Re: A question about the phase difference of the beam splitter.


<dvanbaak@xxxxxxxxxx> wrote in message
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On Aug 15, 2:31 am, "Adam Norton" <AnortonREMOVET...@xxxxxxxxxxxxx>
wrote:
"Neil Bates" <neil_del...@xxxxxxxxxxxxxxx> wrote in message

news:13bq51sougs3873@xxxxxxxxxxxxxxxxxxxxx


. . .

However, your question about a thin metal film got me thinking even more,
and I realized that the above criteria only hold for non-absorbing films.
If the film is absorbing (i.e. has a complex refractive index), then the
phase differences can add up to something other than 180 deg. This
raises
some questions I have never thought about before; for example, when
combining beams with an absorbing beamsplitter, does the total energy
absorbed in the beamsplitter depend on the phase of the two beams
incident
on the beam splitter? When I get more free time, I might look into
questions like these. Right now, I am too busy.

-Adam Norton

Hello all--

There are two issues for all of you to think about:

1) On the vexed matter of the 'conservation of energy' -- please
remember that he beams returning to, and recombining at, the
beamsplitter of a Michelson interferometer generate not one but TWO
output beams. One of them comes off in the 'fourth direction', and
it's the beam which is usually displayed as 'the output' of the
interferometer. And it can indeed drop to zero intensity, for certain
phase conditions. But there is in general another beam, usually
headed straight back to the source, which also in general carries
energy away from the interferometer. No discussion of phases in the
interferometer is complete without analyzing the energy in *both* of
these beams.

Well, the most relevant construction for these questions is a Mach-Zehnder
interferometer instead. Then there's no beam going back to the source
(except for a faint reflection from surfaces if recombiner *cube* used.) For
M-Z, output A = the amplitudes of transmitted input x + refl. input y and
output B = the amplitudes of transmitted input y + refl. input x. You would
naively take what amplitudes x and y would produce if each went in by
itself, add them together, and then square for intensity *if* you could get
away with that (i.e., if there wasn't a need for mutual impedance to prevent
that straightforward superposition, which would violate conservation of
energy if the phase difference between transmitted and reflected beams were
other than 90 degrees, not zero or 180.) You can confirm that 90°
dependency, and that isn't a normal phase difference between transmitted and
reflected beams (think ordinary rules about phase change upon reflection,
either zero or 180°.)


2) A further matter in the applicability of conservation of energy is
the nature of the beamsplitter. If it is wholly dielectric in
character, then it does conserve energy, and the two output beams
together will be found to carry away from the interferometer all the
energy that the input beam brings in. And this has consequences for
phase shifts at the beamsplitter -- suitably defined, they'll have to
be 0 or 180 degrees. (In this respect, akin to the phase shifts in
Fresnel reflection from a dielectric.) It also entails the that
intensities of the two output beams are complementary - when one is
bright, the other is dim, and vice versa.

Well I wasn't saying it wasn't going to conserve energy, only that we
couldn't treat amplitudes "naively." For example, suppose the film reflects
at 180° phase change. If the beams entered at the same phase, then both
reflected beams would be out of phase with both transmitted beams - and
there'd be no output. We can imagine that beams must come out of the
combiner somehow, but I do wonder what phase they'd be in (I mean, the
symmetry selection problem - would the output be the phase of the
transmitted beam, the reflected, or some other?)

But metallic films have been raised in this discussion, and they are
*not* lossless, and the conservation-of-energy argument does NOT apply
in the same way -- some energy can be dissipated in the beamsplitter,
and thus lost to the optical fields. It follows that the two outputs
need *not* be complementary in intensity, and the two output beams are
therefore not redundant in the information-theory sense. In fact the
two output beams can be in 'quadrature' with each other (in intensity
-- a separate matter from optical phase) and this in turn can make
such an interferometer supremely useful.

Sure, but the absorption etc. just introduces complications and corrections
to the outcome. The basic problem is still the same.

It further follows that for a metal-film or other lossy beamsplitter,
the previous argument that phase shifts (suitably defined) have to be
0 or 180 degrees is relaxed, and in fact (by artful selection of the
complex refractive index, and thickness, of the film) a phase shift
intermediate between these two can be achieved.

It can be easily found that 90° phase difference between transmitted and
reflected is required to have the outputs of proper energy under "naive"
amplitude addition, i.e. if you don't want to force the issue by
constraining output in some other way. I must admit I didn't take into
account the issue of complex index etc, however in any case there must be a
symmetrical phase difference for both output channels. Remember always that
it is not a matter of whether a paradigmatic perfect example of the
discrepancy can be had, but only if any discrepancy at all would be found
(*how* wrong of an answer simple addition of amplitudes would give.)

Have a look for 'quadrature interferometry' in the brochure
http://www.teachspin.com/instruments/moderni/ModInter_full.pdf
to see something of the application of these possibilities in a
teaching instrument; that'll also account for how it is that I've
learned something about these issues.

Best regards,
David Van Baak



.

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