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What will happen for single slit diffraction if?

  1. Apr 1, 2013 #1
    If the width of the single slit is much smaller than the wavelength of a monochromatic light, until there is only one point on the wavefront can pass through it?

    Thank you
     
  2. jcsd
  3. Apr 1, 2013 #2

    mfb

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    You get something similar to a spherical wave, expanding in all directions (orthogonal to the slit) behind the slit.
     
  4. Apr 1, 2013 #3
    What has been said is quite true.
    As a good guide you will find that when the slit width = 1 wavelength the angle of the first minimum in the diffraction pattern is 90 degrees.
    This means that a slit width of less than 1 wavelength is essentially a 'point source'
     
  5. Apr 1, 2013 #4

    sophiecentaur

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    Hmm. A bit narrower than that, I think, or you will have a first minimum at +/- 90°, iirc, which is not omnidirectional.
     
  6. Apr 1, 2013 #5
    Thank you guys.
    To double confirm.
    So if I use a slit that is same as the wavelength, then I will get my second maxima at +/-90° after my first?
    If smaller than wavalength , then is just a point source?
     
  7. Apr 1, 2013 #6
    If you use a slit that is 1 wavelength wide the first MINIMA !!!!! (please note MINIMA) occur at +/-90
    There is only the central (0 degrees) maximum.
    If the slit is 2 wavelengths wide then the first MINIMUM is at +/-30, the second MINIMUM is at +/-90 so there is a maximum between 30 and 90 (not half way between !!!)
     
  8. Apr 1, 2013 #7
    I thought the destructive interference only call minima?
    So only the centre is maximum , the rest are called minimum?
     
  9. Apr 1, 2013 #8
    Maxima occur with CONSTRUCTIVE interference.
    minima occur with DESTRUCTIVE interference.
    Are you familiar with 2 source interference....the conditions for the formation of MAX and MIN
     
    Last edited: Apr 1, 2013
  10. Apr 1, 2013 #9

    mfb

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    This is not a yes/no question. If you make the slit smaller and smaller, it will get a better and better approximation of a point source.
     
  11. Apr 1, 2013 #10
    Quite correct, 1 wavelength width should be seen as an upper limit. With this width some of the wave exists out to +/-90 degrees so wave effects can be detected.
    Smaller than 1 wavelength and the intensity decreases, make it small enough to be a better point source and the intensity will be so low that detection of wave effects will be difficult.
    Pay your money and make the choice.
     
  12. Apr 2, 2013 #11
    Is minima , I though is maxima because I mistook that as double slit .thanks

    Smaller than the wavelength, better point source ,lower intensity .

    What do you mean by exists out to +/-90 degrees? Money?
     
  13. Apr 2, 2013 #12
    straight through the slit is 0 degrees.
    +/- 90 degrees means 90 degrees on each side of the straight through direction.

    "pay your money and make the choice"....sorry....that is a colloquialism, more or less means that you can't have it both ways.
    better point source means lower intensity; higher intensity means a less good point source.
     
    Last edited: Apr 2, 2013
  14. Apr 2, 2013 #13

    sophiecentaur

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    Have you looked at the single slit diffraction pattern? The first nulls are at 90 to the straight through direction. Did you not write that you are familiar with Fraunhoffer?
     
  15. Apr 2, 2013 #14
    Thank you technician and mfb
    I have. But I don't have slit that is same as the wavelength of the monochromatic light used.
    I just studied about Fraunhofer's single and double slit. What is wrong?
     
  16. Apr 2, 2013 #15

    sophiecentaur

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    If you have studied the theory for Fraunhofer's single slit pattern then you must be aware that it works for any width you want. If not, then you need to start with the simple, ideal two slit interference' situation and progress from that.
    Do you appreciate that, for two slits of the same width, the resulting pattern can be calculated by 'multiplying' the single slit (wide) pattern by the double slit (fine) pattern? I googled "two slit diffraction pattern" and found an endless supply of hits which give diagrams and 'the maths', plus loads of diagrams - starting with wiki. I am not really sure what extra "visualisation" you are wanting. I can't think of anything simpler than what I have found.
     
  17. Apr 2, 2013 #16

    Andy Resnick

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    You will generate an evanescent field. The geometry you describe is used a lot for near-field scanning microscopy, for example. The details were worked out in the 1940's, as I recall.
     
  18. Apr 8, 2013 #17
    I think I have studied, but I really cant find out why you are so sure it can work for all width? actually last time I wondered how to block light, I thought light is also particle so we can block them if our slit is smaller then their diameter, but now only realise that they are absorbed by opaque not blocked.

    I did try,maybe I am not good at it or can you please guide me your reading material link?
    This one I am not sure too, though there are some math I am not sure.

    http://web.mit.edu/viz/EM/visualizations/notes/modules/guide14.pdf
    I dont know why we can multiply the intensity of the narrow double with the intensity of the wide single slit to get the intensity of the double slit with wide slit.
    Please guide , thank you.
     

    Attached Files:

  19. Apr 8, 2013 #18
    Dont know how the NSMM function, knowing its application makes it more interesting.
    Thank you.
     
  20. Apr 8, 2013 #19

    jtbell

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    One can prove this by starting from scratch and setting up an integral for the total amplitude of the light from the two slits, as in the first equation on this page.

    http://www.phy.davidson.edu/StuHome/grpatterson/Diffraction and Spatial Filtering/double_slit1.htm

    Each of the two integrals, by itself, gives the total amplitude from one of the slits, which turns out to be the "wide single slit" pattern.

    The details of evaluating the integrals are left as an exercise for the student. :wink:

    I'm surprised that the MIT notes that you link to don't use an integral to find the single-slit intensity pattern. Surely MIT students know how to do integrals! :bugeye: :confused:
     
  21. Apr 8, 2013 #20

    mfb

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    That is simple: the derivation of the formula does not need any relevant restriction of the width (well, it cannot be negative...).
     
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