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Michelson Interferometer (Zero Path Difference)

  1. May 9, 2015 #1
    1. The problem statement, all variables and given/known data
    A Michelson interferometer is illuminated with a laser with a wavelength of 514.5nm. A Haidinger fringe pattern is photographed with a lens of focal length 55mm. The diameter of the two adjacent circular fringes in the image are 1.53mm and 2.62mm.

    How far would the mirror that is further away from the beamsplitter need to be moved in order to set the interferometer at zero path difference?

    2. Relevant equations

    Haidinger Fringe, rp

    rp = f √ (pλ/d)

    rp = f [ (( 1 - pλ/d )^-2) - 1 ] ^(1/2)

    cosθp = f / √ ( rp2 + f2 = [ 1 - (pλ / 2d) ]

    2d(1-cosθp) = pλ

    rp = f [ (( 1 - 2d(1-cosθp/2d )^-2) - 1 ] ^(1/2)

    Path Difference = 2dcosθ

    3. The attempt at a solution

    ∴ 0.00131m = 0.055m [ (( 1 - 2*0.05(0.000283516/2*0.05 )^-2) - 1 ]^(1/2)

    Effective path difference
    2d*cosθ

    2*0.05*0.999716484 = 0.0999716484 m

    0.999716484 cm

    I was trying to write an equation for Haidinger's fringes that was independent of p so I could solve the problem. I think I did that correctly, but I'm not sure if the rest of my working is correct. I couldn't rearrange the equation to solve for d, but would like some help doing that.

    I'm a bit confused by the zero path difference question too. Does that mean they want the path difference to = 0. This wouldn't make sense, but I can't really find any literature on what zero path difference is.
     
  2. jcsd
  3. May 11, 2015 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    Doesn't this mean to set the path length between the beam splitter and one mirror equal to the path length between the beam splitter and the other mirror?
     
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