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I Single Slit Fraunhofer Diffraction

  1. Jul 12, 2017 #1
    Just to clarify, this isn't homework. I took an Optics course a few years ago, but in the time between then and now, I've lost all my notes.

    Suppose we have an infinitely tall slit of width [itex]a[/itex] and a parallel screen a distance [itex]r_0[/itex] away. Let the [itex]x[/itex] axis be parallel to the width of the slit, with the origin at the center of the slit (such that the slit extends from [itex]x=-a/2[/itex] to [itex]x=a/2[/itex]). The [itex]z[/itex] axis extends toward the screen from the origin. An angle [itex]\theta[/itex] locates a point [itex]P[/itex] on the screen (with respect to the [itex]z[/itex] axis) a distance [itex]x=X[/itex] along the screen. A ray drawn from the origin to [itex]P[/itex] is length [itex]r[/itex].

    Given a plane wave incident on the slit, what is the formula for the irradiance of the diffracted wave as it encounters the screen?

    I know the answer is similar (if not equal to) [tex]I=I_0\left(\frac{\sin(\beta/2)}{\beta/2}\right)^2[/tex] where [tex]\beta\equiv ka\sin\theta[/tex] I also know that [tex]E=-\frac{ikE_0}{2\pi r_0}\int_{-a/2}^{a/2} e^{ikr} \, dx[/tex] To get from the electric field to the irradiance is not a problem.

    Any recommendations?
     
  2. jcsd
  3. Jul 13, 2017 #2

    sophiecentaur

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    This is the standard formula, I think. The problem is with the limits of the integral. For a real case and a narrow slit, the illumination across the slit is not uniform so the profile should be included in the formula.
     
  4. Jul 13, 2017 #3
    Does saying the incident wave is a (coherent) plane wave not specify the profile? If not, please assume the illumination is uniform across the width of the slit. Also, I know this is just a toy example, but I'd like to get it down before tackling anything more realistic, like a nonuniform incident wave front.

    What I want to know is how to get from the integral to to the [itex]\sin(\beta/2)/(\beta/2)[/itex] function. And if the integral is incorrect, I need some pointers to correct it.
     
  5. Jul 13, 2017 #4

    sophiecentaur

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    If you assume that, then no problem. But a slit in a screen of finite, practical thickness could be several wavelengths deep and, depending on the material (say metal) it will definitely affect the distribution of luminosity at the way out of a narrow slit. It just depends on how far you want to go, I think. Thinking in terms of a microwave slot through a thick sheet of metal, there will be induced currents around the edges of the slit which will alter the effective width.
    Imo, the problem of a narrow slit would probably be best approached not as the limiting case of a wide slit but rather as a microwave slot antenna.
     
  6. Jul 13, 2017 #5
    I found a working of the problem in a book on Optics and was able to replicate the result in notation that was more familiar to me.

    Thank you for your comments.
     
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