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 $\omega_x = 50$ will have a coordinate on the disk $x' = 1.8cm$.

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 $x' = 1.8cm$.