Re: A question about the phase difference of the beam splitter.
Adam Norton
If it were 0 deg. then you could make an interferometer where the light
disappears entirely, and of course that can never happen.
--
Adam Norton
Norton Engineered Optics
Optical design and systems engineering for Silicon Valley and beyond.
www.nortonoptics.com
(Remove antispam feature before replying)
"Chengxian.Zhang" <zhangch...@gmail.com> wrote in message
news:1186542647.9...@m37g2000prh.googlegroups.com...
> Dear all:
> Whether the phase difference between the reflection and transmission
> lights of a 50:50 beam splitter is 180 or a modifiable number that
> depends on the coating scheme?
>
Sa.Sh.
Could yoy please explain your point?
Thank you,
Sasha
"Adam Norton" <AnortonR...@ix.netcom.com> wrote in message
news:13bn268...@corp.supernews.com...
Adam Norton
beams A and B. Recombine the two beams carefully with a second beamsplitter
to produce two more beams C and D whose intensity depends on the relative
phase of A and B. However if the phase is adjusted so C has constructive
interference and is bright, D will be dark with destructive interference.
This is due to the 180 degree phase change between the reflected and
transmitted light of the beamsplitter. If it were possible to make a
beamsplitter with 0 phase difference, then it would be possible to either
make both C and D disappear or to make their combined energy greater than
the input energy. Anything other than a 180 deg. phase shift violates
conservation of energy.
-Adam Norton
"Sa.Sh." <sash...@telus.net> wrote in message
news:gGOui.83126$tB5.54400@edtnps90...
Adam Norton
that the the real restriction is that phase difference between transmitted
and reflected incident from one direction and the phase difference incident
from the other direction have to add to 180 deg.
The example I gave still shows why this is the case.
-Adam Norton
Neil Bates
"Adam Norton" <AnortonR...@ix.netcom.com> wrote in message
news:13bp8a2...@corp.supernews.com...
Well, I think you could create the "impossible" phase scheme with thin
metallic coatings: Recombine beams into a transparent cube with a diagonal
made of thin metal surface. From symmetry, there can't be a basis for
distinguishing/favoring phase, correct? To avoid violating conservation of
energy, I presume the impedance effects between atoms of metal would just
keep the output the same total energy as went in, but it is interesting to
think about. IOW, sometimes a messy outcome, not being able to use direct
phase reasoning? Another interesting example would be to try frustrated
total internal reflection (same symmetry argument) or microwaves with a mesh
diagonal, etc.
Sa.Sh.
"Neil Bates" <neil_...@caloricmail.com> wrote in message
news:13bq51s...@corp.supernews.com...
schemes to emplement constructive inteference, still energy seems to be
conserved.
Did anyone measure those interference effects carefully? Any publications?
S.
redbelly
wrote:
> Take a collimated laser beam and use one beamsplitter to split it into two
> beams A and B. Recombine the two beams carefully with a second beamsplitter
> to produce two more beams C and D whose intensity depends on the relative
> phase of A and B. However if the phase is adjusted so C has constructive
> interference and is bright, D will be dark with destructive interference.
> This is due to the 180 degree phase change between the reflected and
> transmitted light of the beamsplitter. If it were possible to make a
> beamsplitter with 0 phase difference, then it would be possible to either
> make both C and D disappear or to make their combined energy greater than
> the input energy. Anything other than a 180 deg. phase shift violates
> conservation of energy.
>
> -Adam Norton
>
>
> news:gGOui.83126$tB5.54400@edtnps90...
>
> > Adam,
>
> > Could yoy please explain your point?
>
> > Thank you,
>
> > Sasha
>
> >news:13bn268...@corp.supernews.com...
> >> The phase difference is always 180.
>
> >> If it were 0 deg. then you could make an interferometer where the light
> >> disappears entirely, and of course that can never happen.
>
> >> --
> >> Adam Norton
>
> >> Norton Engineered Optics
> >> Optical design and systems engineering for Silicon Valley and beyond.
> >>www.nortonoptics.com
>
> >> (Remove antispam feature before replying)
>
> >>news:1186542647.9...@m37g2000prh.googlegroups.com...
> >>> Dear all:
> >>> Whether the phase difference between the reflection and transmission
> >>> lights of a 50:50 beam splitter is 180 or a modifiable number that
> >>> depends on the coating scheme?
After giving it some thought, what I've come up with is that a 90
degree phase difference is necessary.
Beams A & B are incident on BS #2. Let's say B lags A by 90 degree at
the BS. In one of the output beams, say C, B gets a 90 degree advance
relative to A, so A & B are now in phase for path C. Constructive
interference makes for an actual beam.
In D, the other output beam, it is A that gets a 90 degree advance.
But it already led B by 90, so in path D they are 180 degrees out of
phase, hence no actual beam.
Mark
Neil Bates
But Adam, what about the situation I described adjacently:
Recombine beams of equal phase into a transparent cube with a diagonal made
of thin metal film. From symmetry, there can't be a basis for
distinguishing/favoring phase, correct? (In this case, I think the
reflection inversion of phase would make destructive interference out of
both sides if that were the only consideration. If one beam entered at 180
off, then symmetry about the surface would imply constructive interference
and 2x total original intensity if that were the only consideration.) To
avoid violating conservation of energy, I presume the impedance effects
between atoms of metal would just keep the output the same total energy as
went in. However, I wonder if the phase relations would stay what we expect.
IOW, sometimes a messy outcome, not being able to use direct phase
reasoning?
Another interesting example would be to try frustrated total internal
reflection (same or similar symmetry argument) or microwaves with a mesh
diagonal, etc.
Note: There's a similar discussion up at sci.physics.research, called "Phase
cancellation --where does the energy go?"
Helpful person
> "Adam Norton" <AnortonREMOVET...@ix.netcom.com> wrote in message
I'm afraid that your descriptions are a bit muddled and dificult to
understand. However, Adam's description (after his second message)
are perfectly correct and well explained.
Neil Bates
news:1186965918.3...@j4g2000prf.googlegroups.com...
I was hoping my descriptions weren't muddled, but the claim that the phase
has to be different is simply not true. This isn't too hard: Let's recombine
beams (say, coming together along x and y axes) of same phase (relative to
point of intersection) into a cube with metal-film diagonal. There is
complete symmetry in that arrangement, which is not a traditional beam
splitter with metal coating on one side of a slab of glass.
(BTW, can such beam-splitters be purchased, or are even "laboratory
curiosities"? Could your company Richard Fisher Optical Engineering find or
make such a phase-symmetrical device?)
Hence Nature can't distinguish by say, having light come out along the x
axis and not along the y axis, the way it would in a traditional
Mach-Zehnder interferometer. It also can't distinguish by having phases
different in one output beam than in the other. That is a fundamental
principle of physics, and also the logical principle of sufficient reason.
Hence, energy conservation must be accomplished some other way.
Rene Tschaggelar
> Dear all:
> Whether the phase difference between the reflection and transmission
> lights of a 50:50 beam splitter is 180 or a modifiable number that
> depends on the coating scheme?
>
This shouldn't be too difficult to figute out. Just
solve the maxwell equations to be fullfilled.
The fields have to be continous. and the energy
conserved. Oe of the beams is in one media, the
other beam is in the other media.
Rene
--
Ing.Buero R.Tschaggelar - http://www.ibrtses.com
& commercial newsgroups - http://www.talkto.net
Abe
interferometer, a phase can be changed (distance = phase difference).
In your logic therefore, energy could be created by moving one of the
components or inserting a medium with a slightly different index of
refraction.
On Aug 10, 1:30 pm, "Adam Norton" <AnortonREMOVET...@ix.netcom.com>
wrote:
> Take a collimated laser beam and use one beamsplitter to split it into two
> beams A and B. Recombine the two beams carefully with a second beamsplitter
> to produce two more beams C and D whose intensity depends on the relative
> phase of A and B. However if the phase is adjusted so C has constructive
> interference and is bright, D will be dark with destructive interference.
> This is due to the 180 degree phase change between the reflected and
> transmitted light of the beamsplitter. If it were possible to make a
> beamsplitter with 0 phase difference, then it would be possible to either
> make both C and D disappear or to make their combined energy greater than
> the input energy. Anything other than a 180 deg. phase shift violates
> conservation of energy.
>
> -Adam Norton
>
> "Sa.Sh." <sasha...@telus.net> wrote in message
>
> news:gGOui.83126$tB5.54400@edtnps90...
>
>
>
> > Adam,
>
> > Could yoy please explain your point?
>
> > Thank you,
>
> > Sasha
>
> > "Adam Norton" <AnortonREMOVET...@ix.netcom.com> wrote in message
> >news:13bn268...@corp.supernews.com...
> >> The phase difference is always 180.
>
> >> If it were 0 deg. then you could make an interferometer where the light
> >> disappears entirely, and of course that can never happen.
>
> >> --
> >> Adam Norton
>
> >> Norton Engineered Optics
> >> Optical design and systems engineering for Silicon Valley and beyond.
> >>www.nortonoptics.com
>
> >> (Remove antispam feature before replying)
>
> >> "Chengxian.Zhang" <zhangchengx...@gmail.com> wrote in message
> >>news:1186542647.9...@m37g2000prh.googlegroups.com...
> >>> Dear all:
> >>> Whether the phase difference between the reflection and transmission
> >>> lights of a 50:50 beam splitter is 180 or a modifiable number that
> >>> depends on the coating scheme?- Hide quoted text -
>
> - Show quoted text -
AES
Abe <cambrid...@hotmail.com> wrote:
> This is incorrect. Just by changing a distance anywhere in the
> interferometer, a phase can be changed (distance = phase difference).
> In your logic therefore, energy could be created by moving one of the
> components or inserting a medium with a slightly different index of
> refraction
Guys, I'll not trying to self-promote here, but this is basic, important
and useful stuff, and I tried to explain all these concepts in
reasonably simple fashion in Section 11.1 of my LASERS text.
Neil Bates
"AES" <sie...@stanford.edu> wrote in message
news:siegman-2C7991...@nntp.stanford.edu...
See what I said about special recombiners and equal entering phases. As for
how to get various relative phases, you can just put flat glass in one leg.
Adam Norton
>
> Well, I think you could create the "impossible" phase scheme with thin
> metallic coatings: Recombine beams into a transparent cube with a diagonal
> made of thin metal surface. From symmetry, there can't be a basis for
> distinguishing/favoring phase, correct? To avoid violating conservation of
> energy, I presume the impedance effects between atoms of metal would just
> keep the output the same total energy as went in, but it is interesting to
> think about. IOW, sometimes a messy outcome, not being able to use direct
> phase reasoning? Another interesting example would be to try frustrated
> total internal reflection (same symmetry argument) or microwaves with a
> mesh diagonal, etc.
Neil,
I replied to my own earlier post with a correction, but apparently it did
not get picked up by some news servers or is not easy to spot with some
newsreaders. I will quote it here:
"After pausing to think a little, I have to revise my response. I
realized
that the the real restriction is that phase difference between
transmitted
and reflected incident from one direction and the phase difference
incident
from the other direction have to add to 180 deg.
The example I gave still shows why this is the case."
You are correct that a single-layer film free-standing in air will obviously
have the same phase between incident and transmitted regardless from which
side the light is incident. In that case, to meet the above criteria, the
phase difference simply has to be 90 deg.
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
Neil Bates
"redbelly" <redbe...@yahoo.com> wrote in message
news:1186874749....@l70g2000hse.googlegroups.com...
recombiner. The phase difference you note would violate the requirements of
how the reflective surface works in the case I provided, a thin metal film
diagonal inside a clear block, and its symmetry. What I'm saying is, the
combiner symmetry means that when two beams enter it in-phase, then the
exiting beams must have the same phase and intensity as each other too.
Energy must be conserved by the outputs simply having less energy than we'd
expect from for example two constructively reinforced beams, or more than
from two destructively reinforced beams. IOW, it has to be wrong to imagine
it this way: Consider the individual transmitted and reflected wave
amplitudes and phases of input channel A and ditto for input channel B. Then
combine the amplitudes by the mix from A and B, then square the amplitudes.
What must happen is that interactions in the semi-reflecting film just
require that the output beams have the same energy as input beams, so their
intensities cannot be given by addition of the amplitudes each beam *would*
output separately. It is odd, to wonder what phase light would have, that
would otherwise suffer destructive interference. The symmetry makes it hard
to select out an exit phase. In any case, treating the required phases
simplistically (independently) just gives the wrong answer for total output
power.
Neil Bates
combiner as such, it's the relative phase between transmitted and reflected
beams. Iff (sic) the difference is consistently 90° on both sides, as
redbelly noted, then the added amplitudes squared will indeed give the
correct output power. But that isn't (?) what we expect from a *thin* metal
film diagonal inside a glass cube. We expect a 180° change upon reflection.
Depending on relative input phase, that leads to results varying from both
channels getting constructively reinforced (simplistic amplitude addition
would provide 2x input power) to both destructive, to in between. And if
there was a zero degree change, there'd be an analogous problem. I always
see reference to either zero or 180° phase change, not others (and even if
90° was available, what if we used something that wasn't?) So, either we
find a way to get 90° phase change on reflection - how? What does that? If
not, then we can't just add up amplitudes naively, square them, and get
output powers.
BTW this is interesting, about a mirror that does not change reflected
phase:
http://ieeexplore.ieee.org/Xplore/login.jsp?url=/iel5/10387/33016/01548109.pdf?arnumber=1548109
dvan...@calvin.edu
wrote:
> "Neil Bates" <neil_del...@caloricmail.com> wrote in message
>
> news:13bq51s...@corp.supernews.com...
>
. . .
> 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.
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.
But metallic films have been raised in this discussion, and they are
*not* lossless, and the conservation-of-enrgy 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.
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 slection of the
complex refractive index, and thickness, of the film) a phase shift
intermediate between these two can be achieved.
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
Neil Bates
<dvan...@calvin.edu> wrote in message
news:1187363482....@k79g2000hse.googlegroups.com...
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.)
Neil Bates
recombiner with say metal film inside a cube is used (compare use of
frustrated total internal reflection.) A conventional Mach-Zehnder uses
plates with coatings on the surfaces, and the combiner is not "symmetrical"
like the special kind. In the traditional Mach-Zehnder, if input channels
are in phase, then output channel A is "bright" and output channel B is
"dark" (or vice versa, if adjusted accordingly.) But compare that to the
symmetrical cube recombiner, which *cannot* work like that unless it behaves
in a more complicated way than a simple negligibly thin window/mirror. Even
if does so, is that going to make the exact amount of phase shift to allow
for simple addition of amplitudes giving the correct power output?
redbelly
> "Neil Bates" <neil_del...@caloricmail.com> wrote in message
>
> news:13ca1u9...@corp.supernews.com...
>
>
>
> > "redbelly" <redbell...@yahoo.com> wrote in message
I think, as Prof. Siegman and Rene noted, one needs to review the
details of the math -- or at least I need to. Our arguments may be
simplified and overlooking something. The more I think about it, the
more I realize I'm rather rusty at working out the details without
referring to a book.
Regards,
Mark
Jürgen Appel
[Beam Splitter]
> After pausing to think a little, I have to revise my response. I realized
> that the the real restriction is that phase difference between transmitted
> and reflected incident from one direction and the phase difference
> incident from the other direction have to add to 180 deg.
Proof:
Let E1 = E1in exp(i w t) + c.c. and
E2 = E2in exp(i w t) + c.c be the two incident monochromatic electric
fields and
E3 = E1out exp(i w t) + c.c. and
E4 = E2out exp(i w t) + c.c the two emerging fields.
If the beam splitter is treated as a linear lossless optical element, E3, E4
must be related by a linear relation to E1, E2:
(E3) (E1)
(E4) = A (E2), where A is some 2x2-matrix. That means that
(E1out) (E1in)
(E2out) = A (E2in). (*)
Since the beamsplitter is lossless, the power flowing into it
|E1in|^2 + |E2in|^2 must equal the optical power emanating
|E1out|^2 + |E2out|^2.
Multiplying equation (*) with its conjugate transpose therefore leads to
A * A^+ = 1; so the matrix A has to be unitary.
(a b)
So if A=(c d), that means that c a^* + d b^* =0. Writing
a=|a| exp(i p_a), b=|b| exp(i p_b), ...
one obtains |c||a| + |d||b| exp(i*(p_a-p_b-p_c+p_d)) = 0 and therefore
p_a - p_b - p_c + p_d = pi (=180°).
This is the only restriction on the phases. Assuming real input amplitudes
your claim follows.
For the amplitudes one of course has
|a|=|d|, |b|=|d|, |a|^2 + |b|^2 =1 (Energy conservation).
Cheers,
Jürgen
--
GPG key:
http://pgp.mit.edu:11371/pks/lookup?search=J%FCrgen+Appel&op=get