# User:Tohline/Appendix/CGH/KAH2001

Jump to: navigation, search

# Hologram Reconstruction Using a Digital Micromirror Device

 |   Tiled Menu   |   Tables of Content   |  Banner Video   |  Tohline Home Page   |

## Fresnel Diffraction

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

 Given the intensity immediately in front of the aperture, $~E(x', y', 0)$, this integral expression generates the intensity, $~E(x, y, z)$, on the image plane whose distance from the aperture is, $~z$.
 $~E(x, y, z)$ $~=$ $~ \frac{1}{i \lambda} \iint_{-\infty}^\infty E(x', y', 0) \biggl[ \frac{e^{i k r}}{r}\biggr] \cos\theta~ dx' dy'\, ,$

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

 $~r$ $~\equiv$ $~ \biggl[ z^2 + (x - x')^2 + ( y - y')^2 \biggr]^{1 / 2} = z \biggl[ 1 + \frac{(x - x')^2 + ( y - y')^2}{z^2} \biggr]^{1 / 2} = z\biggl[ 1 + \frac{(x - x')^2 + ( y - y')^2}{2z^2} - \frac{[(x - x')^2 + ( y - y')^2]^2}{8z^4} + \cdots\biggr] \, .$

(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, $~\theta = 0$.

In the Fresnel approximation, the assumption is made that, in the series expansion for $~r$, all terms beyond the first two are very small in magnitude relative to the second term. Adopting this approximation — and setting $~\theta = 0$ — then leads to the expression,

 $~E(x, y, z)$ $~\approx$ $~ \frac{1}{i z \lambda} \iint_{-\infty}^\infty E(x', y', 0) ~\biggl[ 1 - \frac{(x - x')^2 + ( y - y')^2}{2z^2} \biggr] \exp\biggl\{ i k z\biggl[ 1 + \frac{(x - x')^2 + ( y - y')^2}{2z^2}\biggr] \biggr\}~ dx' dy'$ $~=$ $~ \frac{e^{i k z}}{i z \lambda} \iint_{-\infty}^\infty E(x', y', 0) ~\biggl[ 1 - \frac{(x - x')^2 + ( y - y')^2}{2z^2} \biggr] \exp\biggl\{\frac{ i k}{2 z}\biggl[ (x - x')^2 + ( y - y')^2 \biggr] \biggr\}~ dx' dy' \, .$

If "… for the $~r$ 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,

 $~E(x, y, z)$ $~\approx$ $~ \frac{e^{i k z}}{i z \lambda} \iint_{-\infty}^\infty E(x', y', 0) ~ \exp\biggl\{\frac{ i k}{2 z}\biggl[ (x - x')^2 + ( y - y')^2 \biggr] \biggr\}~ dx' dy' \, .$

## Optical Field in the Image Plane

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.

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, $~B(x, y)$, in the image plane at a distance $~d$ from the [aperture]," they write,

 Given the intensity immediately in front of the aperture, $~U(\xi, \eta)$, this integral expression generates the intensity, $~B(x, y)$, on the image plane whose distance from the aperture is, $~d$.
 $~B(x,y)$ $~=$ $~ \frac{e^{i k d}}{i k d} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} U(\xi,\eta) \times \exp\biggl\{ \frac{i \pi}{d \lambda} \biggl[ (x - \xi)^2 + (y-\eta)^2 \biggr] \biggr\} d\xi d\eta$ $~=$ $~ \biggl[\frac{e^{i k d}}{i k d} \biggr] I_\xi(x) \cdot I_\eta(y) \, ,$

with,

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

and where "… the optical field immediately in front of the [aperture]" is assumed to be of the form, $~U(\xi,\eta) = V(\xi)\cdot W(\eta)$. Following KAH2001 — especially the discussion associated with their equations (7) - (10) — if we make the substitutions,

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

the expression for $~I_\xi(x)$ may be written as,

 $~I_\xi(x)$ $~=$ $~ \int_{-\infty}^{\infty} V(\xi) \times \exp\biggl\{ \frac{i \pi}{d \lambda} \biggl[ d \lambda \mu - \frac{\sqrt{d\lambda}}{\sqrt{2}} \biggl( \alpha + \sqrt{2 d\lambda}~\mu \biggr) \biggr]^2 \biggr\} \biggl( \frac{d \lambda}{2}\biggr)^{1 / 2} d\alpha$ $~=$ $~ \int_{-\infty}^{\infty} V(\xi) \times \exp\biggl\{ i \pi d \lambda \biggl[ \mu - \frac{1}{\sqrt{2d \lambda}} \biggl( \alpha + \sqrt{2 d\lambda}~\mu \biggr) \biggr]^2 \biggr\} \biggl( \frac{d \lambda}{2}\biggr)^{1 / 2} d\alpha$ $~=$ $~ \biggl( \frac{d \lambda}{2}\biggr)^{1 / 2} \int_{-\infty}^{\infty} V(\xi) \times \exp\biggl[ \frac{i \pi \alpha^2}{2} \biggr] d\alpha \, .$

The expression for $~I_\eta(y)$ may be rewritten similarly.

## SWMED

In a paper titled, Dynamic holographic 3-D image projection, M. L. Huebschman, B. Munjuluri & H. R. Garner (2003) — Optics Express, vol. 11, no. 5, 437 - 445, hereafter, SWMED03 — describe the experimental 3-D projection system that was developed at UT's Southwestern Medical Center at Dallas. This is the research group that Richard Muffoletto and I visited circa 2004.

In §4 of SWMED03 we find this general description:

The following integral expression is "… 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 Helmholtz form of the wave equation … with additional phase shifts due to a spherical converging lens in the light pathway."

 Given the intensity, $~U(x', y', z')$, on various points across the objects in a 3-D scene, this integral expression generates the resulting intensity, $~U_s(x, y, 0)$, across the hologram plane, i.e., immediately in front of the aperture.
 $~U_s(x, y, 0)$ $~=$ $~ \int_{\mathrm{V}'} U(x', y', z')~ \biggl[\frac{e^{-ikr} }{r}\biggr]~\times \exp\biggl[ \frac{ik(x^2 + y^2)}{2f} \biggr]~dV^'$ $~=$ $~ \int_{\mathrm{V}'} dV^' ~ \frac{U(x', y', z')}{\sqrt{ {z'}^2 + (x - x')^2 + (y - y')^2 }}~ \exp\biggl[-ik \biggl( \sqrt{ {z'}^2 + (x - x')^2 + (y - y')^2 } - \frac{x^2 + y^2}{2f} \biggr) \biggr] \, .$

[The second of these two expressions has the form that appears as equation (1) in 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 SWMED03, "$~U_s(x, y, 0)$ represents the intensity amplitude at a point in the hologram plane — i.e., immediately in front of the aperture — and $~U(x', y', z')$ is the intensity amplitude at a point on the objects in the 3-D scene of volume V' … 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."

"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 … 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 — see T. Kreis, P. Aswendt, & R. Höfling (2001), which is discussed above. 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 Fresnel approximation in the amplitude term in front of the exponential … Experimentally, we observed little change in image quality on reconstruction with this simplification."

## From Our Separate Discussion of CGH

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:

 $~A(y_1)$ $~\approx$ $~ e^{i 2\pi L/\lambda }\biggl[ \frac{w}{2\beta_1} \biggr] \int a_0(\Theta) e^{i\phi(\Theta)} \cdot e^{-i \Theta } d\Theta \, ,$

where,

 $~\frac{1}{\beta_1}$ $~\equiv$ $~\frac{\lambda L}{\pi y_1w} \, ,$ $~L$ $~\equiv$ $~ Z \biggl[1 + \frac{y_1^2}{Z^2} \biggr]^{1 / 2} \, ,$ and, $~\Theta$ $~\equiv$ $~\biggl(\frac{2\pi y_1 Y}{\lambda L} \biggr) \, .$

In other words, making the substitution, $~(2\pi/\lambda) \rightarrow k$, and recognizing that, $~d \leftrightarrow Z$, our expression becomes,

 $~I(y) \equiv \biggl[i k d e^{-i k d} \biggr] A(y_1)$ $~\approx$ $~ \biggl[i k d e^{-i k d} \biggr] e^{i kL }\biggl[ \frac{L}{k y_1} \biggr] \int a_0(\Theta) e^{i\phi(\Theta)} \cdot \exp\biggl[-i \frac{2\pi y_1 Y}{\lambda L} \biggr] \biggl[ \frac{k y_1 }{L} \biggr] dY$ $~=$ $~ (i k Z) e^{i k (L-Z)} \int a_0(\Theta) e^{i\phi(\Theta)} \cdot \exp\biggl[-i 2\pi Y \biggl(\frac{y_1 }{\lambda L}\biggr) \biggr] dY$ $~\approx$ $~ (i k Z) \exp\biggl[i k \biggl( L-Z \biggr)\biggr] \int a_0(\Theta) e^{i\phi(\Theta)} \cdot \exp\biggl[-i 2\pi Y \biggl(\frac{y_1 }{\lambda L}\biggr) \biggr] dY$ $~\approx$ $~ (i k Z) \exp\biggl[\frac{i \pi y_1^2}{Z \lambda} \biggr] \int a_0(\Theta) e^{i\phi(\Theta)} \cdot \exp\biggl\{ -i 2\pi Y \biggl(\frac{y_1 }{\lambda } \biggr) \frac{1}{Z}\biggl[1 - \frac{y_1^2}{2Z^2} \biggr] \biggr\} dY$ $~\approx$ $~ \biggl( \frac{2\pi i Z}{\lambda}\biggr) \exp\biggl[\frac{i \pi y_1^2}{Z \lambda} \biggr] \int a_0(\Theta) e^{i\phi(\Theta)} \cdot \exp\biggl\{ - \biggl(\frac{2\pi i y_1 Y}{Z \lambda } \biggr) \biggl[1 - \cancelto{0}{\frac{y_1^2}{2Z^2}} \biggr] \biggr\} dY \, ,$

where we have used the approximate expressions,

 $~L - Z$ $~\approx$ $~\frac{y_1^2}{2Z}$ and $~\frac{1}{L}$ $~\approx$ $~\frac{1}{Z}\biggl[1 - \frac{y_1^2}{2Z^2} \biggr] \, .$

Next, accounting for the different variable notations that have been adopted in the two separate discussions, namely,

 Notation KAH2001 Our $~x$ $~y_1$ $~\xi$ $~Y$ $~d$ $~Z$ $~\lambda$ $~\frac{2\pi}{k}$ $~\alpha \equiv \frac{\sqrt{2} \xi}{ \sqrt{d \lambda} } - \sqrt{2d\lambda} ~\biggl(\frac{x}{d\lambda}\biggr)$ $~\frac{\sqrt{2} Y}{ \sqrt{Z \lambda} } - \sqrt{2Z \lambda} ~\biggl(\frac{y_1}{Z \lambda}\biggr)$

let's examine how similar this last integral expression is to the key integral from KAH2001 that has been presented above. We have,

 $~I_\xi(x)$ $~=$ $~ \biggl( \frac{d \lambda}{2}\biggr)^{1 / 2} \int_{-\infty}^{\infty} V(\xi) \times \exp\biggl[ \frac{i \pi \alpha^2}{2} \biggr] d\alpha$ $~\rightarrow$ $~ \biggl( \frac{Z \lambda}{2}\biggr)^{1 / 2} \int_{-\infty}^{\infty} V(Y) \times \exp\biggl\{ \frac{i \pi}{2} \biggl[ \frac{\sqrt{2} Y}{ \sqrt{Z \lambda} } - \sqrt{2Z \lambda} ~\biggl(\frac{y_1}{Z \lambda}\biggr) \biggr]^2 \biggr\} \biggl[ \frac{2}{Z\lambda} \biggr]^{1 / 2}dY$ $~=$ $~ \int_{-\infty}^{\infty} V(Y) \times \exp\biggl\{ \frac{i \pi}{2} \biggl[ \frac{2Y^2}{ Z \lambda } - \frac{2\sqrt{2} Y}{ \sqrt{Z \lambda} } \cdot \sqrt{2Z \lambda} ~\biggl(\frac{y_1}{Z \lambda}\biggr) + 2Z \lambda ~\biggl(\frac{y_1}{Z \lambda}\biggr)^2\biggr] \biggr\} dY$ $~=$ $~ \int_{-\infty}^{\infty} V(Y) \times \exp\biggl\{ \frac{i \pi}{Z\lambda} \biggl[ Y^2 - 2y_1Y + y_1^2 \biggr] \biggr\} dY$ $~=$ $~ \exp\biggl[ \frac{i \pi y_1^2}{Z\lambda} \biggr] \int_{-\infty}^{\infty} V(Y) \times \exp\biggl\{ \frac{i \pi}{Z\lambda} \biggl[ - 2y_1Y + Y^2\biggr] \biggr\} dY$ $~=$ $~ \exp\biggl[ \frac{i \pi y_1^2}{Z\lambda} \biggr] \int_{-\infty}^{\infty} V(Y) \times \exp\biggl\{ - \frac{2\pi i y_1 Y}{Z\lambda} + \frac{i \pi Y^2}{Z\lambda} \biggr\} dY \, .$

Therefore, in order that the two expressions to match, we need to ignore the quadratic (Y2) term inside the last exponential, and make the association,

 $~V(Y)$ $~\leftrightarrow$ $~ \biggl( \frac{2\pi i Z}{\lambda}\biggr) a_0(\Theta) e^{i\phi(\Theta)} \, .$

# 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 ]

 © 2014 - 2020 by Joel E. Tohline |   H_Book Home   |   YouTube   | Appendices: | Equations | Variables | References | Ramblings | Images | myphys.lsu | ADS |