Question

In a medium, the electric field in an EM wave is
$E=E_{0} \sin 12 \times 10^{6}\left(z-2 \times 10^{8} t\right) N / C$
The refractive index of the medium is

Answer 1

$E=E_{0} \sin \left[12 \times 10^{6}\left(z-2 \times 10^{8} t\right)\right]$
$=E_{0} \sin \left(12 \times 10^{6} z-24 \times 10^{14} t\right)$
$\therefore k=12 \times 10^{6}, \omega=24 \times 10^{14}$
$\nu=\frac{\omega}{k}=\frac{24 \times 10^{14}}{12 \times 10^{6}}=2 \times 10^{8}$ ;
$ \mu=\frac{c}{v}=\frac{3 \times 10^{8}}{2 \times 10^{8}}=\frac{3}{2}$