Appendix/CGH/ParallelAperturesConsolidate

CGH: Consolidate Expressions Regarding Parallel Apertures

One-dimensional Apertures

From our accompanying discussion of the Utility of FFT Techniques, we start with the most general expression for the amplitude at one point on an image screen, namely,

$A(y_1)$

$=$

$\sum _j{a}_j{e}^{i(2\pi D_j/\lambda +{\varphi }_j)},$

and, assuming that $|Y_j/L|\ll 1$ for all $j$, deduce that,

$A(y_1)$

$\approx$

$\sum _j{a}_j{e}^{i[2\pi L/\lambda +{\varphi }_j]}[\cos (\frac{2\pi y_1{Y}_j}{\lambda L})-i\sin (\frac{2\pi y_1{Y}_j}{\lambda L}\left)\right],$

where,

$L$

$\equiv$

$Z[1+\frac{y_1^2}{Z^2}{]}^{1/2}.$

Note that $L$ is formally a function of $y_1$, but in most of what follows it will be reasonable to assume, $L\approx Z$. Notice, as well, that this last approximate expression for the (complex) amplitude at the image screen may be rewritten in the form that will be referred to as our,

Focal-Point Expression
$A(y_1)$	$\approx$	$e^{i2\pi L/\lambda }\sum _j{a}_j{e}^{i{\varphi }_j}\cdot e^{-i{\mathrm{\Theta }}_j},$

where,

${\mathrm{\Theta }}_j$

$\equiv$

$\left(\frac{2\pi y_1{Y}_j}{\lambda L}\right).$

Case 1

In a related accompanying derivation titled, Analytic Result, we made the substitution,

$a_j$

$\to$

$a_0(Y)dY=a_0(\mathrm{\Theta })\left[\frac{w}{2{\beta }_1}\right]d\mathrm{\Theta },$

where,

$\frac{1}{{\beta }_1}$

$\equiv$

$\frac{\lambda L}{\pi y_{1}w},$

and changed the summation to an integration, obtaining,

$A(y_1)$

$\approx$

$e^{i2\pi L/\lambda }\left[\frac{w}{2{\beta }_1}\right]\int a_0(\mathrm{\Theta })e^{i\varphi (\mathrm{\Theta })}\cdot e^{-i\mathrm{\Theta }}d\mathrm{\Theta }.$

If we assume that both $a_0$ and $\varphi$ are independent of position along the aperture, and that the aperture — and, hence the integration — extends from $Y_2=-w/2$ to $Y_1=+w/2$, we have shown that this last expression can be evaluated analytically to give,

$A(y_1)$	$\approx$	$e^{i[2\pi L/\lambda +\varphi ]}\left[\frac{a_{0}w}{2{\beta }_1}\right]\underset{{\mathrm{\Theta }}_2}{\overset{{\mathrm{\Theta }}_1}{\int }}{e}^{-i\mathrm{\Theta }}d\mathrm{\Theta }$
	$=$	$e^{i[2\pi L/\lambda +\varphi ]}\cdot a_{0}w\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}({\beta }_1).$

We need to explicitly demonstrate that an evaluation of our Focal-Point Expression with $a_j=1$, gives this last sinc-function expression, to within a multiplicative factor of, something like, $j_{\mathrm{m}\mathrm{a}\mathrm{x}}$.

Case 2

In our accompanying discussion of the Fourier Series, we have shown that a square wave can be constructed from the expression,

$f(x)$	$=$	$\frac{c}{L}+\sum _{n=1}^{\mathrm{\infty }}\left(\frac{2}{n\pi }\right)\sin \left(\frac{n\pi c}{L}\right)\cos \left(\frac{n\pi x}{L}\right)$
	$=$	$\frac{2c}{L}\{\frac{1}{2}+\sum _{n=1}^{\mathrm{\infty }}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}(\frac{n\pi c}{L})\cos (\frac{n\pi x}{L}\left)\right\}.$

Can we make this look like our above, Focal-Point Expression?

Let's start by setting

$Y_j$

$=$

$\frac{j\cdot w}{(j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1)}-\frac{w}{2},$

for $0\le j\le (j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1)$, in which case,

${\mathrm{\Theta }}_j$	$\equiv$	$\frac{2\pi y_1}{\lambda L}[\frac{j\cdot w}{(j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1)}-\frac{w}{2}]=\frac{2\pi y_1}{\lambda L}\left[\frac{j\cdot w}{(j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1)}\right]-\frac{2\pi y_1}{\lambda L}\left[\frac{w}{2}\right]$
	$=$	$j\left[\frac{2\pi y_{1}w}{(j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1)\lambda L}\right]-\frac{\pi y_{1}w}{\lambda L}=(\frac{2j}{j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1}-1)\frac{\pi y_{1}w}{\lambda L},$
	$=$	$j\cdot \mathrm{\Delta }\mathrm{\Theta }-\frac{(j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1)}{2}\mathrm{\Delta }\mathrm{\Theta },$

where,

$\mathrm{\Delta }\mathrm{\Theta }\equiv \frac{\pi y_1}{\mathfrak{𝔏}},$     and     $\mathfrak{𝔏}\equiv \left[\frac{(j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1)\lambda L}{2w}\right].$

This means that ${\mathrm{\Theta }}_i=-{\mathrm{\Theta }}_{(j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1-i)}$.

The key expression under the summation therefore becomes,

$a_j{e}^{i{\varphi }_j}\cdot e^{-i{\mathrm{\Theta }}_j}$

$=$

$a_j{e}^{i{\varphi }_j}\cdot [\cos (\frac{j\pi y_1}{\mathfrak{𝔏}}-{\mathrm{\Theta }}_0)-i\sin (\frac{j\pi y_1}{\mathfrak{𝔏}}-{\mathrm{\Theta }}_0\left)\right],$

where,

${\mathrm{\Theta }}_0\equiv \frac{(j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1)}{2}\cdot \pi y_1\left[\frac{2w}{(j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1)\lambda L}\right]=\frac{\pi y_{1}w}{\lambda L}.$

Now, what is the argument of the sinc function? By default, it needs to be something along the lines of,

$\frac{j\pi c}{\mathfrak{𝔏}}$

$=$

$j\pi c\left[\frac{2w}{(j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1)\lambda L}\right].$

Then, as $j$ varies from $0$ to $(j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1)$, the argument goes from $0$ to $[2\pi wc/(\lambda L)]$. In an effort to make the function exhibit reflection symmetry as we move from one side of the aperture to the next, let's subtract half of this upper limit; that is, let's modify the argument of the sinc function to read,

$\frac{j\pi c}{\mathfrak{𝔏}}-\frac{\pi wc}{\lambda L}$

$=$

$j\pi c\left[\frac{2w}{(j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1)\lambda L}\right]-\frac{\pi wc}{\lambda L}=[\frac{2j}{j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1}-1]\left[\frac{\pi wc}{\lambda L}\right].$

This means that in our above, Focal-Point Expression we want to set,

$a_j$

$=$

$\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}\left[\right(\frac{2j}{j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1}-1\left)\frac{\pi wc}{\lambda L}\right].$

This therefore gives the following,

Focal-Point Expression for a Square Wave
$A(y_1)$	$\approx$	$e^{i2\pi L/\lambda }\sum _{j=0}^{j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1}{e}^{i{\varphi }_j}\cdot \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}\left[\right(\frac{2j}{j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1}-1\left)\frac{\pi wc}{\lambda L}\right]\{\cos [(\frac{2j}{j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1}-1)\frac{\pi y_{1}w}{\lambda L}]-i\sin [(\frac{2j}{j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1}-1)\frac{\pi y_{1}w}{\lambda L}\left]\right\}.$

This exhibits a very desirable feature: Both the sinc function and the sine function — and, hence, also their product — have reflection symmetry about the summation index, $j=(j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1)/2$. As a result, if the overall phase factor, $e^{i{\varphi }_j}$, behaves in an appropriately simple way — for example, if it is zero everywhere — then under the summation the sine term will sum to zero and leave only the desired — and real — product, $\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}\times \cos$. Try this out in Excel to see if it works!

This could use a little more manipulation. Let's define the alternate summation index,

$n$

$\equiv$

$\frac{1}{2}[j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1](\frac{2j}{j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1}-1),$

in which case we can write,

$A(y_1)$	$\approx$	$e^{i2\pi L/\lambda }\sum _{n=-(j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1)/2}^{+(j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1)/2}{e}^{i{\varphi }_j}\cdot \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}\left[\right(\frac{2n}{j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1}\left)\frac{\pi wc}{\lambda L}\right]\{\cos [\left(\frac{2n}{j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1}\right)\frac{\pi y_{1}w}{\lambda L}]-i\sin [\left(\frac{2n}{j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1}\right)\frac{\pi y_{1}w}{\lambda L}\left]\right\}$
	$=$	$e^{i2\pi L/\lambda }{e}^{i{\varphi }_{j=0}}+e^{i2\pi L/\lambda }\sum _{n=1}^{+(j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1)/2}2e^{i{\varphi }_j}\cdot \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}\left[\right(\frac{2n}{j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1}\left)\frac{\pi wc}{\lambda L}\right]\cos \left[\right(\frac{2n}{j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1}\left)\frac{\pi y_{1}w}{\lambda L}\right]$
	$=$	$e^{i2\pi L/\lambda }{e}^{i{\varphi }_{j=0}}+e^{i2\pi L/\lambda }\sum _{n=1}^{+(j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1)/2}2e^{i{\varphi }_j}\cdot \mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}\left(\frac{\pi nc}{\mathfrak{𝔏}}\right)\cos \left(\frac{n\pi y_1}{\mathfrak{𝔏}}\right)$
	$=$	$e^{i2\pi L/\lambda }\left(\frac{\mathfrak{𝔏}}{c}\right)\left\{e^{i{\varphi }_{j=0}}\right(\frac{c}{\mathfrak{𝔏}})+\sum _{n=1}^{+(j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1)/2}{e}^{i{\varphi }_j}\cdot (\frac{2}{\pi n})\sin (\frac{\pi nc}{\mathfrak{𝔏}})\cos (\frac{n\pi y_1}{\mathfrak{𝔏}}\left)\right\}.$

Finally, recalling that,

$L$

$\equiv$

$Z[1+\frac{y_1^2}{Z^2}{]}^{1/2}\approx Z[1+\frac{1}{2}\frac{y_1^2}{Z^2}]=Z+\frac{y_1^2}{2Z},$

let's set …

$e^{i{\varphi }_j}$	$=$	$e^{-i2\pi Z/\lambda }$
$\Rightarrow e^{i2\pi L/\lambda }\cdot e^{i{\varphi }_j}$	$=$	$e^{i2\pi (L-Z)/\lambda }\approx e^{i\pi y_1^2/(\lambda Z)}=\cos \left(\frac{\pi y_1^2}{\lambda Z}\right)+i\sin \left(\frac{\pi y_1^2}{\lambda Z}\right).$

As a result, we have,

$A(y_1)$

$\approx$

$[\cos (\frac{\pi y_1^2}{\lambda Z})+i\sin (\frac{\pi y_1^2}{\lambda Z}\left)\right]\left(\frac{\mathfrak{𝔏}}{c}\right)\left\{\right(\frac{c}{\mathfrak{𝔏}})+\sum _{n=1}^{+(j_{\mathrm{m}\mathrm{a}\mathrm{x}}-1)/2}(\frac{2}{\pi n})\sin (\frac{\pi nc}{\mathfrak{𝔏}})\cos (\frac{n\pi y_1}{\mathfrak{𝔏}}\left)\right\}.$

Therefore, a clean square wave will appear only if $[\pi y_1^2/(\lambda Z)]\ll 1$.

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 ]
• Diffraction (Wikipedia)
• Various Google hits:
  • Single Slit Diffraction (University of Tennessee, Knoxville)
  • Diffraction from a Single Slit; Young's Experiment with Finite Slits (University of New South Wales, Sydney, Australia)
  • Single Slit Diffraction Pattern of Light (University of British Columbia, Canada)
  • Fraunhofer Single Slit (Georgia State University)