Double-slit diffraction with thin film interference

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Homework Statement



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?

Homework 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

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.
 

Answers and Replies

  • #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 .
 
  • #3
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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.
 
  • #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.
 
  • #5
SammyS
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number of waves in plastic: = t/(λ/η) = ηt/λ,

number of waves in: = ?
 
  • #6
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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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