\sin(90^\circ) = 1,$ the equation simplifies to:$\\ n_1 \sin(\theta_c) = n_2 \\$Solving for the critical angle $\theta_c: \\ \sin(\theta_c) = \frac{n_2}{n_1} \\$When the angle of incidence is equal to the critical angle, the angle of refraction is $90^\circ. \\$Therefore, the correct answer is Option 4: $90^\circ.$