How Do Lenses Exhibit Fourier Transforming Properties in Optics?

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Lenses exhibit Fourier transforming properties due to their ability to produce the Fourier transform of an object at their back focal plane, unlike Fraunhofer diffraction which requires significant distances to achieve the same result. The lens's phase transfer function effectively cancels out the phase factor present in Fresnel diffraction, allowing for practical applications in optics. While a simple setup for Fraunhofer diffraction can yield a Fourier transform without lenses, incorporating a converging lens focuses the transform at the focal plane, enhancing usability for visible objects. The mathematical basis for this transformation relies on the thin lens approximation, which may not hold for thick lenses or those with large diameter-to-focus ratios. Understanding these principles clarifies why lenses are considered Fourier transform calculators in optical systems.
wellorderingp
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I was reading about this Fourier transforming property of lens,when I came by the experimental setup for Fourier optics(with laser and a 4f correlator system).Part of the setup was that of Fraunhofer diffraction and we get the Fourier transform of the aperture at the focal point of first lens.
But then a simple setup for Fraunhofer diffraction,which need not consist of a lens, could do the Fourier transform,why is said that the lens has the transforming property?
 
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Both Fraunhofer difffraction and lens focusing have Fourier transforming action. The main difference between them is the required distance for the Fourier relation to hold. In the case of positive lens, the Fourier transform of the object can be obtained at the back focal plane. In contrast, for the same object size, the Fraunhofer diffraction pattern which gives the Fourier transform of the object must be observed at distances which satisfy the requirement
$$
L >> \frac{2a^2}{\lambda}
$$
where ##a## is the object size. For daily object of sizes on the order of a few centimeters and red wavelength, the required distance will have to be much larger 1 km.
In this sense, using lens as a Fourier transform calculator for objects with sizes visible to naked eyes is much more realistic. It can be proven that the relief of the Fourier transforming property of positive lens from the Fraunhofer distance requirement is achieved thanks to the phase transfer function of the lens, which cancels the same, but opposite sign phase factor that appears in the Fresnel diffraction formula.

 
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What If I have a simple experimental setup for Fraunhofer diffraction of say a rectangular aperture, with a laser source and get the Fourier transform on a screen(without any kind of lenses).Now I introduce a converging lens in front of the illuminated aperture and I get the same result but at it's focal plane. Isn't this just the focusing property of the lens? I mean all it did was focus the Fourier transform,why is it attributed with the property of transforming?
 
wellorderingp said:
I mean all it did was focus the Fourier transform
Rather than thinking that the lens focuses the Fourier transform, instead it is more correct to think that it's the focusing property of a converging lens which makes it Fourier transform the incident field. You may be able to understand this action through mathematical derivations. The Fourier transforming action was in fact derived using the assumption of thin lens and paraxial approximation. If one uses instead, thick lens and/or lenses with large diameter-to-focus ratio, despite still have the focusing property, the field at the focal plane will deviate from the exact Fourier transform.
 
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