Difference between revisions of "User:Tohline/Appendix/CGH/KAH2001"
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<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' \, . | \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' \, . | ||
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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''', | |||
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<math>~E(x, y, z)</math> | |||
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<math>~\approx</math> | |||
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<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> | </math> | ||
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Revision as of 16:18, 27 March 2020
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.
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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
Labeling it as their equation (5), KAH2001 present the following 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:
<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} U(\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} U(\eta) \times \exp\biggl[ \frac{i \pi}{d \lambda} (y - \eta)^2 \biggr] d\eta \, , </math> |
and where <math>~U(\xi,\eta)</math> is "… the optical field immediately in front of the DMD" — i.e., the aperture. 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} U(\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} U(\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} U(\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.
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:
where,
In other words, making the substitution, <math>~(2\pi/\lambda) \rightarrow k</math>, and recognizing that, <math>~d \leftrightarrow Z</math>, our expression becomes,
|
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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