$E(x,y,z)=\frac{z}{i \lambda}\int E(x',y',0) \frac{e^{ikr}}{r^2}dx'dy'$,

where $E(x',y',0)$ is the electric field of the aperture, and

$r=\sqrt{(x-x')^2+(y-y')^2+z^2}$.

This was an assignment for my E+M class (836) that calculates the diffraction of light around a cross which is expressed exactly by the Fresnel integral:

So, the electric field of a cross would be some amplitude $E(x',y',0) = E_o$ where the cross is located, and $0$ everywhere else. Plug that in, and now you have the diffracted electric field $E(x,y,z)$ everywhere in space...mod-square that to see what our eyes see! In the mathematica file I calculate the intensity on a screen located just 1micron away up to 100 microns away from the aperture screen, and you can gradually see more and more interference. Note that on a screen an "infinite" distance away (called Fraunhofer diffraction) the E-field is just the 2D Fourier transform of the initial electric field!

$E(x,y,z)=\frac{z}{i \lambda}\int E(x',y',0) \frac{e^{ikr}}{r^2}dx'dy'$,

where $E(x',y',0)$ is the electric field of the aperture, and

$r=\sqrt{(x-x')^2+(y-y')^2+z^2}$.

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Great post! Josh.

ReplyDeleteplease drop me an email at hacaoideas@gmail.com.

ReplyDeleteI have many things that I wanna ask you about Mathematica as I'm using mathematica to learn numerical analysis.

I'm following undergraduate naval architecture at my university.