Editing Appendix/CGH/KAH2001 (section)

==From Our Separate Discussion of CGH==

As a point of comparison, in our [[Appendix/CGH/ParallelAperturesConsolidate#Case_1|accompanying discussion of 1D parallel apertures (specifically, the subsection titled, '''Case 1''')]], we have presented the following expression for the y-coordinate variation of the optical field immediately in front of the aperture:


<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~A(y_1)</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~ 
e^{i 2\pi L/\lambda }\biggl[ \frac{w}{2\beta_1} \biggr]  \int a_0(\Theta) e^{i\phi(\Theta)} \cdot  e^{-i \Theta } d\Theta 
\, ,
</math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{1}{\beta_1}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{\lambda L}{\pi y_1w}  \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~L</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
Z \biggl[1 + \frac{y_1^2}{Z^2}  \biggr]^{1 / 2} \, ,
</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\Theta</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{2\pi y_1 Y}{\lambda L} \biggr) \, .</math>
  </td>
</tr>
</table>
In other words, making the substitution, <math>~(2\pi/\lambda) \rightarrow k</math>, and recognizing that, <math>~d \leftrightarrow Z</math>, our expression becomes,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~I(y) \equiv \biggl[i k d e^{-i k d} \biggr] A(y_1)</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~ \biggl[i k d e^{-i k d} \biggr]
e^{i kL }\biggl[ \frac{L}{k y_1} \biggr]  \int a_0(\Theta) e^{i\phi(\Theta)} \cdot  \exp\biggl[-i \frac{2\pi y_1 Y}{\lambda L} \biggr] \biggl[ \frac{k y_1 }{L} \biggr] dY </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ (i k Z) e^{i k (L-Z)} 
\int a_0(\Theta) e^{i\phi(\Theta)} \cdot  \exp\biggl[-i 2\pi Y \biggl(\frac{y_1 }{\lambda L}\biggr) \biggr]  dY 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~ (i k Z) \exp\biggl[i k \biggl( L-Z \biggr)\biggr] 
\int a_0(\Theta) e^{i\phi(\Theta)} \cdot  \exp\biggl[-i 2\pi Y \biggl(\frac{y_1 }{\lambda L}\biggr) \biggr]  dY 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~ (i k Z) \exp\biggl[\frac{i \pi y_1^2}{Z \lambda} \biggr] 
\int a_0(\Theta) e^{i\phi(\Theta)} \cdot  \exp\biggl\{ -i 2\pi Y \biggl(\frac{y_1 }{\lambda } \biggr) \frac{1}{Z}\biggl[1 - \frac{y_1^2}{2Z^2}  \biggr] \biggr\}  dY 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~ \biggl( \frac{2\pi i Z}{\lambda}\biggr) \exp\biggl[\frac{i \pi y_1^2}{Z \lambda} \biggr] 
\int a_0(\Theta) e^{i\phi(\Theta)} \cdot  \exp\biggl\{ - \biggl(\frac{2\pi i y_1 Y}{Z \lambda } \biggr) \biggl[1 - \cancelto{0}{\frac{y_1^2}{2Z^2}}  \biggr] \biggr\}  dY \, ,
</math>
  </td>
</tr>
</table>

where we have used the approximate expressions,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~L - Z</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\frac{y_1^2}{2Z}</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;
  <td align="right">
<math>~\frac{1}{L}</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\frac{1}{Z}\biggl[1 - \frac{y_1^2}{2Z^2}  \biggr] \, .</math>
  </td>
</tr>
</table>

Next, accounting for the different variable notations that have been adopted in the two separate discussions, namely,

<table border="1" align="center" cellpadding="8">
<tr>
  <td align="center" colspan="2">Notation</td>
</tr>

<tr>
  <td align="center" colspan="1">KAH2001</td>
  <td align="center" colspan="1">Our</td>
</tr>

<tr>
  <td align="center" colspan="1"><math>~x</math></td>
  <td align="center" colspan="1"><math>~y_1</math></td>
</tr>

<tr>
  <td align="center" colspan="1"><math>~\xi</math></td>
  <td align="center" colspan="1"><math>~Y</math></td>
</tr>

<tr>
  <td align="center" colspan="1"><math>~d</math></td>
  <td align="center" colspan="1"><math>~Z</math></td>
</tr>

<tr>
  <td align="center" colspan="1"><math>~\lambda</math></td>
  <td align="center" colspan="1"><math>~\frac{2\pi}{k}</math></td>
</tr>

<tr>
  <td align="center" colspan="1"><math>~\alpha \equiv \frac{\sqrt{2} \xi}{ \sqrt{d \lambda} }  - \sqrt{2d\lambda} ~\biggl(\frac{x}{d\lambda}\biggr)  </math></td>
  <td align="center" colspan="1"><math>~\frac{\sqrt{2} Y}{ \sqrt{Z \lambda} }  - \sqrt{2Z \lambda} ~\biggl(\frac{y_1}{Z \lambda}\biggr) </math></td>
</tr>
</table>


let's examine how similar this last integral expression is to the [[#Eq10KAH2001|key integral from KAH2001 that has been presented above]].  We have,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~I_\xi(x)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl( \frac{d \lambda}{2}\biggr)^{1 / 2} 
\int_{-\infty}^{\infty} V(\xi) \times \exp\biggl[ \frac{i \pi \alpha^2}{2} \biggr] d\alpha 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\rightarrow</math>
  </td>
  <td align="left">
<math>~ \biggl( \frac{Z \lambda}{2}\biggr)^{1 / 2} 
\int_{-\infty}^{\infty} V(Y) \times \exp\biggl\{ \frac{i \pi}{2}
\biggl[ \frac{\sqrt{2} Y}{ \sqrt{Z \lambda} }  - \sqrt{2Z \lambda} ~\biggl(\frac{y_1}{Z \lambda}\biggr) \biggr]^2 \biggr\} 
\biggl[ \frac{2}{Z\lambda} \biggr]^{1 / 2}dY 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
\int_{-\infty}^{\infty} V(Y) \times \exp\biggl\{ \frac{i \pi}{2}
\biggl[ \frac{2Y^2}{ Z \lambda } - \frac{2\sqrt{2} Y}{ \sqrt{Z \lambda} }  \cdot \sqrt{2Z \lambda} ~\biggl(\frac{y_1}{Z \lambda}\biggr) + 2Z \lambda ~\biggl(\frac{y_1}{Z \lambda}\biggr)^2\biggr] \biggr\} 
dY 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
\int_{-\infty}^{\infty} V(Y) \times \exp\biggl\{ \frac{i \pi}{Z\lambda}
\biggl[ Y^2 - 2y_1Y + y_1^2 \biggr] \biggr\} 
dY 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \exp\biggl[ \frac{i \pi y_1^2}{Z\lambda} \biggr]
\int_{-\infty}^{\infty} V(Y) \times \exp\biggl\{ \frac{i \pi}{Z\lambda} \biggl[ - 2y_1Y + Y^2\biggr] \biggr\} 
dY 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \exp\biggl[ \frac{i \pi y_1^2}{Z\lambda} \biggr]
\int_{-\infty}^{\infty} V(Y) \times 
\exp\biggl\{ - \frac{2\pi i y_1 Y}{Z\lambda}  + \frac{i \pi Y^2}{Z\lambda} \biggr\} 
dY \, .
</math>
  </td>
</tr>
</table>

Therefore, in order that the two expressions to match, we need to ignore the quadratic (Y<sup>2</sup>) term inside the last exponential, and make the association,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~V(Y)</math>
  </td>
  <td align="center">
<math>~\leftrightarrow</math>
  </td>
  <td align="left">
<math>~ \biggl( \frac{2\pi i Z}{\lambda}\biggr) a_0(\Theta) e^{i\phi(\Theta)} \, .</math>
  </td>
</tr>
</table>