# Why do we get a reflected electromagnetic wave when it hits a perfect conductor?

It is true that the electric field inside a perfect conductor is zero. But consider what is happening on the surface of the conductor.

We can only say that an induced field exists, which is opposite to that of the incident one and lives on the surface of the conduit conductor.

The incident electromagnetic wave moves the free charges on the conductor which produces a current that then creates a radiating field which is the reflected wave.

We can only say that an induced field exists, which is opposite to that of the incident one and lives on the surface of the conduit conductor.

And this field is oscillating charges on the surface of the conductor.

And it doesn't follow that this induced field should should travel out of the surface in the form of a reflected wave.

It is these induced (changing) fields on the surface that then cause the reflected electromagnetic wave.

How can then one deduce that a reflected EM wave exists when a EM wave strikes a perfect conductor?

Since the electric field inside the conductor is zero, there is an infinite impedance to the electromagnetic wave right at the surface. Now if we take the phase of the electromagnetic wave at this interface to be $$0$$ degrees (and since the conductor allows no electric field), there must be an electromagnetic wave with an opposite phase of $$180$$ degrees to cancel the incident electromagnetic wave at the interface.

The presence of a reflected wave is simply a consequence of Maxwell's equations and the boundary conditions imposed on their solutions.

When a wave is incident upon the conducting interface, you are free to try any solution you like for what happens to the electromagnetic fields on either side of the interface. But those fields must be solutions to Maxwell's equations in (I) a good conductor on one side of the interface and (II) vacuum (or whatever) on the incidence side of the interface. Secondly, the components of the electric fields tangential to the interface must be continuous.

The very low impedance in the conductor means that the electric field is almost zero in the conductor, which means it must be almost zero on the incidence side too.

So what solution could we explore for the fields on the incidence side. If we treat the fields as the sum of the incident wave plus some other, unknown, solution to Maxwell's equations, then our options are limited.

We know that the solutions to Maxwell's equations in vacuum (or whatever dielectric it is) must be electromagnetic waves. We know that to maintain the electric field at zero close to the interface, it must have the same frequency and amplitude as the incident wave. We know that this wave must transport almost all the energy in the incident wave away from the interface. And we know that this energy does not travel into the conductor. Finally, we know that this wave must travel at a specific angle (equal to the angle of incidence), because otherwise it would not cancel with the incident wave all along the interface. All that we are left with is a wave of equal amplitude and frequency travelling in the opposite direction (for normal incidence).

The simplest explanation is the following. For a perfect conductor the incoming light makes the conduction electrons move 180 degrees out of phase with it. These electrons create two waves that are 180 degrees out of phase with the incoming field. One is moving along with the incoming wave and cancels it beyond the surface of the metal. The other moves opposite to it and is the reflected wave. This is what Maxwell's equations say if the so called bound charge is used as the source of the field. So Maxwell in vacuum but with the conduction current as a source.

Alternatively the bound charge can be absorbed in an effective medium, characterized by a non-unity relative dielectric constant. To find the answer with this approach you need to use the Fresnel equations.

OP states that "it doesn't follow that this induced field should should travel out of the surface". This follows from Maxwell's equations.