Q 12-09-136JEE MainJEE Main 2022 (25 Jun, Shift 1)Medium
The difference of speed of light in the two media $A$ and $B$, $(v_A - v_B)$, is $2.6\times10^7\ \text{m s}^{-1}$. 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 $c = 3\times10^8\ \text{m s}^{-1}$)
Answer: (C) 1.13
$v_B = \dfrac{3\times10^8}{1.47} = 2.041\times10^8\ \text{m s}^{-1}$, so $v_A = 2.041\times10^8 + 0.26\times10^8 = 2.301\times10^8\ \text{m s}^{-1}$.
$$\frac{n_B}{n_A} = \frac{v_A}{v_B} = \frac{2.301}{2.041}\approx1.13$$
Solution by Sreeraj P, M.Sc Physics