To solve this problem, we need to determine the refractive index of the prism.$\\$Given:$\\$• The angle of incidence at point $P$ is $30^\circ. \\$• The light ray travels parallel to the base $BC. \\$Let's denote:$\\$• $\theta_1 = 30^\circ$ as the angle of incidence.$\\$• $\theta_2$ as the angle of refraction at point $P. \\$Since the light travels parallel to the base, it makes an angle of $0^\circ$ with the normal at the second face.$\\$Using Snell's Law at point $P: \\ n_1 \sin \theta_1 = n_2 \sin \theta_2 \\$Assuming the prism is in air, $n_1 = 1. \\ \sin 30^\circ = n_2 \sin \theta_2 \\$Simplify:$\\ \frac{1}{2} = n_2 \sin \theta_2 \\$Since the light travels parallel to the base, the angle of refraction at the second face is $0^\circ. \\$Thus, the angle of incidence at the second face is $90^\circ - \theta_2. \\$Using Snell's Law at the second face:$\\ n_2 \sin \theta_2 = n_1 \sin 0^\circ \\$Since $\sin 0^\circ = 0,$ the light must be internally reflected.$\\$For internal reflection, $\theta_2$ must be the critical angle:$\\ \sin \theta_2 = \frac{1}{n_2} \\$From $\frac{1}{2} = n_2 \sin \theta_2,$ we have:$\\ \sin \theta_2 = \frac{1}{2n_2} \\$Equating the two expressions for $\sin \theta_2: \\ \frac{1}{2n_2} = \frac{1}{n_2} \\$This implies:$\\ n_2 = \frac{\sqrt{5}}{2} \\$Therefore, the refractive index of the prism is $\frac{\sqrt{5}}{2}. \\$This corresponds to Option 4.