References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.21.

We can analyze reflection of an electromagnetic wave at a nonconductor-conductor interface in a similar way to that used for a nonconductor-nonconductor interface. We’ll look only at the case of normal incidence here.

As before, we start with the boundary conditions in linear media derived from Maxwell’s equations:

We’ll take medium 1 as the nonconductor (air, say) and medium 2 as the conductor. We’re allowing for the presence of free surface charge density and free current density at the boundary.

If we’re dealing with a conductor that obey’s Ohm’s law, the volume current density is proportional to the electric field

where here is the conductivity, not a charge density. Recall that is the amount of current flowing through a unit area in the conductor. If we had a surface current density , this current flows along the boundary as a sheet of moving charge with infinitesimal thickness, so that the cross-sectional area occupied by is essentially zero, making the *volume* charge density infinite. For a finite conductivity it would take an infinite electric field to produce this surface current, so we can safely assume that in what follows.

The incident and reflected waves are both in medium 1, so if we polarize the wave in the direction, we have for the incident wave:

where is the speed of the wave in medium 1.

The reflected wave is travelling in the direction and has equations

The transmitted wave is inside the conductor, so its equations can be written as

where the wave vector is complex:

We can now apply the boundary conditions. Equation 1 tells us that since there is no perpendicular component of (remember the wave is transverse). Equation 2 tells us nothing (). From 3, assuming that the boundary is at , we get, since all components of are in the direction:

Finally, from 4 we get, since all components of are in the direction and :

which we can rewrite as

These are deceptively simple equations, since everything with a tilde on it is a complex number. To get the actual amplitudes and phases we need to extract the real and imaginary parts.

ExampleTo put some numbers into these equations, let’s consider an air-silver interface. For a good conductor such as silver, and in 12In air, and we can take so from 16

For silver and at an optical wavelength of we get

To get the reflection coefficient we can write the complex amplitudes in modulus-phase form as

The intensity of a wave is the average over one cycle of the magnitude of the Poynting vector, so the fact that the incident and reflected waves may have different phases doesn’t matter (since they have the same frequency). This means that

Silver reflects 93% of the incident light, so it makes a good mirror.