Editing Appendix/CGH/KAH2001 (section)

==Fresnel Diffraction==

According to the [https://en.wikipedia.org/wiki/Fresnel_diffraction Wikipedia description of  Fresnel diffraction], "&hellip; the electric field diffraction pattern at a point <math>~(x, y, z)</math> is given by &hellip;" the expression,

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Given the intensity immediately in front of the aperture, <math>~E(x', y', 0)</math>, this integral expression generates the intensity, <math>~E(x, y, z)</math>, on the image plane whose distance from the aperture is, <math>~z</math>.
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<math>~E(x, y, z)</math>
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  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{i \lambda} \iint_{-\infty}^\infty E(x', y', 0) \biggl[ \frac{e^{i k r}}{r}\biggr] \cos\theta~ dx' dy'\, ,
</math>
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where, <math>~E(x', y', 0)</math> is the electric field at the aperture, <math>~k \equiv 2\pi/\lambda</math> is the wavenumber, and,

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  <td align="right">
<math>~r</math>
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<math>~\equiv</math>
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<math>~
\biggl[ z^2 + (x - x')^2 + ( y - y')^2 \biggr]^{1 / 2}
= z \biggl[ 1 + \frac{(x - x')^2 + ( y - y')^2}{z^2} \biggr]^{1 / 2}
= z\biggl[ 1 +  \frac{(x - x')^2 + ( y - y')^2}{2z^2} - \frac{[(x - x')^2 + ( y - y')^2]^2}{8z^4} + \cdots\biggr] \, .
</math>
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(The infinite series in this last expression results from enlisting the [[Appendix/Ramblings/PowerSeriesExpressions#Binomial|binomial theorem]].)  For simplicity, in the discussion that follows we will assume &#8212; as in &sect;2 of [https://ui.adsabs.harvard.edu/abs/2001OptEn..40..926K/abstract KAH2001] &#8212; that the aperture is illuminated by a monochromatic plane wave that is impinging normally onto the aperture, in which case, the angle, <math>~\theta = 0</math>.

In the '''Fresnel approximation''', the assumption is made that, in the series expansion for <math>~r</math>, all terms beyond the first two are very small in magnitude relative to the second term.  Adopting this approximation &#8212; and setting <math>~\theta = 0</math> &#8212; then leads to the expression,

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<tr>
  <td align="right">
<math>~E(x, y, z)</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
\frac{1}{i z \lambda} \iint_{-\infty}^\infty E(x', y', 0) ~\biggl[ 1 -  \frac{(x - x')^2 + ( y - y')^2}{2z^2} \biggr] \exp\biggl\{ i k z\biggl[ 1 +  \frac{(x - x')^2 + ( y - y')^2}{2z^2}\biggr] \biggr\}~ dx' dy'
</math>
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&nbsp;
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  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{e^{i k z}}{i z \lambda} \iint_{-\infty}^\infty E(x', y', 0) ~\biggl[ 1 -  \frac{(x - x')^2 + ( y - y')^2}{2z^2} \biggr] \exp\biggl\{\frac{ i k}{2 z}\biggl[ (x - x')^2 + ( y - y')^2 \biggr] \biggr\}~ dx' dy' \, .
</math>
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If "&hellip; for the <math>~r</math> in the denominator we go one step further, and approximate it with only the first term &hellip;", then our expression results in the '''Fresnel diffraction integral''', 

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<tr>
  <td align="right">
<math>~E(x, y, z)</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
\frac{e^{i k z}}{i z \lambda} \iint_{-\infty}^\infty E(x', y', 0) ~ \exp\biggl\{\frac{ i k}{2 z}\biggl[ (x - x')^2 + ( y - y')^2 \biggr] \biggr\}~ dx' dy' \, .
</math>
  </td>
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