# Can non-monochromatic light be elliptically or circularly polarised?

We all know that monochromatic light is always polarised. Non-monochromatic light can obviously also be linearly polarised. However, can non-monochromatic light be elliptically or circularly polarised?

• all satellite communications and some radar signals are circularly polarized, these are always modulated and not monochromatic. Apr 13, 2018 at 23:34
• There is no such thing as monochromatic light and so the comment that is it always polarised doesn't matter. Also linearly polarised light can be expressed as a superposition of circularly polarised light and visa versa so fundamentally they are the same.
– MJC
Apr 19, 2018 at 9:31

Yes. The two images in a 3D cinema are left- and right-handed circularly polarized.

If you have a pair of such 3D-glasses, it is fun to look in a mirror, and close your right eye or your left eye.

• Apr 13, 2018 at 23:00

There's some subtleties around the edges of this question, but the short answer is yes.

Non-monochromatic light is made up of some combination (which might be coherent or incoherent) of light of a range of different frequencies. If each component of that combination is circularly polarized, then we say that the combination is circularly polarized.

• If the combination is coherent, then it is a linear superposition that forms some form of pulse, in which case the field shape rotates with time; see e.g. this paper for a recent example.

• If the combination is incoherent, then it becomes quite hard to say anything about the detailed time dependence of the pulse, which will generally be a mess, but it will tend to rotate in the specified direction.

In either case, to actually make a non-monochromatic circular polarization, you need a quarter-wave plate as usual -- except that the plate will need to add exactly a quarter-wavelength of optical path to each wavelength, which can't be achieved with a plate of constant birefringence $\Delta n$. Instead, what's needed is an achromatic waveplate whose birefringence $\Delta n$ changes with wavelength in precisely the right way; this is mostly doable over limited frequency ranges, but it gets more and more challenging (and expensive) as the bandwidth of interest increases.