MHTCET 18 Apr 2026 Shift 2 Paper

Concept:Snell's law relates the refractive indices of two media to the sines of the angles of incidence and refraction.Explanation:Let the common length of the air-water interface be $L$.The angle of incidence $i$ equals the angle between the incident wavefront and the interface in air.From the right-angled triangle formed in air:$$\sin i = \frac{z}{L}$$Similarly, the angle of refraction $r$ equals the angle between the refracted wavefront and the interface in water.From the right-angled triangle formed in water:$$\sin r = \frac{w}{L}$$Applying Snell's law at the air-water interface:$$\mu_a \sin i = \mu_w \sin r$$Substituting the expressions for $\sin i$ and $\sin r$:$$\mu_a \left(\frac{z}{L}\right) = \mu_w \left(\frac{w}{L}\right)$$Cancelling $L$ from both sides:$$\mu_a z = \mu_w w$$Rearranging gives the refractive index of air with respect to water:$$\mu_{aw} = \frac{\mu_a}{\mu_w} = \frac{w}{z}$$Answer:$$\frac{w}{z}$$, which is option A.