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Light diffraction

  1. Feb 13, 2009 #1
    Two isotropic (radiating the same intensity of EM wave in all directions) point sources S1and S2 emit light in phase at wavelength λ and at the same amplitude. The sources are separated by distance 2d=6.00λ. They lie on the axis that parallel to the x-axis, which runs along a viewing screen at distance D=2.00λ. The origin lies on the perpendicular bisector between the sources. The figure below shows two rays reaching point P on the screen, at position xp

    At what value of xp do the rays have the minimum possible phase difference?


    http://www.wisheyebio.com/images/screen.png [Broken]

    From question, the minimum possible phase difference, does it mean that the first dark point, or not?

    I tried like this...
    [tex]\Delta L=d\sin \theta) = (m+0.5)\lambda[/tex]
    [tex]\frac{2dx_{p}}{D} = 0.5\lambda[/tex]
    [tex]\frac{6 \lambda x_{p}}{20\lambda}[/tex]
    [tex]x_{p} = 1.66\lambda[/tex]

    Am I correct or not?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 13, 2009 #2

    Delphi51

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    Your attachment hasn't been approved yet, so I can't see it.
    Are the sources a distance 2 wavelengths above the screen?
    If the sources are that close to the screen, I would forget the formulas and just work out the distance from each source to P - expressions involving Xp, d and D. Choose Xp so the difference between the distances is one wavelength, to get the first constructive interference spot.

    The formulas are approximate and they work well when the distance is large, but not close up.
     
  4. Feb 14, 2009 #3
    Delphi is correct, noppawit. The formula you use involves the approximation that d is much smaller than D. That's not true here. So you have gotta write down the formulas for the path lengths direct : take the distances between each slit and the point x_p. Constructive interference occurs first when the two path lengths differ by a wavelength.

    You have to solve a quadratic equation to get the answer, I reckon.
     
  5. Feb 14, 2009 #4

    Delphi51

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    For the S1 to Xp path, draw a line straight up from S1 to make a right triangle. Use the pythagorean formula to find the length of the path. Do the same for the other path. Write that the longer one is equal to the shorter one plus one wavelength. Solve for the distance Xp.
     
  6. Feb 14, 2009 #5

    Redbelly98

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    Does one wavelength represent the minimum possible phase difference?

    Sorry to jump in here, but it isn't quite clear to me what is being asked in the original problem statement.
     
  7. Feb 14, 2009 #6

    Delphi51

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    The minimum phase difference is zero. It will occur when the path difference is 0, 1, 2, 3 ... wavelengths.
     
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