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Cause of interference pattern in the Michelson Interferometer.

by Zanychap
Tags: interferometer, michelson
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Zanychap
#1
May4-11, 07:53 AM
P: 7


The books say that interference happens basically between the light waves from the two mirrors. But it seems to me that the path difference between the pairs remains constant for all the waves (as long as both the mirrors are perpendicular to each other). So why do circular rings form? How does the path difference vary?
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Feldoh
#2
May4-11, 10:24 AM
P: 1,345
You need to vary the length of one of the mirrors over very small differences. Typically micrometers will do it for the visible light spectra.
schip666!
#3
May4-11, 11:24 AM
P: 595
The point of the Interferometer is to see _very_ small differences in length and changes thereof. In general, I would challenge anyone to set one up such that it doesn't have fringes...

Zanychap
#4
May4-11, 11:43 AM
P: 7
Cause of interference pattern in the Michelson Interferometer.

The point that I am not getting is why are the fringes circular.
Like in Newton's rings I know that the thickness of air is constant over concentric circles causing circular rings. I just can't understand what causes this in Michelson's.
schip666!
#5
May4-11, 11:47 AM
P: 595
Ah...

I think it's because the wave-front of both beams expands in a cone, so the travel path is different at each point on the detector screen.
jtbell
#6
May4-11, 11:54 AM
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P: 11,778
The usual simple analysis gives you the interference condition when the light goes through exactly perpendicular to the mirrors. This is the center of the actual interference pattern. At other points on the screen, the light reflects off the mirrors at a small angle. This makes a longer path which is generally slightly different in length for the two mirrors.

http://www.phy.davidson.edu/stuhome/.../Michelson.htm
Zanychap
#7
May4-11, 12:29 PM
P: 7
Thanks.


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