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Basic doubt on diffraction

  1. Oct 15, 2007 #1
    In single slit diffraction, my professor says that only 2 rays get diffracted. Both at the 2 ends of the slit. no other rays are diffracted. Is this true??:confused:
    What i think is all rays in the slit get diffracted and not only the 2 rays at the 2 ends of the slit.
     
    Last edited: Oct 15, 2007
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  3. Oct 15, 2007 #2

    jtbell

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    Last edited: Oct 15, 2007
  4. Oct 15, 2007 #3

    vanesch

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    You must (a) have misunderstood him or (b) he expressed himself badly or (c) he wanted to illustrate something peculiar and you missed somehow the point and got this out of context or (d) he doesn't know what he's talking about.
     
  5. Oct 15, 2007 #4
    He was right. Only the photons that "see" or "touch" the slit are diffracted. There is no physical basis for diffraction of a photon that passes through the slits without touching one or both of the edges of the slits.
     
  6. Oct 15, 2007 #5
    no i didn't misunderstood him, i even questioned him when the lecture was over. He emphasized on his point that in single slit only 2 rays at the end of slit get diffracted..
     
  7. Oct 15, 2007 #6
    hey but diffraction deals with the wave nature of light. Then how can u explain it using photons??:confused:
     
  8. Oct 15, 2007 #7

    vanesch

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    Because photons are of course mechanical bullets....
     
  9. Oct 15, 2007 #8
    Think of diffraction as a change in the path of rays. They do not reflect (i.e., they do not obey Snell's law); they simply "diffract", following different paths of propagation. On the other side, each diffraction point gives rise to a secondary wavefront that interfere constructivelly and destructivelly with each other according to their phases at the slit. And when they are measured, each interference component of the wavefield gives rise to a pattern. That is what the diffraction experiment tell us.
     
  10. Oct 15, 2007 #9

    Claude Bile

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    You're absolutely correct, Diffraction is a wave phenomenon and can be explained perfectly well without invoking the particulate nature of light.

    I think I understand the OP's question and I think it involves a particular derivation of the minima in a single slit diffraction pattern that involves calculating the path difference between the centre of the slit and one edge.

    This particular derivation works because of symmetry. If you divide the slit into a number of "slitlets" (essentially invoking the Huygen's principle) and look at the phase contribution of each "slitlet", you can see that, in the case of a minimum, the phase contributions from the "slitlets" can be canceled in a pairwise fashion.

    I have a feeling something was lost in translation between you and your professor. No professor would use the phrase "rays get diffracted", because ray optics implicitly ignores wave-like behaviour like diffraction.

    Claude.
     
  11. Oct 16, 2007 #10
    Absolutely right. I know that optics is an approximation, considering light travels in straight lines and not like wave.
    But what i meant to ask was that, "do only the 2 light "waves" arriving at the 2 ends of the slit get diffracted? i.e. do no other light wave arriving at the slit get diffracted?"
    As diffraction is just flaring of light, which takes place at every point of the slit and not just only at the 2 ends of the slit.
     
  12. Oct 16, 2007 #11
    2 light waves

    I think it is possible that only 2 light waves get diffracted at the ends of the slit, since diffraction is an interference measure. But i state this based on the scalar diffraction theory, where the Fresnel-Kirchhoff integral operator (or any other approximation that describes diffraction) "sums" all light waves -- mathematically speaking, it sums all its oscillatory sin and cosine terms -- incident upon the slit and "puts" it on a observation point. If, by chance, the geometry of the experiment results in amplitudes at the observation point of only 2 light waves diffracted at the ends of the slit, i see no problem in this. But it means that all other contributions from the slit, except from the edges, interfere destructively. Also, this observation is valid depending where the diffraction pattern is observed, and the latter is directly related to the geometry of the experiment. But rather than being a law, i see it as an exception.
     
  13. Oct 16, 2007 #12

    ZapperZ

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    Consider the scenario that this is correct. Then it shouldn't matter how wide I make the slit because only light that is at the end of the slit will get diffracted. The degree of diffraction should remain the same if I keep the intensity constant.

    Yet, we know that isn't true. I get a larger diffraction effect as I make the slit smaller and smaller.

    Secondly, I can also get the identical diffraction pattern with electrons, and not just having them go through a slit, but rather in a superconducting circuit such as a SQUID. Unless one is willing to contradict the physics of superconductivity, there are no "scattering" off the edge of anything here because the state that the supercurrent has condensed to has long-range coherence that has been "protected" from scattering off the material (quantum protectorate).

    Finally, how does one get the diffraction pattern, especially when this is done one photon at a time? If one considers a classical distribution of "ballistic photons" in terms of angle of impact with the side of the slit, there should be a continuous spread of impact on the screen after many photons have passed through the slit one at a time. One can expect this from both sides of the slit opening. There should be a continuous distribution and no cancellation (why would the detector care that a future photon that will hit it in the exact location has an opposite "phase"?).

    So no. That explanation is not correct.

    Zz.
     
  14. Oct 16, 2007 #13

    Claude Bile

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    Diffraction is the evolution of a wavefront as it propagates. I think if you apply this definition, many of your questions will be resolved.

    In the case of the single-slit, being able to calculate the minima by only considering the phase contribution of 2 Huygen's wavelets is pure happenstance, and occurs because of how the phase contribution from each wavelets can be cancelled in a pairwise fashion. This can ONLY be done if the observation point is at a minimum, and can't be done for any other observation points (As carlos_asarf pointed out, it is the exception rather than the rule).

    Anyone that uses this method to explain single slit diffraction should explain that, because the phase contribution from those two Huygens wavelets can be cancelled, you can extend the same argument to every pair of Huygens wavelets. So in actuality, you are regarding the entire wavefront as you should be doing.

    Claude.
     
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