Editing Appendix/CGH/ParallelApertures (section)

===Analytic Result===

If we consider the limit where the aperture shown in Figure I.1 is divided into an infinite number of divisions, then we can convert the summation in the above equations into an integral running between the limits, Y<sub>2</sub> and Y<sub>1</sub>.  Specifically, linearized expression immediately above becomes,
<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>A_0~a_0 \int
e^{-i[2\pi y_1 Y/(\lambda L)]} dY
\, ,
</math>
  </td>
</tr>
</table>

where, again, we have returned to the case where the aperture is assumed to be uniformly bright and set a(Y) = a<sub>0</sub>dY.  (It should be understood that a<sub>0</sub> is the brightness of the aperture per unit length.)  If we now make the substitution,
<div align="center">
<math>\Theta \equiv \frac{2\pi y_1 Y}{\lambda L} \, ,</math>
</div>

this integral expression takes the form,

<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>A_0 \biggl[ \frac{a_0 w}{2\beta_1} \biggr] \int_{\Theta_2}^{\Theta_1}
e^{-i \Theta} d\Theta
\, ,
</math>
  </td>
</tr>
</table>
where the limits of integration are, respectively,

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

<tr>
  <td align="right">
<math>\Theta_2</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\frac{2\pi y_1 Y_2}{\lambda L} = \vartheta_1 - \beta_1
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Theta_1</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\frac{2\pi y_1 Y_1}{\lambda L} = \vartheta_1 + \beta_1
\, ,
</math>
  </td>
</tr>
</table>
and,

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

<tr>
  <td align="right">
<math>\beta_1</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\frac{\pi y_1}{\lambda L} \biggl[ Y_1 - Y_2 \biggr] = \frac{\pi y_1 w}{\lambda L} 
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\vartheta_1</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\frac{\pi y_1}{\lambda L} \biggl[ Y_1 + Y_2 \biggr] 
\, .
</math>
  </td>
</tr>
</table>
This definite integral can be readily evaluated, giving,

<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>
(-i)^{-1} A_0 \biggl[ \frac{a_0 w}{2\beta_1} \biggr] \biggl[e^{-i\Theta_1} - e^{-i \Theta_2}  \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
i A_0 \biggl[ \frac{a_0 w}{2\beta_1} \biggr] e^{-i\vartheta_1} \biggl[e^{-i\beta_1} - e^{+i \beta_1}  \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ A_0 ~a_0 ~w ~e^{-i\vartheta_1} \biggr]\frac{\sin\beta_1}{\beta_1} 
= \biggl[ A_0 ~a_0 ~w ~e^{-i\vartheta_1} \biggr] \mathrm{sinc}(\beta_1) \, .
</math>
  </td>
</tr>
</table>
The last two lines have been derived by realizing that, for any angle <math>\chi</math>,

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

<tr>
  <td align="right">
<math>e^{-i\chi} - e^{+i\chi}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-2i\sin(\chi) \, ;
</math>
  </td>
</tr>
</table>
and the "sinc" function,

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

<tr>
  <td align="right">
<math>\mathrm{sinc}(\chi)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\sin\chi}{\chi} \, .
</math>
  </td>
</tr>
</table>
Notice that the center of the aperture is used to define the origin of the Y (and y) coordinate axis so that Y<sub>2</sub> = -Y<sub>1</sub>, as illustrate in Figure I.1; then <math>\vartheta_1 = 0</math> and the amplitude, A(y<sub>1</sub>), has no complex component.  If the aperture is not centered as depicted in Figure I.1, the angle <math>\vartheta_1</math> simple serves to introduce a phase shift in the evaluation of the amplitude.

Although this problem was solved for a specific location, y<sub>1</sub>, on the image screen, we see that the sinc function solution can readily be evaluated for any other location on the screen without having to redo the integral.