Question

A uniform plane wave in free space with $\overline{E}=8cos(\omega t-4x-3z){\hat{a}}_{y}V/m$ is incident on a dielectric slab $(z\ge 0)$ with ${\mu }_r=1,{ϵ}_r=2.5$, then reflected electric field ${\overline{E}}_r$ is

• A. ${\overline{E}}_r=-3.112cos(\omega t-4x+3z){\hat{a}}_{y}V/m$
• B. ${\overline{E}}_r=8cos(\omega t-4x+3z){\hat{a}}_{y}V/m$
• C. ${\overline{E}}_r=-8cos(\omega t+4x+3z){\hat{a}}_{y}V/m$
• D. ${\overline{E}}_r=-8cos(\omega t+4x-3z){\hat{a}}_{y}V/m$

Answer 1

The correct option is A ${\overline{E}}_r=-3.112cos(\omega t-4x+3z){\hat{a}}_{y}V/m$



$\overline{E}=8e^{-[4x+3z]}{\hat{a}}_y$

$k_1=\frac{4{\hat{a}}_x+3{\hat{a}}_z}{5}$

${\theta }_i={\theta }_r$

${\overline{k}}_i={\beta }_1(x\sin {\theta }_i+z\cos {\theta }_i)$

$\tan {\theta }_i=\frac{4}{3}$

${\theta }_i={53.13}^{\circ }$

${\overline{k}}_t={\beta }_1(x\sin {\theta }_i-z\cos {\theta }_i)$

${\overline{k}}_t=4x-3z$

$\Gamma =\frac{E_{01}}{E_{01}}=\frac{{\eta }_2\sec {\theta }_t-{\eta }_1\sec {\theta }_i}{{\eta }_2\sec {\theta }_t+{\eta }_1\sec {\theta }_i}$

$\frac{\sin {\theta }_1}{\sin {\theta }_t}=\sqrt{\frac{{ϵ}_{r_2}}{{ϵ}_{r_1}}}$

$\sin {\theta }_i=\frac{\sin 53.13}{\sqrt{2.5}}$

${\theta }_t=30.{39}^{\circ }$

Now we will get

$\Gamma =\frac{E_{or}}{E_{oi}}=-0.389$

$E_{01}=\Gamma E_{oi}=-3.112$

${\overline{E}}_r=-3.112\cos (\omega t-4x+3z){\hat{a}}_{y}V/m$