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Q. The electromagnetic waves travel in a medium at a speed of $2.0 \times 10^{8} \,m / s$. The relative permeability of the medium is $1.0$. The relative permittivity of the medium will be:

JEE MainJEE Main 2022Electromagnetic Waves

Solution:

Given,
Speed of electromagnetic waves, $v=2 \times 10^8 m / s$
Relative permeability of the medium $=1$
Speed of light, $C=3 \times 10^8 m / s$
Using the formula of speed of electromagnetic wave in a medium,
$ \begin{array}{l} v=\frac{1}{\sqrt{\mu \varepsilon}} \\ v=\frac{1}{\sqrt{\mu_0 \mu_r\left(\varepsilon_0 \varepsilon_r\right)}} \\ v=\frac{1}{\sqrt{\mu_0 \varepsilon_0}} \times \frac{1}{\sqrt{\mu_{ r } \varepsilon_{ r }}} \end{array} $
Therefore,
Relative permittivity,
$ \begin{aligned} \varepsilon_r & =\frac{c^2}{v^2 \mu_r} \\ & =\frac{\left(3 \times 10^8\right)^2}{\left(2 \times 10^8\right)^2 \times 1} \\ & =2.25 \end{aligned} $