Exercise 6.4.1: Reading an oblique wave from its phasor

${\beta }_1=\sqrt{(12\pi {)}^2+(16\pi {)}^2}=20\pi rad/m$, so ${\lambda }_1=10\mathrm{cm}$ and $f={\beta }_{1}c/(2\pi )=3\mathrm{GHz}$. $\tan {\theta }_i=16/12$: ${\theta }_i=53.13^{\circ }$ ($\cos {\theta }_i=0.6$, $\sin {\theta }_i=0.8$). Perpendicular (S) polarisation.

${\beta }_2=2{\beta }_1=40\pi rad/m$. $\sin {\theta }_t={\beta }_y/{\beta }_2=16\pi /40\pi =0.4$, ${\theta }_t=23.58^{\circ }$, $\cos {\theta }_t=0.9165$, and

Along the interface both waves vary as ${\mathrm{e}}^{-\mathrm{j}16\pi y}$: the trace wavelength is $2\pi /16\pi =12.5\mathrm{cm}$ on both sides, longer than ${\lambda }_1=10\mathrm{cm}$ and ${\lambda }_2=5\mathrm{cm}$, since ${\lambda }_1/\sin {\theta }_i={\lambda }_2/\sin {\theta }_t=12.5\mathrm{cm}$.

In units of ${\eta }_0$, ${\eta }_1=1$, ${\eta }_2=0.5$:

$R=0.507^2=0.257$, $T=\frac{{\eta }_1\cos {\theta }_t}{{\eta }_2\cos {\theta }_i}{\tau }_{\perp }^2=\frac{0.9165}{0.5\times 0.6}\times 0.243=0.743$, and $R+T=1$.

Tangential $H_y$ at $x=0$: incident $-\frac{5}{377}\times 0.6=-7.96mA/m$, reflected $+\frac{-2.53}{377}\times 0.6=-4.03mA/m$, sum $-11.99mA/m$ $=H_y^t$.