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Why is the bending of waves related to wavelength?

  1. Jan 26, 2010 #1
    Why is the "bending" of waves related to wavelength?

    Hi!
    As stated in the title, my question concerns the "bending" of waves. (I'm quite sure that "bending" is not the proper term, so if you know the correct one, please tell me.)
    Let's say that we investigate waves in a ripple tank. We create a plane wave and put a blockage, parallell to the wavefront, with a small slit in it. If the slit is small enough, we will get almost circular waves on the other side of the blockage, but if the slit is large, the wave on the other side will not be as much bent. This seems natural to me, but I don't understand how it is related to wavelength. My questions are:

    1. In our physics textbook, there was an inequality for "bending" of waves in an experiment like the one described above: D[tex]\leq[/tex][tex]\lambda[/tex], where D is the width of the slit and lambda is the wavelength. What do they mean by this? The bending of the waves gets less and less, the wider the slit is, but it never disappears. How then, can you give such a condition for "bending".

    2. Why does the wavelength affect the "bending" of the waves?

    Thanks in advance!
     
  2. jcsd
  3. Jan 28, 2010 #2

    Claude Bile

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    Re: Why is the "bending" of waves related to wavelength?

    1. The condition you describe is where you get the diffracted wave propagating through 180 degrees; it only occurs when the width of the slit is comparable to (or less than) the wavelength/2. At wider slit widths, "bending" still occurs, just not through the whole 180 degrees.

    2. It is due to the invariance of scale, which basically says that if you halve the wavelength and halve the aperture size, the wave diffracts in exactly the same way. If you halve the wavelength without halving the aperture size, this is equivalent to doubling the aperture size without changing the wavelength. To summarise, the wavelength matters, because the diffraction you get depends on the ratio of the wavelength to the aperture size, not just the wavelength.

    Claude.
     
  4. Jan 28, 2010 #3

    Andy Resnick

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    Re: Why is the "bending" of waves related to wavelength?

    As Claude states, you are talking about *diffraction*. The origin of diffraction is truncating a wavefront. Perhaps you have seen Huyghens' principle: each point on a wavefront can be considered a source point, which will radiate in all directions.

    So, how to construct a flat wavefront? That occurs when all the different source points interfere in such a way that the total wavefront is a flat surface, and this can only occur if the wavefront is infinitely large (or, the original point source is infinitely far away). So, by truncating the flat wavefront (by passing through an aperture, for example), the small piece of wavefront diffracts outward ('spreads', 'curves', 'bends'..), and the rate at which it diffracts is related to the ratio of the wavelength and size of aperture. Even if the aperture is much larger than a wavelength, if the wavefront is not a complete sphere, there will be diffraction.

    One small note of caution: surface waves, in a ripple tank, are significantly different that electromagnetic waves- both are waves, but that's the only real similarity.
     
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