Huygen's principle states that every plane wave is made up of an infinite number of point sources, and that the constructive interference between each wavelet forms the next wave front, and so the cycle continues.

Diffraction, then, can be explained by the entirety of a wave front being blocked out, but with a slit just wide enough so that exactly one point source can "fit", producing a spherical wave on the other side of the slit.

However, that raises some questions:

  • The effect of diffraction is at its greatest when the slit is the exact length of the wavelength. Hence, can one assume that each point source on the wave front is spaced exactly one wavelength apart?
  • Why is the diffraction effect reduced but not completely eliminated if the slit is wider than the wavelength? Surely if there is more than one wavelet that can "fit" into the slit width, then all that should be produced on the other side is another, albeit shorter, wave front. Why is it slightly bent on the sides when the original wave front is not?
  • When one wavelet can fit into the slit, what happens to the rest? Do they reflect back at the boundary?
  • Why are single slit diffraction patterns produced? If there is only one wavelet in the slit width, what is it interfering with to produce an interference pattern?
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    $\begingroup$ I think you may be a little confused about Huygen's principle. The diffraction doesn't occur only due to one wavelet; instead, at the wavefront meeting the slits, each point on the wavefront behaves as a point source to emit wavelets. Hence you get an infinite number of point sources emitting wavelets at the slits, not just one, and the superposition of the wavelets forms the wave front. $\endgroup$ Jun 22, 2017 at 18:50
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    $\begingroup$ Oh. In that case why is the diffraction effect at its greatest when the slit width is equal to the wavelength? If there are an infinite number of wavelets regardless of the width of the slit then surely how much the wave bends should be independent of the width of the slit. $\endgroup$ Jun 22, 2017 at 18:56
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    $\begingroup$ "The effect of diffraction is at its greatest when the slit is the exact length of the wavelength." Can you elaborate more on this? What do you mean by "the effect of diffraction"? $\endgroup$
    – Mostafa
    Jun 22, 2017 at 20:18
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    $\begingroup$ The bending is at its maximum when the aperture is a point (not a $\lambda$ wide) and the diffracted field is a spherical wave . $\endgroup$
    – Mostafa
    Jun 22, 2017 at 21:27
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    $\begingroup$ You don't even need a slit to get diffraction. A Single edge has diffraction and a slit has two edges. You can derive any fringe pattern from one or two edges. A double slit experiment has four edges. You don't need an infinite Number of point sources in between. $\endgroup$ Jun 22, 2017 at 21:32

1 Answer 1


In order to understand Huygens principle in this context clearly, one needs to resort to the mathematical formulation of the scalar diffraction theory for diffraction from an aperture. According to the Rayleigh-Sommerfeld formula, the diffracted field at a point in space in front of the aperture can be written as

$$U_P(x,y) = \frac{1}{j\lambda}\iint_{\text{aperture}}U_I(x',y')\frac{\exp{(jkr)}}{r}\cos \theta \,\,ds$$

                                    enter image description here

As you see from the above equation, the observed field $U_P$ is a sum of diverging spherical waves in the form of $\dfrac{\exp{(jkr})}{r}$ located at each and every point in the aperture (as stated in the Huygens principle), multiplied by a factor of $\dfrac{1}{j\lambda}\,U_I(x',y') \cos \theta$. Therefore, the fictitious source located at $(x',y')$ has the complex amplitude proprtional to the incident field at that point, $U(x',y')$. This seems reasonable considering the linearity of the problem. (The presence of the remaining multiplicative factors $1/j\lambda$ and $\cos \theta$ may be explained in some other ways but not very intuitively.)

To summarize, your statement that "each point source on the wave front is spaced exactly one wavelength apart" is wrong. The single slit problem is usually treated within the scope of the Fraunhoffer (far-field) approximation of the more general formula above, where the observed diffraction pattern is the Fourier transofrm of the aperture. It means the width of of observed pattern is inversely proportional to the width of the aperture.

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    $\begingroup$ I thank you for the effort in your answer and for trying to help, but I'm afraid I'm only a high school student and this has gone completely over my head! I have, however, been pondering on the issue for a while now and I believe most of my reservations can simply be put to rest by understanding in high school terms why light bends when it diffracts. If you could perhaps answer me this, I would be very grateful. $\endgroup$ Jun 22, 2017 at 22:29
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    $\begingroup$ Actually, upon further though it's probably for the best that nothing more happens on this thread. The entire question was based on a fundamental misunderstanding of Huygen's principle, so therefore I will post a new and less flawed question tomorrow that will allow for more concise answers. $\endgroup$ Jun 22, 2017 at 23:20
  • $\begingroup$ @Pancake_Senpai I dis agree. If one wavelength fits in the slit. What is the angular dependence of the phase difference between the two ends of the slit. As seen after propagating to a distance point at some angle? $\endgroup$
    – JEB
    Nov 27, 2022 at 18:15

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