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I am getting really confused about the experiment of Michelson and Morley. As I understand it, we should be able to observe a destructive interference because light beams go through different optical paths. Destructive interferences form minimums of interference whereas constructive interferences form maximums. Then I translate here what my book says: "After the rotation of $90^\circ$ degrees of the interferometer, we should see a number of fringes equal to the times the width of the phase shift, which is $$\delta\ =\ 4\pi \frac{L_0}{λ} \frac{v^2}{c^2}$$ (which is the sum in absolute value of the longitudinal and transversal phase shift, as I understand it) contains $2\pi$, since according to the following equation, Intensity $$I\ =\ I_0\ cos^2 \frac{\delta}{2}$$ and $$\delta\ =\ \frac{2\pi}{λ} \Delta R,$$ where $2 \pi$ is the phase shift between 2 consecutive maximums of interference"

So what I don't understand is the last line. If a destructive interference is expected, shouldn't we be looking for minimums(that is what destructives interference forms)? Why here are we assuming that the expected result is a constructive interference? Please don't delete this post since I really need help on this.

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A pair of light beams that are not correlated will add in intensity. The oscillating electric fields add in random fashion, two 1 V/meter fields summing to 1.4 V/meter. The intensity of the sum, being proportional to the square of the field, is then 1.4^2 = 2. If the beams ARE correlated, as in the Michaelson-Morley apparatus, the sum can be 2 V/meter if they are in phase (constructive interference) giving intensity 4, or can be 0 V/meter if they are opposed in phase (destructive interference) giving intensity 0. They can also be anything inbetween, but that's the entire expected range.

By conservation of energy, though, the full area on which the beams shine will have not just 0 or 4 intensity, it will have bands, the so-called fringes, with intensity 4 peaks (bright band) and intensity 0 troughs (dark band) and averaging the same intensity of 2 as the uncorrelated light beams.

The experimenters expected to watch a set of fringes that shifted according to the orientation, because (1) the fringes are large enough, and bright enough to observe on a screen, and (2) light was expected to travel differently in different directions (and a VERY SMALL difference makes a big fringe movement). So, the observers were seeing both constructive and destructive interference, before, after, and during the rotation of the apparatus.

What they didn't see, was movement of the fringes when orientation was changed.

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  • $\begingroup$ But if both constructive and destructive were expected to be measured why exactly in the calculation above are we taking into account only the phase 2pi ? Shouldn't we also calculate the number of fringes for destructive interference which has a phase of pi? $\endgroup$ – Maths64 Sep 7 '16 at 15:31
  • $\begingroup$ I believe the phrase 'maximums of interference' means bright bands, i.e. maxima of intensity of the interference pattern. The bands are uniformly spaced, when the ridges move so do the valleys, so there is no second observation for motion of the valleys. $\endgroup$ – Whit3rd Sep 8 '16 at 4:51

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