# Do interference rings disappear in an interferometer if the path lengths are identical?

Consider a standard Michelson Interferometer (I took the picture from http://en.wikipedia.org/wiki/Interferometer)

The incident beam is split into two parts, where the two parts travel on different paths to the screen. If the paths are of different length and the incident wave is a spherical one, one sees the interference ring pattern at the detector. I now have three questions:

1.) What happens if the path lengths are exactly identical. Do I then still see the interference pattern?

2.) How does the patter on the screen change if I then slowly change one of the path lengths?

3.) What happens if I use plane waves instead of spherical ones? If the difference between the two paths corresponds to a phase difference of $\Delta \phi= \pi$ do the plane waves cancel out at the screen?

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So why didn't you get the answers from Wikipedia while you were at it ? – user26165 Oct 26 '13 at 5:40
I didn't find an explanation about what happens if the path lengths are identical or almost identical (i.e. only a difference in the order of the wavelength of the incident radiation). From Wikipedia I got the impression that one always sees these interference rings, which I can't understand in the case when the path lengths are almost identical. That's why I'm asking what one sees in that case. If you think the answers are trivial I would nevertheless very much appreciate it if you could answer my question. – Peter Oct 26 '13 at 9:17

The answer is no, you don't get any interference rings if both arms of the Michelson are identical and the beamsplitter is perfectly flat.

The answer with real optics is somewhat more complicated. Real laser beams have curved wavefronts (as opposed to plane) whose curvature changes as they propagate. If you take two laser beams of the same frequency but different curvatures and overlap them on a camera, then you will get interference rings like shown in your wikipedia diagram, two laser beams of the same curvature give an interference pattern with no rings.

In a Michelson interferometer the things which could cause the two returning beams to have different curvatures are numerous but the most common ones are: path length differences between the arms, curvature of the two end mirrors, or curvature of the beamsplitter.

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The curvature of wavefront comes from diffraction, so if beam is narrow, it always will have a curved edges, so it will interfere differently on different distances from the center. – user2622016 Jan 10 '14 at 17:12
The curvature of the wavefront is indeed due to diffraction, but the symmetry of a perfect Michelson ensures that the beams at the output have the same curvature. Two overlapped fields of identical curvature still give an interference pattern without any rings. – Chris Mueller Jan 10 '14 at 17:22

First consider only the center point of the screen.

• Identical lengths: red dot.
• Length differ by $\lambda/2$: black dot.

Light source is a laser then you can add any natural number of wavelengths to one distance, and still result will be the same. If light source is for example lightbulb with lens making parallel rays and a "pupil" making the beam narrow, then the effect will disappear when difference of lengths is around or grater than length of a single wave packet. If I recall correctly one wave packet from lightbulb has length around $100 {\mu}m$.

Ad 2) Separation of interference fringe will not change, but fringes will gradually move from center or to center.

Ad 3) There will be no interference fringes. Whole screen will have the same intensity as the center point.

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Whether the centerline path length is identical or differs by a whole number of wavelengths does not matter. You will still get a bright spot in the center. If the setup is perfect as you show, with the beamsplitter exactly at $45^\circ$ and incidence on the mirrors and screen perfectly perpendicular, you will not have any fringes because all lengths do not change. Fig 1 in the Wikipedia article has a sketch that shows the angles not perfect. In that case, the path length will vary over the area and you will see fringes.

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