NEET 2020 Physics Question 140 Solution

The ray emerges normally from the second surface of the prism. This means the ray is perpendicular to the second interface ($$\theta = 90^\circ$$ to the surface, hence along the normal): $$r_2 = 0^\circ \quad \text{and} \quad e = 0^\circ$$

For any triangular prism, geometry dictates that the prism angle $$A$$ equals the sum of the two internal refraction angles: $$r_1 + r_2 = A$$ Since $$r_2 = 0$$: $$r_1 + 0 = A \implies r_1 = A$$

At the first refracting surface, light passes from air ($$\mu_1 = 1$$) into the prism medium ($$\mu_2 = \mu$$) at incident angle $$i$$: $$\sin i = \mu \sin r_1$$ Substituting $$r_1 = A$$: $$\sin i = \mu \sin A$$

Because the question explicitly specifies a "small angle prism" (typically $$A \le 10^\circ$$), we utilize the standard paraxial small-angle trigonometric approximations: $$\sin i \approx i \quad \text{and} \quad \sin A \approx A$$ Substituting these approximations yields: $$i \approx \mu A$$ This directly gives Option (3).