Beam Splitter: were is the power?

[lpfr]

If the path difference is not a rational factor of lambda, I don't think that a 50:50 beam splitter ratio is even possible.
It is not clear to me, how there could be an arbitrary phase difference, if interferences are supposed to occur.

There is the optical path for constructive multiple reflections and the phase lag of the transmitted beam due to the thickness of the layer and a variable parameter: the refraction index of the layer.
I have not made a program to calculate this, so I'm not sure if you can satisfy the two conditions. But then, I do not know how real BS are really made. Maybe you can introduce others parameters if you do (say) a sandwich of dielectrics instead just a layer.

 

[Cthugha]

There is the optical path for constructive multiple reflections and the phase lag of the transmitted beam due to the thickness of the layer and a variable parameter: the refraction index of the layer.
I have not made a program to calculate this, so I'm not sure if you can satisfy the two conditions. But then, I do not know how real BS are really made. Maybe you can introduce others parameters if you do (say) a sandwich of dielectrics instead just a layer.

Ok, I see what you mean.
In the postings before I assumed, that the discussion was just about symmetric BS. Asymmetric BS can of course have any arbitrary phase difference.
Sorry if there were any misunderstandings. I am not as used to reading english texts as I would like to be.

However, this link might be interesting for Alexander:

http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=AJPIAS000068000002000186000001&idtype=cvips&gifs=yes"

This is a calculation of several BS designs, including phase shifts, coatings and such stuff. I haven't read it completely, but it might be of use.

 

[Edgardo]

Hello Alexander-1,

I once read an explanation for the Mach-Zehnder interferometer on why one output has zero intensity while the other has the full intensity of the original lightbeam. (The Mach-Zehnder interferometer is the the setup in your post #5, or see also here)
Instead of beamsplitter cubes, half-silvered mirrors are used.

Unfortunately, the website (http://www.upscale.utoronto.ca/GeneralInterest/Harrison/MachZehnder/MachZehnder.html) is not available anymore but you can still view the website in the google cache:
Website from google cache http://209.85.129.104/search?q=cach...ehnder+how+it+works&hl=de&ct=clnk&cd=10&gl=de (takes a minute to load)
The image belonging to the website can be found here (type in google "mach-zehnder upscale" and do an image search, see the first image)

Although the paths in the interferometer look symmetric, they are not. They only look symmetric if you assume "infinitely thin mirrors", but you have to take into account that the mirrors have a thickness. See here, the top mirror on the right side. The paths at that mirror show that the situation is asymmetrical for the two beams, so you would have to change your picture http://physics.nad.ru/img/bs2.gif.

You may have a look at this paper ("How does a Mach-Zehnder interferometer work"): http://www.iop.org/EJ/abstract/0031-9120/35/1/308
It could give some explanation of the phase shifts.

You said about "reflections at both edges".. What edges? If we consider "Beamsplitter Plate" (semireflecting mirorr), then the glass plate is covered by reflecting metal or dielectric coating from one side, while the other side is covered by an anti-reflection coating. So reflection occurs just at one side of the plate. In cubic splitter two right-angle prisms are cemented together at their hypotenuse faces; one of such faces are covered by reflecting layer. So there is no two edges too.

For your beamsplitter cube, I first thought that both sides of the triangle shaped pieces are reflecting. But if only one side is reflecting as you said, then the situation is similar to the half-silvered mirrors.

I think you should definitely have a look at Cthugha's links and why the phase difference between the reflected and the transmitted beam at a half-silvered mirror is pi/2.

 

[lpfr]

I once read an explanation for the Mach-Zehnder interferometer...

I think you should definitely have a look at Cthugha's links and why the phase difference between the reflected and the transmitted beam at a half-silvered mirror is pi/2.

This thread is NOT concerned with interferometers and still less with the Mach-Zehnder interferometer.

I did have a look to the only of the Cthugha's links on the web. Have you read it? I doubt.
It has no more to do with this thread than the Mach-Zehnder interferometer.

 

[Cthugha]

This thread is NOT concerned with interferometers and still less with the Mach-Zehnder interferometer.

I did have a look to the only of the Cthugha's links on the web. Have you read it? I doubt.
It has no more to do with this thread than the Mach-Zehnder interferometer.

This must be the reason, why alexander, who started this thread, posted a picture of a Mach-Zehnder interferometer in post #5...

However, the other picture is about a symmetric BS design with a phase shift of pi. Mach-Zehnder interferometers use symmetric BS and are therefore absolutely not off topic.

You should also mention, that waveguides, resonators and single dielectric interfaces have absolutely nothing to do with the original topic. Especially single interfaces are strictly asymmetric.
The most important point here is, that it is not possible to achieve a phase difference different from pi/2 with a symmetric design. I calculated some designs and found all symmetric designs to show a phase shift of pi/2 and all other designs to be asymmetric.
Of course that is not a guarantee. If you find a different design, post it.

 

[Manchot]

Anyhow, if you have a conductor without ohmmic losses, but with few free electrons, the wave created by the oscillating electrons will not have the amplitude needed to "reflect" totally the incoming wave.

Well then, that's not a perfect conductor, is it? (Typically, a perfect conductor is defined as an object with infinite conductance, such that any applied fields are canceled out immediately. Both the mobility and free electron density factor into conductance, so an object with high mobility and a low free electron density will still have a mediocre conductance.)

As for dielectric splitters, keep in mind that the formulas you posted earlier only apply to single interfaces. Multiple interfaces require the use of propagation matrices to fully solve.

 

[Alexander-1]

Links, which "Cthugha" sent in the beginning of this thread, shows that in a lossless symmetric beamsplitter the phase shift between the reflected and transmitted beams equals pi/2 for any value of the splitting ratio. In the meanwhile the physical mechanism of this effect was not described. Maxwell's Equations give this phase shift equal to pi if the light is incident from the air on the medium with refraction index n>1 and if the angle of incidence is less than Brewster's angle. So, if we consider the BS in the form of a simple glass plate at angle 45 deg to the incident beam, then the reflected beam must be in antiphase to the incident beam and the transmitted beam must be in phase to it. There is a contradiction. The light propagating in the opposite direction can not influence the phase of reflected beam because of linear properties of the medium. The reason may be in a design of BS. The matter is that the BS in the form of glass plate will have a thickness and two reflecting surfaces. Careful calculations may give a correct result. So, I shall try to think in this direction.

 

[Cthugha]

So, if we consider the BS in the form of a simple glass plate at angle 45 deg to the incident beam, then the reflected beam must be in antiphase to the incident beam and the transmitted beam must be in phase to it. There is a contradiction. The light propagating in the opposite direction can not influence the phase of reflected beam because of linear properties of the medium. The reason may be in a design of BS. The matter is that the BS in the form of glass plate will have a thickness and two reflecting surfaces. Careful calculations may give a correct result. So, I shall try to think in this direction.

That's right. Even a simple thin glass plate shows two reflections. Try triggering a photodiode with a part of a laser pulse divided by a thin glass plate. Both reflections will occur and trigger, if you are want to trigger in a matching timerange.

My calculations showed me the following:

A single interface is not a model for a symmetric BS. Strictly speaking a single interface would mean, that there is just air on one side and just dielectric on the other side, which is clearly asymmetric. In this case Fresnel's formula can be applied as shown before by lpfr.

An antireflection coating just on one edge of the BS does almost the same. There is no symmetry for beams coming from different directions. You would need antireflection coatings on more than one edge to get a symmetric design.

Any symmetric design leads to a phase shift of pi/2.
You can calculate that very easily. Just keep in mind, that there is a phase shift of pi, when an reflection happens at an air-dielectric interface, that there is no phase shift, when an reflection happens at an dielectric-air-interface, that antireflection coatings work using destructive interference and usually have a thickness of lambda/4, that the thickness of the bs is very important and that reflections from all edges should show constructive or destructive interference (depends on the design and defines the needed thickness).
I found no symmetric design, which satisfies all of these boundary conditions and has a different phase shift.

 

[Alexander-1]

An antireflection coating just on one edge of the BS does almost the same. There is no symmetry for beams coming from different directions. You would need antireflection coatings on more than one edge to get a symmetric design.

Any symmetric design leads to a phase shift of pi/2.
You can calculate that very easily. Just keep in mind, that there is a phase shift of pi, when an reflection happens at an air-dielectric interface, that there is no phase shift, when an reflection happens at an dielectric-air-interface, that antireflection coatings work using destructive interference and usually have a thickness of lambda/4, that the thickness of the bs is very important and that reflections from all edges should show constructive or destructive interference (depends on the design and defines the needed thickness).
I found no symmetric design, which satisfies all of these boundary conditions and has a different phase shift.

One approach consists in consideration of the plate one side of which is covered by reflecting layers and the oposite side is covered by anti-reflection layers. Reflection layers will divide a beam strictly in a half, while anti-reflection layers will suppres reflection and amplify transmission. After careful calculations I expect to see pi/2 in reflected beam and 0 in transmitted beam. If such a system of reflected and antireflected layers are close to ideal, then there will be no interferens between the beams reflected by opposite sides of the plate. For the time being these are only the words and I hope to prove them by calculations.

The other approch is consideration of totally symetrical interferometer. This can be done if we use simple glass plate at 45° without any coatings and the source with the coherence length less that the optical thickness of the plate. In this case interference will be observed between the beems with the same optical length. So, beams reflected by the front surface of the plate (phase shift is pi) will interfere only with the beams reflected by the rear surface of the plate (phase shift is 0). This gives MAX in one arm and MIN in other arm. It would be better to draw a figure here.

 

[Edgardo]

Here is a paper that mentions the pi/2 phase difference between reflected and transmitted beam:
1) C.H. Holbrow, E.J. Galvez and M.E. Parks, "Photon Quantum Mechanics and beam splitters," Am. J. Phys. 70, 260-265 (2002)
http://departments.colgate.edu/physics/research/Photon/root/photon_quantum_mechanics.htm

Other papers about beam splitters:
2) A. Zeilinger, “General Properties of Lossless Beam Splitters in Interferometry,” Am. J. Phys. 49, 882-883 (1981)

3) "A quantum description of the beam splitter"
A Luis et al 1995 Quantum Semiclass. Opt. 7 153-160
http://www.iop.org/EJ/abstract/1355-5111/7/2/005