The rays reflected by mirrors M1 and M2 meet at the glass G. For each ray(specifically 1 and 2 in the figure which may go on 1,2,3... for an extended object as is required for the experiment) as the splitted beams meet at G the path difference for each ray(1,2,3..) would be the same (x2-x1) according to the figure, 
then why do we see circular fringes at the screen?
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$\begingroup$ Not sure what you mean, the path difference of one of the two arms can be adjusted so they aren't the same between leg 1 and leg 2 (like in the case of gravitational wave detection). There is interference only when the path length difference isn't an integer number of wavelengths. $\endgroup$– TriatticusCommented Jan 7, 2021 at 20:02
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$\begingroup$ Maybe I am not able to frame my question properly. I am trying to say that if for instance we take beam 2(in the figure) then the split beam that reaches mirror M2(vertical) and returns back to G and the one that reaches M1 and returns to G, the path difference between the two beams(split parts of Beam 2) is same (x2-x1) and same is true for all the other beams(1,2,....), then how do we see circular fringes? $\endgroup$– Devesh VaishCommented Jan 7, 2021 at 20:44
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1$\begingroup$ See physics.stackexchange.com/a/317114/81224 $\endgroup$– Apoorv PotnisCommented Apr 24 at 18:27
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2 Answers
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Circular fringes imply that at least one of the beams is not collimated.
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$\begingroup$ This shouldn't be an answer, just a comment. $\endgroup$ Commented Apr 24 at 18:28
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It is similar to the interference pattern of a slits or particularly a round openings. The perspective is in line with the center of the rings but perpendicular to the surface of the mirrors. Consider the light beam from the source to the center of the pattern as straight. Then every ring out from there corresponds to a particular value of theta. These are fringes of equal inclination.
answered Jan 8, 2021 at 5:54