# Imaging fidelity in a 4F setup

In Fourier optics, a 4$f$ setup is an arrangement of 4 lenses like so:

The idea is that the beam waist $\omega_f$ at the last position (at $x = 2f_1 + 2f_2$) is equal to the waist $\omega_i$ at x = 0 (times the magnification, given by the ratio of the two focal lengths). Does this only happen when the lenses are separated by $f_1 + f_2$ ?

Ideally the distance between the lenses can be infinite, since the beam is collimated. Obviously the beam is Gaussian so one cannot do that, but still.
Is the image at $x = 2f_1 + 2f_2$ only focussed (at a waist) if the inter-lens distance is $f_1 + f_2$?

## 1 Answer

Yes, the lenses have to be separated by $f_1+f_2$. To see why, consider in the ray optics picture what happens when the distance is different and the input rays are parallel instead of originating at a point in the focus. Or consider in the Fourier optics picture where the Fourier transform plane would be in such a system.

• But in real space the beam is collimated between the lenses, why does it care about propagating for a fixed and not infinite length>? – SuperCiocia Sep 19 '17 at 8:44
• You pointed out yourself that the beam isn't collimated, it's Gaussian. You have to consider the Fourier optics picture. The position of the waist will change if the propagation length changes. – ptomato Sep 19 '17 at 8:51
• What would the Fourier transform of this beam look like? – SuperCiocia Sep 19 '17 at 10:38
• The Fourier transform of a Gaussian is another Gaussian. The first lens does a Fourier transform, that's why the plane where the focal lengths overlap is called the Fourier plane; then the second lens does the inverse Fourier transform and you get your original beam back. If the distance isn't the same, then you don't get your original beam back, you get a Fourier transform of a propagated version of it. That will also be a Gaussian, but the waist will not be in the same position. – ptomato Sep 19 '17 at 18:27