Programming a Diffraction Pattern by the (pseudo?) method of images

[PhDeezNutz]

The last picture was from a rotating dipole. Anyone have any ideas how I can mess around with frequencies to get rid of the spiral? Maybe introduce a rotating magnetic dipole in the orthogonal plane?

 

[PhDeezNutz]

Getting so close but doing so by arbitrarily making electric and magnetic dipole moments complex in different ratios. If I get a good result I'll have to reconcile it physically.

 

[PhDeezNutz]

@hutchphd @etotheipi @TSny

I had some serious misconceptions about the dipole approach to diffraction (or more generally "aperture coupling") . The image dipoles on either side of the infinite (closed off) conducting sheet are supposed to yield total power through the aperture, it is not supposed to recover the Bessel function. Apparently for many applications total power is the most important parameter as opposed to its distribution.

The first two sections of this paper offer an excellent explanation if you're interested.

https://www.tandfonline.com/doi/full/10.1080/02726343.2011.590960?scroll=top&needAccess=true&

The values of the dipoles can be chosen such that the structure on the left absorbs power and the structure on the right emits the same amount of power thus simulating power through an aperture.

 

[hutchphd]

@hutchphd @etotheipi @TSny

I had some serious misconceptions about the dipole approach to diffraction (or more generally "aperture coupling") . The image dipoles on either side of the infinite (closed off) conducting sheet are supposed to yield total power through the aperture, it is not supposed to recover the Bessel function. Apparently for many applications total power is the most important parameter as opposed to its distribution.

The first two sections of this paper offer an excellent explanation if you're interested.

https://www.tandfonline.com/doi/full/10.1080/02726343.2011.590960?scroll=top&needAccess=true&

The values of the dipoles can be chosen such that the structure on the left absorbs power and the structure on the right emits the same amount of power thus simulating power through an aperture.

Thanks for the link. I see that his primary reference is to

Theory of Diffraction by Small Holes
H. A. Bethe
Phys. Rev. 66, 163 – Published 1 October 1944

which is, as always, a nice treatment. It was never clear to me the range of applicability of Bethe but he does say it is for holes small compared to wavelength but then he generalizes some. I think your assessment is on point.

 

[PhDeezNutz]

Thanks for the link. I see that his primary reference is to

Theory of Diffraction by Small Holes
H. A. Bethe
Phys. Rev. 66, 163 – Published 1 October 1944

which is, as always, a nice treatment. It was never clear to me the range of applicability of Bethe but he does say it is for holes small compared to wavelength but then he generalizes some. I think your assessment is on point.

I think Bethe deals with the small hole limit whereas the Kirchhoff Integral deals with the large hole limit. In the Bethe limit I think the hole is so small that the electric field is essentially constant over its extent thus resembling a linear relationship $\vec{p} = (constant) \vec{e}$ or some diagonal polarizability tensor.

Bethe mentions in his paper that the total radiation/power from the Kirchhoff Integral to the his results should differ on the order of $\frac{a^2}{\lambda^2}$ but reading his results more carefully I think there is a slight difference.

He says the Kirchhoff Integral yields $\vec{H}$ on the order $k a^2$ whereas his solution yields $\vec{H}$ on the order of $k^2 a^3$.I presume the same goes for $\vec{E}$If so the radiation intensity is related to the square of these values

$\frac{Kirchhoff}{Bethe} = \frac{\left(ka\right)^2}{\left( k^2 a^3\right)^2} = \frac{k^2 a^4}{k^4 a^6} = \frac{1}{k^2 a^2} = \frac{\lambda^2}{4 \pi^2 a^2}$

Basically I think the Kirchhoff Integral should differ from the Bethe solution by the following

$Bethe = \frac{4 \pi^2 a^2}{\lambda^2} \left( Kirchhoff \right)$

I hope I did that math right.