If an optical medium possesses a relative permeability of $$\frac{10}{\pi}$$ and relative permittivity of $$\frac{1}{0.0885}$$, then the velocity of light is greater in vacuum than that in this medium by ______ times.
$$(\mu_{\circ}=4\pi \times 10^{-7} H/m, \in_{\circ} = 8.85 \times 10^{-12} F/m, c = 3 \times 10^{8} m/s)$$

The speed of an electromagnetic wave in any medium is

$$v = \frac{1}{\sqrt{\mu \, \varepsilon}}$$

where $$\mu = \mu_r \mu_0$$ and $$\varepsilon = \varepsilon_r \varepsilon_0$$. Putting these into the expression, the speed in the medium becomes

$$v = \frac{1}{\sqrt{\mu_r \mu_0 \, \varepsilon_r \varepsilon_0}} = \frac{1}{\sqrt{\mu_r \varepsilon_r}\;\sqrt{\mu_0 \varepsilon_0}} = \frac{c}{\sqrt{\mu_r \varepsilon_r}}$$

Here $$c = \dfrac{1}{\sqrt{\mu_0 \varepsilon_0}}$$ is the speed of light in vacuum. Therefore the factor by which light travels faster in vacuum than in the given medium is

$$\frac{c}{v} = \sqrt{\mu_r \varepsilon_r}$$

The medium has

$$\mu_r = \frac{10}{\pi}, \qquad \varepsilon_r = \frac{1}{0.0885}$$

First find the product $$\mu_r \varepsilon_r$$:

$$\mu_r \varepsilon_r = \frac{10}{\pi}\;\times\;\frac{1}{0.0885} = \frac{10}{\pi \times 0.0885}$$

Using $$\pi \approx 3.1416$$:

$$\pi \times 0.0885 \approx 3.1416 \times 0.0885 = 0.2780$$

Hence

$$\mu_r \varepsilon_r \approx \frac{10}{0.2780} \approx 35.96$$

Now take the square root:

$$\sqrt{35.96} \approx 5.996 \approx 6$$

Thus

$$\frac{c}{v} \approx 6$$

Therefore, the velocity of light in vacuum is greater than that in this medium by 6 times.