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, if we evaluate the square and substitute <math>~\mu = x/(d \lambda)</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 x^2}{d \lambda} \biggl(1 - \frac{2 \xi}{x} + \frac{\xi^2}{x^2} \biggr) \biggr] d\xi </math>
	<math>~=</math>	<math>~ \int_{-\infty}^{\infty} V(\xi) \times \exp\biggl[ \frac{i \pi x^2}{d \lambda} \biggr] \times \exp\biggl[- \frac{i \pi x^2}{d \lambda} \biggl(\frac{2 \xi}{x} \biggr) \biggr] \times \exp\biggl[ \frac{i \pi x^2}{d \lambda} \biggl( \frac{\xi^2}{x^2} \biggr) \biggr] d\xi </math>
	<math>~=</math>	<math>~ \exp( i \pi d \lambda \mu^2 ) \int_{-\infty}^{\infty} V(\xi) \times \exp (- i 2\pi \mu \xi ) \times \exp \biggl[\biggl( \frac{i \pi }{d \lambda}\biggr) \xi^2 \biggr] d\xi \, . </math>

Note that all three of the exponential terms in this expression can be found in equation (7) of KAH2001. Notice, as well, that the product of exponentials that appears under the integral may be rewritten as,

<math>~ \exp (- i 2\pi \mu \xi ) \times \exp \biggl[\biggl( \frac{i \pi }{d \lambda}\biggr) \xi^2 \biggr] </math>	<math>~=</math>	<math>~ \exp \biggl[i \pi \biggl( \frac{\xi^2}{d \lambda} - 2\mu \xi \biggr) \biggr] </math>
	<math>~=</math>	<math>~\exp(- i\pi d \lambda \mu^2) \times \exp \biggl[i \pi \biggl( \frac{\xi^2}{d \lambda} - 2\mu \xi + d\lambda \mu^2 \biggr) \biggr] </math>
	<math>~=</math>	<math>~\exp(- i\pi d \lambda \mu^2) \times \exp \biggl[i \pi \biggl( \frac{\xi}{ \sqrt{d \lambda} } - \sqrt{d\lambda} ~\mu \biggr)^2 \biggr] </math>
	<math>~=</math>	<math>~\exp(- i\pi d \lambda \mu^2) \times \exp \biggl[\frac{ i \pi \alpha^2 }{2} \biggr] \, , </math>

where — see the discussion accompanying equation (9) of KAH2001 —

<math>~\alpha \equiv \frac{\sqrt{2} \xi}{ \sqrt{d \lambda} } - \sqrt{2d\lambda} ~\mu \, .</math>

Hence — see equation (10) of KAH2001 — we may also write,

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

<math>~=</math>

<math>~ \biggl(\frac{d \lambda}{2}\biggr)^{1 / 2} \int_{-\infty}^{\infty} V(\xi) \exp \biggl[ \frac{i \pi \alpha^2}{2} \biggr] d\alpha \, . </math>

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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