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Michelson Interferometer - Fringe Shape

  1. Dec 30, 2011 #1
    Hi all

    Edit: The reflecting side is the right side of left slab

    Please look at the pic above ... The light from source splits at left slab and goes in 2 directions and come back to interfere.

    From figure, it is obvious that path diff. in slabs for both rays is 0
    and only path diff. would be because of the difference in distance of 2 mirrors from left slab, which is d

    so, i found the path diff, as [itex]\Delta = 2\mu d cos \theta \pm \frac{\lambda}{2}[/itex]

    [itex]\frac{\lambda}{2}[/itex] as green ray is reflected by denser medium ...

    But i need to know how the circular or straight fringes are formed and how it depends on alignment of M1 and M'2

    Some help please ... :)

    EDIT: μ would be the refractive index of stuff filled in whole system ... like air for example
    and θ is incidence angle angle at slabs ... :}

    Attached Files:

    • MI.png
      File size:
      3.3 KB
    Last edited: Dec 30, 2011
  2. jcsd
  3. Jan 1, 2012 #2
    Some help please ???
  4. Jan 6, 2012 #3
    Oh come on, someone please help me !!!!!!
  5. Jan 6, 2012 #4
    Perfectly parallel mirrors results in circles if the optics are clean. Else you get what appears to be straight lines but i believe they are just the edges of a circle.
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