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Double-Slit Interference

  1. Apr 17, 2005 #1
    Two slits are separated by 0.180 mm. An interference pattern is formed on a screen 80.0 cm away by 656.3-nm light. Calculate the fraction of the maximum intensity 0.600 cm above the central maximum

    I was using the equation I=Imaxcos^2(pie(d)sintheta/wavelength and I don't seem to be getting anywhere is there another equation because this one doesn't seem right and it's the only one in the chapter.....
  2. jcsd
  3. Apr 17, 2005 #2
    You are indeed using the right equation. But you are probably getting stuck trying to find [itex]\theta[/itex]. The intensity as a function of the phase [itex]\Delta\phi[/itex] is:

    [tex]I = I_{max}\cos^2\frac{\Delta\phi}{2}[/itex]

    Now, [itex]\Delta\phi = \frac{2\pi}{\lambda}\Delta x[/itex]

    This should see you through...you don't need to compute [itex]\theta[/itex] if you observe that

    [itex]d\sin\theta[/itex] = Path difference = [itex]\Delta x[/itex]

  4. Apr 18, 2005 #3


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    Looks like the correct equation except that you do not have a closing parenthesis. Make sure you have the units right. You have mm, cm, and nm in the problem. Are you making the necessary conversions?
  5. Apr 18, 2005 #4
    OOps..I didn't see you have the y-coordinate of the fringe as well...in that case, for small theta,

    [itex]\sin\theta = \frac{y}{D}[/itex] and that should do it.
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