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(D) The speed of an electromagnetic wave in a medium is given by $v = \frac{1}{\sqrt{\mu \varepsilon}}$,where $\mu = \mu_0 \mu_r$ and $\varepsilon = \

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$2$

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(D) The speed of an electromagnetic wave in a medium is given by $v = \frac{1}{\sqrt{\mu \varepsilon}}$,where $\mu = \mu_0 \mu_r$ and $\varepsilon = \varepsilon_0 \varepsilon_r$.Substituting these,we get $v = \frac{1}{\sqrt{\mu_0 \mu_r \varepsilon_0 \varepsilon_r}} = \frac{c}{\sqrt{\mu_r \varepsilon_r}}$,where $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = 3 	imes 10^8 \ m/s$.Rearranging the formula,we have $\sqrt{\mu_r \varepsilon_r} = \frac{c}{v}$.Squaring both sides,$\mu_r \varepsilon_r = \left(\frac{c}{v}ight)^2$.Given $\mu_r = 2.0$ and $v = 1.5 	imes 10^8 \ m/s$,we have $2.0 	imes \varepsilon_r = \left(\frac{3 	imes 10^8}{1.5 	imes 10^8}ight)^2$.$2.0 	imes \varepsilon_r = (2)^2 = 4$.Therefore,$\varepsilon_r = \frac{4}{2} = 2$.

Electromagnetic waves travel in a medium with a speed of $1.5 \times 10^8 \ m/s$. The relative permeability of the medium is $2.0$. The relative permittivity will be:

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