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A simple Michelson interferometer is generally presented like the one bellow

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If we observes through the telescope, we see fringes like these enter image description here

Is it the lens of the telescope that form theses fringes? Because if we use a monochromatic light source, say a laser, the result we are supposed to see is simply a spot that would be bright or dim depending on the kind of interference, not fringes. So is there some other optical equipment between the final beam and the telescope? Or is it just the lens of the telescope that causes these fringes, as opposed to a spot, to appear?

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You can produce either one depending on the angle and the focus point etc. http://labman.phys.utk.edu/phys222core/modules/m9/interferometers.htm

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If the beam in the first figure ihas a small diameter, only a small spot will apoear. The spot may have fringes in it if the two beams are not perfectly aligned. If the beam is large, there will be a large spot and you are more likely to see fringes.

If the beams are perfectly collimated and perfectly aligned, there will be no fringes visible; just a spot whose brightness varies as the phase difference varies between the two beams. If fringes are visible, their spacing corresponds to the angle between the two beams.

In general, an eyepiece does not cause fringes to appear in an interferometer; it just helps you see the fringes.

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  • $\begingroup$ I am honestly still confused about why there should be any fringes at all if mirror 2, for example, is $\lambda/2$ away from the beam splitter than mirror 1 is, then the resulting optical path difference applies to all the rays of light that hit the screen, thus all the rays should either constructively interfere or destructively interfere, and so we should have either a bright spot, or a dark spot, regardless of the beam's "thickness". $\endgroup$
    – Hilbert
    Feb 6, 2020 at 22:27
  • $\begingroup$ If there is a slight misalignment between two collimated beams, so that they are not exactly parallel, the phase of one beam relative to the other will vary across the region in which the beams overlap. If you draw a diagram showing the two wavefronts, you will probably see what I mean. The fringes, which correspond to same-phase surfaces, run along the bisector of the angle between the beams. $\endgroup$
    – S. McGrew
    Feb 6, 2020 at 23:39

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