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Light Diffraction

  1. Dec 19, 2008 #1
    For diffraction patterns, the simple relation for the angle corresponding to dark fringes is sinθd = mλ/a, in which a is the width of the slit and m = {all integers excluding 0}. From this, it appears that a slit only allows certain wavelengths (a slit width corresponding to a length equal to an integer multiple of the wavelength). Is this true...and if so...why (the direction of propagation of the wave is perpendicular to the slit, so how would the width of the slit have such an influence)?
     
    Last edited: Dec 19, 2008
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  3. Dec 19, 2008 #2

    vanesch

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    I don't know where you get this from, or I misunderstand your question.

    If you send light on a single slit, then you get a an outgoing light distribution with a continuously varying intensity as a function of the angle

    The diffraction by a single slit is different. It doesn't give you specific Bragg angles, but a specific continuum of intensity as a function of angle given by a formula of the kind:

    I(theta) = I0 sinc^2(a sin(theta)/ lambda)

    (or something of the kind, I didn't check).

    Now, that's a function with zeros at specific angles, and those angles are given by the formula you stated. Note that m=0 is not a solution: the direct beam always gets through (eh, as you noted yourself).

    What makes you think that no light of an arbitrary wavelength can get through ?
     
  4. Dec 19, 2008 #3

    ZapperZ

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    Er... try doing this with an incandescent light bulb and then look at the resulting diffraction. You'll see for yourself how what you see falsifies what you think here.

    Zz.
     
  5. Dec 19, 2008 #4

    jtbell

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    This would be true if [itex]\theta_d[/itex] were constant. But it's not. It varies from 0 to 90 degrees, with [itex]\sin \theta_d[/itex] varying correspondingly from 0 to 1, depending on the position the light arrives at your viewing screen. The source where you got that equation should have a diagram showing the geometry of the situation and how [itex]\theta_d[/itex] fits in.

    For given values of [itex]\lambda[/itex] and a, different values of m give you different values of [itex]\theta_d[/itex], which give you the positions on the screen at which the diffracted light is at a minimum.
     
  6. Dec 20, 2008 #5
    Looks like I misunderstood the content. But, I'm still not understanding how the equation came up. Could anyone go ahead and post a proper derivation of the equation sinθd = mλ/a??
     
  7. Dec 20, 2008 #6

    Doc Al

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    Look up single slit diffraction (Fraunhofer diffraction) in any textbook. Or try this: Fraunhofer Single Slit
     
  8. Dec 20, 2008 #7
    Thanks...I looked into it more...

    I think I found the source of my misunderstanding. For a*sinθd = mλ, if a < λ, θd does not have a solution within a single domain, whereas a > λ would hold proper solutions. Earlier in the text, it was stated that (or I read it as) proper diffraction occurs for slits comparable to or smaller than the wavelength, which seems to contradict the equation above.
     
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