Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Diffraction of a circular aperture

  1. Feb 11, 2010 #1


    User Avatar

    It is quite typical example for a text to mention Airy disc (a diffraction patten for a circular aperture), also in wiki http://en.wikipedia.org/wiki/Airy_disc. But what wiki confusing me is , in the mathematical details section, the intensity is given by J1(x)/x, where J1 is the first order Bessel function, it is ZERO around x=0. But the Airy disc has a maximum in the center, so how can on use J1(x)/x to describe an Airy disc?

    And I saw some introduction on using a circular slit (annulus) to produce a diffraction pattern and the it is said that annulus will produce a bessel beam on the screen. Any one know if this is true and where can I find some information on this?
  2. jcsd
  3. Feb 11, 2010 #2

    Andy Resnick

    User Avatar
    Science Advisor
    Education Advisor

    The intensity should be J0(x)/x: also called a 'sombrero' function.

    edit: oops... yep, it's J1(x)/x.
    Last edited: Feb 12, 2010
  4. Feb 12, 2010 #3
    Any idea what the diffraction pattern of a rectangular slit is? Hint: It's based on Sin(x)/x. Go look up that function. Notice how when X = 0 that sin (x) = 0? But 0/0 is undefined. Hence one must look at the limit of the sin(x)/x function as x -> 0. It's NOT zero! Go Google the sinx/x function and see what it looks like!

    well, when you shift from rectangular coordinates to circular coordinates the calculation of the diffraction patterns shifts from Fourier-style results (with sin and cos) to Bessle function results. There you end up with J1(x)/x which is analogous to the sinx/x above. And the same ideas apply at x = 0. OK?

    For more information go look at any advanced optics book such as Born and Wolf.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook