A $4F$-configuration is usually utilized for imaging with $2$ lenses. An object is placed in the focus of the first lens ($f_{1}$), then the second lens is located at a distance $f_{1}+f_{2}$ from the first, where $f_{2}$ is the focal length of the second lens. Finally, an image is formed in the focal plane of the second lens. I know that the advantage of this system is to have access to the Fourier plane exactly in between these 2 lenses. But let's say I don't need it and just interested in imaging. What will happen if I don't keep the distance between lenses $f_{1}+f_{2}$, make it larger or smaller? From the perspective of Fourier optics, the image would be formed in the wrong way since the second lens would form an image not from the Fourier distribution but from some propagative form of it. Does it affect the image very much?


Does it affect the image very much?

Well, yes.

With a $4f$ setup, at $+f_2$ you have exactly the same image you had at $-f_1$, just (de)magnified by a factor $f_2/f_1$. You can also put a spatial filter in the middle of the two lenses to clean up high spatial frequency modes.

If you don’t keep the $f_1+ f_2$ distance between the lenses, and you still look at the same position at $+f_2$, then you won’t get the same exact relationship between the images before and after the $4f$ setup. You will get something else.

Your “very much” is hard to estimate because Fourier transforms aren’t very intuitive. Best bet it to do the simulation.

You could also do a simulation with the ABCD matrix to see how much the focus (image) after the $4f$ setup shifts depending on your variations from $f_1+f_2$.

  • $\begingroup$ Thank you for your answer! I think the ray-tracing wouldn't help since the rays are just collimated by the first lens and then focused by the second. No matter what is the distance between two lenses, the second lens just focuses the collimated beam at the focal plane giving exactly the same magnification. So I think, only the Fourier transform could give the answer. $\endgroup$ May 21 '20 at 7:12
  • $\begingroup$ Sorry, not ray tracing (I edited the answer). ABCD matrices. They take divergence and waist into account, hence are good for Fourier optics. $\endgroup$
    – SuperCiocia
    May 21 '20 at 7:24

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