On Reflection Radar Theory 101

From the post "Turning With The Rest Of Us"  and "More Bending Of Light" both dated 13 Aug 2015,  when

$\theta_1-\alpha\lt0$

For $\mu_2\lt\mu_1$,

$\alpha_{2s}\gt-180^o+\alpha$

and $\mu_2\gt\mu_1$,

$\alpha_{2s}\lt-180^o+\alpha$

The reflected ray, $\alpha_{2s}$ can be steered by changing $\mu_2$,

$tan(\theta_2-\alpha_{2s})=\cfrac{\mu_2}{\mu_1}tan(\theta_1-\alpha)$

$\alpha_{2s}$ being reflected by rotating $-180^o$.

$\theta_1=\theta_2$

$tan(\theta_1-\alpha_{2s})=\cfrac{\mu_2}{\mu_1}tan(\theta_1-\alpha)$

This is important in the case of $EMW$ where a reflector opposite to an emitting source focuses the radiation behind the source.

In the case of a perfect conductor, $\alpha_{2p}$ does not exist as there cannot be magnetic fields inside the conductor.  $\alpha_{2p}$ is absorbed by the conductor; half of the energy in the wave is absorbed.  If $\theta$ is kept small, the vertical component of  the $E$ field is small compared to the horizontal component in the direction of travel of the radiation.  This horizontal component attenuates as the wave move into medium of varying $\varepsilon$; the vertical component remained unchanged.  $\theta$ should be kept large $\theta\to90^o$ as $\alpha_{2s}$ only depends on $\alpha$, $\mu_2$ and $\mu_1$, up to the range over which the horizontal component of $E$ attenuates completely.  As the wave attenuates, $\theta$ decreases and is more readily reflected given the incident angle $\alpha$.

Apart from keeping polarization high, $\Delta\theta$ dose not effect $\alpha_{2s}$, the reflected EMW.