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I get the slit version of diffraction, with the Huygen's principle. But I don't find the same principle useful.

I keep thinking that a obstacle like a boat is the same as two large slits on both sides. So on this basis, the wave is never gonna diffract.

edit: I am wondering why the length of the wave shown here is so small compared to that of the ship, but yet it diffract?! enter image description here

2nd edit: Huygen's principle: enter image description here

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  • $\begingroup$ That about the segments of a wave: peaks and troughs. $\endgroup$
    – Jimmy360
    Jun 4, 2015 at 3:31
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    $\begingroup$ The statement is incorrect to begin with. Waves always diffract, no matter how large the object. $\endgroup$
    – CuriousOne
    Jun 4, 2015 at 3:41
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    $\begingroup$ @Courious One is right about diffraction being ubiquitous, so a better way to ask the question why the diffracted amplitude drops so rapidly for angles directed into the vessel's shadow. $\endgroup$ Jun 4, 2015 at 3:47
  • $\begingroup$ "Every point of a wave front may be considered the source of secondary wavelets that spread out in all directions with a speed equal to the speed of propagation of the waves." $\endgroup$
    – Jimmy360
    Jun 4, 2015 at 3:48
  • $\begingroup$ I thought that the above was Huygen's Principle. $\endgroup$
    – Jimmy360
    Jun 4, 2015 at 3:48

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The water wave will diffract at the edges , but with respect to the size of the boat it will be like a small shadow.

It is analogous to what happens to water waves going through a slit, demonstrated here at about 2'30", one sees that the shorter the wavelength the more narrow the diffraction pattern. If the wave length were made very small then it would pass through with only a small dispersion at the edge. If the wave length is very large the slit acts as a point source for the dispersion of energy.

It all depends on how a plane wave reacts to and obstacle. Think of the two limits:

1) if wavelength is very much larger than obstacle, a piece of wood in the water, the wood is essentially a part of the mass of the water that is undergoing the up and down motion, not distinguishable from water and allows the energy to go through without effect.

2)if the wavelength is very small with respect to the obstacle, the part of the wave hitting the obstacle will give up all its energy on the obstacle and be absorbed. Only at the edges there will be a continuation of the plane wave plus an interference from the part of the wave that scattered at the edge, a bit of difraction, which, as the wavelength is small, will be of small extent in the shadow of the obstacle.

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  • $\begingroup$ Are you saying obstacle diffraction is the complete opposite of slit diffraction? (The smaller the wavelength compared to that of the obstacle, the more refraction. Sorry maybe my understanding is problematic) $\endgroup$
    – Valkyrie
    Jun 4, 2015 at 14:11
  • $\begingroup$ No the smaller the wavelength the less diffraction at the edges , same as the smaller the wavelength the narrower the diffraction pattern through a slit. The edge will behave as the edge of the slit $\endgroup$
    – anna v
    Jun 4, 2015 at 15:34
  • $\begingroup$ the wave diffracts in the picture on the left, but the shadow is much larger. there is always diffraction at the edge. $\endgroup$
    – anna v
    Jun 4, 2015 at 15:36
  • $\begingroup$ What is shadow? $\endgroup$
    – Valkyrie
    Jun 4, 2015 at 15:37
  • $\begingroup$ But there shouldn't be diffraction b/c the length of ship is several times the wavelength.?! $\endgroup$
    – Valkyrie
    Jun 4, 2015 at 15:40

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