Home exercise 7.2.2: The TE_1 mode as two plane waves

With $h=\pi /b$ and $\sin u=({\mathrm{e}}^{\mathrm{j}u}-{\mathrm{e}}^{-\mathrm{j}u})/2\mathrm{j}$,

two plane waves with $k_{1,2}=\tau a_z\mp h{a}_y$, one travelling obliquely downwards, the other upwards. Numerically $h=\pi /0.03=104.7rad/m$, $k=2\pi \times 6\times 10^9/c=125.7rad/m$ and $\tau =\sqrt{k^2-h^2}=69.5rad/m$.

$|k_{1,2}|=\sqrt{{\tau }^2+h^2}=k$, so each is an ordinary plane wave in air, and the angle to the axis satisfies

The wavefronts of each plane wave are spaced $\lambda$ apart along $k$. Along the $z$ axis, which makes the angle $\theta$ with $k$, they are spaced $\lambda /\cos \theta$: this is ${\lambda }_g$, consistent with $\tau =k\cos \theta$. The energy of each wave travels along its ray at $c$, and its component along the axis is $c\cos \theta =v_g$. Since $\cos \theta =\sqrt{11}/6=0.553$,

$2\pi /\tau =2\pi /69.5=9.05\mathrm{cm}$ and $v_p{v}_g=9.0\times 10^{16}=c^2$.