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

Single Slit diffraction question

  1. Nov 1, 2006 #1
    Im trying to get a good physical picture of what is causing the interference pattern in single slit diffraction (Im using a water wave that goes through a slit and then hits a wall Im also ignoring reflections)...

    Let me know if I am on the right track...

    I am picturing a wave front approaching the slit and Im using Huygens model to picture the "point" sources for the plane wave...

    I picture waves spreading out from the slit....some from the middle "source" (meaning the middle of the slit) and some from the edge "sources" (meaning from the edge of the slit). The waves created by the different sources will reach various points on the wall at different times due to path length differences.

    My question is can I say that the waves created by the "edge sources" will interact at a certain point on the wall with, say, the waves created by the "middle source" of the next wave front?
  2. jcsd
  3. Nov 1, 2006 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Sounds good to me. Waves spreading out from all points on the wave front reach all points of the screen. Their path length differences create phase differences which lead to various amounts of interference.

    For more insight into single slit diffraction, read this: http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslitd.html#c1"
    Last edited by a moderator: Apr 22, 2017
  4. Nov 1, 2006 #3

    Thanks for the reply and the link.
    Last edited by a moderator: Apr 22, 2017
  5. Nov 1, 2006 #4


    User Avatar

    Staff: Mentor

    You've got the right idea. Just keep in mind that every point across the width of the slit is a "source". You have "sources" at the two edges, at the midpoint, at the 1/4 point, the 3/4 point, etc. etc. The width of the slit contains an infinite number of sources of waves, separated by an infinitesimal distance from their neighbors.

    At the point on the screen where you want to find the resultant light intensity, you have to add all those waves using integral calculus, in general. From this you can get a graph of intensity versus position on the screen.

    For certain positions on the screen (certain angles away from the central axis of the setup), the waves are related symmetrically enough that you can "cancel" them pairwise and deduce that the intensity is zero, without having to do the integral. This is the way that many introductory books derive the locations of the minima.
  6. Nov 19, 2006 #5

    Right but the waves created by the "edge sources" might end up interfering with (just as an example) the waves created by the "middle source".

    I would think that the path length differences would ensure that there is interaction taking place between waves created by two consecutive wave fronts (albeit having different point sources along the slit).

    I did read it...good link...thanks.

    BTW...I realize that there are theoretically (in reality we are limited by the number of water molecules) an infinite number of point sources along the slit...I am only talking about "middle" and "edge" to simplify the discussion.

    Last edited by a moderator: Apr 22, 2017
  7. Nov 19, 2006 #6
    Right...for simplicity I was speaking as if there werent an infinite number.

    But when we say "infinite" we are speaking mathematically...not physically because the number of water molecules would limit the number of possible sources.

    I assume though that the number involved wouldnt create much difference in terms of the mathematics or physcial results from an infinite number of sources...

    Interesting....never saw this calculation done before...most textbooks I have seen dont develop the method in detail...

    Yeah that what Ive seen also...any online references for the more in depth treatment?

  8. Nov 19, 2006 #7


    User Avatar

    Staff: Mentor

    Most general physics textbooks probably don't do it, but most optics textbooks probably do it. I'm pretty sure all the ones in my office do it. Two that I can remember specifically off the top of my head are Pedrotti and Pedrotti, "Introduction to Optics" (which I use in my optics course) and Hecht, "Optics".

    I found the following online page which sketches the derivation. You'll have to fill in some of the purely mathematical steps.

  9. Nov 19, 2006 #8

    Thanks...thats a nice website...
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook