User:Tohline/Appendix/CGH/KAH2001

Hologram Reconstruction Using a Digital Micromirror Device

In a paper titled, Hologram reconstruction using a digital micromirror device, T. Kreis, P. Aswendt, & R. Höfling (2001) — Optical Engineering, vol. 40, no. 6, 926 - 933), hereafter, KAH2001 — present some background theoretical development that was used to underpin work of the group at UT's Southwestern Medical Center at Dallas that Richard Muffoletto and I visited circa 2004.

Fresnel Diffraction

According to the Wikipedia description of Fresnel diffraction, "… the electric field diffraction pattern at a point <math>~(x, y, z)</math> is given by …" the expression,

<math>~E(x, y, z)</math>

<math>~=</math>

<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>

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,

<math>~r</math>

<math>~\equiv</math>

<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>

(The infinite series in this last expression results from enlisting the binomial theorem.) For simplicity, in the discussion that follows we will assume — as in §2 of KAH2001 — 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 — and setting <math>~\theta = 0</math> — then leads to the expression,

<math>~E(x, y, z)</math>	<math>~\approx</math>	<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>
	<math>~=</math>	<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>

If "… for the <math>~r</math> in the denominator we go one step further, and approximate it with only the first term …", then our expression results in the Fresnel diffraction integral,

<math>~E(x, y, z)</math>

<math>~\approx</math>

<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>

Optical Field in the Image Plane

This same integral expression — with a slightly different leading normalization factor — appears as equation (5) of 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,

<math>~B(x,y)</math>	<math>~=</math>	<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>
	<math>~=</math>	<math>~ \biggl[\frac{e^{i k d}}{i k d} \biggr] I_\xi(x) \cdot I_\eta(y) \, , </math>

with,

<math>~I_\xi(x)</math>	<math>~=</math>	<math>~ \int_{-\infty}^{\infty} V(\xi) \times \exp\biggl[ \frac{i \pi}{d \lambda} (x - \xi)^2 \biggr] d\xi \, , </math>
<math>~I_\eta(y)</math>	<math>~=</math>	<math>~ \int_{-\infty}^{\infty} W(\eta) \times \exp\biggl[ \frac{i \pi}{d \lambda} (y - \eta)^2 \biggr] d\eta \, , </math>

and where "… 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 KAH2001 — especially the discussion associated with their equations (7) - (10) — if we make the substitutions,

<math>~\mu \equiv \frac{x}{d\lambda} \, ,</math>

and,

<math>~\alpha \equiv \frac{\sqrt{2} \xi}{ \sqrt{d \lambda} } - \sqrt{2d\lambda} ~\mu </math>      <math>~\Rightarrow ~~~ d\xi = \biggl(\frac{d \lambda}{2}\biggr)^{1 / 2} d\alpha \, ,</math>

the expression for <math>~I_\xi(x)</math> may be written as,

<math>~I_\xi(x)</math>	<math>~=</math>	<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>
	<math>~=</math>	<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>
	<math>~=</math>	<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>

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

As a point of comparison, in our 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: <math>~A(y_1)</math> <math>~\approx</math> <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> where, <math>~\frac{1}{\beta_1}</math> <math>~\equiv</math> <math>~\frac{\lambda L}{\pi y_1w} \, ,</math>       <math>~L</math> <math>~\equiv</math> <math>~ Z \biggl[1 + \frac{y_1^2}{Z^2} \biggr]^{1 / 2} \, , </math>       and,       <math>~\Theta</math> <math>~\equiv</math> <math>~\biggl(\frac{2\pi y_1 Y}{\lambda L} \biggr) \, .</math> In other words, making the substitution, <math>~(2\pi/\lambda) \rightarrow k</math>, and recognizing that, <math>~d \leftrightarrow Z</math>, our expression becomes, <math>~I(y) \equiv \biggl[i k d e^{-i k d} \biggr] A(y_1)</math> <math>~\approx</math> <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>   <math>~=</math> <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>   <math>~\approx</math> <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>

See Also

• Updated Table of Contents
• Tohline, J. E., (2008) Computing in Science & Engineering, vol. 10, no. 4, pp. 84-85 — Where is My Digital Holographic Display? [ PDF ]

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