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CGH: Consolidate Expressions Regarding Parallel Apertures

Whitworth's (1981) Isothermal Free-Energy Surface
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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 + \phi_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 + \phi_j]}\biggl[ \cos\biggl(\frac{2\pi y_1 Y_j}{\lambda L} \biggr) - i  \sin\biggl(\frac{2\pi y_1 Y_j}{\lambda L} \biggr) \biggr]
\, ,

where,

~L

~\equiv

~
Z \biggl[1 + \frac{y_1^2}{Z^2}  \biggr]^{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^{i 2\pi L/\lambda }  \sum_j  a_j e^{i \phi_j}   \cdot e^{-i \Theta_j } 
\, ,

where,

~\Theta_j

~\equiv

~\biggl(\frac{2\pi y_1 Y_j}{\lambda L} \biggr) \, .

Case 1

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

~a_j

~\rightarrow

~a_0(Y) dY = a_0(\Theta) \biggl[ \frac{w}{2\beta_1} \biggr] d\Theta \, ,

where,

~\frac{1}{\beta_1}

~\equiv

~\frac{\lambda L}{\pi y_1w}  \, ,

and changed the summation to an integration, obtaining,

~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 
\, .

If we assume that both ~a_0 and ~\phi 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 + \phi] }\biggl[ \frac{a_0 w}{2\beta_1} \biggr]  \int_{\Theta_2}^{\Theta_1} e^{-i \Theta } d\Theta

 

~=

~ 
e^{i [2\pi L/\lambda + \phi] } \cdot a_0 w ~\mathrm{sinc}(\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{max}.

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}^{\infty} \biggl( \frac{2}{n\pi} \biggr) \sin \biggl( \frac{n\pi c}{L} \biggr) \cos \biggl(\frac{n\pi x}{L}\biggr)

 

~=

~
\frac{2c}{L}\biggl\{\frac{1}{2} + \sum_{n=1}^{\infty} \mathrm{sinc} \biggl( \frac{n\pi c}{L} \biggr) \cos \biggl(\frac{n\pi x}{L}\biggr) \biggr\} \, .

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

Let's start by setting

~Y_j

~=

~\frac{j\cdot w}{(j_\mathrm{max}-1)} - \frac{w}{2} \, ,

for ~0 \le j \le (j_\mathrm{max}-1), in which case,

~\Theta_j

~\equiv

~
\frac{2\pi y_1}{\lambda L} \biggl[ \frac{j\cdot w}{(j_\mathrm{max}-1)} - \frac{w}{2} \biggr] 
= \frac{2\pi y_1}{\lambda L} \biggl[ \frac{j\cdot w}{(j_\mathrm{max}-1)} \biggr] - \frac{2\pi y_1}{\lambda L} \biggl[ \frac{w}{2} \biggr]

~=

~
j \biggl[ \frac{2\pi y_1 w}{(j_\mathrm{max}-1) \lambda L} \biggr] - \frac{\pi y_1 w }{\lambda L} 
= \biggl( \frac{2j}{j_\mathrm{max} - 1} - 1 \biggr) \frac{\pi y_1 w }{\lambda L}
\, ,

~=

~
j \cdot \Delta\Theta - \frac{(j_\mathrm{max} -1)}{2} \Delta\Theta
\, ,

where,

~\Delta\Theta \equiv  \frac{\pi y_1}{\mathfrak{L}}  \, ,     and     ~\mathfrak{L} \equiv \biggl[ \frac{(j_\mathrm{max}-1) \lambda L}{2w} \biggr] \, .

This means that ~\Theta_{i} = - \Theta_{( j_\mathrm{max} - 1 - i )}.

The key expression under the summation therefore becomes,

~a_j e^{i \phi_j}   \cdot e^{-i \Theta_j }

~=

~~a_j e^{i \phi_j}   \cdot \biggl[ \cos \biggl( \frac{j\pi y_1}{\mathfrak{L}}- \Theta_0 \biggr) - i \sin \biggl( \frac{j\pi y_1}{\mathfrak{L}}- \Theta_0  \biggr) \biggr] \, ,

where,

~\Theta_0 \equiv  \frac{(j_\mathrm{max} - 1)}{2} \cdot \pi y_1 \biggl[ \frac{2w}{(j_\mathrm{max}-1) \lambda L} \biggr] = \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 \biggl[ \frac{2w}{(j_\mathrm{max}-1) \lambda L} \biggr] \, .

Then, as ~j varies from ~0 to ~(j_\mathrm{max} - 1), the argument goes from ~0 to ~[2\pi w c/(\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 w c}{\lambda L}

~=

~
j \pi c \biggl[ \frac{2w}{(j_\mathrm{max}-1) \lambda L} \biggr]  - \frac{\pi w c}{\lambda L}
= \biggl[ \frac{2j}{j_\mathrm{max}-1}  - 1\biggr]\biggl[ \frac{\pi w c}{\lambda L} \biggr]  \, .

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

~a_j

~=

~
\mathrm{sinc} \biggl[ \biggl( \frac{2j}{j_\mathrm{max}-1}  - 1 \biggr) \frac{\pi w c}{\lambda L} \biggr] \, .

This therefore gives the following,

Focal-Point Expression for a Square Wave

~A(y_1)

~\approx

~ e^{i 2\pi L/\lambda }  \sum_{j=0}^{j_\mathrm{max}-1}  e^{i \phi_j}  \cdot~
\mathrm{sinc} \biggl[ \biggl( \frac{2j}{j_\mathrm{max}-1}  - 1 \biggr) \frac{\pi w c}{\lambda L} \biggr] 
\biggl\{ \cos \biggl[ \biggl( \frac{2j}{j_\mathrm{max} - 1} - 1 \biggr) \frac{\pi y_1 w }{\lambda L} \biggr] - i \sin \biggl[ \biggl( \frac{2j}{j_\mathrm{max} - 1} - 1 \biggr) \frac{\pi y_1 w }{\lambda L}  \biggr] 
\biggr\} \, .

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{max}-1)/2. As a result, if the overall phase factor, ~e^{i \phi_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{sinc} \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} \biggl[ j_\mathrm{max}-1 \biggr] \biggl( \frac{2j}{j_\mathrm{max}-1}  - 1 \biggr) \, ,

in which case we can write,

~A(y_1)

~\approx

~ e^{i 2\pi L/\lambda }  \sum_{n~=~-(j_\mathrm{max} - 1)/2}^{+(j_\mathrm{max} - 1)/2}  e^{i \phi_j}  \cdot~
\mathrm{sinc} \biggl[ \biggl( \frac{2n}{ j_\mathrm{max}-1} \biggr) \frac{\pi w c}{\lambda L} \biggr] 
\biggl\{ \cos \biggl[ \biggl( \frac{2n}{ j_\mathrm{max}-1} \biggr) \frac{\pi y_1 w }{\lambda L} \biggr] - i \sin \biggl[ \biggl( \frac{2n}{ j_\mathrm{max}-1} \biggr) \frac{\pi y_1 w }{\lambda L}  \biggr] 
\biggr\}

 

~=

~ e^{i 2\pi L/\lambda }   e^{i \phi_{j=0} }
~+~e^{i 2\pi L/\lambda }  \sum_{n~=~1}^{+(j_\mathrm{max} - 1)/2}  2e^{i \phi_j}  \cdot~
\mathrm{sinc} \biggl[ \biggl( \frac{2n}{ j_\mathrm{max}-1} \biggr) \frac{\pi w c}{\lambda L} \biggr] 
~ \cos \biggl[ \biggl( \frac{2n}{ j_\mathrm{max}-1} \biggr) \frac{\pi y_1 w }{\lambda L} \biggr]

 

~=

~ e^{i 2\pi L/\lambda }   e^{i \phi_{j=0} }
~+~e^{i 2\pi L/\lambda }  \sum_{n~=~1}^{+(j_\mathrm{max} - 1)/2}  2e^{i \phi_j}  \cdot~
\mathrm{sinc} \biggl(\frac{\pi n c}{\mathfrak{L} } \biggr) 
~ \cos \biggl(  \frac{n \pi y_1 }{\mathfrak{L} } \biggr)

 

~=

~ e^{i 2\pi L/\lambda } \biggl(\frac{\mathfrak{L}}{c} \biggr) \biggl\{   e^{i \phi_{j=0} }  \biggl(\frac{c}{\mathfrak{L}} \biggr)
~+~ \sum_{n~=~1}^{+(j_\mathrm{max} - 1)/2}  e^{i \phi_j}  \cdot~
\biggl(\frac{ 2  }{\pi n } \biggr) \sin \biggl(\frac{\pi n c}{\mathfrak{L} } \biggr) 
~ \cos \biggl(  \frac{n \pi y_1 }{\mathfrak{L} } \biggr)  \biggr\}
\, .

Finally, recalling that,

~L

~\equiv

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

let's set …

~e^{i\phi_j}

~=

~
e^{-i2\pi Z/\lambda}

~\Rightarrow ~~~ e^{i2\pi L/\lambda} \cdot e^{i\phi_j}

~=

~
e^{i2\pi (L-Z)/\lambda} \approx e^{i\pi y_1^2/(\lambda Z)} = \cos\biggl( \frac{\pi y_1^2}{\lambda Z} \biggr) + i \sin \biggl( \frac{\pi y_1^2}{\lambda Z} \biggr) \, .

As a result, we have,

~A(y_1)

~\approx

~ \biggl[ \cos\biggl( \frac{\pi y_1^2}{\lambda Z} \biggr) + i \sin \biggl( \frac{\pi y_1^2}{\lambda Z} \biggr) \biggr] \biggl(\frac{\mathfrak{L}}{c} \biggr) \biggl\{  \biggl(\frac{c}{\mathfrak{L}} \biggr)
~+~ \sum_{n~=~1}^{+(j_\mathrm{max} - 1)/2}  
\biggl(\frac{ 2  }{\pi n } \biggr) \sin \biggl(\frac{\pi n c}{\mathfrak{L} } \biggr) 
~ \cos \biggl(  \frac{n \pi y_1 }{\mathfrak{L} } \biggr)  \biggr\}
\, .

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

See Also


Whitworth's (1981) Isothermal Free-Energy Surface

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