# How can we show that a lens is a low pass filter?

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?

• 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. Apr 14, 2022 at 13:38
• 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.
– Ian
Apr 14, 2022 at 14:24
• 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. Apr 14, 2022 at 15:16
• 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). Apr 14, 2022 at 18:40