Editing Appendix/CGH/ParallelApertures (section)

===Two Slits===

Starting with just a pair of slits (as illustrated in frame ''b'' of [[#Figure5|Figure 5]]) and setting the x-direction zero point exactly midway between the two, we can expand upon the [[Appendix/Ramblings/FourierSeries#Example_.232|accompanying ''Example #2'' discussion]] to obtain,

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

<tr>
  <td align="right">
<math>~a_n</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{L} \biggl\{ \int_{\nu}^{\nu +2c} \beta_{+1} \cos\biggl( \frac{n\pi x}{L} \biggr) dx
+ \int^{-\nu}_{-(\nu +2c)} \beta_{-1}\cos\biggl( \frac{n\pi x}{L} \biggr) dx \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{n\pi} \biggl\{ \int_{n\pi \nu/L}^{n\pi(\nu +2c)/L} \beta_{+1} \cos\theta  ~d\theta
+ \int^{-n\pi \nu/L}_{-n\pi(\nu +2c)/L} \beta_{-1}\cos\theta ~d\theta \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{n\pi} \biggl\{\beta_{+1} \biggl[ \sin\theta \biggr]_{n\pi \nu/L}^{n\pi(\nu +2c)/L} 
+ \beta_{-1}\biggl[ \sin\theta \biggr]^{-n\pi \nu/L}_{-n\pi(\nu +2c)/L} \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{n\pi} \biggl\{\beta_{+1} \biggl[ \sin\biggl( \frac{n\pi \nu }{L} + \frac{2n\pi c}{L}\biggr) - \sin\biggl( \frac{n\pi \nu}{L} \biggr) \biggr] 
+ \beta_{-1}\biggl[\sin\biggl( -~\frac{n\pi \nu}{L} \biggr)  - \sin\biggl( -~ \frac{n\pi \nu }{L} - \frac{2n\pi c}{L}\biggr)  \biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\beta_{+1} +  \beta_{-1}}{n\pi} \biggl[ \sin\biggl( \frac{n\pi \nu }{L} + \frac{2n\pi c}{L}\biggr) - \sin\biggl( \frac{n\pi \nu}{L} \biggr) \biggr]  \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~b_n</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{n\pi} \biggl\{ \int_{n\pi \nu/L}^{n\pi(\nu +2c)/L} \beta_{+1} \sin\theta  ~d\theta
+ \int^{-n\pi \nu/L}_{-n\pi(\nu +2c)/L} \beta_{-1}\sin\theta ~d\theta \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{n\pi} \biggl\{ - \beta_{+1} \biggl[ \cos\theta\biggr]_{n\pi \nu/L}^{n\pi(\nu +2c)/L} 
- \beta_{-1} \biggl[ \cos\theta \biggr]^{-n\pi \nu/L}_{-n\pi(\nu +2c)/L}  \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{n\pi} \biggl\{ - \beta_{+1} \biggl[ \cos\biggl( \frac{n\pi \nu}{L} + \frac{2n\pi c}{L} \biggr) - \cos\biggl( \frac{n\pi\nu}{L} \biggr)\biggr] 
- \beta_{-1} \biggl[ \cos\biggl( - \frac{n\pi\nu}{L} \biggr) - \cos\biggl( -~\frac{n\pi\nu}{L} - ~\frac{2n\pi c}{L} \biggr) \biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~
\frac{1}{n\pi} \biggl\{ \beta_{+1} \biggl[ \cos\biggl( \frac{n\pi \nu}{L} + \frac{2n\pi c}{L} \biggr) - \cos\biggl( \frac{n\pi\nu}{L} \biggr)\biggr] 
- \beta_{-1} \biggl[ \cos\biggl( \frac{n\pi\nu}{L} + \frac{2n\pi c}{L} \biggr) - \cos\biggl( \frac{n\pi\nu}{L} \biggr)  \biggr]  \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\beta_{-1} -  \beta_{+1}}{n\pi} \biggl[ \cos\biggl( \frac{n\pi \nu}{L} + \frac{2n\pi c}{L} \biggr) - \cos\biggl( \frac{n\pi\nu}{L} \biggr)\biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

Note that,

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

<tr>
  <td align="right">
<math>~a_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{L} \biggl\{ \int_{\nu}^{\nu +2c} \beta_{+1} dx
+ \int^{-\nu}_{-(\nu +2c)} \beta_{-1} dx \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{L} \biggl\{ \beta_{+1} \biggl[ \nu +2c - \nu \biggr]
+ \beta_{-1} \biggl[- \nu + (\nu +2c) \biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2c(\beta_+ + \beta_-)}{L} \, .
</math>
  </td>
</tr>
</table>
</div>

Considering, first, the case where the two slits are illuminated equally <math>(~\beta_- = \beta_+)</math>, we see that <math>~b_n = 0</math>, so the amplitude is,
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<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~|a_n|</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{n\pi} \biggl| \sin\biggl( \frac{n\pi \nu }{L} + \frac{2n\pi c}{L}\biggr) - \sin\biggl( \frac{n\pi \nu}{L} \biggr) \biggr|
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2c}{L\chi_n} \biggl| \sin\biggl( \frac{\chi_n \nu }{c} + 2\chi_n \biggr) - \sin\biggl( \frac{\chi_n\nu}{c} \biggr) \biggr| 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2c}{L\chi_n} \biggl| \sin\biggl[ \biggl( \frac{\chi_n\nu }{c} + \chi_n \biggr) + \chi_n \biggr] - \sin\biggl[ \biggl( \frac{\chi_n\nu }{c} + \chi_n \biggr) - \chi_n \biggr] \biggr| 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{4c}{L} \biggl| \cos\biggl[ \chi_n \biggl( 1 + \frac{\nu }{c} \biggr) \biggr] \frac{\sin \chi_n }{\chi_n} \biggr| \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<math>~\chi_n \equiv \frac{n\pi c}{L} \, .</math>
</div>

In the following animated figure, the red-dotted curve shows how <math>~\tfrac{L}{4c}|a_n|</math> varies with <math>~\chi_n/\pi</math> for a variety of values of the dimensionless parameter, <math>~0.01 \le \nu/c \le 5</math>, as recorded in the upper-right-hand corner of the plot.

<table border="1" align="center" cellpadding="8">
<tr>
  <th align="center"><font size="+1">Figure 4:</font>&nbsp; Double-Aperture Diffraction Pattern &#8212; Variable <math>~\nu/c</math></th>
</tr>
<tr>
  <td align="center">[[File:TwoSlit01.gif|center|Diffraction pattern for two-slit aperture]]</td>
</tr>
</table>

The static, dashed-black curve in the figure displays the function,
<div align="center">
<math>~\mathrm{sinc}(2\chi_n) = \frac{\sin(2\chi_n)}{2\chi_n} \, ,</math>
</div>
which is the proper amplitude behavior in the limit of <math>~\nu/c \rightarrow 0</math>.  The static, ''solid'' black curve in the figure displays the function, <math>~\mathrm{sinc}(\chi_n)</math>; by inspection, this curve serves as the upper envelope to the amplitude function for all values of the dimensionless parameter, <math>~\nu/c</math>.