Transmission and reflectivity in a Michelson interferometer.

[thomas19981]

Homework Statement

Sketch and explain the operation of the Michelson interferometer.
How is the transmission of the interferometer modified if the amplitude transmission t and reflectivity r of the

partially-reflecting beamsplitter are not equal? What happens to the fraction of light that is not transmitted by the instrument?

Homework Equations

Not really any maths[/B]

The Attempt at a Solution

So I have managed to explain the Michelson setup and I said that the light not transmitted it sent back to the source so it isn't received at the detector after the two rays combine. What I'm unsure about is the second part where it asks about the setup modification. I assumed that since r doesn't equal t then the intensity received at the detector isn't as strong so the only thing that they'd change is to use a stronger source. But I feel as though they would have used the strongest source avalible in the first place. Is this right?
Thanks in advance.

 

[BvU]

Not really any maths

Well, an expression for the on-axis amplitude at the observer, perhaps ?

 

[Charles Link]

From the looks of it, they are just trying to give you a quick introduction to the Michelson interferometer. For its operation, the beamsplitter is used twice, and the same processes could actually be performed by using two beamsplitters. In its first use, the beamsplitter splits the wavefront into two separate, but mutually coherent beams. The beams then travel a short distance on two different paths, ( each to a mirror and back), that can have different lengths resulting in a phase difference. The two beams are then incident on the beamsplitter again (and here is where a second beamsplitter could be used instead). The emerging beams ( each of which results from a partial reflection and partial transmission of the two incident beams), can take two directions, (back to the source or to the receiver), and there is interference in both cases. $\\$ It is possible to use two different types of sources in the Michelson interferometer= 1) A plane wave so that the beam seen at the receiver is a uniform plane wave, but varies in intensity as the path difference between the beams is changed by moving one of the mirrors. 2) A diffuse source=a scatterer (a frosted glass) is put in the path of the source (before it is incident onto the beamsplitter the first time) so that the interference is between point sources from the two images that appear=the result is a series of bright and dark rings, where the location of the bright rings depends upon the difference in path distance that the light from the point sources needs to travel before it is recombined by the second pass onto the beamsplitter. $\\$ In the case of the plane waves, if there is constructive interference for the two partial emerging beams(wavefronts) in the direction of the receiver, then there is necessarily destructive interference between these two wavefronts in the direction of the source. The result is that energy is conserved, and what doesn't show up at the receiver winds up going back to the source. By changing the path length, destructive interference can be made to occur at the receiver, and constructive interference at the source. One question you may have is , why the asymmetry? This is a more advanced topic, but the answer is the beamsplitter normally has an AR (ant-reflection) coating on one face (so that one face of the beamsplitter is 100% transmissive), and the other face is where the beamsplitting occurs. For the beam that gets split by being incident from the air (lower index) side, there is a $\pi$ phase change. The result is there is an asymmetry to the problem with the result that constructive interference occurs at the receiver when destructive interference occurs in the direction that is back to the source. (Probably more advanced than necessary for the introductory level, but necessary to try to make the explanation semi-complete) . One more item: In the case of the diffuse source, the ring patterns in the two directions will be opposites of each other=a bright ring at the receiver has a dark ring when viewed in the direction of the source and visa versa. Once again, energy is conserved in the Michelson interferometer interference with the result that the energy in the source is equal to the sum of the energies that goes to the receiver plus that which returns to the source. $\\$ (In general, the homework helpers are not supposed to give complete solutions, but this is a somewhat specialized homework problem, and I don't know of any textbook, (even any advanced Optics text), that really gives a thorough and complete treatment of this subject, either at the introductory level, or at the advanced level).