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With expanding beam I understand I will get circular fringes, and I understand if you move the mirror by d the path difference change of the interfering beam is $2dcos\theta$. But I don't understand why in all those experiments they bsay if you move the mirror by half of a wavelength, the pattern will repeat itself as the path difference is shifted by 2d and thus one whole wavelength. But isn't this true only for the central spot where $\theta$ is zero? For other circular fringes with the $cos\theta$ term, wouldn't you have to move the mirror by $(0.5\lambda)/cos\theta$ for the fringe to shift by $2\pi$? That's why I'm confused when they ask you to move the mirror while counting the fringes, any fringe, not the central fringe to determine the wavelength of the light.

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You are not noting the position of a particular fringe what you are doing is counting the fringes and it is much easier to do this whilst observing fringes "sharper" than the central fringe.
The count for every fringe is the same and then use that count for the central, $\theta=0$, fringe.

As you watch the video Interference pattern from a Michelson interferometer put your mouse cursor over different parts of the image including the centre and count the fringes as they pass the cursor.
Note that if the cursor is kept in one position the $\theta$ is constant and one fringe is cycled through when the mirror separation is by the same amount as for the central fringe.

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  • $\begingroup$ I get it now! Thanks! $\endgroup$
    – Cosmo
    Nov 15, 2023 at 2:33

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