Plane Wave Incidence of Material Boundaries at Normal Incidence

Incident, Reflected and Transmitted

Boundary Conditions

If medium 2 is a perfect conductor

If medium 2 is a perfect conductor, we note that $\vec E = \vec H = 0$ inside the conductor

• Written as a boundary condition, we have $\tilde{\vec E_x}(z=0)=0$, and the only way we can satisfy this boundary condition is if we get a perfect reflection: $E_{0,i}e^{-jk0} + \Gamma E_{0,i}e^{+jk0} = 0 \ \rightarrow \ \Gamma = -1$
• Therefore $\tilde{\vec E_x}(z<0) = \hat xE_{0,i}e^{-jk_1z}-\hat xE_{0,i}e^{+jk_1z} = -\hat x2jE_{0,i}\sin(k_1z)$
  • And we see the same standing wave pattern we saw in transmission lines
• We can also determine that $\tilde{\vec H}(z<0) = \hat y\frac{2}{\eta_1}E_{0,i}\cos(k_1z)$

  • From this and magnetic field boundary condition $\hat z \times(\vec H_2 - \vec H_1) = \vec J_s$, we can find the surface current on the conductor

    $$ \vec J_s = \hat z \times (0-\hat y\frac{2}{\eta_1}E_{0,i}) \ \rightarrow \ \vec J_s = \hat x \frac{2E_{0,i}}{\eta_1}\text{ A/m} $$

General $\vec E$ in normal incidence

In general, the electric field can have both x and y components:

$$ \tilde{\vec{E_i}} = (\hat xE_{0,x} + \hat yE_{0,y})e^{-jk_1z} $$

• Note that the electric field remains tangential to the interface, so the boundary conditions would be exactly the same as the previous case where we only had x components.
• Therefore, $\Gamma, \Tau$ is the same.

Plane Wave Incidence of Material Boundaries at Oblique Incidence