Electromagnetic wave travels with speed of \( 2.0\times10^{8}\,\mathrm{m/s} \) in any medium of relative permeability \( (\mu_r) \) 1.5. Find the relative permittivity of the medium \( (\epsilon_r) \)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1.50
The speed of an electromagnetic wave in a medium is \( v = 1/\sqrt{\mu\varepsilon} \). Writing \( \mu = \mu_r\mu_0 \), \( \varepsilon = \epsilon_r\varepsilon_0 \) and \( c = 1/\sqrt{\mu_0\varepsilon_0} \):
\( \displaystyle v = \frac{c}{\sqrt{\mu_r\,\epsilon_r}} \;\Rightarrow\; \epsilon_r = \frac{c^{2}}{\mu_r\,v^{2}} \)
\( \displaystyle \epsilon_r = \frac{(3\times10^{8})^{2}}{1.5 \times (2.0\times10^{8})^{2}} = \frac{9}{1.5 \times 4} = 1.50 \)
The refractive index here is \( c/v = 1.5 \), and \( \mu_r\epsilon_r = 1.5^{2} = 2.25 \).