How does the fractional Fourier transform apply to an out-of-focus imaging system? Do we use the fractional distance to the focal plane?

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?

Gaussian beam optics might be the simplest approach to answering here as well.