Answer to Question #171472 in Electricity and Magnetism for Bhesha

Question

A linearly polarized wave with Ē in the xy plane is incident from x > 0 on an interface at x = 0. In the region x > 0 is free space and in the region x< 0 is a medium with dielectric constant e,

a) If the incidence angle is 45 degrees and angle of transmitted(refracted wave) is 90 degree find the dielectric constant ?

b)Find E and B for z>0 and calculate the time average of Poynting vector component along x (i.e. S_x) flowing through x = 0.

Answer 1

Let's draw the diagram for the given condition,

For the region (i), x<0

For the region (ii), x>0

Permittivity for the free space $(\epsilon_o)=1$

Here y-z is the interface between the two medium.

Now, using the electric field

For the boundary condition,

$E_I''=E_{II}'' ...(i)$

The perpendicular component of D is noncontinuous function,

$D_{I}^{\perp}-D_{II}^{\perp}=\sigma_f$

Electric displacement vector $\overrightarrow{D}=\epsilon \overrightarrow{E}$

$D_{I}^{\perp}=D_{II}^{\perp}$

$E_{I}''=E_{II}\sin(\pi/4)=\frac{E_{II}}{\sqrt{2}}$

Now, substituting the values,

$\epsilon_{I} \overrightarrow{E}_{I}^{\perp}=\epsilon_{II} \overrightarrow{E}_{II}^{\perp}$

$\epsilon=\infty$

b) $\overrightarrow{E}$ is linearly polarized in x-y plane

We know that poynting vector $\overrightarrow{S}=\frac{\overrightarrow{E}\times \overrightarrow{B}}{\mu_o}$

$\overrightarrow{E}=\frac{E}{\sqrt{2}}(\hat{i}+\hat{j})$

$B=\frac{E}{\sqrt{2}c}$

$S=<|\overrightarrow{S}|>$

$=\frac{E^2}{\sqrt{2}\mu c}$

$=0.002E^2$