What are the dimensions of huygen wavelets

  • Thread starter RedX
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  • #1
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According to Wikipedia the formula for the field created by an aperture (the Kirchoff-Fresnel Integral) is:

[tex]\Psi(r)\propto \int\!\!\!\int_\mathrm{aperture} E_{inc}(x',y')~ \frac{e^{ik | \bold r - \bold r'|}}{4 \pi | \bold r - \bold r' |} \,dx'\, dy', [/tex]

http://en.wikipedia.org/wiki/Huygen's_principle#Diffraction_by_a_general_aperture

How do the units work out for this? The right hand side has units of electric field times length, so don't we need to get rid of the length somehow? The only thing I can think of is to divide by the wavelength, but if that's the case, then they would have added it to the right hand side to emphasize it, because that seems important. The wavelength can already appear in equations due to the k in the integral.
 

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  • #3
970
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The Wikipedia article is completely wrong. The "field" ψ(x) is supposed to be a scalar, and the Eint on the right hand side is supposed to be its normal derivative. Take a look somewhere else, such as http://www.physics.byu.edu/faculty/berrondo/wt642/diffraction.pdf
Those are some pretty good notes.

Indeed, the right hand side should be the normal derivative of the scalar field, and even if that's changed, that's only half the equation. There should also be a term that is the field times the normal derivative of the Green's function. These two terms almost are the same, but one gives an inclination from the slit point to the field point, and the other gives an inclination from the source point to the slit point.

I checked out an optics book from the library and they seem to neglect the inclination from the source point to the slit point, and only consider the inclination from the slit point to the field point.
 

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