MHTCET 26 Apr 2025 Shift 1 Paper

Concept:A thin prism causes small deviations, so Snell’s law can be applied with the small-angle approximation $\sin\theta \approx \theta$ for both the refracted ray and the internally reflected ray.Explanation:The ray enters normally through the first face, so it undergoes no deviation at the first face.At the second face, the angle of incidence inside the prism is $A$.For the ray that emerges after refraction, the deviation is $\delta = 1.15^\circ$.Applying Snell’s law at the second face:$$\mu \sin A = \sin(A + \delta)$$So,$$\mu = \frac{\sin(A+\delta)}{\sin A}$$The ray reflected internally from the second face strikes the first face at an angle $2A$.Let $e = 6.3^\circ$ be the angle this emerging ray makes with the incident ray.Applying Snell’s law at the first face:$$\mu \sin(2A) = \sin e$$Thus,$$\mu = \frac{\sin e}{\sin 2A}$$Equating the two expressions for $\mu$:$$\frac{\sin(A+\delta)}{\sin A} = \frac{\sin e}{\sin 2A}$$Using the small-angle approximation:$$\frac{A+\delta}{A} = \frac{e}{2A}$$This gives:$$2A + 2\delta = e$$$$A = \frac{e - 2\delta}{2}$$Substitute $e = 6.3^\circ$ and $\delta = 1.15^\circ$:$$A = \frac{6.3 - 2(1.15)}{2} = 2^\circ$$Now,$$\mu = \frac{A + \delta}{A} = \frac{2 + 1.15}{2} = 1.575$$Answer:The refractive index of the prism is $\mu = 1.575$, which matches option B.