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Thickness of film for maximum interference

  1. Oct 28, 2016 #1
    1. The problem statement, all variables and given/known data
    As shown in the figure below, beams of light whose wavelength in air is 600 nm are perpendicularly incident upon two transparent thin films (A and B) in the same phase. The thickness of A and B is l. The refractive indices of A and B are, respectively, nA and nB. The difference of these two indices (nA - nB) is 6.0 x 10-3, and the refractive index of air is 1.00.
    What value should be assigned to film thickness l in order that the two beams of light directly passing through A and B, respectively, without reflection at the film – air boundaries are of opposite phase (i.e. phase difference corresponding to half a wavelength)?
    1234.png

    2. Relevant equations
    phase difference = (2 x n x thickness) / λ - 1/2

    3. The attempt at a solution
    What is the meaning of "in order that the two beams of light directly passing through A and B, respectively, without reflection at the film – air boundaries are of opposite phase (i.e. phase difference corresponding to half a wavelength)"?

    The light will be reflected when it travels from air to the film and from the film to air so there will be 4 reflected rays, from left and right side of A and B.

    Let say:
    reflected ray from lef side of A = light 1
    reflected ray from right side of A = light 2
    reflected ray from left side of B = light 3
    reflected ray from right side of B = light 4

    So the question means that light 1 and 2 are in phase and light 3 and 4 are in phase, or light 1 is in phase with light 3 and light 2 is in phase with light 4? Or is it just the same for the two cases?
     
  2. jcsd
  3. Oct 28, 2016 #2

    andrevdh

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    The two beams enter in phase and passes straight through the media. When they exit they will not be in phase anymore due to the optical path difference between the two media.
     
  4. Oct 28, 2016 #3
    So the question compares the phase of light that passes through A and B, not comparing the light that reflected at the boundary of thin film - air?
     
  5. Oct 28, 2016 #4

    andrevdh

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    Correct. The words in the question
    seems to point to that, also the arrows going straight through the media.
     
  6. Oct 28, 2016 #5
    I thought the phase change is due to reflection and no phase change due to refraction.

    And I also interpret the question a little bit different:

    I interpret it as: there is reflection at the film - air boundaries but the reflected rays are not opposite phase, i.e. they are in phase.

    Do I misinterpret the question?

    Then, the formula I write above can't be used because it is for reflected waves and if we compare the light that passes through A and B, won't they be in phase since there is no phase change due to refraction?
     
  7. Oct 28, 2016 #6

    andrevdh

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    Due to the different refractive indices the beams travel at different speeds through A and B and do not exit in phase anymore.
    Have a (close :) ) look at the concept of optical path.
     
  8. Oct 28, 2016 #7
    optical path difference = length (na - nb)

    phase difference = optical path difference / wavelength

    What value should I put for the phase difference? They are not in phase anymore so it can not be an integer. If the two waves are out of phase, the phase difference will be (n + 1/2) but how can we determine whether they will be out of phase or not and what will be the value of n if they are really out of phase?
     
  9. Oct 31, 2016 #8

    andrevdh

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    They will be completely out of phase if the optical path difference between the two beams is half a wavelength.
    This is the first occurance, of cause you will get that the beams are again completely out of phase if the otical path difference is 3/2 lambda ....
     
  10. Nov 3, 2016 #9
    How can we determine whether they will be out of phase or not? They won't be in phase but can the phase difference be, like, 1/4 ?
     
  11. Nov 3, 2016 #10

    andrevdh

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    The two beams are in phase when they enter the films, but due to the fact that they travel at different speeds in the films we get that the optical distances or paths differ for the two beams. That means that not the same amount of wavelengths fit in the two optical paths of the beams. Surely the phase difference between the two beams is dependent on the thickness of the films.
     
  12. Nov 3, 2016 #11
    Do I post the correct formula?

    optical path difference = length (na - nb)

    phase difference = optical path difference / wavelength

    If yes, then the question asks about the value of length (which is equal to the thickness of the film). I have the value of (na - nb) and also the value of wavelength so I need the value of phase difference to calculate the thickness. What should I put for the value of the phase difference? Is it 1/2? If yes, I don't know why I should put 1/2 instead of other value
     
  13. Nov 3, 2016 #12

    andrevdh

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  14. Nov 3, 2016 #13
    Let us try using number:

    phase difference = optical path difference / wavelength

    1/2 = L x 6.0 x 10-3 / (600 x 10-9)

    L = 5 x 10-5

    Is that correct?
     
  15. Nov 4, 2016 #14

    andrevdh

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    Yes but give the answer to 3 significant digits plus units: L = 5.00 x 10-5 m
    PB040364.JPG

    At (a) the beams enter the films and at (b) they exit at the same time.
    The optical path for beam A traveling through the film is ΔA and for B it is ΔB.
    They are not in phase when they exit due to the fact that the refractive indices and thus their speeds differ in the two films. Also their wavelengths will differ in the two films leading to different amount of wavelengths fitting in the films.
    For A 1 and 3/4 wavelengths fit in L and for B 1 and 1/4 wavelengths. So when they exit B lags with 1/2 wavelength with respect to A.
     
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