1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Michelson Interferometer - Fringe Shape

  1. Dec 30, 2011 #1
    Hi all

    attachment.php?attachmentid=42306&stc=1&d=1325256305.png
    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
      MI.png
      File size:
      3.3 KB
      Views:
      135
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Michelson Interferometer - Fringe Shape
Loading...