The difference of speed of light in the two media A and B (v A - v B ) is 2.6 × 10⁷ m/s. If the refractive index of medium B is 1.47, then the ratio of refractive index of medium B to medium A is : (Given : speed of light in vacuum c = 3 × 10⁸ ms⁻¹)

Solution

Physics sir · 3 steps

Get v_B from n_B, add the difference to get v_A, then the speed ratio gives the index ratio.

1. Speed in B. Use n = c/v. So v_B = (3 × 10⁸)/1.47 = 2.041 × 10⁸ m s⁻¹.
2. Speed in A. Since v_A - v_B = 0.26 × 10⁸ m s⁻¹, we get v_A = 2.301 × 10⁸ m s⁻¹. Light is faster in A, so A is the optically rarer medium.

   n_B/n_A = (c/v_B)/(c/v_A) = v_A/v_B = (2.301 × 10⁸)/(2.041 × 10⁸) ≈ 1.127

3. So. The ratio of the refractive index of B to that of A is about 1.13, option (c).

The idea to keep. Refractive index is inversely proportional to speed of light, so n_B/n_A = v_A/v_B. The ratio flips; c cancels out.

Answer from the NTA final answer key. Solution by PrepWiser.