Kostya answered a question that was asking what the diffraction pattern looks like for a hexagonal aperture in front of a lens. He lists an equation which was derived using a Heaviside function to describe the shape of the aperture.

I was wondering if anyone knows what units were used for the display of his Fourier Transform. He mentions he plotted that from -100 to 100 for (what I am assuming) is $\omega_x$ and $\omega_y$. The units seem to be degrees but I wanted to make sure, as sometimes these problems regarding FTs are treated as being unitless.

I have tried other approaches to predict what that diffraction pattern would look like, and so far Kostya's answer agrees by far the best with my own experimental results and with graphical Fourier Transforms of similarly shaped apertures.


1 Answer 1


What I've calculated is just a Fourier transform of the aperture $h(x,y)$:

$$f(\omega_x,\omega_y)=\int dx\, dy\, h(x,y)e^{-i(\omega_xx+\omega_yy)}$$

(And I was plotting $|f|^2$ as a function of $\omega_{x,y}$ each changing from -100 to 100.)
Already here one can see that $\omega_{x,y}$ both have a dimension of inverse length. So it is not an angle.

Now, the actual expression for Fraunhofer diffraction is something like: $$u(x',y') = \int dx\,dy\,h(x,y)e^{-i\frac{k}{Z}(xx'+yy')}$$ Where $x'$ and $y'$ are coordinates on the screen, where you observe the diffraction pattern, $k$ is a wave-vector $k=\frac{2\pi}{\lambda}$, and $Z$ is the distance to the screen.

As you can see these formulae are very similar. Namely you get the Fourier transform by renaming: $$\frac{kx'}{Z}\leftrightarrow \omega_x\quad\frac{ky'}{Z}\leftrightarrow \omega_y$$ So, in practice $\omega_{x,y}$ denote a position on the screen -- given $\omega_x$, you get an $x'$ by rescaling: $$x'=\frac{\lambda Z}{2\pi}\omega_x$$ Finally let us put some numbers. Let's say that $\lambda=600nm$ and $Z = 2m$. Then a point with $\omega_x = 50cm^{-1}$ will have a coordinate on the disk $x' = 1.8cm$.


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