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Source wikipedia

It's given that CA - CB is d, the angle CAE is $\theta$. (E is not on the normal of mirror M1). I want to find the condition for maxima at E (as a function of $\theta$).

I tried finding the path difference but it turned out to be very complicated, assuming that light after passing through beamsplitter continues traveling along the same straight line. How does one find the radius of rings formed on the screen, if there's a screen instead of detector?

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enter image description here It is much easier to "fold" the interferometer. If you imagine that there is a lens of short focal length at the exit of the laser, (otherwise, the rays do not diverge), you have a point source S which after reflection on the separator and the mirrors has two images S1 and S2 at the top, distant from $2d$. (When you move a mirror by $d$, the image move by $2d$)

If the screen is at a sufficiently large distance, the rays emitted by S1 and S2 which interfere at E are practically parallel and the difference in the path is $\delta =2d\cos (\theta )$ (if there is no additional phase shift due to reflections on the mirrors ).

The bright rings correspond to $\delta =2d\cos ({{\theta }_{n}})=n\lambda $ and the rings on the screen are obtained by writing $\tan ({{\theta }_{n}})\approx {{\theta }_{n}}={{R}_{n}}/D$ if $D$ is the distance of the sources on the screen.

Be careful : the order of interference is maximum in the center and decreases when one moves away from the center.

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  • $\begingroup$ Can you please add a picture with labels, I can't easily 'fold' the interferometer in my mind? $\endgroup$ – user183683 Feb 20 '19 at 15:44
  • $\begingroup$ I have added my old optic courses, in French but may be the pictures can help ? $\endgroup$ – Vincent Fraticelli Feb 20 '19 at 15:54
  • $\begingroup$ So D is $S'_1S'_2$ here. is that right ? $\endgroup$ – user183683 Feb 20 '19 at 16:04
  • $\begingroup$ No. S'1S'2 is 2D. D is the distance between the mirrors (e in my pictures) But if you move a mirror by D, the image move by 2D. (Sorry for my english !) $\endgroup$ – Vincent Fraticelli Feb 20 '19 at 16:06
  • $\begingroup$ @VincentFraticelli Your diagrams shows the set up for fringes of equal thickness ie the sort of fringes formed when there is a wedge. The OP’s diagram looks as though the mirrors are at right angles to one another and so the two surfaces (mirror M1 and virtual mirror M2) causing the reflections are parallel to one another and the fringes which are formed are fringes of equal inclination. These fringes will be circular. Have a look at this answer. $\endgroup$ – Farcher Feb 20 '19 at 17:06

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