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A glass mirror (with metal backing layer) will reverse the polarisation of circularly polarised light upon reflection.

A polished piece of metal will also reverse the polarisation of circularly polarised light upon reflection. (I have tested and confirmed this for myself).

wikipedia states the reason a mirror will reverse the polarisation of circularly polarised light is:

...[A]s a result of the interaction of the electromagnetic field with the conducting surface of the mirror, both orthogonal components are effectively shifted by one half of a wavelength.

However, my understanding of mirrors is that only a polished piece of metal will phase shift a wavelength by half a wavelength, whereas a glass mirror (with metal backing layer) will not produce a phase shift. For example wikipedia which states:

According to Fresnel equations there is only a phase shift if n2 > n1 (n = refractive index). This is the case in the transition of air to reflector, but not from glass to reflector

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What about from reflector to glass? – jhobbie Jul 3 '14 at 0:21
Yes, both a polished metal mirror and a glass mirror with metal backing ('reflector through glass') both reverse the polarisation of circularly polarised light. The simple test is to look through the circular polariser at either type of mirror; the reflection of the circular polariser appears completely opaque (ie. you cannot see your eye in reflection) – user52673 Jul 3 '14 at 2:50

From a quantum-mechanical perspective, circularly-polarized light is made of photons with their spins parallel to their momentum. The mirror reverses the photons' momentum but does not affect their spins, so the dot product $\sigma\cdot p$ changes sign.

Both the quantum and classical approaches are examined in Beth's 1936 measurement of the angular momentum of light, one of my favorite underrated classic papers.

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Thanks Rob, I'll go to the library and get a copy of that paper later today. I think this explanation should be added to the Wikipedia entry. – user52673 Jul 3 '14 at 2:53
I went to add it and someone had beaten me to it. :-) – rob Jul 3 '14 at 15:43

While rob is correct about the quantum mechanical picture I think that this case is at least as easy to understand in the classical description.

Classically circular polarization can be described in terms of a time-varying linear polarization, so lets just look at two points on a wave.

I'm going to chose a beam in the $+z$ direction at examine two points on the wave: one where the polarization currently points along $+\hat{x} - \epsilon \hat{y}$, and a very short time later where the polarization is in the $+\hat{x} + \epsilon \hat{y}$ direction. The wave has right-handed circular polarization.

Now we let the beam bounce off a mirror in the $x\text{--}y$ plane. This reverse the direction of propagation but leaves the time-order in which are two points of interest pass any given point unaffected. A little though suffices to show that the reflected wave has left-handed circular polarization.

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