My friend and I recently did an experiment on Lasers where we shot a laser beam through a Michelson interferometer and observed the interference pattern on the wall. Here is the basic setup:


One basic thing we don't understand is why when the two arms of the Michelson interferometer are at the same distance, there is a spherical interference pattern. When we block one of the paths, we see a single dot of red light on the wall. It seems like when you unblock both paths, the two beams should constructively interfere, and a dot of red light of higher intensity would appear. But our observations show the circular interference pattern we mentioned earlier. So what is actually the correct model for the laser light?

I read the following link: Do interference rings disappear in an interferometer if the path lengths are identical? and I don't completely understand it.

The main thing I don't understand about the link is even if the two wave fronts have different curvatures, if you look at a particular point on the wall, shouldn't the waves have traveled the same distance on the two paths? Or perhaps I'm misunderstanding where the wave front on each path first gains some curvature.


marked as duplicate by ACuriousMind, Sofia, Kyle Kanos, WetSavannaAnimal, Danu Mar 19 '15 at 12:53

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    $\begingroup$ Can you explain what you don't understand? If one of the mirrors has a slight curvature, you're getting a spherical output wavefront which differs from the other path. $\endgroup$ – Carl Witthoft Mar 10 '15 at 12:07
  • $\begingroup$ possible duplicate of Do interference rings disappear in an interferometer if the path lengths are identical? $\endgroup$ – Sofia Mar 10 '15 at 14:34
  • $\begingroup$ @CarlWitthoft I've edited my post to contain an explanation. $\endgroup$ – user35734 Mar 11 '15 at 20:00
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    $\begingroup$ May I suggest that you draw your setup rather than use an image that isn't actually your configuration? It would make things clearer... $\endgroup$ – Floris Mar 11 '15 at 20:25
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    $\begingroup$ Here it is: imgur.com/mibpzDH $\endgroup$ – user35734 Mar 11 '15 at 20:36

The interference pattern you are thinking of relies on the interfering waves being plane waves. In real life, laser beams have phase fronts which have some curvature. If your Michelson interferometer is less than ideal, then the returning beams have slightly different curvatures. When you interfere two wavefronts with different curvatures, you get the interference pattern you witnessed.

As a helpful tool for understanding why curved wavefronts produce an interference pattern with concentric rings, have a look at the Newton's rings experiment. The resulting interference pattern is the same for the same reason.


The laser beams expand (slowly) as they travel. If you now consider the distance from the laser aperture (assumed to be a point) to the wall through the two arms (assumed to result in paths of unequal length $\ell$) then there will be a phase shift as a function of lateral position $d$ that goes as $(1-\cos(\frac{d}{\ell}))/\lambda$. If $\ell$ is different for the two arms, then there will be a phase shift between the two beams.

When the two arms are the same length, the pattern should disappear.

  • $\begingroup$ The main thing I'm confused about is we still saw the wave pattern when the two arms were the same length. $\endgroup$ – user35734 Mar 11 '15 at 20:25
  • $\begingroup$ How do you know they were the same length? $\endgroup$ – Floris Mar 11 '15 at 20:26
  • $\begingroup$ I don't, but I moved one of the arms to a distance that was closer than the other arm to one that was further away. Every time I stopped moving the arm, I still saw a similar pattern. $\endgroup$ – user35734 Mar 11 '15 at 20:28
  • $\begingroup$ Did the ring spacing change with position of the mirror? Did you see the center go bright and dark as you moved the mirror? I would expect the answer to both questions to be "yes". If not, then maybe you are looking at the diffraction pattern of your aperture! How big was it? $\endgroup$ – Floris Mar 11 '15 at 20:45
  • $\begingroup$ I can't remember for sure but I'm fairly sure the answer to both questions is yes. $\endgroup$ – user35734 Mar 11 '15 at 21:04

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