A wave of light is polarized along a plane…it vibrates at a particular angle perpendicular to the direction of the light beam.
A light beam is also made up of an electric field AND a magnetic field…both of which are perpendicular to each other (right angle) as well as the light beam itself.
If you’re asking which field gets acted upon by a polarizer, then it’s the electric field, I believe. As for number 2 it is neither a coincindence nor a specific function of the electric field. It’s a property of the field itself in the way it interacts with the electron structure of the polarizer.
The equation for the electromagnetic wave has a solution with an electric field in one direction perpendicular to the direction of propagation. To this you can add another wave, with the E-field perpendicular to that direction, if you wish, perhaps with a phase shift (so that one wave starts going “up” when the other goes “down”, for instance). The polarization state is simply the relationship between these two electric-field vectors, their relative amplitudes and phase. If both are precisely in-phase, then you have linearly polarized light. (In fact, you can choose a different coordinate system, if you wish, and you can thus get an E-field all in one plane) If there’s a quarter phase difference, and the amplitudes are equal, you have circularly polarized light. In general, with randomly related amplitudes and phase, you get elliptically polarized light.
I always think of this as “bookkeeping”, because all the physics has already been done in solving the wave equation. All you’re doing with polarization math is reshuffling the amplitudes as you change your coordinate system. There are two elegant methods of treating polarization – the Jones Calculus and the Mueller Calculus. Both are matrix methods. The Jones Calculus uses complex numbers, but it easy to remember. The Mueller calculus uses all real numbers , but they’re not so trivial to remember.
See a good book on polarization, like Shurcliffe and Ballard’s.
Anything that effects polarization is acting on the whole electromagnetic field, BTW, so it doesn’t seem to me to ask whether something affects the electric field or the magnetic field. It’s a property of both.
I just want to clarify that the two methods are not equivalent. The Jones vector can only describe coherent (perfectly polarized) light. An individual photon is perfectly polarized so a Jones vector can describe its polarization status, and the Jones matrix can describe what happens to it.
But a real-life light beam contains a mixture of differently polarized photons. It may even be unpolarized, meaning it contains all polarizations in equal proportions. You need a 4-dimentional Stokes vector to describe the polarization state of such a beam, and a Mueller matrix to describe what happens to it.
I’ll beg to differ with you on this. There are methods for converting Mueller matrices into Jones matrices, and vice versa. The Stokes vector has four parameters, the same as the Jones vector (which has two components, but each has a real and imaginary part). The two are perfectly equivalent, the only real difference being that Mueller caslculus contains all real values, while the Jones matrix is complex. For handling randomly polarized light, you can as easily average Jones values over all directions. And Mueller calculus with the Stokes vector can be used for – is normally used for – well-defined states of polarization. Again, see S&B.
I don’t think so. A single Stokes vector can describe an unpolarized light - it’ll look like (1,0,0,0). Any light described by a single Jones vector is perfectly polarized. The Jones vector does have 4 parameters, but that’s because it has other information that a Stokes vector does not: the phase of the photon. So you can’t use a Stokes vector to calculate interference effects.
Depends upon your definitions. If you’re time-averaging the coefficients, then, yeah, you can have a state like that. But for a monochromatic wave, you have to have the relationship that (S0)**2 = (S1)**2 +(S2)**2+(S3)**2 (see Born and Wolf, equation 1.4.1 #44) The sum of the squares of the last three Stokes’ parameters is equal to the square of the first parameter. In fact, the last three parameters give the orientation of the Stokes vector within the Poincare sphere, so they’d better not all be zero.
Some folks do perform a time average over times longer than the coherence time of the source, getting the result above. But most folks don’t do that, because it throws away useful information. The Stokes vector certainly can represent a unique polarizationm state (B&W p. 31)