This question has been asked in a similar way but no satisfactory answers came up.

Huygen's principle relies on the fact that if we consider an aperture, every element dS on the surface of the aperture is emitting as a secondary wavelet.

Though it is also mentioned that these wavelets are actually fictitious.Indeed if we were to carry the experiment in vacuum, there would not be anything in the aperture that could be responsible for generating these secondary wavelets. Though diffraction would actually happen.

So the question is :

Is there a physical explanation that can explain diffraction in a intuitive manner that does not rely on an artificial hypothesis based on fictitious emitters??


Original response: There is a perfectly valid physical explanation: use Maxwell's equations to find how the wave propagates beyond the aperture. Use as boundary conditions the idea that the wave is completely killed off beyond the aperture. Sound reasonable? This results in an integral. Turns out that the integral can be interpreted, at least to a very good approximation, as Huygens's principle!

So it's a clean principle - derivable from Maxwell's equations which are the foundation of electrodynamics, and is not some contrived idea.

October 2016 addition: Looking at my answer after so long, I see that the original question asked for an "intuitive" picture of the origin of wavelets. The fact is that even the basic Kirchoff integral is scalar in nature while the fields are vectorial; the reference given in my comments below (Jackson) does attempt to extend this to the full vector case. However, this is non-trivial to do and also to understand.

Nevertheless, the basic idea is that the simplest "Green function" needed to solve the equation has a form implying "spherical wavelets emanating from every source point". Unfortunately, the simplest Green function is not universally applicable to all problems, but needs to be different for and must be adapted to every particular boundary condition.

Reading Jackson carefully it is evident that there are several different approximations that can be made under various circumstances, and the popular idea of Huygens's principle is the best simple summary of a mathematically complicated situation. And now I admit I've exhausted my own capacity to clarify this problem and its solution.

  • 1
    $\begingroup$ This isn't much of an "intuitive" picture. $\endgroup$ – Señor O Mar 6 '15 at 16:56
  • $\begingroup$ This is interesting since it does not use the idea of secondary wavelets but I do not get how you find the wave beyond the aperture from Maxwell equation unless you consider the aperture as a widely spread(and not a point source) source emitting spherical radiation. Unfortunately, it is not indeed very intuitive .. $\endgroup$ – Ronan Tarik Drevon Mar 6 '15 at 17:25
  • $\begingroup$ @RonanTarikDrevon The point is that the secondary wavelet is the intuitive picture. The full classical picture is to solve Maxwell's equations with the appropriate boundary conditions. $\endgroup$ – Chris Mueller Mar 6 '15 at 18:32
  • $\begingroup$ Indeed but it is artificial since if the aperture is in the vacuum you can not consider vacuum element dS as emitting wavelets. To me Huygen's principle is a nice artifact to explain that everything works AS IF there were secondary wavelets. On the other hand, saying that diffraction corresponds to the solution of Maxwell's equation is not satisfactory either but I will have a look at the complete demonstration. $\endgroup$ – Ronan Tarik Drevon Mar 6 '15 at 20:22
  • $\begingroup$ In fact saying $x=x_0cos(\omega t+\phi)$ is the solution of the classical harmonic oscillator does not explain the underlying physical phenomenon that the mass is driven by the force from the spring and that an exchange between kinetic and potential energy is constantly occurring but thanks for your help. Maybe my question is not very relevant .. $\endgroup$ – Ronan Tarik Drevon Mar 6 '15 at 20:24

It's not quite as fictitious as you think! Imagine a circular wave pool. To make perfect circular waves of wavelength 5m, you would need a cylinder of radius 5m to rapidly go in and out of the water, displacing a lot of water for a brief period of time, periodically.

Now imagine ocean waves of wavelength 5m going through a slit 5m wide. Inside that 5m wide slit, water is getting displaced at the exact same rate and same fashion that it is above. So as far as the calm water inside the slit sees it, there is a wave machine making radial waves. Think as the wave peak travels through the slit into calm water, it has no preference as to which direction it "falls" to.

If the slit was wider than 5m, then the wave peak in the middle has forces to its left and right keeping it moving in the same direction.

If the slit was narrower than 5m, the wave getting reflected on either side of the slit would have enough momentum to carry the wave trying to go through the slit back in the opposite direction.

  • $\begingroup$ I kind a feel what you say but I am still a bit confused. What do you mean when you say there is a wave machine making radial waves ? And for the picture of aperture wider than wavelength, I do not get why this argument holds only when the wavelength is small compared to the size of the aperture.Moreover you invoke some mechanical aspects(force) which does not stand for EM waves. $\endgroup$ – Ronan Tarik Drevon Mar 6 '15 at 17:18

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