Editing Arago spot (section)

== Theory ==

[[File:huygensfresnelintegral.jpg|300px|thumb|right|Notation for calculating the wave amplitude at point P<sub>1</sub> from a spherical point source at P<sub>0</sub>.]]

At the heart of Fresnel's wave theory is the [[Huygens–Fresnel principle]], which states that every unobstructed point of a wavefront becomes the source of a secondary spherical wavelet and that the amplitude of the optical field ''E'' at a point on the screen is given by the superposition of all those secondary wavelets taking into account their relative phases.<ref name="sommerfeld"/> This means that the field at a point P<sub>1</sub> on the screen is given by a surface integral:
<math display="block">U(P_1) = \frac{A e^{\mathbf{i} k r_0}}{r_0} \int_S \frac{e^{\mathbf{i} k r_1}}{r_1} K(\chi) \, dS,</math>
where the inclination factor <math>K(\chi)</math> which ensures that the secondary wavelets do not propagate backwards is given by
<math display="block">K(\chi) = \frac{\mathbf{i}}{2 \lambda} (1 + \cos(\chi))</math>
and
* ''A'' is the amplitude of the source wave
* <math display="inline">k = \frac{2\pi}{\lambda}</math> is the [[wavenumber]]
* ''S'' is the unobstructed surface.

The first term outside of the integral represents the oscillations from the source wave at a distance ''r''<sub>0</sub>. Similarly, the term inside the integral represents the oscillations from the secondary wavelets at distances ''r''<sub>1</sub>.

In order to derive the intensity behind the circular obstacle using this integral one assumes that the experimental parameters fulfill the requirements of the [[Fresnel diffraction|near-field diffraction]] regime (the size of the circular obstacle is large compared to the wavelength and small compared to the distances ''g'' = P<sub>0</sub>C and ''b'' = CP<sub>1</sub>). Going to [[polar coordinates]]{{dubious|reason=Going to polar coordinates, from what?? This is an example of where someone switched to a different textbook without reading over the whole thing. For the paraxial regime (as just implored) a simpler construction for above would assume collimated light (as Arago used) so g=∞; in fact this is what the result written below appears to have assumed: a simpler model.|date=October 2022}} then yields the integral for a circular object of radius ''a'' (see for example Born and Wolf<ref name="bornwolf"/>):
<math display="block">U(P_1) = - \frac{\mathbf{i}}{\lambda} \frac{A e^{\mathbf{i} k (g + b)}}{g b} 2\pi \int_a^\infty e^{\mathbf{i} k \frac{1}{2} \left(\frac{1}{g} + \frac{1}{b}\right) r^2} r \, dr.</math>

[[File:poissonspotintensity.jpg|300px|thumb|right|The on-axis intensity at the center of the shadow of a small circular obstacle converges to the unobstructed intensity.]]

This integral can be solved numerically{{dubious|reason=Actually it can be solved analytically! This is the case for the ''on-axis'' amplitude (center of the spot) which is easier.|date=October 2022}} (see below). If ''g'' is large and ''b'' is small so that the angle <math>\chi</math> is not negligible{{dubious|reason=Actually it was just stipulated above that a ≤≤ b, so that chi IS small for r ≤ a. But of course not for r ≥≥ a, could never be if you're integrating to ∞. I think the book this was copied from might have said "if .... chi at r=a IS negligible....."|date=October 2022}} one can write the integral for the on-axis case (P<sub>1</sub> is at the center of the shadow) as (see Sommerfeld<ref name="sommerfeldp186"/>):
<math display="block">U(P_1) = \frac{A e^{\mathbf{i} k g}}{g} \frac{b}{\sqrt{b^2 + a^2}} e^{\mathbf{i} k \sqrt{b^2 + a^2}}.</math>

The source [[intensity (physics)|intensity]], which is the square of the field amplitude, is <math display="inline">I_0 = \left|\frac{1}{g} A e^{\mathbf{i} k g}\right|^2</math> and the intensity at the screen <math>I = \left| U(P_1) \right|^2</math>. The on-axis intensity as a function of the distance ''b'' is hence given by:
<math display="block">I = \frac{b^2}{b^2 + a^2} I_0.</math>

This shows that the on-axis intensity at distances ''b'' much greater than the diameter of the circular obstacle is the same as the source intensity, as if the circular object was not present at all. However at larger distances ''b'', it turns out that the ''size'' of the bright spot (as can be seen in the simulations below where ''b/a'' is increased in successive images) is larger therefore making the spot easier to discern.

=== Calculation of diffraction images ===

To calculate the full diffraction image that is visible on the screen one has to consider the surface integral of the previous section. One cannot exploit circular symmetry anymore, since the line between the source and an arbitrary point on the screen does not pass through the center of the circular object. With the aperture function <math>g(r,\theta)</math> which is 1 for transparent parts of the object plane and 0 otherwise (i.e. It is 0 if the direct line between source and the point on the screen passes through the blocking circular object.) the integral that needs to be solved is given by:
<math display="block">U(P_1) \propto \int_0^{2\pi} \int_0^\infty g(r,\theta) e^{\frac{\mathbf{i} \pi \rho^2}{\lambda} \left( \frac{1}{g} + \frac{1}{b} \right)} \rho \, d\rho \, d\theta.</math>

Numerical calculation of the integral using the [[trapezoidal rule]] or [[Simpson's rule]] is not efficient and becomes numerically unstable especially for configurations with large [[Fresnel number]]. However, it is possible to solve the radial part of the integral so that only the integration over the azimuth angle remains to be done numerically.<ref name="dauger"/> For a particular angle one must solve the line integral for the ray with origin at the intersection point of the line P<sub>0</sub>P<sub>1</sub> with the circular object plane. The contribution for a particular ray with azimuth angle <math>\theta_1</math> and passing a transparent part of the object plane from <math>r = s</math> to <math>r = t</math> is:
<math display="block">R(\theta_1) \propto e^{\frac{\pi}{2} \mathbf{i} s^2} - e^{\frac{\pi}{2} \mathbf{i} t^2}.</math>

So for each angle one has to compute the intersection point(''s'') of the ray with the circular object and then sum the contributions <math>I(\theta_1)</math> for a certain number of angles between 0 and <math>2\pi</math>. Results of such a calculation are shown in the following images.

[[File:poissonspot simulation d4mm.jpg|200px]]
[[File:poissonspot simulation d2mm.jpg|200px]]
[[File:poissonspot simulation d1mm.jpg|200px]]

The images are simulations of the Arago spot in the shadow of discs of diameter 4&nbsp;mm, 2&nbsp;mm, and 1&nbsp;mm, imaged 1&nbsp;m behind each disc. The disks are illuminated by light of wavelength of 633&nbsp;nm, diverging from a point 1&nbsp;m in front of each disc. Each image is 16&nbsp;mm wide.

The Arago spot can also be visualized using lines of average energy flow calculated numerically by averaging the [[Poynting vector]] of the electromagnetic field.<ref name=Gondran-2010/><ref>{{Cite book |last=Born |first=Max |title=Principles of optics: electromagnetic theory of propagation, interference and diffraction of light |last2=Wolf |first2=Emil |date=1993 |publisher=Pergamon Press |isbn=978-0-08-026481-3 |edition=6. ed., reprinted (with corrections) |location=Oxford}}</ref>{{rp|575}}
[[File:Arago spot.jpg|thumb|left|700px|Numerical simulation of the intensity of monochromatic light of wavelength λ&nbsp;=&nbsp;0.5&nbsp;μm behind a circular obstacle of radius {{nowrap|1=R = 5 μm = 10λ}}.<ref name=Gondran-2010>{{Cite journal |last=Gondran |first=Michel |last2=Gondran |first2=Alexandre |date=2010-06-01 |title=Energy flow lines and the spot of Poisson–Arago |url=https://pubs.aip.org/aapt/ajp/article-abstract/78/6/598/1042061/Energy-flow-lines-and-the-spot-of-Poisson-Arago?redirectedFrom=fulltext |journal=American Journal of Physics |volume=78 |issue=6 |pages=598–602 |doi=10.1119/1.3291215 |issn=0002-9505|arxiv=0909.2302 }}</ref>]]

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