QJEEM22Ph5e99: JEE Main 2022 Physics - Class 12 JEEM22Ph5e99 Solution: The electromagnetic waves travel in a medium at a speed of 2.0…

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

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

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

Given that $\mu = \mu_0 \mu_r$ and $\varepsilon = \varepsilon_0 \varepsilon_r$ , we can rewrite the above equation as :

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

which simplifies to :

$v = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \times \frac{1}{\sqrt{\mu_r \varepsilon_r}}$

Substituting $c$ for $\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$ (the speed of light in a vacuum), we get:

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

We can rearrange this equation to solve for $\varepsilon_r$ :

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

So, the relative permittivity of the medium is 2.25.