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I understand from the derivation in Goodman Chapter 6 that a lens Fourier transforms light from the front focal plane onto the back focal plane, ignoring aperture effects. I've also read that a lens acts as a low pass filter, effectively masking the image in the back focal plane. It seems then that for an object located in the front focal plane, the back focal plane should reveal the Fourier transform of the object multiplied by a masking function which sets the image to zero at radii outside the mask that correspond to high frequencies in the object.

Simple multiplicative masking in the back focal plane to account for aperture effects rooted in the finite size of the lens seems strange to me. It would make more sense to me that a lens operates by masking the field present at the plane of the lens. Then this masked field would be further propagated from the lens plane to the back focal plane via the Fresnel diffraction formula which yields a more complicated expression than simply the Fourier transform of the object multiplied by a mask.

Can someone help me resolve the confusion? How is masking at the objective plane equivalent to masking at the back focal plane? In essence how is it possible that a lens acts as a low pass filter between the focal planes?

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  • $\begingroup$ I do not understand your exact problem. High frequency components mean high diffraction, so at the position of a lens, you will "undersample" high frequencies because your aperture is small. Fourier optics is a simple plane to plane transformation considering the properties of the transformer while fresnel diffraction will be more on the propagation type of solving. I guess both will give you the same answer. $\endgroup$
    – José Andrade
    Commented Apr 14, 2022 at 13:38
  • $\begingroup$ My exact problem is that I want a derivation of how in a 2f system the aperture in the objective plane acts as a multiplicative mask in the back focal plane, rather than as a multiplicative mask in the objective plane. $\endgroup$
    – Ian
    Commented Apr 14, 2022 at 14:24
  • $\begingroup$ I don't understand that claim about a lens being "low-pass" . Take a look, for example, at the FFT response curve for a thin ring lens (central obscuration) , which loses the low spatial frequency information. A lens only limits the maximum spatial frequency as a result of its entrance aperture. $\endgroup$
    – Carl Witthoft
    Commented Apr 14, 2022 at 15:16
  • $\begingroup$ AH! I understand. I'm a bit rusty, but in this case, its multiplicative because at the lens position, the "filtering" is a convolution (in frequency space). Or alternatively, the filter is indeed a multiplication at the lens, but now in conjugate of frequency (so spatial-space). $\endgroup$
    – José Andrade
    Commented Apr 14, 2022 at 18:40

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A lens transforms the phase front at its entrance plane (not the object plane off in the distance) into an intensity pattern at the focal plane. It looks like you're confusing the principal planes of the lens with the object and image planes.

So, yes, to the extent that the size of the lens system's entrance aperture limits the high-frequency components, the lens is "low-pass." I don't think that's a useful way to view the system -- other than the equivalent analysis which shows the size of the Airy Disk as a function of entrance aperture (and lens focal length).

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Edited Answer - Initially I mentioned that its due to the paraxial approximation. But that is not the correct reasoning for this question; because the lens will anyway image all the rays hitting its surface, and the focal lengths may change for rays beyond the paraxial limit. But this alone doesn't imply that lens is a low pass filter. A better reasoning would be - the finite size of the lens diameter (or aperture). If the lens diameter is very small, then only rays which are parallel or near parallel to the optic axis would pass through the lens. If the lens diameter is infinitely large, rays with angles upto 89.99° from the optic axis would be able to pass through the lens. So, therefore it's the aperture of the lens that sets the limit for the ray with the maximum angle that can enter the lens. Larger the angle of the ray, higher is the spatial frequency that it would carry.

Original Answer - It's a consequence of the paraxial approximation which is used to derive the focal length of the lens. Only those rays with small angles (sinθ = θ) are imaged by the lens. Hence lens can be considered a low pass filter.

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