Editing Appendix/CGH/ParallelAperturesConsolidate

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=CGH:  Consolidate Expressions Regarding Parallel Apertures=

==One-dimensional Apertures==

From our accompanying discussion of the [[Appendix/CGH/ParallelApertures#Utility_of_FFT_Techniques|''Utility of FFT Techniques'']], we start with the most general expression for the amplitude at one point on an image screen, namely,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~A(y_1)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\sum_j
a_j e^{i(2\pi D_j/\lambda + \phi_j)} 
\, ,
</math>
  </td>
</tr>
</table>
and, assuming that <math>~|Y_j/L| \ll 1</math> for all <math>~j</math>, deduce that,
<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>~\sum_j
a_j e^{i[ 2\pi L/\lambda + \phi_j]}\biggl[ \cos\biggl(\frac{2\pi y_1 Y_j}{\lambda L} \biggr) - i  \sin\biggl(\frac{2\pi y_1 Y_j}{\lambda L} \biggr) \biggr]
\, ,
</math>
  </td>
</tr>
</table>
where, 
<table border="0" cellpadding="5" align="center">

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

<span id="FocalPoint">Note that</span> <math>~L</math> is formally a function of <math>~y_1</math>, but in most of what follows it will be reasonable to assume, <math>~L \approx Z</math>.  Notice, as well, that this last approximate expression for the (complex) amplitude at the image screen may be rewritten in the form that will be referred to as our,

<table border="0" cellpadding="5" align="center">
<tr>
  <th align="center" colspan="3">Focal-Point Expression</th>
</tr>

<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 }  \sum_j  a_j e^{i \phi_j}   \cdot e^{-i \Theta_j } 
\, ,
</math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Theta_j</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{2\pi y_1 Y_j}{\lambda L} \biggr) \, .</math>
  </td>
</tr>
</table>

===Case 1===
In a related accompanying derivation titled, [[Appendix/CGH/ParallelApertures#Analytic_Result|''Analytic Result'']], we made the substitution,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~a_j </math>
  </td>
  <td align="center">
<math>~\rightarrow</math>
  </td>
  <td align="left">
<math>~a_0(Y) dY = a_0(\Theta) \biggl[ \frac{w}{2\beta_1} \biggr] 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>
</tr>
</table>

and changed the summation to an integration, obtaining,

<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>

If we assume that both <math>~a_0</math> and <math>~\phi</math> are independent of position along the aperture, and that the aperture &#8212; and, hence the integration &#8212; extends from <math>~Y_2 = -w/2</math> to <math>~Y_1 = +w/2</math>, we have shown that this last expression can be evaluated analytically to give,

<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 + \phi] }\biggl[ \frac{a_0 w}{2\beta_1} \biggr]  \int_{\Theta_2}^{\Theta_1} e^{-i \Theta } d\Theta 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
e^{i [2\pi L/\lambda + \phi] } \cdot a_0 w ~\mathrm{sinc}(\beta_1) \, . 
</math>
  </td>
</tr>
</table>

We need to explicitly demonstrate that an evaluation of our [[#FocalPoint|Focal-Point Expression]] with <math>~a_j = 1</math>, gives this last sinc-function expression, to within a multiplicative factor of, something like, <math>~j_\mathrm{max}</math>.

===Case 2===

In our accompanying discussion of the [[Appendix/Ramblings/FourierSeries|Fourier Series]], we have shown that a square wave can be constructed from the expression,

<div align="center" id="StandardExpression">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~f(x)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{c}{L} + \sum_{n=1}^{\infty} \biggl( \frac{2}{n\pi} \biggr) \sin \biggl( \frac{n\pi c}{L} \biggr) \cos \biggl(\frac{n\pi x}{L}\biggr) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2c}{L}\biggl\{\frac{1}{2} + \sum_{n=1}^{\infty} \mathrm{sinc} \biggl( \frac{n\pi c}{L} \biggr) \cos \biggl(\frac{n\pi x}{L}\biggr) \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>

Can we make this look like our [[#FocalPoint|above, Focal-Point Expression]]?  

Let's start by setting
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~Y_j</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{j\cdot w}{(j_\mathrm{max}-1)} - \frac{w}{2} \, ,</math>
  </td>
</tr>
</table>

for <math>~0 \le j \le (j_\mathrm{max}-1)</math>, in which case,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Theta_j</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\frac{2\pi y_1}{\lambda L} \biggl[ \frac{j\cdot w}{(j_\mathrm{max}-1)} - \frac{w}{2} \biggr] 
= \frac{2\pi y_1}{\lambda L} \biggl[ \frac{j\cdot w}{(j_\mathrm{max}-1)} \biggr] - \frac{2\pi y_1}{\lambda L} \biggl[ \frac{w}{2} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
j \biggl[ \frac{2\pi y_1 w}{(j_\mathrm{max}-1) \lambda L} \biggr] - \frac{\pi y_1 w }{\lambda L} 
= \biggl( \frac{2j}{j_\mathrm{max} - 1} - 1 \biggr) \frac{\pi y_1 w }{\lambda L}
\, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
j \cdot \Delta\Theta - \frac{(j_\mathrm{max} -1)}{2} \Delta\Theta
\, ,</math>
  </td>
</tr>
</table>
where,
<div align="center">
<math>~\Delta\Theta \equiv  \frac{\pi y_1}{\mathfrak{L}}  \, ,</math> &nbsp; &nbsp; and &nbsp; &nbsp; <math>~\mathfrak{L} \equiv \biggl[ \frac{(j_\mathrm{max}-1) \lambda L}{2w} \biggr] \, .</math>
</div>
This means that <math>~\Theta_{i} = - \Theta_{( j_\mathrm{max} - 1 - i )}</math>.

The key  expression under the summation therefore becomes,

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

<tr>
  <td align="right">
<math>~a_j e^{i \phi_j}   \cdot e^{-i \Theta_j }  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~~a_j e^{i \phi_j}   \cdot \biggl[ \cos \biggl( \frac{j\pi y_1}{\mathfrak{L}}- \Theta_0 \biggr) - i \sin \biggl( \frac{j\pi y_1}{\mathfrak{L}}- \Theta_0  \biggr) \biggr] \, ,</math>
  </td>
</tr>
</table>
where,
<div align="center">
<math>~\Theta_0 \equiv  \frac{(j_\mathrm{max} - 1)}{2} \cdot \pi y_1 \biggl[ \frac{2w}{(j_\mathrm{max}-1) \lambda L} \biggr] = \frac{\pi y_1 w}{\lambda L} \, .</math>
</div>

Now, what is the argument of the sinc function?  By default, it needs to be something along the lines of,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{j \pi c}{\mathfrak{L}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~j \pi c \biggl[ \frac{2w}{(j_\mathrm{max}-1) \lambda L} \biggr] \, .</math>
  </td>
</tr>
</table>
Then, as <math>~j</math> varies from <math>~0</math> to <math>~(j_\mathrm{max} - 1)</math>, the argument goes from <math>~0</math> to <math>~[2\pi w c/(\lambda L)]</math>.  In an effort to make the function exhibit reflection symmetry as we move from one side of the aperture to the next, let's subtract half of this upper limit; that is, let's modify the argument of the sinc function to read,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{j \pi c}{\mathfrak{L}} - \frac{\pi w c}{\lambda L}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
j \pi c \biggl[ \frac{2w}{(j_\mathrm{max}-1) \lambda L} \biggr]  - \frac{\pi w c}{\lambda L}
= \biggl[ \frac{2j}{j_\mathrm{max}-1}  - 1\biggr]\biggl[ \frac{\pi w c}{\lambda L} \biggr]  \, .
</math>
  </td>
</tr>
</table>
This means that in our [[#FocalPoint|above, Focal-Point Expression]] we want to set,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~a_j</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\mathrm{sinc} \biggl[ \biggl( \frac{2j}{j_\mathrm{max}-1}  - 1 \biggr) \frac{\pi w c}{\lambda L} \biggr] \, .
</math>
  </td>
</tr>
</table>

This therefore gives the following,

<table border="0" cellpadding="5" align="center">
<tr>
  <th align="center" colspan="3">Focal-Point Expression for a Square Wave</th>
</tr>

<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 }  \sum_{j=0}^{j_\mathrm{max}-1}  e^{i \phi_j}  \cdot~
\mathrm{sinc} \biggl[ \biggl( \frac{2j}{j_\mathrm{max}-1}  - 1 \biggr) \frac{\pi w c}{\lambda L} \biggr] 
\biggl\{ \cos \biggl[ \biggl( \frac{2j}{j_\mathrm{max} - 1} - 1 \biggr) \frac{\pi y_1 w }{\lambda L} \biggr] - i \sin \biggl[ \biggl( \frac{2j}{j_\mathrm{max} - 1} - 1 \biggr) \frac{\pi y_1 w }{\lambda L}  \biggr] 
\biggr\} \, .
</math>
  </td>
</tr>
</table>

This exhibits a very desirable feature:  Both the sinc function and the sine function &#8212; and, hence, also their product &#8212; have reflection symmetry about the summation index, <math>~j = (j_\mathrm{max}-1)/2</math>.  As a result, if the overall phase factor, <math>~e^{i \phi_j}</math>, behaves in an appropriately simple way &#8212; for example, if it is zero everywhere &#8212; then under the summation the sine term will sum to zero and leave only the desired &#8212; and ''real'' &#8212; product, <math>~\mathrm{sinc} \times \cos</math>.  Try this out in Excel to see if it works!

This could use a little more manipulation.  Let's define the alternate summation index,

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

<tr>
  <td align="right">
<math>~n</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>\frac{1}{2} \biggl[ j_\mathrm{max}-1 \biggr] \biggl( \frac{2j}{j_\mathrm{max}-1}  - 1 \biggr) \, ,</math>
  </td>
</tr>
</table>
in which case we can write,

<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 }  \sum_{n~=~-(j_\mathrm{max} - 1)/2}^{+(j_\mathrm{max} - 1)/2}  e^{i \phi_j}  \cdot~
\mathrm{sinc} \biggl[ \biggl( \frac{2n}{ j_\mathrm{max}-1} \biggr) \frac{\pi w c}{\lambda L} \biggr] 
\biggl\{ \cos \biggl[ \biggl( \frac{2n}{ j_\mathrm{max}-1} \biggr) \frac{\pi y_1 w }{\lambda L} \biggr] - i \sin \biggl[ \biggl( \frac{2n}{ j_\mathrm{max}-1} \biggr) \frac{\pi y_1 w }{\lambda L}  \biggr] 
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ e^{i 2\pi L/\lambda }   e^{i \phi_{j=0} }
~+~e^{i 2\pi L/\lambda }  \sum_{n~=~1}^{+(j_\mathrm{max} - 1)/2}  2e^{i \phi_j}  \cdot~
\mathrm{sinc} \biggl[ \biggl( \frac{2n}{ j_\mathrm{max}-1} \biggr) \frac{\pi w c}{\lambda L} \biggr] 
~ \cos \biggl[ \biggl( \frac{2n}{ j_\mathrm{max}-1} \biggr) \frac{\pi y_1 w }{\lambda L} \biggr]  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ e^{i 2\pi L/\lambda }   e^{i \phi_{j=0} }
~+~e^{i 2\pi L/\lambda }  \sum_{n~=~1}^{+(j_\mathrm{max} - 1)/2}  2e^{i \phi_j}  \cdot~
\mathrm{sinc} \biggl(\frac{\pi n c}{\mathfrak{L} } \biggr) 
~ \cos \biggl(  \frac{n \pi y_1 }{\mathfrak{L} } \biggr)  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ e^{i 2\pi L/\lambda } \biggl(\frac{\mathfrak{L}}{c} \biggr) \biggl\{   e^{i \phi_{j=0} }  \biggl(\frac{c}{\mathfrak{L}} \biggr)
~+~ \sum_{n~=~1}^{+(j_\mathrm{max} - 1)/2}  e^{i \phi_j}  \cdot~
\biggl(\frac{ 2  }{\pi n } \biggr) \sin \biggl(\frac{\pi n c}{\mathfrak{L} } \biggr) 
~ \cos \biggl(  \frac{n \pi y_1 }{\mathfrak{L} } \biggr)  \biggr\}
\, .
</math>
  </td>
</tr>
</table>

Finally, recalling that,
<table border="0" cellpadding="5" align="center">

<tr>
  <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} \approx Z \biggl[1 + \frac{1}{2}\frac{y_1^2}{Z^2}  \biggr] = Z + \frac{y_1^2}{2Z}  \, ,
</math>
  </td>
</tr>
</table>

let's set &hellip;
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~e^{i\phi_j}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
e^{-i2\pi Z/\lambda}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ e^{i2\pi L/\lambda} \cdot e^{i\phi_j}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
e^{i2\pi (L-Z)/\lambda} \approx e^{i\pi y_1^2/(\lambda Z)} = \cos\biggl( \frac{\pi y_1^2}{\lambda Z} \biggr) + i \sin \biggl( \frac{\pi y_1^2}{\lambda Z} \biggr) \, .
</math>
  </td>
</tr>
</table>

As a result, we have,

<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>~ \biggl[ \cos\biggl( \frac{\pi y_1^2}{\lambda Z} \biggr) + i \sin \biggl( \frac{\pi y_1^2}{\lambda Z} \biggr) \biggr] \biggl(\frac{\mathfrak{L}}{c} \biggr) \biggl\{  \biggl(\frac{c}{\mathfrak{L}} \biggr)
~+~ \sum_{n~=~1}^{+(j_\mathrm{max} - 1)/2}  
\biggl(\frac{ 2  }{\pi n } \biggr) \sin \biggl(\frac{\pi n c}{\mathfrak{L} } \biggr) 
~ \cos \biggl(  \frac{n \pi y_1 }{\mathfrak{L} } \biggr)  \biggr\}
\, .
</math>
  </td>
</tr>
</table>
Therefore, a clean square wave will appear only if <math>~[\pi y_1^2/(\lambda Z)] \ll 1</math>.

=See Also=
* Updated [[Appendix/Ramblings#Computer-Generated_Holography|Table of Contents]]
* Tohline, J. E., (2008) Computing in Science &amp; Engineering, vol. 10, no. 4, pp. 84-85 &#8212; ''Where is My Digital Holographic Display?'' [ [http://www.phys.lsu.edu/~tohline/CiSE/CiSE2008.Vol10No4.pdf PDF] ]
* [https://en.wikipedia.org/wiki/Diffraction Diffraction] (Wikipedia)
* Various Google hits:
** [http://labman.phys.utk.edu/phys222core/modules/m9/diffraction.htm Single Slit Diffraction] (University of Tennessee, Knoxville)
** [http://www.animations.physics.unsw.edu.au/jw/light/single-slit-diffraction.html Diffraction from a Single Slit; Young's Experiment with Finite Slits] (University of New South Wales, Sydney, Australia)
** [http://www.math.ubc.ca/~cass/courses/m309-03a/m309-projects/krzak/ Single Slit Diffraction Pattern of Light] (University of British Columbia, Canada)
** [http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslit.html Fraunhofer Single Slit] (Georgia State University)

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