MHTCET 19 Apr 2026 Shift 2 Paper

Concept:At the critical angle, light travels from the denser medium to air, and $\sin C$ equals the ratio of the speed of light in the medium to the speed in air.Explanation:The speed of light in air is obtained by dividing distance by time:$$v_{\text{air}} = \frac{x}{t_0}$$The speed of light in the denser medium is:$$v_{\text{medium}} = \frac{4x}{t_1}$$For total internal reflection at the critical angle $C$, Snell's law gives:$$\sin C = \frac{n_{\text{rarer}}}{n_{\text{denser}}}$$Since refractive index is inversely proportional to the speed of light, this can be written as:$$\sin C = \frac{v_{\text{medium}}}{v_{\text{air}}}$$Substituting the two speed expressions:$$\sin C = \frac{\frac{4x}{t_1}}{\frac{x}{t_0}} = \frac{4t_0}{t_1}$$Taking the inverse sine on both sides:$$C = \sin^{-1}\left(\frac{4t_0}{t_1}\right)$$Answer:The correct option is B: $C = \sin^{-1}\left(\frac{4t_0}{t_1}\right)$.