Question

uniform plane wave in air is normally incident on a dielectric material $(\sigma =0,{\mu }_r=1,{\in }_r=3)$. The transmitted wave is given by $7.32cos(\omega t-\sqrt{3}Z)a_y$V/m. The incident wave expression is _______.

• A. $5.8cos(\omega t+z)a_{y}V/m$
• B. $10cos(\omega t+z)a_{y}V/m$
• C. $10cos(\omega t-z)a_{y}V/m$
• D. $5.8cos(\omega t-z)a_{y}V/m$

Answer 1

The correct option is C $10cos(\omega t-z)a_{y}V/m$

${\eta }_1={\eta }_0$

$\Gamma =\frac{{\eta }_2-{\eta }_1}{{\eta }_2+{\eta }_1}=\frac{1-\sqrt{{\in }_r}}{1+\sqrt{{\in }_r}}$

$\Gamma =-0.268$

$\tau =1+\Gamma =0.732$

$\tau =\frac{E_{Transmitted}}{E_{incident}}$

$also{\beta }_{incident}=1$

$E_{incident}=\frac{7.32cos(\omega t-z)}{0.732}$

$E_{incident}=10cos(\omega t-z)a_{y}V/m$