The difference of speed of light in the two media $$A$$ and $$B$$ ($$v_A - v_B$$) is $$2.6 \times 10^7$$ 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 $$c = 3 \times 10^8$$ m s$$^{-1}$$)

We are given that $$v_A - v_B = 2.6 \times 10^7$$ m/s, $$n_B = 1.47$$, and $$c = 3 \times 10^8$$ m/s. We need to find $$\frac{n_B}{n_A}$$.

First, the speed of light in medium B is $$v_B = \frac{c}{n_B} = \frac{3 \times 10^8}{1.47} = 2.0408 \times 10^8 \text{ m/s}$$. Since $$v_A - v_B = 2.6 \times 10^7$$ m/s, we have $$v_A = v_B + 2.6 \times 10^7 = 2.0408 \times 10^8 + 0.26 \times 10^8 = 2.3008 \times 10^8 \text{ m/s}$$.

The refractive index of medium A is then $$n_A = \frac{c}{v_A} = \frac{3 \times 10^8}{2.3008 \times 10^8} = 1.3039$$, and hence the ratio is $$\frac{n_B}{n_A} = \frac{1.47}{1.3039} = 1.1274 \approx 1.13$$. The correct answer is Option C: 1.13.