Interferometer and Interference.

[binbagsss]

I have some questions regarding a fabry perot interferometer. (they aren’t problems as such however I have had a good look around the internet and am struggling to find what I’m looking for).

1) Why is it that a circular fringe pattern is observed. How does a single source interfere with itself and is there variation in the pattern for a given wavelength as the distance between the plates is varied.

2) Am I correct in thinking that the term fringe visibility can only be used when the source is not monochromatic – e.g if there are two wavelengths present, at a given distance between the two plates, two independent ring systems are produced and fringe visibility refers to whether the bright and dark bands of each system coincide or not, and this is dictated by the distance between the plates. But then my next question is…

3) If the source is monochromatic and so changing the distance can not influence the fringe visbiity, what oservable affects does it give arise to?

Thanks a lot for any assistance on this. Really Appreciated.

 

[Simon Bridge]

Your question seem to be answered in wikipedia.

1) Why is it that a circular fringe pattern is observed. How does a single source interfere with itself and is there variation in the pattern for a given wavelength as the distance between the plates is varied.

Interference can happen from a single source if there are at least two paths that light (or whatever) could be travelling. The interference comes from the phase difference between the paths.
http://en.wikipedia.org/wiki/Fabry–Pérot_interferometer
... see the diagrams of the possible paths.

2) Am I correct in thinking that the term fringe visibility can only be used when the source is not monochromatic – e.g if there are two wavelengths present, at a given distance between the two plates, two independent ring systems are produced and fringe visibility refers to whether the bright and dark bands of each system coincide or not, and this is dictated by the distance between the plates.

http://en.wikipedia.org/wiki/Interferometric_visibility
I'm reading that as a strict "no": for some interference patters the intensity is not zero at every minima.
This happens when the the interfering waves are not equally bright.

In the fabry perot system - there is attenuation (due to reflection/transmission) in each path - so the waves which travel farthest will be dimmer.

 

[binbagsss]

Your question seem to be answered in wikipedia.
Interference can happen from a single source if there are at least two paths that light (or whatever) could be travelling. The interference comes from the phase difference between the paths.
http://en.wikipedia.org/wiki/Fabry–Pérot_interferometer
... see the diagrams of the possible paths.

From what I can see this does not explain why it is a circular pattern that is produced.


[/QUOTE]In the fabry perot system - there is attenuation (due to reflection/transmission) in each path - so the waves which travel farthest will be dimmer.[/QUOTE]

But does this have negligible influence on the ring pattern obtained, which is primarily caused by the different path lengths traveled by the light.?

 

[Simon Bridge]

From what I can see this does not explain why it is a circular pattern that is produced.

Circular lenses.

But does this [waves which travel farthest will be dimmer] have negligible influence on the ring pattern obtained, which is primarily caused by the different path lengths traveled by the light.?

It does not change the pattern of light and dark fringes - it just affects the brightness of the fringes.
Imagine you have 2-source interference between a bright and a dim source - everything else the same. Where you have destructive interference to give a minima, the dim source will not be able to completely cancel the bright one... but the minima are still in the same places. But if you changed the separation of the two sources, the pattern would change.

 

[binbagsss]

Circular lenses

Ahhh okay, does anyone know where I can get more detail on this?

.It does not change the pattern of light and dark fringes - it just affects the brightness of the fringes.
Imagine you have 2-source interference between a bright and a dim source - everything else the same. Where you have destructive interference to give a minima, the dim source will not be able to completely cancel the bright one... but the minima are still in the same places. But if you changed the separation of the two sources, the pattern would change.

Ahh thanks, that makes sense. My confusion stemmed from 'which travel farthest will be dimmer', rather than the initial conditions being a dim and brighter source. And therefore the intensity at a given point on the detecting medium would be influenced in two ways, the first being the above - whether it interferes destructively or constructively with other rays, the second being my interpretation of this, its intensity once reaching the screen due to attenuation.

In which case, the major contribution of the intensity at a given point would be due to the first reason?

 

[binbagsss]

Also you mentioned the circular pattern being due to the shape of lens, however I have read on a few sources that if the plates are not exactly parallel the fringe pattern is not circular. Are the two connected?

 

[binbagsss]

Sorry one more question, does the interference occur in the gap between the plates or once the rays have emerged?

 

[Simon Bridge]

Circular lenses

Ahh thanks, that makes sense. My confusion stemmed from 'which travel farthest will be dimmer', rather than the initial conditions being a dim and brighter source.

In the interferometer you are looking at, you have a single uniform source. The interference comes from the path difference as there are several ways light can get from the source to the screen, each involving a different number of reflections. At each reflection, some of the light is lost (and so cannot contribute to the interference). The more reflections the longer the path the light has to take and the more of the light is lost ... so: the light which travels farther will be dimmer.

And therefore the intensity at a given point on the detecting medium would be influenced in two ways, the first being the above - whether it interferes destructively or constructively with other rays, the second being my interpretation of this, its intensity once reaching the screen due to attenuation.

That's the one.

In which case, the major contribution of the intensity at a given point would be due to the first reason?

For this interferometer, yes. Interferometers are usually engineered to make sure that there are well-defined fringes so they want to minimize the attenuation difference between paths. (They also like to minimize the attenuation period.)

Also you mentioned the circular pattern being due to the shape of lens, however I have read on a few sources that if the plates are not exactly parallel the fringe pattern is not circular. Are the two connected?

It is due to cylindrical symmetries in the optical components used to build the interferometer. One of these components is the lens.

You seem to be looking at the crossection diagrams in resources and thinking they are rectangular in the third dimension. You should try sketching what happens for overhead and side views for different paths. Explore.

The change in pattern with misaligned plates is related to Newton's rings.
If you have two flat sheets of glass parallel to each other and shine monochromatic light on them you don't get anything interesting, but if you tilt one of them slightly you get a row of interference fringes.

I think you need to get hold of the equipment and try this stuff.

Sorry one more question, does the interference occur in the gap between the plates or once the rays have emerged?

In the wave model being used for this discussion, interference appears everywhere there are waves to interfere with each other .. look at waves in a ripple-tank. What the screen does is intercept part of the overall 3D interference. The interference pattern is at the screen.