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Assume two obstacles are complementary. For example, a circular aperture and a circular stop of same radii. My theoretical analysis leads to contradictory conclusions about such situations.

1) Assuming incident light is a plane wave consisting of N sources of Huygens wavelets, obstacle 1 keeps a subset of those wavelets (blocking the others) while the obstacle 2 keeps the complementary subset of wavelets. In other words, the SUM of the diffracted patterns is expected to be a uniform intensity on the screen (all the wavelets = same as what we have when there is NO obstacle). In other words, when the Fraunhofer criteria are met, complementary obstacles produce complementary diffraction patterns.

2) Experiment shows that a circular stop and a circular aperture indeed produce opposite fringes, but the circular stop's diffraction pattern has the Poisson-Arago spot in the middle, while the circular aperture doesn't have a corresponding dark spot in the middle. Therefore these patterns are NOT complementary.

What am I missing ? I somehow have the feeling that patterns are complementary "everywhere except the center" because there is something ill-defined about using Huygens wavelets to predict the center of the pattern... Anyone has input on this ?

David Reishi
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Teacher77
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    As far as I know (I may be wrong) the Arago spot shows up only in the Fresnel regime. I'm going to guess that Fraunhofer diffraction from a circular stop does not exhibit a bright spot at the center. Alternatively, it may simply be the case that one is never truly in the Fraunhofer limit, and that a faint bright spot on a dark background is easier to observe than a small dimming against a bright background. – garyp Feb 24 '16 at 18:54
  • You aren't missing anything. The apertures are boundary conditions of the wave equation and the functional that maps the boundary conditions to the solutions is NOT linear and additive. That's part of the reason why boundary value problems are seriously hard, much harder than initial value problems that can be solved with a simple Green's function. – CuriousOne Feb 24 '16 at 20:35
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    @garyp you are exactly right, and if the OP had bothered to look at the wikipedia page, he'd have seen that right off. – Carl Witthoft Feb 24 '16 at 20:47
  • @CarlWitthoft: please don't assume I didn't research this point before asking. I did. Besides, can you please tell us which of Garyp's 3 hypothesis seems "exactly right" to you ? – Teacher77 Feb 24 '16 at 20:59
  • @CuriousOne : The point that you are making is most troubling to me. Can you please elaborate on this issue : why exactly two patterns obtained through the linear superposition of linear wavelets wouldn't be allowed to add ? – Teacher77 Feb 24 '16 at 21:00
  • You aren't adding waves. You are adding solutions of the wave equation. If you were to actually "superimpose" the two complimentary masks physically, what would the result be? 100% light? Of course not. The result would be no light because the boundary condition is now a solid wall. Does this mean there are no scenarios where solutions are complimentary for complimentary boundary conditions? No, it doesn't mean that, either. Phase masks can do just that. Another example would be interferometer patterns (at least near the center of the beam). – CuriousOne Feb 24 '16 at 21:07
  • Actually no, I am adding waves. In fact, I'm adding Huygens-Fresnel wavelets, assuming linear superposition. That's how the Arago spot was predicted by Poisson in 1818 (not through solutions of a 3D wave equation which didn't exist at the time).

    Even then... Isn't the wave equation a linear differential equation ? Solutions should be additive. I still don't see your point.

     – Teacher77 Feb 24 '16 at 21:14
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    Look up Babinet's principle; there are many different ways to prove it. It is covered in all of the standard optics texts. – Peter Diehr Feb 24 '16 at 21:20
  • Consider a simple scenario: You have a mirror in front of you - that's reflective boundary conditions. What do you see? You see yourself. Now consider a mirror in your back. What do you see? You can see your back. Now put the two mirrors together... what do you see? An infinite number of reflections of yourself. Where did they come from... there was only you and your back with the individual mirrors, right? If boundary conditions were additive, shouldn't you just see yourself front and back exactly once? – CuriousOne Feb 24 '16 at 21:30
  • @PeterDiehr : Ah thanks ! I studied that as an undergrad about 15 years ago and the concept was still in my mind but not the name. Nevertheless, I just went through 3 sources which don't seem much clearer than me : Hecht claims that the principle holds for the entire pattern but only in the Fraunhofer limit, wikipedia claim in works in any limit but not at the center of the screen, etc. None provide a proof. Do you have any idea why complementariness wouldn't work at the center ? – Teacher77 Feb 24 '16 at 22:13
  • @CuriousOne : thanks for this good example. But according to your boundary condition approach, which I'm not familiar with (that's from Jackson's, right ?), how can one derive Babinet's principle and in which limit is it valid ? – Teacher77 Feb 24 '16 at 22:14
  • That's a good question... I thought I was taught that it was valid for scalar diffraction in the Frauenhofer limit (i.e. far away from the screen), but now I have found an article that claims that that's not correct, either. Unfortunately, I don't have access to that journal... "On the Validity of Babinet's Principle for Fraunhofer Diffraction", H. Lipsona & K. Walkleya, Optica Acta: International Journal of Optics, Volume 15, Issue 1, 1968, pages 83-91... so I can't give you an answer to that... – CuriousOne Feb 24 '16 at 22:32
  • @Teacher77: Hecht states that Babinet's Principle is strictly true in the Fraunhofer limit, and is almost true in the Fresnel range - almost true because the irradiances don't add up right. Hecht mostly gives examples. JD Jackson's "Classical Electrodynamics ", 2nd edition, pp. 432-441, derives Babinet's Principle from the "approximate Kirchoff integrals" for the scalar theory, and then again for polarized light. It seems quite thorough as I skim through it. – Peter Diehr Feb 24 '16 at 23:31
  • Thanks to both of you. Your answers are amazing. I'll check into the Itnl Jnl of Optics and into Jackson. I'm sure I'll find both on campus. – Teacher77 Feb 25 '16 at 01:53
  • @PeterDiehr : I looked up the article cited by CuriousOne and I'm surprised to read that Babinet principle seems to have 2 versions : I've always read that complementary obstacles produce complementary patterns, but it seems that other sources claim that complementary obstacles produce identical patterns, except for a small portion in the center. What the h... ? – Teacher77 Feb 26 '16 at 05:32
  • More on this : the 2 arguments differ on what the "addition" of two patterns should be : "uniform intensity on the screen" or "total black except for a small spot of light in the center". The 1st leads to the idea that if an obstacle gives intensity I than the complementary obstacle gives the intensity 1 - I, therefore inverted bright/dark fringes. But if "black except at the center" is assumed on the screen, than an obstacle giving amplitude A leads to the complementary obstacle giveing 0 - A (therefore identical intensities). – Teacher77 Feb 26 '16 at 05:37
  • But Fraunhofer diffraction assumes an incident PLANE WAVE, which should fill the screen with a uniform intensity in the absence of an obstacle... So I'm assuming my reasoning is correct. I'm curious about @CuriousOne 's opinion too. – Teacher77 Feb 26 '16 at 05:38
  • To be honest... I though I understood this myself, but the more I am thinking about (and the more of the abstracts about the topic I see), the less sure I am. – CuriousOne Feb 26 '16 at 05:42
  • @CuriousOne : There may be two limits : one where the incident wave really is a plane wave (ie. infinite wave front). And the other one where it's a laser beam, which indeed has quasi-plane wave fronts, but radially decreasing intensity (gaussian beam).

    The article you cited seems to say that : they tested a pattern of "stops" enclosed in a wide cicular aperture. The details of the obtained diffraction pattern depended on the diameter of the circular aperture, even though the stops remained identical.

    I feel I should make some experimental testing before I write anything in my lecture notes.

     – Teacher77 Feb 26 '16 at 06:04
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    @The record, last night I gave an answer which I now think is wrong. I said that the rings of interference around a disk's shadow in the Arago spot are caused by the light source (a point source in the original experiment), as though, if you remove the disk, the outer rings would stay as part of an Airy disk. But I just watched a vid of the coin moving, and with it the outer rings. I now see that the area of diffracted light from the disk is greater than I thought, and that the outer rings are caused by interference between light-waves from the source and from the opposite edge of the disk. – David Reishi May 05 '16 at 16:28

3 Answers3

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According to wikipedia's article about the Poisson-Arago spot it is a phenomenon occurring in Fresnel diffraction, which is a different limit from Fraunhofer diffraction.

From the question:

In other words, when the Fraunhofer criteria are met, complementary obstacles produce complementary diffraction patterns.

If I believe the OP, and also this article seems to assume the Fraunhofer limit, the contradiction is resolved. However I don't completely understand which limits Babinet's principle applies in. Therefore I will only claim that this is a possible answer, which relies upon Babinet's principle only applying in the Fraunhofer limit.

We don't even need to consider the applicability of Babinet's principle though. I think the confusion might simply come from that the OP used an argument about the interference pattern in Fraunhofer diffraction and added the Arago spot into the mix, which does not actually appear there, i.e. the diffraction pattern in the Fraunhofer limit does not have a spot in the middle. So there is certainly no contradiction in this limit.

Wolpertinger
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I think you are missing things in both points.

1) There is no reason why the sum of diffracted patterns would give a uniform intensity. Why? Because waves are made of amplitudes, but patterns are made of intensities, that is: squared amplitudes. Two waves that would interfere destructively, give non-zero intensities when taken individually. In fact, they give the same intensity pattern when taken individually, because the only way to interfere destructively is to be in phase opposition (same absolute value of amplitude, but opposite signs).

This observation is the gist of Babinet's principle in Fraunhofer conditions: in any direction other than that of the incident plane wave, the intensity is zero, so the complementary obstacles yield the same pattern in those directions. In the direction of the plane wave, the intensity is non-zero; when one obstacle is finite and the other is infinite (as is the case here), the finite one (circular stop) yields full intensity, and the infinite one (hole) yields zero intensity: the energy passing through the hole is negligible as compared to that passing around the stop (we are in Fraunhofer conditions, at infinity, a finite energy gives infinitesimally small intensity).

2) The Poisson-Arago spot is a specific phenomenon observed in Fresnel regime.

About the Fraunhofer regime: it is not only a limiting case. You can produce experimentally a Fraunhofer diffraction pattern. How? Use a convergent lens. The convergent lens conjugates infinity in its object space to the focal plane in the image space. In other words: put a lens after a diffraction obstacle, and you will observe the Fraunhofer diffraction pattern on a screen in its focal plane, where coordinate $x$ on the screen matches direction $θ$ according to $x=f'\,\tan θ$.

With the circular stop, you will observe a bright dot at the center. This is not the Poisson-Arago spot. This dot (of theoretically zero width, unlike the Poisson-Arago spot) is the conjugate of the $θ=0$ direction in the Fraunhofer pattern, and as I explained above in this direction the finite obstacle gives maximal intensity (of course experimentally the lens is finite, has aberrations, the incident wave doesn't have infinite width... so the intensity of the dot is not infinite).

L. Levrel
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You're mostly correct, about how you expect both patterns together to produce uniform intensity corresponding to a plane wave (this is described as Babinet's principle. What you're forgetting is that it's the sum of the amplitude, including a phase. A circular hole and the complementary shape will produce identical patterns but opposite phase, except for additional background plane wave.

Peter Diehr
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David
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