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Light - Thin Film interference

  1. Sep 25, 2008 #1
    1. The problem statement, all variables and given/known data
    A thin film of MgF2 with n = 1.38 coats a piece of glass. Constructive interference is observed for the reflection of light with wavelengths of 500 nm and 625 nm.

    Part A -
    What is the thinnest film for which this can occur?
    2. Relevant equations
    t = lambda/2n *(m + .5)

    3. The attempt at a solution
    Not sure what I am supposed to do, because of the two wavelengths. I know what to do for a single wavelength...

    So this is my work for solving the problem with each wavelength


    500 = 2(1.38)(t)/(m+.5)

    500(0+.5) = 2.76(t)

    t = 90.6


    625(0+.5) = 2.76(t)
    t = 113

    Now that I have done this, I am not sure how to proceed. Any help is appreciated.
  2. jcsd
  3. Sep 26, 2008 #2


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    Hi Foxhound101,

    I don't think these are correct. I believe these equations would be for destructive interference; isn't the index of refraction of glass greater than 1.38?

    By setting m=0, you have found a minimum thickness for each wavelength (for destructive interference with the equations you are using). But in this problem you don't want the minimum for each; you want a thickness t that is true for both wavlengths. So use m1 for one wavelength and m2 for another wavelength.

    After eliminating t, you can find a relationship between the two m values. What do you get?
  4. Sep 26, 2008 #3
    Thanks for the help, alphysicist

    <i>isn't the index of refraction of glass greater than 1.38?</i> Index of refraction in glass is greater than 1.38...the value is sometimes 1.55

    So...another formula I have is t=(m*lambda)/(2n)

    lambda = 500

    t= 181

    t= (2*500)/(2*1.38)
    t= 362

    lambda = 625
    t = (1*625)/(2*1.38)
    t = 226

    t = (2*625)/(2*1.38)
    t = 453
  5. Sep 26, 2008 #4
    So the relationship between these values...

    181/226 = .8

    362/453 = .8

    Interesting...so what do I do, now that I know this relationship?
  6. Sep 26, 2008 #5


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    That's the right formula for this problem. (If, for example, the thin film was on something that with an index of refraction less than 1.38, then your original problem would have been correct for constructive interference.)

    I don't think this is the way you want to do it. You are setting the m's to be specific numbers; but in this problem they are unknown. Just leave them as m1 and m2:


    t = (m2*625)/(2*1.38)

    So you have these two equations; now eliminate t to find out how the two m values are related. For example, if you end up with:

    m1 = 1.7 m2

    then you know that m1=17, m2=10 would be the smallest integer values that m1 and m2 could have that still obey that relationship.

    (The ratios you found in your last post are related to this, but I think it rather hides the process.)
  7. Sep 26, 2008 #6

    (m1*500)/2.76 = (m2*625)/2.76
    m1 * 500 = m2 * 625

    m1 = .8m2

    m1 = 10 and m2 = 8
  8. Sep 26, 2008 #7
    Plugging those m's in...

    t = (10*500)/(2.76)
    t = 1812

    t = (8*625)/(2.76)
    t = 1812

    Does this seem like a good answer? I typed 1810 into the program and it told me I was still wrong.
  9. Sep 26, 2008 #8


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    I think you have those backwards (m1 is 8 and m2 is 10).

    However, those are not the smallest integers that satisfy the relationship. (Rewrite 0.8 as a fraction and then simplify.)

    Once you get the set of smallest integers, what do you get for the minimum thickness?
  10. Sep 26, 2008 #9
    If I change m1 and m2, then they won't equal each other.

    If I leave m1 = 5 and m2 = 4 then I get...

    t = 906

    t = 906

    This seems like a decent answer...answers for other problems were t=1000ish, so this is decent.

    Does this seem like a good answer now?
  11. Sep 26, 2008 #10


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    Okay, I see what happened. You wrote

    m1 = .8m2

    a few posts ago but you meant

    m2 = .8m1

    So your numbers for m1 and m2 are fine.

    That looks right to me (if they want the answer in nanometers).
  12. Sep 26, 2008 #11
    That was the correct answer. Thank you for your help.
  13. Sep 26, 2008 #12


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    Sure, glad to help!
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