Question

A perpendicularly polarized plane wave impinges on a very thick sheet of plastic at an angle ${\theta }_i$ = 30° from free space. The relative permittivity of the plastic is ${\epsilon }_r=4$ and its relative permeability is ${\mu }_r=1$. If the amplitude of the incident electric field intensity is $E_i$ = 100V/m, then the magnitude of time averaged power density transmitted into the plastic is

• A. 10.1$W/m^2$
• B. 53.7$Mw/m^2$
• C. 23.2$W/m^2$
• D. 62.8$W/m^2$

Answer 1

The correct option is A 10.1$W/m^2$

Given

${\theta }_1\left(or\right){\theta }_1=30{}^{\circ }$

from Snell's low

$\frac{\sin {\theta }_t}{\sin {\theta }_1}=\sqrt{\frac{{\epsilon }_1}{{\epsilon }_2}}\Rightarrow \sin {\theta }_t=\sin 30{}^{\circ }\sqrt{\frac{1}{4}}$

${\theta }_2\left(or\right){\theta }_t=14.47{}^{\circ }$

${\eta }_1=120\pi \Omega$

${\eta }_2=120\pi \sqrt{\frac{1}{4}}=60\pi \Omega$

$E_i=100V/m$

Transmission coefficient:

$T_1=\frac{2{\eta }_2\cos \theta }{{\eta }_2\cos {\theta }_1+{\eta }_1\cos {\theta }_2}=\frac{2\times 60\pi \cos 30{}^{\circ }}{60\pi \cos 30{}^{\circ }+120\pi \cos 14.47{}^{\circ }}$

$T_1=0.618$

$T_1=\frac{E_{t0}}{E_{t0}}=0.618$

$E_{t0}=0.618\times 100$

$E_{t0}=61.87V/m$

Magnitude of time-averaged power density of transmitted wave is

$W_t=\frac{E_{t0}^2}{2{\eta }_2}=\frac{{\left(61.8\right)}^2}{2\times \left(60\pi \right)}$

$=10.130W/m^2\left(if{}^{\prime }{\pi }^{\prime }taken\right)$

$\left(or\right)10.135W/m^2(if\pi =3.14taken)$