Exercise 5.1.1: Plane wave in a dielectric

Repeating the derivation of Home exercise 5.1.1 with $\epsilon ={\epsilon }_r{\epsilon }_0$ and $\mu ={\mu }_0$, the two curl equations are consistent only if $\omega /k=1/\sqrt{\mu \epsilon }$:

$\nabla \times E=(\partial E_x/\partial z)a_y=20k\sin (\omega t-kz)a_y=-\partial B/\partial t$. Integrating in time,

$\nabla \times H=-(\partial H_y/\partial z)a_x=-(20k/\eta )\sin (\omega t-kz)a_x$ and $\partial D/\partial t=-\epsilon \omega 20\sin (\omega t-kz)a_x$. They are equal if $k/\eta =\epsilon \omega$, that is $\eta =k/(\epsilon \omega )=1/(\epsilon v)=\sqrt{{\mu }_0/\epsilon }$, the same $\eta$ as in (b).

Units: ${\mu }_{0}v$ is $(H/m)(m/s)=H/s=\Omega$. In a dielectric the wave is slower by $\sqrt{{\epsilon }_r}=n$, the refractive index ($n=1.5$ here, a typical glass), and shorter by the same factor. $H$ is larger for the same $E$ because $\eta$ is smaller. Both results are used in the language of media of Section 3.4 and in every optics course.