JEE Main 2022 — Electromagnetic Waves Question with Solution

The speed of electromagnetic waves in a medium is given by the formula:

$$v=\frac{1}{\sqrt{\mu \epsilon }}$$

where $\mu$ and $\epsilon$ are the absolute permeability and absolute permittivity of the medium, respectively.

Given that $\mu ={\mu }_0{\mu }_r$ and $\epsilon ={\epsilon }_0{\epsilon }_r$, we can rewrite the above equation as :

$$v=\frac{1}{\sqrt{{\mu }_0{\mu }_r{\epsilon }_0{\epsilon }_r}}$$

which simplifies to :

$$v=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}\times \frac{1}{\sqrt{{\mu }_r{\epsilon }_r}}$$

Substituting $c$ for $\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$ (the speed of light in a vacuum), we get:

$$v=\frac{c}{\sqrt{{\mu }_r{\epsilon }_r}}$$

We can rearrange this equation to solve for ${\epsilon }_r$ :

$${\epsilon }_r=\frac{c^2}{v^2{\mu }_r}=\frac{(3\times 10^{8}m/s{)}^2}{(2\times 10^{8}m/s{)}^2\times 1}=2.25$$

So, the relative permittivity of the medium is 2.25.