29

Over on Photography, a question was asked as to why (camera) lenses are always cylindrical. Paraphrasing slightly, one of the answers and follow-up comments asserted that quantum effects are significant and that you need to understand QED in order to understand lens design. My intuition tells me that this probably isn't the case, and that modern lens design can be handled via classical optical theory. Is my intuition correct, or are quantum effects really significant in modern lenses?

Qmechanic
  • 201,751
Philip Kendall
  • 486
  • My first thought was that QED is necessary to explain the refraction index to begin with. There was a recent video that covered how QED is needed for this, and how it's actually quite complicated and non-intuitive https://www.youtube.com/watch?v=CiHN0ZWE5bk – Alan Rominger Jul 15 '13 at 19:09
  • 8
    @AlanSE: There are good classical explanations for the refraction of light. See http://physics.stackexchange.com/q/65812/ . You only need quantum mechanics if you want a detailed explanation of how the index of refraction varies with frequency, and even then, you don't need the full machinery of QED. –  Jul 15 '13 at 19:24
  • From the point of view of lens design, I think that varying refractive index with frequency is more of an input to the problem than anything else (I have these materials with these dispersion curves, how can I minimize chromatic aberrations?) - it's not something the lens designer needs to understand, just something that is. Still a worthwhile note that classical optics can't explain varying refractive index though. – Philip Kendall Jul 15 '13 at 19:59
  • That is total nonsense. How did Steinheil or Abbe design their lenses? – Georg Jul 15 '13 at 21:07
  • @Georg two possible answers: 1) The usual combination of experience, intuition, trial and error and a little bit of blind luck. 2) It could be the case that while the lenses of 100 years ago were firmly in the classical realm, the lenses of today have started getting into quantum behaviour. Note I don't believe either of those answers, but what I believe is pretty much irrelevant :-) – Philip Kendall Jul 15 '13 at 21:29
  • @PhilipKendall Welcome to Physics, by the way! Feel free to come by with any more physics questions arising in photography. –  Jul 16 '13 at 02:33
  • It was appropriate if someone could estimate the error of not using quantum mechanics to design a lens.(it's not that hard to do it, and it shatters all the clouds) – Ali Jul 16 '13 at 06:12
  • ""@Georg two possible answers:"" No, there is a third: You do not know who Ernst Abbe was. – Georg Jul 16 '13 at 08:58
  • @Ali That would be an absolutely wonderful answer to the question :-) – Philip Kendall Jul 16 '13 at 09:01
  • @Georg Actually, I do. My point is that saying "that is total nonsense" without backing that up with facts isn't a helpful contribution to the discussion. – Philip Kendall Jul 16 '13 at 09:03
  • Philip, why dont You ask the guy in "photography" for facts first? – Georg Jul 16 '13 at 09:15
  • @Georg largely because it seemed better as a separate question, and because the discussion on that answer had already degenerated into a sequence of "yes, it is", "no, it's not" posts. – Philip Kendall Jul 16 '13 at 09:25
  • Using QED to calculate lenses would be like starting out with GR to calculate the trajectory of an apple falling from a tree onto a head:=) – Georg Jul 16 '13 at 10:31

9 Answers9

31

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.

  • 4
    +1 - I think this should be an answer to that question as well, as it's better than any answer currently there. – BlueRaja - Danny Pflughoeft Jul 15 '13 at 19:01
  • 2
    So, if the lens is not round, the world according to Stan Rogers explodes, because the single photon ends up out of focus with itself. :) Thus, lenses must be round to protect the world. QED, by QED. – Kaz Jul 15 '13 at 22:43
  • I've seen it stated that the aperture when to small cause blur due to refraction. One source for this statement is here: http://www.kenrockwell.com/tech/macro.htm This would be an effect to be considered, that can't be explained by the ray model of light. – Jens Schauder Jul 16 '13 at 05:55
  • @JensSchauder I don't know how much of a camera person you are, but the general recommendation is to ignore Ken Rockwell entirely. He's very good at SEO and writing content which looks at first glance to be valuable, but not so good at actual correct information. – Philip Kendall Jul 16 '13 at 09:01
  • @PhilipKendall I'm aware of that, at least to some extend. But I've seen similar statements in other places. Obviously I haven't checked if they copied it from Ken Rockwell, nor if the statement is plausible. Although I should be able to do the later. – Jens Schauder Jul 16 '13 at 11:10
  • @JensSchauder: I've seen it stated that the aperture when to small cause blur due to refraction. This makes sense (assuming you mean diffraction rather than refraction), but with a small aperture, the noncircular shape of the lens would be irrelevant. I'll edit the answer appropriately. –  Jul 16 '13 at 15:42
  • 2
    @Jens: The diffraction of light is explained by Maxwell's equations and the Huygens–Fresnel principle. There is no need for QM. – BlueRaja - Danny Pflughoeft Jul 16 '13 at 18:37
16

Your intuition is correct, you don't need quantum electrodynamics to explain/model/engineer camera lenses. When considering the propagation of light, the results of geometric optics can be interpreted in terms of path integrals, as Feynman does in his QED: The Strange Theory of Light and Matter, but this is not necessary for lens design. Geometric optics itself suffices in most cases, but there are some design tasks which require an incorporation of the wave nature of light (or the application of empirical rules). (Thanks @Matt J for this info)

twistor59
  • 16,746
  • 2
    This is mostly right, but geometric optics is not sufficient for designing a high quality lens (such as is routine in photography) with minimizes spherical aberration, chromatic aberration, coma, etc. These have to be explained with either phenomenological rules or (for a deeper explanation) with wave optics. That is why the last several editions of a very popular optics text (Born) have started out with wave optics based on Maxwell's Equations, leaving little or no time for geometric optics at all. But really, the rules suffice, there is only very rare need for QED in military applications. – Matt J. Jul 16 '13 at 01:14
  • 1
    @Matt J OK thanks for the input - I've modified the last sentence to reflect this. But I'm curious - which military applications require the application of QED in optical design? – twistor59 Jul 16 '13 at 06:27
  • 1
    The example I was thinking of was with microwave optics -- and I didn't have the security clearance to hear much about it:( But you got the important point: such needs for QED in design are quite rare. Rare enough that we can be sure it is not needed for lens design. But I can't rule out the idea that reference to QED was needed to design the materials that go into designing stealth-technology either, since I don't know enough about how that is done. QED is the basic means of understanding the interaction of light and matter, after all. Yet phenomenological rules suffice for most engineers. – Matt J. Jul 16 '13 at 23:54
4

Well Phillip, I guess you ARE the original questioner on this site, and your real question seems to be is QED or any form of quantum mechanics needed in "modern" lens design. So let me try again.

First some history. Significant lens design goes back certainly to the start of the 20th century; but the foundations go back much earlier.

Computation was an expensive chore, so "tracing geometrical rays" was time consuming. Certainly, wave based diffraction theory was prohibitively expensive for "design" as distinct from research.

So much work was done on purely mathematical theoretical calculation of ray paths and aberration theory. The Seidel aberration theory, was a great leap forward,

Now the actual math used is ordinary Algebra, and ordinary Trigonometry. Euclidean Geometry of course.

Since ray tracing was time consuming; aberration theory was very important, so fewer rays needed to be traced.

Between WW-I and WW-II lens design was such that one could design an achromatic cemented doublet objective lens for a telescope or binocular, completely to manufacturing specs, by ray tracing just three special rays in two different colors. That allowed correction for primary chromatic aberration, primary spherical aberration, and primary coma.

Eventually, this theory was condensed into a classic text book, that no optics designer should be without. That is "Applied Optics and Optical Design." By A. E. Conrady. It is available in two paperback Penguin book for a few dollars. The first volume material dates from his lecture notes from 1926. It is an intensely equation filled text but as I said, simple algebra trig, and geometry, and he developed a formalism that many designers adopted even using it today (I do)

Now of course we have high speed computers in our pockets, and ray tracing is cheap.

Geometrical optics is a mathematical simplification that basically assumes that the wavelength is zero so diffraction wave effects don't occur. So it breaks down when real practical wavelengths are used. But the difference between the geometrical and the diffraction based calculations, doesn't affect many real cases, so the basic design is still done with geometrical ray optics, and the diffraction optics is used to polish things up, once a reasonable solution is reached. And the mathematical aberration theory is till often used to establish a likely successful architecture.

The Seidel Aberrations; spherical aberration, coma, astigmatism, Petzval curvature, distortion, and then chromatic aberration, are all related, and you can specifically control each of them (given enough variables), but often, in modern design, the software can optimize spot sizes, or modulation transfer functions, paying no attention to the Seidel theory. The successful result (if you get one) doesn't care whether you used aberration theory to achieve it, or whether you simply told the computer to find the best result.

BUT ! there are some serious "you can't get there from here" restrictions. Little nuisances like the second law of thermodynamics, and failure to heed those will keep you from finding your answer, because you are asking for something that physics doesn't permit.

But I can't ever prove that you don't need QED to design any lens. But as Einstein told us, you only have to identify A SINGLE CASE that does require QED, to prove me wrong.

Lens design is a very heierarchical discipline; so simple lens formulae, Seidel theory, geometrical ray tracing, and full diffraction theory, are used interactively.

It is well known at the university entry level, that geometrical optics predicts a different spot size from the true diffraction result, and moreover, the best spot image is in a different location. But that is just icing the cake.

But get the Conrady books if you really are interested; I keep it at my elbow at all times. Of course , Born and Wolfe is a well known text, and Warren J. Smith's "Modern Optical Design" is a must have text book. I'll let you know if I ever have to use QED

2

Well camera lenses are NOT always cylindrical. 99.99999999 % of them are NEVER cylindrical.

They are usually rotationally symmetrical at least; but for the most part, it is the shape of the lens surfaces (which would be dead flat on a cylinder) which determines the optical properties. Lens thicknesses are not unimportant, but the are secondary, to the curvature (1/r) of the surfaces.

As for QED, trying to use that for lens design, would be akin to making a hole in the garden to plant a seed, using a small nuclear bomb.

Geometrical ray optics suffices for 99 % of all camera lens design. Very few camera lenses come anywhere near being diffraction limited. A lot of them don't even get within a factor of ten of the diffraction limited spot size. Telescope and microscope optics are a different story, and near diffraction limits is typical, specially for large telescopes.

  • 2
    I think you confused two meanings of lens: the more or less simple piece of glass vs the device attached to the front of a camera which is made up of lenses of the first kind + various stuff of metal, plastic, electronics. The later is in first approximation cylindrical. – Jens Schauder Jul 16 '13 at 05:47
  • What @JensSchauder said. I think everyone is aware that the actual bits of glass in a camera lens have curved surfaces. – Philip Kendall Jul 16 '13 at 09:07
1

You dont need QED because lenses can be any shape you want for example rectangular: http://www.anchoroptics.com/catalog/product.cfm?id=20

Most important is a surface form not shape.

Asphir Dom
  • 665
  • 2
    Sorry, I don't follow this. Why does the existence of rectangular lenses mean that quantum effects are not significant? – Philip Kendall Jul 16 '13 at 09:12
  • 2
    There is no restriction on the lens shape. That is why you do not need any theory for it. Only restricted parameter is form of the front and back surfaces. They are well described by classical optics. – Asphir Dom Jul 16 '13 at 15:18
  • But how do I know (other than your assertion) that the front and back surfaces are well described by classical optics? That's really the question here :-) – Philip Kendall Jul 16 '13 at 15:53
  • @PhilipKendall You do not know. There are tons of books on optics, you can create your own lens (there are many of enthusiasts in diy astronomy) and see that everything is fine :) Validity of the theory is about size of its objects -- "size" of visible light quanta is about 700nm so here you go. If you want lense that is 700nm small then you need to ask here again :) – Asphir Dom Jul 16 '13 at 16:17
1

On the one hand the answer is an emphatic no (in the sense that ray and wave propagation is valid). But ray optics and all the rest of it in a sense ARE quantum optics - Maxwell's equations are the first quantized description of one photon propagation through a lens system and raytracing / wave simulation are different approximations for solving this.

Although the equations are the same, it is helpful to think about this in designing things that transmit one photon states (or a sequence thereof) such as fluorescence confocal microscopy, see

  1. J. Opt. Soc. Am. B/Vol. 24, No. 4/ April 2007 p 928
  2. J. Opt. Soc. Am. B/Vol. 24, No. 4/ April 2007 p 942
  3. J. Opt. Soc. Am. B/Vol. 24, No. 6/ June 2007 p 1369

These are paywalled, email me if you're interested.

Selene Routley
  • 88,112
  • 1
    Sure, everything is quantum if you look small enough. But if the classical approximations are good enough to design a modern camera lens, then it's not something the designer needs to know or worry about. – Philip Kendall Jul 16 '13 at 09:18
  • @PhilipKendal As someone who does actual design in this field, I would say that depends what you mean by a "camera". If you are imaging one photon sources or fluorophores, then it is very helpful to keep the basic physics in mind. In design of such things, it impacts more on the specification of imaging systems, rather than the model. Also, from a curiousity standpoint - haven't you ever wondered why raytracing is valid for one photon systems - not only is it a good approximation - but for some quantum states it is as good as anything else. – Selene Routley Jul 16 '13 at 10:12
  • Fair point. I guess I'm primarily interested in what "most people" would think of as photography - visual spectrum imaging at focal lengths from a few mm up to maybe a metre or so. But I guess there's another interesting question along the lines of "are there any sorts of photography where quantum effects are significant?" – Philip Kendall Jul 16 '13 at 16:02
0

Besides classical optics, some high-end lens exploit (or suppress) wave effects, but no designs are influenced by considerations on quantum effects.

Taiben
  • 432
0

If you want to design lenses then no But if you want to understand how lenses actually work you would need to understand the QED theory of light. Also it is indeeed non intuitive but not very difficult.

Kushagra Dixit
  • 1
0

The answer there seems to be a bit confused. It is true that you need to account for wave-like effects (i.e. interference and diffraction) to understand lens design.

However, these effects appear in both the classical field theory of light (Maxwell's equations/wave optics) and the quantum field theory of light (QED). Both classes of theories contain the basic phenomena of interference and diffraction, and the more subtle additional effects in QED don't play a role.

knzhou
  • 101,976