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Double-slit diffraction with thin film interference

  1. Nov 4, 2011 #1
    1. The problem statement, all variables and given/known data

    A very thin sheet of plastic (n=1.70) covers one slit of a double-slit apparatus illuminated by 630 nm light. The center point on the screen, instead of being a maximum, is dark. What is the (minimum) thickness of the plastic?

    2. Relevant equations

    Constructive interference d*sinθ=m*λ where m=0,1,2,3,...

    Destructive interference d*sinθ=(m+1/2)*λ where m=1,2,3,...

    c=λ*f

    c=c0/n

    λ1*n1=λ2*n2

    3. The attempt at a solution

    Obviously this is a case of destructive interference so the sheet of plastic needs to shift the phase of that wave by λ/2. I am unsure of how to determine the thickness of the plastic necessary to make that happen though.
     
  2. jcsd
  3. Nov 4, 2011 #2

    SammyS

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    The thickness, t, of the plastic must be such that there is an extra 1/2 a wavelength in the plastic, compared to the number of wavelengths in the same thickness of air.

    Index of refraction for air is ≈ 1 .
     
  4. Nov 4, 2011 #3
    As stated above in my attempted solutions, I understand that the wavelength needs to be shifted by λ/2 but I don't understand how to use this to find out how thick the glass needs to be to shift the wave by that amount.
     
  5. Nov 5, 2011 #4

    SammyS

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    If λ is the wavelength in air, then λ/n is the wavelength in the plastic, right?

    How many waves of wavelength, λ/n, will fit in a layer of plastic having a thickness, t ?

    How many waves of wavelength, λ, will fit in a layer of air having a thickness, t ?

    The difference the the above two quantities must be 1/2.
     
  6. Nov 5, 2011 #5

    SammyS

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    number of waves in plastic: = t/(λ/η) = ηt/λ,

    number of waves in: = ?
     
  7. Nov 5, 2011 #6
    Ok, thanks, that helped a lot. For some reason I was blanking on how to get the number of waves in the thickness of the plastic, but I figured it out.
     
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