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Calculate Interference in thin films

  1. Dec 18, 2006 #1
    1. The problem statement, all variables and given/known data

    Calculate the minimum thickness of a layer of manesium flouride(n=1.38) on flint glass (n=1.66) in a lens system, if light wavelength 5.50x10^2 nm in air undergoes destructive interference.

    λ1=5.5*10^2 nm

    2. Relevant equations

    n2/n1 = λ1/λ2

    3. The attempt at a solution

    Destructive interference is at λ/4, 3λ/4, 5λ/4 ....

    So the λ in flint flass is λ2=(n1*λ)/n2
    = 4.57 * 10^-7

    Therefore the shortest thicknss should be λ/4

    The answer however is wrong the answer is suppose to be 199 nm.

    Can someone please help! Thanku
  2. jcsd
  3. Dec 18, 2006 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    The interference is between the incident wave at the air (n1=1) and wave reflecting from the first layer (n2=1.38). Because the second layer has a higher index of refraction, the reflection at the first/second layer surface has a phase shift of [itex]\pi[/itex]. So in order to have destructive interference (a phase difference of [itex]\pi[/itex]), the path length through the first layer to the reflecting surface and back to the air has to be a full wavelength. So the thickness has to be [itex]\lambda_2/2[/itex].

    The wavelength in the magnesium fluoride layer is [itex]\lambda_2 = n_1\lambda_1/n_2 = 5.5\times 10^{-7}/1.38 = 3.98\times 10^{-7}m = 398 nm[/itex]

  4. Dec 18, 2006 #3


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    Staff: Mentor

    I get 99.6nm, not 199nm. Remember to use the n of air, not the flint glass.
  5. Dec 18, 2006 #4
    Thank You!
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