# Beam Splitter: were is the power?

by Alexander-1
Tags: beam, power, splitter
 P: n/a 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. Attached Images
Sci Advisor
P: 1,414
 Quote by Alexander-1 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.
P: n/a
 Quote by Cthugha 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 devide 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.
 P: 681 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/physi..._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

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