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Thin Film Interference Problems

  1. Apr 20, 2005 #1

    404

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    There's two problems I can't seem to figure out...

    I sort of got the answer, except I used m as 28 instead of 27. So my question is why are you suppose to use m as 27 to multiply instead of 28? Also What do they mean by incident normally, is it like striking the glass at a right angel?

    and the other one is ...
     
  2. jcsd
  3. Apr 22, 2005 #2

    OlderDan

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    The key is to understand that two of the 28 dark lines are at the ends of the glass. The dark line where the two flats touch is dark because there is phase reversal at one surface, but not at the other. The next dark line occurs when the path length across the space and back is exactly one wavelength, the one after that exactly 2 wavelengths, etc. Since there are 28 lines, and one is at the edge, the space at the first line from the touching edge is 1/27 times the space at the last line, which is the thickness of the spacer. So 1/27 times the spacer thickness is 1/2 wavelength. The "normal" incidence does mean perpendicular to the glass flats. It has to be normal so that the extra path length of the light is two times the space. If the light were at an angle, the path length difference would be more than twice the space. You would still see lines, but not as many and the calculation would be more complicated.
     
  4. Apr 22, 2005 #3

    OlderDan

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    First you need to reduce the ratio of the wavelenths to a ratio of integers. Then you need to understand what conditions cause phase reversal for some reflections and not others, and decide if both surfaces reflect the same phase, or differently. When you do that, you will realize that those integers are connected to the number of wavelengths of light that will "fit" into the thin film layer. There will be more of the sorter wavelengths than longer wavelengths in the film. Once you know how many of each wavelength are in the film, you can calculate the film thickness as a multiple of the wavelength. Since the film has an index of refraction other than 1, the wavelength in the film will be shorter than in the air. You will have to deal with that.

    Edit

    I think this problem may be flawed. For this combination of refractive indices, the wavelength ratio may be impossible. With the ratio given, even if the materials were reversed I'm not getting the answer given. I seem to be getting twice the thickness. Anyone else have an opinion?
     
    Last edited: Apr 22, 2005
  5. Apr 22, 2005 #4

    Andrew Mason

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    The first glass/air surface causes the inner surface of the first glass to reflect without a phase shift. At the second air/glass surface, the second glass reflects with a pi phase shift so it provides constructive interference if the space pi/2 ([itex]\lambda/4[/itex]) thick and destructive interference if the space is 0 or pi ([itex]2\lambda/4[/itex]) thick. Thereafter the dark band (destructive interference) occurs at half wavelength intervals: [itex]4\lambda/4, 6\lambda/4, 8\lambda/4... 54\lambda/4[/itex] and light (constructive) at [itex]3\lambda/4, 5\lambda/4, 7\lambda/4... 53\lambda/4[/itex]. So the maximum thickness is 54*500e-9/4 = 6.75e-6 m.

    Normal means right angle.

    The second one is a little more complicated. Both the surfaces (air/alcohol, alcohol/glass) reflect with a pi phase shift since the index of refraction of the subsequent medium is higher at both surfaces. So there will be minimal reflection when the alcohol is pi/2 or [itex]\lambda/4[/itex]thick. The wavelength changes with the index of refraction of the alcohol (not of the glass). The wavelength in alcohol of the first light is about 376 nm and 471 nm for the second light. Can you work it out from that?

    AM
     
  6. Apr 22, 2005 #5

    OlderDan

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    Given that both surfaces have the same phase at reflection, your [itex]\lambda/4[/itex] condition is certainly correct for the minimum thickness. The next thickness that will result in destructive interference is an additional [itex]\lambda/2[/itex], and the one after that still another [itex]\lambda/2[/itex]. The sequence being

    [itex]\lambda/4[/itex], [itex]3\lambda/4[/itex], [itex]5\lambda/4[/itex], etc

    The thickness of the film is the same for both wavelengths, so we should be looking for a solution to the equation

    [tex]L = (2n+1)\lambda_1/4 = (2m+1)\lambda_2/4[/tex]

    for integer values of m, n. Solving for the ratio of the wavelengths gives

    [tex](2n+1)/(2m+1) = \lambda_2/\lambda_1 = 640/512 = 5/4[/tex]

    which has no solution for integer n and m. The only way you could have a ratio of wavelengths equal to consecutive integers is if the path length through the film was a multiple of a wavelength, which would be the case if there was phase reversal at only one surface, which would happen if the film layer had a higher index of refraction than the air or the bottom layer. The possible thicknesses would then be

    [itex]\lambda/2[/itex], [itex]2\lambda/2[/itex], [itex]3\lambda/2[/itex], etc., leading to

    [tex]L = n\lambda_1/2 = m\lambda_2/2[/tex]

    for integer values of m, n. Solving for the ratio of the wavelengths gives

    [tex]n/m = \lambda_2/\lambda_1 = 640/512 = 5/4[/tex]

    This equation has the solution n = 5, m = 4. The thickness of the film must be 5/2 times the shorter wavelength and 4/2 = 2 times the longer wavelength. Using your correct adjustment for the index of refraction of the film, this gives

    [tex]L = 5\lambda_1/2 = 4\lambda_2/2 = 5*376 nm/2 = 4*471 nm/2 = 941 nm[/tex]

    So, it appears there is no solution to problem as stated, and if you try to salvage the problem by changing the index of refraction of the bottoom layer to less than 1.36, you get a different answer from the one given.

    Have I gone wrong somewhere?
     
    Last edited: Apr 22, 2005
  7. Apr 22, 2005 #6

    Andrew Mason

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    I agree with you. On the other hand, if the wavelength was at a maximum at 640 and a minimum at 512, it would work. If the question said: "at a maximum at wavelength = 640nm", the answer would be d=471 nm.

    AM
     
  8. Apr 22, 2005 #7

    OlderDan

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    Yes. Maybe that's what it was supposed to say. Nice catch.
     
  9. Jun 5, 2010 #8
    thin coating (t=340) is placed on glass .which visible wavelength
    (400 <wavelength<700 nm )
    will be absent in the reflected beam if the glass has an index of refraction n =1.35
     
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