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Haidinger fringe (interferometer at zero path difference)

  1. May 8, 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*√[(p*λ)/d]

    where
    rp = Haidinger Fringe
    f = focal length
    p = order of interference
    λ = wavelength
    d = difference in distance between the two mirrors and beamsplitter

    Effective path difference
    2d*cosθ = pλ

    3. The attempt at a solution
    Trying to derive an equation for the two path differences from the Haidinger Fringe equation, which is independent of the p value, but I'm struggling...

    rp=f*√[(p*λ)/d]

    ( rp / f )2 = (p*λ)/d

    ( rp / f )2 = 2d*cosθ/d

    d = 2d*cosθ / ( rp / f )2

    d = 2d*cos1.3644 / 1.31 mm/55mm

    d = 2*cos1.3644 / 0.02381818181

    d = 17.2082026858

    2 * 17.2082026858 * cos1.3644 = pλ

    p = 2 * 17.2082026858 * cos1.3644 / 5.145*10^-7

    p = 7.05309334257 / 5.145*10^-7

    p = 13708636.2344

    5.145*10^-7 * 13708636.2344 = 7.05309334257 metres = difference in distance between the two mirrors and the beam splitter

    So would the mirror need to be moved 7.05309334257 metres in order to set the interferometer at zero path difference?
     
  2. jcsd
  3. May 8, 2015 #2
    I made a new attempt. Bare with me!

    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


    3. The attempt at a solution
    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)

    A lot of math later and I get a value of...

    p = 55
    d = 5cm

    ∴ Effective path difference
    2d*cosθ

    2*0.05*0.999716484 = 0.0999716484 m

    0.999716484 cm

    The mirror would have to be moved 0.999716484 cm in order to set the interferometer at zero path difference.

    I'm not sure if this is correct. I've not had much experience with interferometers or the theory behind them.
     
    Last edited: May 8, 2015
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