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Optical path difference of air wedge

  1. Sep 1, 2014 #1
    1. The problem statement, all variables and given/known data

    i cant understand why the optical path diffrence is 2nt+0.5λ. why is it so? the 0.5λ is due to the path difference change of 180 degree of when it is reflected lower ray when it is reflected from the lower glass surface... why there's also phase change of 2nt. since both ray pass thru the glass surface twice , so it should cancel each other , leaving only phase change of 0.5λ . Am i right?
    here's the link :
    http://astarmathsandphysics.com/ib-physics-notes/optics/ib-physics-notes-wedge-films.html [Broken]
    2. Relevant equations

    3. The attempt at a solution
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 1, 2014 #2


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    The 2nt refers to the path difference in the space between the plates. You see that one ray travels an extra distance of 2t relative to the other, right? Here, n is the index of refraction of the material wedged between the two glass plates, not the index of refraction of the glass (so n is approximately 1 if it is air).
    Last edited by a moderator: May 6, 2017
  4. Sep 1, 2014 #3


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    n is not index of refraction here http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon6.gif [Broken]

    n is any general integer, and it's used to account for cancellation if path difference is λ/2, or λ+λ/2, or 2λ+λ/2, or 3λ+λ/2, or....

    λ is the wavelength in the material comprising that wedge
    Last edited by a moderator: May 6, 2017
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