Are quantum effects significant in lens design?

Livia Cardenas
2022-07-23
Answered

Are quantum effects significant in lens design?

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slapadabassyc

Answered 2022-07-24
Author has **21** answers

Stan Rogers' answer on photography.SE seems to be claiming that QED is not just sufficient but also necessary to explain the the effect of the lens's shape. This is wrong. Ray optics suffices at ordinary magnifications, and even at high magnifications, classical wave optics suffices.

Let's say you use a rectangular lens rather than a cylindrical one. First off, the shape of the lens won't matter at all unless you have the aperture all the way open; on any slower setting, the approximately circular shape of the diaphram will be the determining factor. Assuming that you do have the aperture all the way open, the main effect, which is purely a geometrical optics (ray optics) effect will be as follows. You will have a certain depth of field. If object point A is at the correct distance to produce a pointlike image, then this point is still a point regardless of the rectangular shape of the lens. However, if object point B is at some other distance, we get a blur as the image of that point. The blur occurs because there is a bundle of light rays, and the bundle has some finite size where it intersects the film or chip. Since the lens is rectangular, this bundle is pyramidal, and the blur will be a rectangular blur rather than the usual circular one. For example, say you're photographing someone's face with a starry sky in the background. You focus on the face. The stars will appear as little fuzzy rectangles.

At very high magnifications (maybe with a very long lens that's effectively a small telescope), it's possible that you would also see diffraction patterns. In the example of the face with the starry background, suppose that we change the focus to infinity, putting the face out of focus. Wave optics would now predict that (in the absence of aberrations), the diffraction pattern for a star would be a central (order 0) fringe surrounded by a ring (first-order fringe) if you used a circular aperture, but a rectangular aperture would give a different pattern (more like a rectangular grid of fringes). In practice, I don't think a camera would ever be diffraction-limited with the aperture all the way open. Diffraction decreases as the aperture gets wider, while ray-optical aberrations increase, so aberration would dominate diffraction under these conditions.

Quantum effects are totally irrelevant here.

Stan Rogers says:

It's difficult to explain without launching into a complete explanation of quantum electrodynamics, but all of the light that reaches the sensor "goes through" all of the lens, at least in a sense, even if we're just talking about a single photon. A photon doesn't take just one path (unless you make the mistake of trying to figure out which path it took), it takes all possible paths. Weird, but true.

This is an OK description of why QED suffices for a description of the phenomenon, but it's completely misleading in its implication that QED is necessary. The word "photon" can be replaced with the word "ray" wherever it occurs in this quote, and the description remains valid.

Let's say you use a rectangular lens rather than a cylindrical one. First off, the shape of the lens won't matter at all unless you have the aperture all the way open; on any slower setting, the approximately circular shape of the diaphram will be the determining factor. Assuming that you do have the aperture all the way open, the main effect, which is purely a geometrical optics (ray optics) effect will be as follows. You will have a certain depth of field. If object point A is at the correct distance to produce a pointlike image, then this point is still a point regardless of the rectangular shape of the lens. However, if object point B is at some other distance, we get a blur as the image of that point. The blur occurs because there is a bundle of light rays, and the bundle has some finite size where it intersects the film or chip. Since the lens is rectangular, this bundle is pyramidal, and the blur will be a rectangular blur rather than the usual circular one. For example, say you're photographing someone's face with a starry sky in the background. You focus on the face. The stars will appear as little fuzzy rectangles.

At very high magnifications (maybe with a very long lens that's effectively a small telescope), it's possible that you would also see diffraction patterns. In the example of the face with the starry background, suppose that we change the focus to infinity, putting the face out of focus. Wave optics would now predict that (in the absence of aberrations), the diffraction pattern for a star would be a central (order 0) fringe surrounded by a ring (first-order fringe) if you used a circular aperture, but a rectangular aperture would give a different pattern (more like a rectangular grid of fringes). In practice, I don't think a camera would ever be diffraction-limited with the aperture all the way open. Diffraction decreases as the aperture gets wider, while ray-optical aberrations increase, so aberration would dominate diffraction under these conditions.

Quantum effects are totally irrelevant here.

Stan Rogers says:

It's difficult to explain without launching into a complete explanation of quantum electrodynamics, but all of the light that reaches the sensor "goes through" all of the lens, at least in a sense, even if we're just talking about a single photon. A photon doesn't take just one path (unless you make the mistake of trying to figure out which path it took), it takes all possible paths. Weird, but true.

This is an OK description of why QED suffices for a description of the phenomenon, but it's completely misleading in its implication that QED is necessary. The word "photon" can be replaced with the word "ray" wherever it occurs in this quote, and the description remains valid.

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