Time taken by light to travel in two different materials $$A$$ and $$B$$ of refractive indices $$\mu_A$$ and $$\mu_B$$ of same thickness is $$t_1$$ and $$t_2$$ respectively. If $$t_2 - t_1 = 5 \times 10^{-10} \text{ s}$$ and the ratio of $$\mu_A$$ to $$\mu_B$$ is $$1:2$$. Then the thickness of material, in meter is: (Given $$v_A$$ and $$v_B$$ are velocities of light in $$A$$ and $$B$$ materials respectively).

Two materials A and B have refractive indices $$\mu_A$$ and $$\mu_B$$ with $$\mu_A : \mu_B = 1 : 2$$ and both have the same thickness $$d$$. The time difference between light traversing them is given as $$t_2 - t_1 = 5 \times 10^{-10} \text{ s}$$, and the individual times can be expressed as $$t_1 = \frac{d}{v_A}$$ and $$t_2 = \frac{d}{v_B}$$.

The refractive index is related to the speed of light in a medium by $$\mu = \frac{c}{v}$$, so that $$v = \frac{c}{\mu}$$. Given $$\mu_A : \mu_B = 1 : 2$$, it follows that $$\mu_B = 2\mu_A$$, hence $$v_A = \frac{c}{\mu_A}$$ and $$v_B = \frac{c}{\mu_B} = \frac{c}{2\mu_A}$$, which implies $$v_B = \frac{v_A}{2}$$.

Substituting these velocities into the expression for the time difference yields $$t_2 - t_1 = \frac{d}{v_B} - \frac{d}{v_A} = d\left(\frac{1}{v_B} - \frac{1}{v_A}\right) = d\left(\frac{2}{v_A} - \frac{1}{v_A}\right) = \frac{d}{v_A}\,.$$

Equating this to the given time difference, $$\frac{d}{v_A} = 5 \times 10^{-10}$$, gives $$d = 5 \times 10^{-10} \times v_A \text{ m}\,.$$

Hence, the correct answer is Option A.