B Rayleigh Interferometer: Impact of Slip Placement on Interference Pattern

AI Thread Summary
The discussion centers on the placement of slits in a Rayleigh Interferometer and its effect on the interference pattern. Participants argue that moving the slits in front of the converging lens instead of behind the collimating lens should not change the interference pattern, as the geometric and optical path lengths remain unaffected. However, real-world factors like diffraction and the size of the slits can influence the resulting pattern, particularly if the slits are small. The conversation also touches on the need for clearer references and images of interferograms to better understand the outcomes. Ultimately, the consensus leans toward minimal impact on the interference pattern under ideal conditions, but acknowledges potential variations in practical scenarios.
HotFurnace
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What will change if we place the slips in front of the converging lens instead of behind the collimating lens? My teacher say nothing will change in the interference pattern but I doubt his answer.
 

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I don't think this is like Young's double slit apparatus, where the distance between the slits and the screen affects the size of the diffraction pattern.

Moving the slips here does not change the geometric path lengths. Only the optical path lengths are changed, by the material in the tubes.

If you consider any point on the screen, neither the geometric nor the optical path length from the pinhole to the screen is affected by the position of the slips.

Looking at the ray diagram the light is a parallel beam through the tube region between the two possible slip positions. So exactly the same rays reach the screen irrespective of which position the slips are in.
 
HotFurnace said:
My teacher say nothing will change in the interference pattern but I doubt his answer.
I have my suspicions too, although there is a sort of reciprocity argument that justifies the statement. The slips (slits?) are, presumably there to suppress off axis beams entering the two gas tubes and causing crosstalk between the two paths. If they are downstream of the tubes, would they achieve the same effect? I am not familiar with the Rayleigh Interferometer and I may have misunderstood what they do. The problem I am referring to would be eliminated if the two tubes were run through holes in an opaque block.
Google hasn't found a link for me that's detailed enough but not too detailed to help me. (Ain't that always the way?)
 
Sorry everbody, I almost forget the thread!
If the slits are large relative to light wavelength (sorry it's not slips) are to eliminate the off axis beam then yes the interference will be the same, but if the slips is small then what will happen? I guess if the slits are placed in front of the converging lens, then we will observe a interference pattern similar to the one we observed with a large slits, however if the slits are place behind the collimating lens then I think the only visible interference will be a small spot at the center, bright or dark depending on the gas contained inside the tubes.
According to wikipedia, the slips are there to split the incoming beams into two beams?
 
HotFurnace said:
View attachment 236575
What will change if we place the slips in front of the converging lens instead of behind the collimating lens? My teacher say nothing will change in the interference pattern but I doubt his answer.

In the ideal case, your teacher is correct- I am assuming that the slits are actually masks. In the ideal case, the pinhole and collimating lens work to generate a perfect plane wave and the masks do not create significant diffraction within the apparatus. If the gas tubes have plane-parallel sides at normal incidence to the wavefront, there will be no change in the direction of the beam as it propagates through the gas tubes. Then, the apertures are simply present to mask off that part of the wavefront that does not encounter either gas tube.

Now, in real life the incident beam is not a perfect plane wave, there is not a single wavevector that characterizes the illumination. The masks will slightly diffract the incident beam; the specific location of the masks matter- not just if they are on either side of the gas tubes but also how far they are from either the collimating or converging lens.

Note- as drawn, the interfereogram will be the same size as the pinhole unless the lenses have differing focal lengths.
 
Andy Resnick said:
Note- as drawn, the interfereogram will be the same size as the pinhole unless the lenses have differing focal lengths.
That seems to imply that the interferogram would be a sinx/x type pattern - due to the lens aperture- or perhaps the aperture of the gas tubes. A google search doesn't seem to reveal the shape of the actual pattern but it does seem to suggest from the plan and elevation diagrams in some links that each gas tube is above one axis and that there is an air path reference 'under' each tube so that two gases can be compared with air.
I need a noddy reference - but not too noddy as to omit the most important bits!
 
sophiecentaur said:
That seems to imply that the interferogram would be a sinx/x type pattern - due to the lens aperture- or perhaps the aperture of the gas tubes. A google search doesn't seem to reveal the shape of the actual pattern but it does seem to suggest from the plan and elevation diagrams in some links that each gas tube is above one axis and that there is an air path reference 'under' each tube so that two gases can be compared with air.
I need a noddy reference - but not too noddy as to omit the most important bits!

If the masks are square holes (and the gas tubes are square prisms), then each path will result in, as you say, a sinc(αx) type diffraction pattern. However, each path also has a phase factor, so the overlap of each path will result in an overall intensity pattern that varies as I0cos2(Δ)sinc(αx), where Δ is the relative phase between the two paths. That relative phase Δ will only depend on the two gases *if* the gas cells are symmetric about the optical axis.
 
Andy Resnick said:
That relative phase Δ will only depend on the two gases *if* the gas cells are symmetric about the optical axis.
That makes sense to me but some of the diagrams on Google Images seem to divide the field into quadrants with two dummy air channels. I guess you are suggesting that the diagonally opposite air path would be used as reference for each gas tube. I must say, I was very disappointed not to find some actual images of the interferograms with a search. After all, that's what it's all about.
 
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