Re: Which Polarizer?

In article <1154792677.076625.165860@xxxxxxxxxxxxxxxxxxxxxxxxxxx>, achilleaslazarides@xxxxxxxxxxx writes
David Littlewood wrote:
>
You can have a simple answer to that question, but not a simple
explanation.

Hello,

Could you please explain what is wrong with or complicated about the
following explanation:

Consider a completely polarised wave, the axis of polarisation of which
is vertical (along the z-axis). That is, the electric field is always
in the direction of the z axis. It now passes through a plate with the
property that, along a direction at 45 degrees clockwise from the z
axis and perpendicular to the direction of propagation, it passes light
with no change; along a direction at 45 degrees anticlockwise from the
z-axis and in the same plane, it retards the phase of this wave by 1/4
wavelength.

So, assume that the moment our wave arrives at this plane, the electric
field is at its maximum (top) position (this amounts to choosing the
origin of time, in this case). Its component along the +45 degree axis
is transmitted unchanged, but the one along the -45 degree axis is
retarded by 1/4 wavelength, ie, since it was at the top of its cycle,
it ends up at exactly zero and going up (of the previous cycle). So
there is a phase difference of pi/2.

After the plane, the +45 component starts decreasing from the max,
while the -45 increasing from zero, and, in general, they just continue
evolving as before. Some thought will show that their resultant will
rotate (clockwise). Hence circularly polarised light.

I know this is wordy, but if it's drawn, it's pretty simple really. Of
course, I may misunderstand what a "quarter wave plate is".

It's fine, provided it is understood that the retardation is proportional to the thickness of the material. It is, BTW, conventional to refer to the first of your two directions as the ordinary (O) ray and the one retarded relative to it as the extraordinary (E) ray. Also, it might be slightly confusing to say the O ray is "transmitted unchanged". Both the O and E rays pass through according to the usual laws, but the refractive index of the material differs in the two directions, so their velocities are different, and hence their wavelengths are different. The E ray is slower than the O ray, and thus the two get out of phase.


The circular polariser works exactly the same for coherent and
incoherent light. That's the easy bit.

The problem with explanations is that they usually ignore the
wave/particle dual nature of light. Explanations of polarisation always
focus on the wave part and ignore the photons. The simplest way to think
of it is that each photon behaves like a little burst of waves, with a
wavelength, a polarisation direction etc. In random light, each photon
will have its polarisation direction and the phase of its light
different. It will pass through polarisers, or not, according to the
relative orientations.

But why is it a problem? That light is made of particles is irrelevant
to this, as far as I can see. Could you explain why you think it is?

Well, the question arose as to "coherent" (in-phase) light, and it seemed that confusion was arising between the phase coherence between different photons, and the phase coherence of the two orthogonal vector directions (the "O and E rays") of any given photon. If you weren't confused, then obviously the clarification was not necessary for you.

Coherent light consists of photons whose polarisation direction and
phase are all the same. However, each one follows the same rules when it
passes through a polariser; it's just that the results will all be the
same for each photon because of their identical polarisation/phase.

However, the "phase" in the description of retardation refers, not to
the phase of one photon relative to its neighbours, but to the phase
relationship between its O and E components as it passes through a
birefringent material. It is, if you like, its internal phase, not its
phase relative to its, er, relatives.

Well, I don't see how it is either useful or meaningful to talk of
photons passing through birefringent materials. Since, as far as I
(vaguely) remember, such materials are formally ones for which the
refractive index is tensorial, it seems the whole thing can be
understood using nothing but Maxwell's eqns.

As I said, I only introduced the photon bit to disentangle a possible confusion between external coherence (i.e. between different photons/quanta) and the phase difference between different field vector angles in a single photon/quantum of light.

Furthermore, you talk of the phase of a photon relative to others. But,
if you are going to think of this at such a microscopic level, photons
don't pass through matter, they get absorbed and reemitted and so on,
and in general, do very complicated things in there (if we insist of
thinking of photons in more or less classical terms; we don't insist,
the picture is completely different). So, which photon, and which
relative?

Or maybe you meant something completely different and I misunderstood?

Well, I cannot claim to be a particle physicist, only someone who did chemistry many years ago. Most of us need to think of such things in terms of analogues which our brains can grasp, even though we know it is merely a feeble attempt to describe a more complex reality. I find it helpful to think of photons as tiny bundles of energy which have a discrete existence (quantisation) but also behave to an extent like energy waves. As long as one remembers the wave/particle duality, and that one's nice, easy mental pictures are just stories-for-children, then one can mostly get by in real life.

However, it is undoubtedly true that different "bits" of light (call them photons, quanta, or what you will) can be either coherent with respect to each other, or not. How you choose to envisage this is up to you.

BTW, I like the idea that "the whole thing can be understood using nothing but Maxwell's equations". I think if I tried to get my brain to handle it using nothing but Maxwell's equations (rather than its nice comfortable little mental pictures) I would struggle.

David
--
David Littlewood
.

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