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In descriptions of the experiment, the two arms of the interferometer have the same length. There's an interference pattern which was expected to be shifted when the system was rotated 90 degrees. But there was no shift in the interference pattern, which led to the conclusion there's no ether.

My question is why is there an interference pattern in the first place? The path lengths are the same, and there's no ether. The two waves should be in phase right? Is this because in practice it is impossible to have the two arms of the interferometer be the same length?

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    $\begingroup$ You need to analyze the optical path with a slight angular offset of the mirrors, which breaks the symmetry that you are correctly expecting if the actual instrument was built like it's being described in the simplified analysis. Look at the Wikipedia page to see the actual light path, which uses multiple mirrors under slight angles. $\endgroup$
    – CuriousOne
    Feb 8, 2016 at 16:39
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    $\begingroup$ why the "relativity" tag? $\endgroup$
    – scrx2
    Feb 8, 2016 at 19:49

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This is only true if the source is perfectly collimated, and that is never the case. If the source is not perfectly collimated, then it will produce a fan of rays and each of these rays will appear at a different place of the detector.

One such ray, for example, will interfere like this:

Image source.

A ray sent slightly downwards as shown in the image will arrive, after the interference, at a detector placed to the left of E. Moreover, unless the placement of the mirrors is exactly right, these two rays will accumulate slightly different phases along the beams, so they are unlikely to both have the same relative phase between the beams and therefore the same type and amount of interference. The interference conditions will therefore change across the detector screen, causing the interference pattern.

The interferometer can therefore be adjusted so that there is a clearly-defined interference pattern on the screen, or such that the interference is perfect throughout (which requires much more careful alignment). In particular, it is often advantageous to keep the alignment such that a number of fringes show on the screen, which can make it easier to count these fringes as they slide past the screen.

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    $\begingroup$ Does S have to be a coherent light source? $\endgroup$
    – Farcher
    Feb 8, 2016 at 19:34
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    $\begingroup$ @Farcher - it has to have a coherence length greater than the total path length. In simplest terms, yes. $\endgroup$
    – WhatRoughBeast
    Feb 8, 2016 at 20:12
  • $\begingroup$ You can see white light fringes with such a device so it is the path difference which is important. For white light that path difference would be less than a few wavelengths? $\endgroup$
    – Farcher
    Feb 8, 2016 at 20:18

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