In Fourier optics it is sometimes convenient to think of lenses as "Fourier transformers". For an imaging system between two planes with a pupil in the center, the amplitude in the pupil is the FT of the image.
A simple way to think about this is that a lens swaps angles and positions (for a plane at the focal distance); all rays from a given position on the focal plane will have the same angle at the lens pupil and vice-versa.
Mathematically there is the fractional Fourier transform $F_{\alpha}(x)$ and when $\alpha = \pi/2$ this is the normal or "continuous" Fourier transform.
Question: How does the fractional Fourier transform apply to an out-of-focus imaging system? For example, if I measure the amplitude at 90% of the distance to the focal plane, would it be reproduced by taking the fractional Fourier transform of the amplitude in the pupil using $\alpha = 0.9 \pi/2$? Is it as simple as that?
Related and potentially helpful:
- Fractional Fourier Transform and Fresnel Propagation
- What is an intuitive explanation of Gouy phase?
- Wikipedia: Gaussian beam optics might be the simplest approach to answering here as well.
- JOSA A: Fractional Fourier optics
- Optik: Fractional Fourier transform in optics – a new perspective
- Optica Applicita: Fractional Fourier transform in optical setups
