KCET 2026 Physics Solved Paper

Concept:At the critical angle, light travels from a denser to a rarer medium, and the angle of refraction becomes $90^\circ$.Explanation:Light goes from medium A to medium B, with critical angle $\theta$.So, at the critical angle, the angle of refraction in medium B is $90^\circ$.Using Snell's law:$$n_A \sin \theta = n_B \sin 90^\circ$$Since $\sin 90^\circ = 1$,$$n_A \sin \theta = n_B$$The refractive index of a medium is $n = \frac{c}{v}$.For medium A, speed is $V$, so:$$n_A = \frac{c}{V}$$Let the speed in medium B be $v_B$. Then:$$n_B = \frac{c}{v_B}$$Substituting these into the equation:$$\frac{c}{V} \sin \theta = \frac{c}{v_B}$$Cancelling $c$ from both sides:$$\frac{\sin \theta}{V} = \frac{1}{v_B}$$Therefore:$$v_B = \frac{V}{\sin \theta}$$Answer:The speed of light in medium B is $\displaystyle \frac{V}{\sin \theta}$.Correct option: C.