Setting up Michelson Interferometer

[mmwave]

I was setting up a Michelson Interferometer the other day & was trying to adjust it so the movable & fixed paths are indentical in length. I could not really tell when they were identical so I asked a colleague. He told me that as the path lengths become equal, the spacing between the rings in the interference pattern get larger.

I don't believe this can be true. The rings come from the spherical mirror that changes the (approx.) planewave beam into a varying phase wavefront so you get a bullseye pattern instead of a single light or dark spot. The spacing is determined by the curvature of the lens.

Does anyone agree with this?

Also, is there an easy way to tell when the two path lengths are equal? I found it's not reliable to judge by the "darkest dark" or "brightest bright" in the bullseye.

 

[Integral]

I think you need to listen to your colleague. If the apparatus is set up correctly you should be seeing interference fringes, these are due to the difference in optical path length between the two arms, not due to the shape of the source lens.

I do not think that you can ever know when the arms are the same length, only whether they differ by a integral number of wave lengths or not. If there is to much difference between the arm lengths you will not get fringes, this is determined by the coherence length of the light source. There is a range where you will get fringes, the arms are the same length somewhere in this range. It is not clear to me that there is a guaranteed method of finding the point of equal length.

 

[mmwave]

Originally posted by Integral
I think you need to listen to your colleague. If the apparatus is set up correctly you should be seeing interference fringes, these are due to the difference in optical path length between the two arms, not due to the shape of the source lens.

The phase difference across the beam cannot depend on the difference in path length in the two arms. That is the same at all points across the beam. For easier viewing, a spherical lens is placed in the beam so that as you move from the beam center to the outside edges there is a phase shift. This turns the interference pattern from spot of light or dark to a bullseye pattern. In a Fabry-Perot system multiple reflections give the bullseye pattern w/o a spherical lens.

 

[mmwave]

Originally posted by mmwave
The phase difference across the beam cannot depend on the difference in path length in the two arms. That is the same at all points across the beam. For easier viewing, a spherical lens is placed in the beam so that as you move from the beam center to the outside edges there is a phase shift. This turns the interference pattern from spot of light or dark to a bullseye pattern. In a Fabry-Perot system multiple reflections give the bullseye pattern w/o a spherical lens.

If I had drawn a diagram you would have laughed me off this forum.

Since the spherical lens is placed before the splitter its phase shift is not the cause of the bullseye pattern, it just enlarges the beam to make it easier to see.

The cause of the bullseye pattern (never explained in any book I've seen)is multiple reflections that do not perfectly coincide. For example, reflection off the outer surface of the glass backing on the splitter. Like the infinite chain of images you see in two mirrors. If you look closely at your beam (w/o spherical lens) you can make it small but never a single round dot. You'd think careful adjustment of the system could eliminate this but in practice not.

 

[krab]

mmwave, you are right; the fringe separation does not depend on the difference in length of the 2 arms. In order to get them equal, use white light. The fringes wash out if the path lengths are more than a few wavelengths different, because there is a range of wavelengths present.

 

[mmwave]

Originally posted by krab
mmwave, you are right; the fringe separation does not depend on the difference in length of the 2 arms. In order to get them equal, use white light. The fringes wash out if the path lengths are more than a few wavelengths different, because there is a range of wavelengths present.

This makes sense. I think the coherence length of white light is around 100nm? The resolution of my micrometer with reduction gearing is 2 um. It would be nice to have a source with a coherence length of about about 20 um. Thanks for the suggestion.