# Interferometer Question

Please correct me if/where I'm wrong in my understanding of the following (laymans terms) explanation and question:

A michelson interferometer like used in 1887 works off a basic principle of having the same beam of light arrive at the same place slightly behind/ahead of itself. We are able to see interference (rings) by making the light's direction strike a flat surface at differing angles thereby causing both addative and subtractive interaction to be seen across that surface.

The assumption for this system to work is that both beams of light are of exactly the same wavelength but because of the differing initial arrival times there is a constant and consistant phase (arrival) difference. The peaks are constantly the same distance apart.

If used in a certain way, an interferometer (of a little different design than michelson 1887) can also be used to determine the wavelength of an unknown beam of light when interfering with a known lightsource. However, michelson's interferometer was not used/designed to detect this.

The Queston:
If a michelson interferometer of the design and methodology of use of the 1887 model received 2 beams of slightly unlike wavelengths, what would be seen? (For example, a beam of 600nm and a beam of 600.06nm)

From what I understand, you could still establish visual interference rings but not as well defined because of the oscillation action that would be caused. Is this true?
Without the equipment I could mentally see reasons why this could be true and reasons why this could be false.
True because I know interference rings are seen when using an interferometer to determine an unknown wavelength diferent than the reference beam, But I don't know if they stand still.
False because oscillation would seem to cause blurring...
I lean toward believing that the rings become averages of sets of waves instead of representative of single waves.

If true...
At this point since the two beams are not the same, there seems to no longer be any such thing a a phase shift since the lineup of peaks and troughs is constantly moving.
Would it be oscillating slowly enough to be able to visually set up a temporary lineup of peaks or does it oscillate too fast for the human eye? (I would guess that even as small of a difference as I mentioned, the ocillation rate would be far above the 30somthingish times per second that a human eye can detect.)
Now if the instrument was calibrated to show rings with two different wavelengths, is it true that you could still detect a phase shift but only on an order of magnitude smaller in proportion to the number of waves averaged to create each fringe?

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pervect
Staff Emeritus
Without getting into the specifics, it seems to me that you'd see constantly moving interference fringes if the two frequencies were not the same.

Think of a wave tank - two waves of the same frequency will produce a standing wave pattern, while two waves of a slightly different frequency will exhibit a moving wave pattern.

That was what I was thinking at first but then I started reading about the method of determinng an unknown wavelength and was no longer sure about what would happen.

My questions leads into a thought that perhaps the MMX was null in more than one way...

This is precisely what you see with a sodium light viewing the sodium "D" lines which are very close in wavelength. I have used a small Michelson interferometer extensively. With the "D" lines as the mirror moves first you see well defined rings then they get blurry until they all but disappear. Then they they get well defined again. This repeats at the "beat wavelength."
If you don't move the stage, the fringes do not move.

By the way the white light fringes are so pretty, don't miss them. I want my own interferometer.