Q 12-08-020NEETJEE MainMedium
The speed of an electromagnetic wave in a medium of relative permittivity $4$ and relative permeability $1$ is
Answer: (A) $1.5 \times 10^8$ m/s
$v = \dfrac{c}{\sqrt{\mu_r\epsilon_r}} = \dfrac{3 \times 10^8}{2} = 1.5 \times 10^8$ m/s. (Refractive index $2$.)
Solution by Sreeraj P, M.Sc Physics