Editing Appendix/CGH/ParallelApertures (section)

===Two (or more) Slit Pairs===

Now, let's examine four identical slits.

<div align="center">
<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_{(3\nu +2c)}^{(3\nu +4c)} \beta_{+2} \cos\biggl( \frac{n\pi x}{L} \biggr) dx + 
\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 +
\int^{-(3\nu +2c)}_{-(3\nu + 4c)} \beta_{-2} \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(3\nu +2c)/L}^{n\pi(3\nu +4c)/L} \beta_{+2} \cos\theta ~ d\theta + 
\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 +
\int^{-n\pi (3\nu +2c)/L}_{-n\pi (3\nu + 4c)/L} \beta_{-2} \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_{+2} \biggl[ \sin\theta \biggr]_{n\pi(3\nu +2c)/L}^{n\pi(3\nu +4c)/L} + 
\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} +
\beta_{-2}\biggl[ \sin\theta \biggr]^{-n\pi (3\nu +2c)/L}_{-n\pi (3\nu + 4c)/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_{+2} \biggl[ \sin\biggl( \frac{3n\pi \nu }{L} + \frac{4n\pi c}{L}\biggr) - \sin\biggl( \frac{3n\pi \nu }{L} + \frac{2n\pi c}{L} \biggr) \biggr] 
+ \beta_{-2}\biggl[\sin\biggl( -~ \frac{3n\pi \nu }{L} - \frac{2n\pi c}{L}\biggr)  - \sin\biggl( -~ \frac{3n\pi \nu }{L} - \frac{4n\pi c}{L}\biggr)  \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \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{1}{n\pi} \biggl\{
(\beta_{+2} + \beta_{-2}) \biggl[ \sin\biggl( \frac{3n\pi \nu }{L} + \frac{4n\pi c}{L}\biggr) - \sin\biggl( \frac{3n\pi \nu }{L} + \frac{2n\pi c}{L} \biggr) \biggr] 
+ (\beta_{+1}+ \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] 
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(\beta_{+2} + \beta_{-2})}{n\pi} \biggl\{ \sin\biggl[ \frac{n\pi }{L} \biggr( 3\nu  + 4c \biggr) \biggr] - \sin\biggl[ \frac{n\pi }{L} \biggr( 3\nu  + 2c \biggr) \biggr] \biggr\} 
+ \frac{(\beta_{+1}+ \beta_{-1})}{n\pi} \biggl\{ \sin\biggl[ \frac{n\pi }{L} \biggr( \nu  + 2c \biggr) \biggr] - \sin\biggl[ \frac{n\pi }{L} \biggr( \nu \biggr)\biggr] \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>


Note that,

<div align="center">
<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_{ (3\nu +2c)}^{ (3\nu +4c)} \beta_{+2} dx + 
\int_{\nu}^{\nu +2c} \beta_{+1}  dx + 
\int^{-\nu}_{-(\nu +2c)} \beta_{-1} dx +
\int^{-(3\nu +2c)}_{-(3\nu + 4c)} \beta_{-2} 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_{+2} \biggl[ (3\nu +4c) -  (3\nu +2c)  \biggr]
+ \beta_{+1} \biggl[ \nu +2c - \nu \biggr]
+ \beta_{-1} \biggl[- \nu + (\nu +2c) \biggr] 
+ \beta_{-2} \biggl[-(3\nu +2c) + (3\nu + 4c) \biggr] 
\biggr\}
</math>
  </td>
</tr>

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

Therefore, for an arbitrary number of slit ''pairs'', <math>~j_\mathrm{slits}</math>, it appears as though,

<div align="center">
<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>~\sum_{j=1}^{j_\mathrm{slits}}
\frac{(\beta_{+j} + \beta_{-j})}{n\pi} \biggl\{ \sin\biggl\{ \frac{n\pi }{L} \biggr[ (2j-1)(\nu + c)  + c \biggr] \biggr\} - \sin\biggl\{ \frac{n\pi }{L} \biggr[ (2j-1)(\nu + c) - c \biggr] \biggr\} \biggr\} 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\sum_{j=1}^{j_\mathrm{slits}}
\frac{2c(\beta_{+j} + \beta_{-j})}{L} \biggl\{ \cos \biggl[ \chi_n(2j-1)\biggl( 1 + \frac{\nu}{c} \biggr)  \biggr] \frac{\sin\chi_n}{\chi_n}  \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2c}{L} \cdot \frac{\sin\chi_n}{\chi_n}  ~\sum_{j=1}^{j_\mathrm{slits}} (\beta_{+j} + \beta_{-j})
\cos \biggl[ \chi_n(2j-1)\biggl( 1 + \frac{\nu}{c} \biggr)  \biggr] \, ,
</math>
  </td>
</tr>
</table>
</div>
where, as before, <math>~\chi_n \equiv (n\pi c/L)</math>, with the special case,

<div align="center">
<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{2c}{L} \sum_{j=1}^{j_\mathrm{slits}}
\biggl[ \beta_{+j} + \beta_{-j} \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

<table border="1" width="85%" cellpadding="8" align="center">
<tr>
  <td align="center">
Behavior of <math>~a_n</math> when <math>~\beta_{\pm j} = 1</math> for all <math>~j</math>, and in the limit, <math>~\frac{\nu}{c} \rightarrow 0</math>.
  </td>
</tr>
<tr><td align="left">
When <math>~j_\mathrm{slits} = 1</math>:

<div align="center">
<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{2c}{L} \cdot \frac{\sin\chi_n}{\chi_n}  \cdot (2)
\cos \chi_n 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4c}{L} \cdot \frac{\sin(2\chi_n)}{(2\chi_n)}  
</math>
  </td>
</tr>
</table>
</div>
When <math>~j_\mathrm{slits} = 2</math>:

<div align="center">
<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{2c}{L} \cdot \frac{\sin\chi_n}{\chi_n}  ~\biggl\{ \sum_{j=1}^{2} (2)
\cos \biggl[ \chi_n(2j-1) \biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4c}{L} \cdot \frac{\sin\chi_n}{\chi_n}  ~\biggl\{ \cos \chi_n  +  \cos (3\chi_n ) \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4c}{L} \cdot \frac{\sin\chi_n}{\chi_n}  ~\biggl\{ \cos \chi_n  +  4\cos^3\chi_n -3\cos\chi_n \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4c}{L} \cdot \frac{\sin\chi_n \cos\chi_n}{\chi_n}  ~\biggl\{ 4\cos^2\chi_n - 2 \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4c}{L} \cdot \frac{\sin\chi_n \cos\chi_n}{\chi_n}  ~\biggl\{ 2 - 4\sin^2\chi_n \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{8c}{L} \cdot \frac{\sin(4\chi_n )}{(4\chi_n)}  \, .
</math>
  </td>
</tr>
</table>
</div>
When <math>~j_\mathrm{slits} = 3</math>:

<div align="center">
<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{4c}{L} \cdot \frac{\sin\chi_n}{\chi_n}  ~\biggl\{ \sum_{j=1}^{3}
\cos \biggl[ \chi_n(2j-1) \biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4c}{L} \cdot \frac{\sin\chi_n}{\chi_n}  ~\biggl\{ 
\cos \chi_n  + \cos (3\chi_n ) + \cos (5\chi_n) 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4c}{L} \cdot \frac{\sin\chi_n}{\chi_n}  ~\biggl\{ 
\cos \chi_n  
+ [ 4\cos^3\chi_n - 3\cos\chi_n ] 
+ [ 16\cos^5\chi_n -20\cos^3\chi_n + 5\cos\chi_n  ]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4c}{L} \cdot \frac{\sin\chi_n \cos\chi_n}{\chi_n}  ~\biggl\{ 
3   -16\cos^2\chi_n + 16\cos^4\chi_n
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{12c}{L} \cdot \frac{\sin(6\chi_n) }{6\chi_n}  
</math>
  </td>
</tr>
</table>
</div>
Therefore, it ''appears'' as though, for arbitrary values of <math>~j_\mathrm{slits} \ge 1</math>,

<div align="center">
<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{4 j_\mathrm{slits}c}{L} \cdot \frac{\sin(2j_\mathrm{slits}\chi_n) }{(2 j_\mathrm{slits}\chi_n)}  
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{(2c)(2 j_\mathrm{slits})}{L} \cdot \mathrm{sinc} (2j_\mathrm{slits}\chi_n)  
</math>
  </td>
</tr>
</table>
</div>

----

I have not yet proven that the above, generalized expression for <math>~a_n</math> works for all values of <math>~j_\mathrm{slits}</math>.  To do so will likely involve enlisting the following two generalized trigonometric (''Multiple-angle'') relations that appear on p. 190 of my CRC handbook of ''Standard Mathematical Tables.''
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\sin(n\alpha)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\sin[(n-1)\alpha] \cos\alpha - \sin[(n-2)\alpha] \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\cos(n\alpha)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\cos[(n-1)\alpha] \cos\alpha - \cos[(n-2)\alpha] \, .
</math>
  </td>
</tr>
</table>
</div>

</td></tr>
</table>

VARIOUS SCALINGS:

<ol>
<li>
Given that <math>~\beta</math> has units of brightness per unit length, with <math>~\beta_0</math> referring to the brightness per unit length that is incident on the single slit; and, given that in the above formulation each separate slit has a width of <math>~2c</math>, the ''total'' brightness emerging from a uniformly illuminated multi-slit configuration is, 
<div align="center">
<math>~b_\mathrm{tot} = 2j_\mathrm{slits}(2c\beta_0) \, .</math>
</div> 
Hence, in order for every configuration's total brightness to be the same &#8212; that is, to be equal to <math>~2c_0\beta_0,</math> where <math>~c_0</math> is the ''half''-width of the single slit &#8212; in each case we need to set <math>~c \rightarrow c_0/(2j_\mathrm{slits})</math>.
</li>
<li>As is illustrated by frame ''a'' of [[#Figure5|Figure 5]], below, the total width of the ''single'' slit is <math>~2c_0</math>, while &#8212; see [[#Figure5|frames ''b'' and ''c'' of Figure 5]] &#8212; the total width of the combined slit(s) is <div align="center">
<math>~C_\mathrm{tot} = (2j_\mathrm{slits})2c + (2j_\mathrm{slits}-1)2\nu \, .</math>
</div>
Hence, if we want the ''ratio'' of the slit width to the ''Fourier'' width, <math>~2L</math>, to be the same for each multi-slit example, in each case we need to set,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{C_\mathrm{tot}}{2L}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{c_0}{L_0}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ L \rightarrow \frac{L_0 C_\mathrm{tot}}{2c_0} </math>
  </td>
  <td align="center">
<math>~=</math>
  <td align="left">
<math>~\frac{L_0 }{c_0} \biggl[ (2j_\mathrm{slits})c + (2j_\mathrm{slits} -1) \nu \biggr]</math>
  </td>
</tr>
</table>
</div>
Or, put together, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~ L </math>
  </td>
  <td align="center">
<math>~\rightarrow</math>
  <td align="left">
<math>~L_0 \biggl[ 1 + \frac{(2j_\mathrm{slits} -1) \nu}{ (2j_\mathrm{slits})c}\biggr]</math>
  </td>
</tr>
</table>
</div>
</li>
</ol>

<span id="Figure5">&nbsp;</span>
<table border="1" cellpadding="8" align="center">
<tr><th align="center"><font size="+1">Figure 5</font></th><tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <th align="center">Frame a:&nbsp; Single Aperture</th>
  <th align="center">Frame b:&nbsp; Double Aperture <math>~(j_\mathrm{slits} = 1)</math></th>
  <th align="center">Frame c:&nbsp; Multiple Apertures <math>~(j_\mathrm{slits} = 3)</math></th>
</tr>
<tr>
  <td align="left">[[File:SingleSlit03.png|250px|center|Single Aperture]]</td>
  <td align="left">[[File:DoubleSlit04.png|center|250px|DoubleSlit]]</td>
  <td align="left">[[File:MultiSlit04.png|270px|center|Multiple Apertures]]</td>
</tr>
</table>
</td>
</tr></table>

Incorporating these normalizations, we have,

<div align="center">
<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{c_0}{j_\mathrm{slits} L_0} \biggl[ 1 + \frac{(2j_\mathrm{slits} -1) \nu}{ (2j_\mathrm{slits})c}\biggr]^{-1} \frac{\sin\chi_n}{\chi_n}  ~\sum_{j=1}^{j_\mathrm{slits}} (\beta_{+j} + \beta_{-j})
\cos \biggl[ \chi_n(2j-1)\biggl( 1 + \frac{\nu}{c} \biggr)  \biggr] \, ,
</math>
  </td>
</tr>
</table>
</div>
and,

<div align="center">
<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{c_0}{j_\mathrm{slits} L_0} \biggl[ 1 + \frac{(2j_\mathrm{slits} -1) \nu}{ (2j_\mathrm{slits})c}\biggr]^{-1} \sum_{j=1}^{j_\mathrm{slits}}
\biggl[ \beta_{+j} + \beta_{-j} \biggr] \, ,
</math>
  </td>
</tr>
</table>
</div>

where,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\chi_n \equiv \frac{n\pi c}{L}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{n \pi c_0}{ 2 j_\mathrm{slits} L_0} \biggl[ 1 + \frac{(2j_\mathrm{slits} -1) \nu}{ (2j_\mathrm{slits})c}\biggr]^{-1} 
\, .
</math>
  </td>
</tr>
</table>
</div>

<table border="1" align="center" cellpadding="5">
<tr>
  <th align="center">
<font size="+1">Figure 6</font>
  </th>
</tr>
<tr>
<th align="center">
<math>~\frac{c_0}{L_0} = \frac{1}{20} </math> &nbsp; &nbsp; and &nbsp; &nbsp; <math>~\frac{\nu}{c} = \frac{1}{10}</math>&nbsp; &nbsp; for various <math>~j_\mathrm{slits}</math>
</th>
</tr>
<tr><td align="center">
[[File:MultiSlits01.gif|center|Diffraction pattern for multi-slit aperture]]
</td></tr>
</table>

----

By generalizing, we also see that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<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>

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,
<div align="center">
<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} \biggl| \sin\biggl( \frac{\chi\nu }{c} + 2\chi \biggr) - \sin\biggl( \frac{\chi\nu}{c} \biggr) \biggr| \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<math>~\chi \equiv \frac{n\pi c}{L} \, .</math>
</div>