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Why is wave diffraction around object better at longer wavelength?

  1. Sep 17, 2010 #1
    Does the wave theory itself (the mathematical formulation) imply that diffraction is more efficient at longer wavelengths? I just want the logical connection of why (for example) radio waves diffract and bend around objects better than visible light. :confused:

    Thank you.
  2. jcsd
  3. Sep 17, 2010 #2


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    I dont think radio waves diffract any "better" than visible light. I can shine a light through a lens and have an enormous bend in the light. Radio waves wouldnt do that. The frequency of the light (And hence its wavelength) will cause it to diffract differently than any other frequency. Does this help and make sense?
  4. Sep 18, 2010 #3
    I'm not talking about refraction I'm talking about diffraction. Lens is something completely different.
    However, if a longer wavelength encounters an obstacle it will become a wave "as if no obstacle was present" faster than shorter wavelength.
    As illustrated here: http://hyperphysics.phy-astr.gsu.edu/hbase/sound/diffrac.html

    But, I don't understand why.
  5. Sep 18, 2010 #4
    By Planck's law we get wavelength is inversely proportional to the energy....so if wavelength increases the energy decreases.....diffracts more because energy is less an so it cannot pass it.............
  6. Sep 18, 2010 #5
    This Planck's Law? http://en.wikipedia.org/wiki/Planck's_law
    I don't see what's the connection.

    However I know that E=hf for photons, and [tex]f=c/\lambda[/tex], so [tex]E=hc/\lambda[/tex] . Energy is inversely proportional to wavelength. So what?
  7. Sep 18, 2010 #6


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    well it depends on your system. In acoustics, long wavelengths travel the farthest with the least distortion. The bigger the wavelength, the less likely something on Earth will be big enough to stop it. This is why the CTBT is using infrasound sensors around the world to detect low frequency signals from explosions.

    Acoustic waves still can experience diffraction over mountains, though.
  8. Sep 18, 2010 #7


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    And you're not using diffraction correctly, I just noticed. Radio waves simply pass through objects that dont distort them. A diffraction would distort the wave signal.
  9. Sep 18, 2010 #8
    Ok fine, then radio waves passing through a single slit (and the rest of the wall is radio-proof), or a ripple tank with a single slit. Longer waves will diffract better through the slit. Why is that?

    I'm thinking shorter wavelength create a superposition that creates destructive interference thorough the slit (because the slit can't be a single dot, it has width, and therefore you have several "new" circular waves starting at the slit).
  10. Sep 18, 2010 #9


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    I don't know what you mean by "diffract better" but the point is the same as with acoustics. Larger wavelengths simply don't interact with the slit because they're too big. As the u derlined sentence you quoted states, diffraction is more pronounced when the object is around the same size. Bigger and smaller waves won't be affected by the slit.
  11. Sep 18, 2010 #10
    To pass trough an object without much bending a wave requires certain energy.....if the wavelength is longer then.............it will have to do more efforts which it cannot.....and so it bends.....(energy inversely proportional to wavelength)
  12. Sep 18, 2010 #11
    So if a wave has more energy it will have higher tendency to continue in a straight line? and a less energetic one will bend, as if due to lower inertia?
  13. Sep 18, 2010 #12
    yes i think............on my side..........
  14. Sep 18, 2010 #13


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    Does this whole thing with diffraction of sound work with light as well?
  15. Sep 18, 2010 #14


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    I think by "diffract better" he means that the widening of the wave's pattern, as compared to a "ray" description, is through a wider angle.
    Yes, and water waves too.

    I was amazed to learn that the diffraction pattern by a given geometry (a single slit, for example) depends on the wavelength but not on the wave's speed. So whether it's light, sound, or water, you just need to know the wavelength and the size of the slit, and you can calculate the diffraction pattern. (Ignoring any nonlinear or dissipative effects.)
  16. Sep 19, 2010 #15


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    Ah, that makes more sense.

    Then, etamorph, you're referring to large and small waves that are on the order of the slot's magnitude already?

    In the classical view, it's pretty much the same as with water waves. As you tighten the slit, you pinch wave forms, making them bellow more (explicitly with acoustic/water waves, it's because the particles near the edges of the slit are getting slowed down by the barrier, having a rigid object (the non slit part of the slit) to push against. The particles near the middle of the grating are less encumbered because they're surrounded by more easily movable particles.

    In the quantum view:

    Well....not really. The diffraction grating is just in the molecular structure in refractive materials.
  17. Sep 19, 2010 #16
    Thanks for all the answers, one more question:

    If the slit is say width 5x, and then you let waves diffract through it.
    wave of lambda=x will create a visible single-slit interference as explained by the Huygens Principle, whilst lambda=10x will diffract as if it is a point-slit, without seeing interference. How is this explained?
  18. Sep 19, 2010 #17


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    It all has to do with the electrical size of the scatterer. All waves will diffract, the difference between the diffraction of visible light versus RF is the electrical scale that we are viewing the problem. That is, we typically view the diffraction of light over an area of millions of wavelengths visually but this corresponds to subwavelengths when we talk about radio waves. If you scaled the problem and the area over which you observe it to the scale of the wavelength, then the amount of diffraction for radio waves would be same as for visible light.

    So the amount of diffraction for an infinite object, like a wedge, looks smaller for visible light because we are not scaling the problem size to see the diffraction. The diffraction occurs over a distance on the order of wavelengths and we can't see that if we view it on the order of meters when the wavelength is on the order of nanometers.

    Likewise, with scatterers of finite size, like a cylindrical coumn, the impact on the incident waves has to do with the electrical size (size in terms of wavelength) of the object. So if you decrease the wavelength of your radiation, you need to scale the size of the scatterer to see the same physics. The primary reason for this is that for small scatterers, the phase shift over the distance of the scatterer is small. Diffraction occurs as a result of the interference of the incident wave with the scattered wave. If the size of the scatterer is electrically small, then the size in "phase," so to speak, is small too and the resulting interference is short ranged and subtle.

    Slits are a bit different. If we have a very large slit then there is still diffraction byt it occurs at the edges of the slit and is short lived. We do not see any effects on the large scale because these diffractions die out quickly. But when the slits are small, then the diffraction from the edges are close enough to gether that the diffractions interfere with each other. This interference is what produces the fringing patterns that we see with small slits.
    Last edited: Sep 19, 2010
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