Suppose you have a uniform plane wave travelling from a dielectric medium towards a perfect conductor (suppose the interface is planar as well).

From my understading of the theory behind EM waves and general EM, the following should happen:

The conductor acts as a "mirror", i.e. it reflects all the wave (in case of a real conductor, I'd expect it to reflect most of the wave). This would imply a "load reflection coefficient" of -1. Why you say, let's consider the perfect conductor as a short circuit:

$\Gamma_L = \frac{Z_L-Z_0}{Z_L+Z_0} = -1$ where $Z_L=0$ is the load impedance and $Z_0 > 0$ is the reference impedance.

Considering transmission lines theory, the voltage at the load would then be:


which is consistent with expecting a short circuit at the load end. This also means that the reflected wave $V^- = - V^+$, that is, it is equal to the forward travelling wave but has a 180 degrees phase shift.

I would also expect the tangential component of the E field to be zero at the interface between dielectric and perfect conductor since the tangential component of E is conserved and E must be zero inside a perfect conductor at steady state.

Is my reasoning correct?

  • $\begingroup$ Yep, that makes sense. $\endgroup$ – my2cts Jun 22 '18 at 18:13

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