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I Huygens-Fresnel principle of diffraction

  1. Jan 3, 2018 #1
    1. The problem statement, all variables and given/known data

    This question does not concern a homework problem but I don't really understand the Huygens-Fresnel principle of diffraction. My book states that an assumption is made that a wavefront acts as a source of secondary wavelets. They continue with the following derivation. This derivation concerns a single slit experiment with a source point and an observation point at either side of the slit.

    Let ##r'## be the distance between the source point and the slit. The wave will reach the slit at ##t=0## so the amplitude becomes $$E_A = \frac {E_0} {r'} e^{ i(kr')} $$ Now let the distance from the slit to the observation point equal ##r##. They state the amplitude at this observation point caused by a single wavelet will then equal $$dE_p = \frac {E_A} {r} e^{ i(kr - w*t)}$$ This is where I get confused. Why does the source amplitude of such a wavelet equal the amplitude of the entire original wave? The way I understand it the number of wavelets should go to infinity. so how come they all have the amplitude of the original wave. Wouldn't you be creating energy out of nowhere?

    Thanks!
     
  2. jcsd
  3. Jan 3, 2018 #2

    Charles Link

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    To show energy conservation, the problem really needs to be written with an aperture that covers two dimensions. The complete diffraction equation is ## E(r)=\frac{i}{\lambda} \int\int E_{inc}(r') \frac {e^{-ik|r-r'|}}{|r-r'|} dA ##. (Note ## r ## and ## r' ## represent vectors here). For a square aperture of side ## b ##, this makes a pattern that has an effective angle ## \theta \approx \lambda/b ## so that the effective solid angle of the pattern ## \Omega \approx (\lambda/b)^2=\lambda^2/A ##. Meanwhile, the on-axis (peak) irradiance (watts/m^2) will be proportional to ## E^2 ##. The electric field ## E ## on- axis, (by evaluating the integral), is proportional to ## \frac{E_{inc} A }{ \lambda r} ##, so that ## E^2 ## is proportional to ## \frac{E_{inc}^2 A^2 }{\lambda^2 r^2} ##. The total power in the pattern is proportional to ## P_{inc}= E_{inc}^2 A ##. The power in the far field is computed to be ## P=[\frac{E_{inc}^2 A^2}{r^2 \lambda^2}][\frac{\lambda^2}{A}]r^2 ## is consistent with the incident power. ## \\ ## Note: Depending on the system of units, the power ## P ## will contain some additional constant ## k ##, so that more correctly ## P_{inc}=k E_{inc}^2 A ##, and ## P=k E^2 \Omega r^2 ##. ## \\ ## Conservation of energy can be shown in more detail by actually performing the integral of ## P=\int\int E^2 (r) \, dA ## of the complete diffraction pattern. Just a simple "effective" solid angle calculation shows to some degree that the results are consistent. ## \\ ## Note: When the single slit diffraction equation is derived, they often simply compute ## E(\theta, t)=E_o \int\limits_{0}^{b} \cos(\frac{2 \pi x \sin{\theta}}{\lambda}-\omega t) \, dx ##, with ## I(\theta)=E^2(\theta) ##, and the primary concern is on the basic shape of the diffraction pattern as a function of ## \theta ##, rather than trying to demonstrate energy conservation.
     
    Last edited: Jan 3, 2018
  4. Jan 3, 2018 #3

    Ibix

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    At the plane where you switch from describing the wave as a plane to describing it as a sum of wavelets the amplitudes must be equal. Otherwise the descriptions are inconsistent.

    Regarding your concerns about energy - each wavelet comes from an infinitesimal area of the incoming wave. So it carries an infinitesimal amount of energy, whatever its amplitude. Summing the infinitesimal energy times the infinite wavelets gets you the original energy.
     
  5. Jan 3, 2018 #4

    vanhees71

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    Well, be careful. E.g., a plane wave (or even a spherical wave as written in the OP) has an infinite energy and thus doesn't really exist in nature. A true em. field always is a wave packet, for which energy, momentum, and angular momentum are finite.

    In practice you can use Kirchhoff's theory and be pragmatic, leaving the overall normalization of the intensity a free parameter. The most simple approximation is to condider Fraunhofer observation. Then it boils down to the finding that the diffraction picture is the Fourier transform of the openings. This you can find even on Wikipedia:

    https://en.wikipedia.org/wiki/Diffraction_formalism

    Exact diffraction theory was worked out by Sommerfeld in his habilitation thesis, and it's a pretty complicated problem. Of course, Sommerfeld has given a pedagogical exposition of his theory in the famous "Lectures on Theoretical Physics" (vol. 4, optics). I can only recommend this 6-volume series, which I consider still the best classical-physics textbooks ever written (although being about >50 years old).
     
  6. Jan 3, 2018 #5
    Huge thanks to all the answers in this thread! My understanding of the topic is a lot better now.
     
  7. Jan 3, 2018 #6

    Charles Link

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    @Decimal It may interest you that ## \int\limits_{-\infty}^{+\infty} \frac{\sin^2{x}}{x^2} \, dx=\pi ##. This result can be used to show that energy is conserved for the single slit diffraction pattern in two dimensions, at least for the case of a somewhat narrow pattern ## \Delta \theta ##, i.e. where ## \phi=\frac{2 \pi b sin(\theta)}{\lambda} \approx \frac{2 \pi b \theta}{\lambda} ##. (This requires a somewhat wide slit ## b ##, but I think it is a useful result). ## \\ ## Mathematically, Kirchhoff's diffraction theory is not exact for very narrow apertures, as was discussed previously on PF. https://www.physicsforums.com/threa...offs-diffraction-formula.919105/#post-5799277 This is a very fine detail though, and for reasonably wide slits, energy is conserved reasonably well with the Kirchhoff diffraction theory formulation.
     
    Last edited: Jan 3, 2018
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