Editing Appendix/CGH/KAH2001

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=Hologram Reconstruction Using a Digital Micromirror Device=

==Fresnel Diffraction==

According to the [https://en.wikipedia.org/wiki/Fresnel_diffraction Wikipedia description of  Fresnel diffraction], "&hellip; the electric field diffraction pattern at a point <math>~(x, y, z)</math> is given by &hellip;" the expression,

<table border="1" align="center" cellpadding="10" width="65%"><tr><td align="center">
Given the intensity immediately in front of the aperture, <math>~E(x', y', 0)</math>, this integral expression generates the intensity, <math>~E(x, y, z)</math>, on the image plane whose distance from the aperture is, <math>~z</math>.
</td></tr></table>

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

<tr>
  <td align="right">
<math>~E(x, y, z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{i \lambda} \iint_{-\infty}^\infty E(x', y', 0) \biggl[ \frac{e^{i k r}}{r}\biggr] \cos\theta~ dx' dy'\, ,
</math>
  </td>
</tr>
</table>

where, <math>~E(x', y', 0)</math> is the electric field at the aperture, <math>~k \equiv 2\pi/\lambda</math> is the wavenumber, and,

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

<tr>
  <td align="right">
<math>~r</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\biggl[ z^2 + (x - x')^2 + ( y - y')^2 \biggr]^{1 / 2}
= z \biggl[ 1 + \frac{(x - x')^2 + ( y - y')^2}{z^2} \biggr]^{1 / 2}
= z\biggl[ 1 +  \frac{(x - x')^2 + ( y - y')^2}{2z^2} - \frac{[(x - x')^2 + ( y - y')^2]^2}{8z^4} + \cdots\biggr] \, .
</math>
  </td>
</tr>
</table>

(The infinite series in this last expression results from enlisting the [[Appendix/Ramblings/PowerSeriesExpressions#Binomial|binomial theorem]].)  For simplicity, in the discussion that follows we will assume &#8212; as in &sect;2 of [https://ui.adsabs.harvard.edu/abs/2001OptEn..40..926K/abstract KAH2001] &#8212; that the aperture is illuminated by a monochromatic plane wave that is impinging normally onto the aperture, in which case, the angle, <math>~\theta = 0</math>.

In the '''Fresnel approximation''', the assumption is made that, in the series expansion for <math>~r</math>, all terms beyond the first two are very small in magnitude relative to the second term.  Adopting this approximation &#8212; and setting <math>~\theta = 0</math> &#8212; then leads to the expression,

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

<tr>
  <td align="right">
<math>~E(x, y, z)</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
\frac{1}{i z \lambda} \iint_{-\infty}^\infty E(x', y', 0) ~\biggl[ 1 -  \frac{(x - x')^2 + ( y - y')^2}{2z^2} \biggr] \exp\biggl\{ i k z\biggl[ 1 +  \frac{(x - x')^2 + ( y - y')^2}{2z^2}\biggr] \biggr\}~ dx' dy'
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{e^{i k z}}{i z \lambda} \iint_{-\infty}^\infty E(x', y', 0) ~\biggl[ 1 -  \frac{(x - x')^2 + ( y - y')^2}{2z^2} \biggr] \exp\biggl\{\frac{ i k}{2 z}\biggl[ (x - x')^2 + ( y - y')^2 \biggr] \biggr\}~ dx' dy' \, .
</math>
  </td>
</tr>
</table>
If "&hellip; for the <math>~r</math> in the denominator we go one step further, and approximate it with only the first term &hellip;", then our expression results in the '''Fresnel diffraction integral''', 

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

<tr>
  <td align="right">
<math>~E(x, y, z)</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
\frac{e^{i k z}}{i z \lambda} \iint_{-\infty}^\infty E(x', y', 0) ~ \exp\biggl\{\frac{ i k}{2 z}\biggl[ (x - x')^2 + ( y - y')^2 \biggr] \biggr\}~ dx' dy' \, .
</math>
  </td>
</tr>
</table>

==Optical Field in the Image Plane==

In a paper titled, ''Hologram reconstruction using a digital micromirror device'', [https://ui.adsabs.harvard.edu/abs/2001OptEn..40..926K/abstract T. Kreis, P. Aswendt, &amp; R. H&ouml;fling (2001)] &#8212; Optical Engineering, vol. 40, no. 6, 926 - 933, hereafter, KAH2001 &#8212; present some background theoretical development that was used to underpin work of the group at [[Appendix/CGH/ZebraImaging#UT_Southwestern_Medical_Center_at_Dallas|UT's Southwestern Medical Center at Dallas]] that Richard Muffoletto and I visited circa 2004.

This same integral expression &#8212; with a slightly different leading normalization factor &#8212; appears as equation (5) of [https://ui.adsabs.harvard.edu/abs/2001OptEn..40..926K/abstract KAH2001].  Referring to it as the ''Fresnel transform'' expression for the "optical field, <math>~B(x, y)</math>, in the image plane at a distance <math>~d</math> from the [aperture]," they write,

<table border="1" align="center" cellpadding="10" width="65%"><tr><td align="center">
Given the intensity immediately in front of the aperture, <math>~U(\xi, \eta)</math>, this integral expression generates the intensity, <math>~B(x, y)</math>, on the image plane whose distance from the aperture is, <math>~d</math>.
</td></tr></table>

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

<tr>
  <td align="right">
<math>~B(x,y)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{e^{i k d}}{i k d} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} U(\xi,\eta) \times \exp\biggl\{ \frac{i \pi}{d \lambda} \biggl[ (x - \xi)^2 + (y-\eta)^2 \biggr] \biggr\} d\xi d\eta
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{e^{i k d}}{i k d} \biggr] I_\xi(x) \cdot I_\eta(y) \, ,
</math>
  </td>
</tr>
</table>
with,
<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>~
\int_{-\infty}^{\infty} V(\xi) \times \exp\biggl[ \frac{i \pi}{d \lambda} (x - \xi)^2 \biggr] d\xi \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~I_\eta(y)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_{-\infty}^{\infty} W(\eta) \times \exp\biggl[ \frac{i \pi}{d \lambda} (y - \eta)^2 \biggr] d\eta \, ,
</math>
  </td>
</tr>
</table>
and where "&hellip; the optical field immediately in front of the [aperture]" is assumed to be of the form, <math>~U(\xi,\eta) = V(\xi)\cdot W(\eta)</math>.  Following [https://ui.adsabs.harvard.edu/abs/2001OptEn..40..926K/abstract KAH2001] &#8212; especially the discussion associated with their equations (7) - (10) &#8212; if we make the substitutions,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mu \equiv \frac{x}{d\lambda} \, ,</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~\alpha \equiv \frac{\sqrt{2} \xi}{ \sqrt{d \lambda} }  - \sqrt{2d\lambda} ~\mu </math>&nbsp; &nbsp; &nbsp;
<math>~\Rightarrow ~~~ d\xi = \biggl(\frac{d \lambda}{2}\biggr)^{1 / 2} d\alpha \, ,</math>
  </td>
</tr>
</table>

<span id="Eq10KAH2001">the expression for <math>~I_\xi(x)</math> may be written as,</span>
<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>~
\int_{-\infty}^{\infty} V(\xi) \times \exp\biggl\{  \frac{i \pi}{d \lambda} \biggl[ d \lambda \mu - \frac{\sqrt{d\lambda}}{\sqrt{2}} \biggl( \alpha + \sqrt{2 d\lambda}~\mu \biggr) \biggr]^2 \biggr\} \biggl( \frac{d \lambda}{2}\biggr)^{1 / 2} d\alpha
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </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>
</table>

The expression for <math>~I_\eta(y)</math> may be rewritten similarly.

==SWMED==

In a paper titled, ''Dynamic holographic 3-D image projection'', [https://ui.adsabs.harvard.edu/abs/2003OExpr..11..437H/abstract M. L. Huebschman, B. Munjuluri &amp; H. R. Garner (2003)] &#8212; Optics Express, vol. 11, no. 5, 437 - 445, hereafter, SWMED03 &#8212; describe the experimental 3-D projection system that was developed at [[Appendix/CGH/ZebraImaging#UT_Southwestern_Medical_Center_at_Dallas|UT's Southwestern Medical Center at Dallas]].  This is the research group that Richard Muffoletto and I visited circa 2004.

In &sect;4 of [https://ui.adsabs.harvard.edu/abs/2003OExpr..11..437H/abstract SWMED03] we find this general description:

The following integral expression is "<font color="orange">&hellip; the mathematical transform containing the wave physics of monochromatic light emanating from each object point, passing through the optical system and being superimposed at each point in the holographic plane.  It represents the integration over the object of spherical wave solutions of the [https://en.wikipedia.org/wiki/Helmholtz_equation Helmholtz form of the wave equation] &hellip; with additional phase shifts due to a spherical converging lens in the light pathway.</font>"

<table border="1" align="center" cellpadding="10" width="70%"><tr><td align="center">
Given the intensity, <math>~U(x', y', z')</math>, on various points across the objects in a 3-D scene, this integral expression generates the resulting intensity, <math>~U_s(x, y, 0)</math>, across the ''hologram plane'', i.e., immediately in front of the aperture.
</td></tr></table>

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

<tr>
  <td align="right">
<math>~U_s(x, y, 0)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_{\mathrm{V}'} U(x', y', z')~ \biggl[\frac{e^{-ikr} }{r}\biggr]~\times  \exp\biggl[ \frac{ik(x^2 + y^2)}{2f} \biggr]~dV^'
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_{\mathrm{V}'} dV^' ~ \frac{U(x', y', z')}{\sqrt{ {z'}^2 + (x - x')^2 + (y - y')^2 }}~
\exp\biggl[-ik \biggl( \sqrt{ {z'}^2 + (x - x')^2 + (y - y')^2 } - \frac{x^2 + y^2}{2f} \biggr) \biggr] \, .
</math>
  </td>
</tr>
</table>

[The second of these two expressions has the form that appears as equation (1) in [https://ui.adsabs.harvard.edu/abs/2003OExpr..11..437H/abstract SWMED03]; the first has been rewritten here in a form that can more easily be compared with related expressions that are found in the above subsections of this chapter.]  As is explained in [https://ui.adsabs.harvard.edu/abs/2003OExpr..11..437H/abstract SWMED03], "<font color="orange">&hellip; <math>~U_s(x, y, 0)</math> represents the intensity amplitude at a point in the hologram plane</font> &#8212; i.e., immediately in front of the aperture &#8212; <font color="orange"> and <math>~U(x', y', z')</math> is the intensity amplitude at a point on the objects in the 3-D scene of volume V' &hellip; The z'-axis is normal to the center of the hologram plane and extends through the center of the reconstructed 3-D scene volume.  The wave number of the light is given by k and f is the focal length of the converging lens.</font>"

"<font color="orange">The second term in the exponential represents the phase change due to the differing lengths of material the light traverses in the converging lens in our system &hellip;  Recall, a function of the converging lens is to produce the 3-D image reconstruction over a relatively near field.  Operating in the near field spatially disperses the depth information which is lost for reconstruction images that converge at infinity</font> &#8212; see [https://ui.adsabs.harvard.edu/abs/2001OptEn..40..926K/abstract T. Kreis, P. Aswendt, &amp; R. H&ouml;fling (2001)], which is [[#Optical_Field_in_the_Image_Plane|discussed above]]. <font color="orange">The phase information due to depth is conveyed via the first term in the exponential which is the radial distance from an object point to a hologram point.  We do not approximate this distance in our calculation, but we have incorporated the [https://en.wikipedia.org/wiki/Fresnel_diffraction#The_Fresnel_approximation Fresnel approximation] in the amplitude term in front of the exponential &hellip; Experimentally, we observed little change in image quality on reconstruction with this simplification.</font>"

==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>

=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] ]

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