# Far field intensity distribution

1. Nov 24, 2012

### gboff21

1. The problem statement, all variables and given/known data
Describe the far-field intensity distribution if

$f(x',y') = circ(\frac{\rho}{D}) cos(\frac{k \theta x'}{2}$

where $\rho = \sqrt{x^{2} + y^{2}}$.

2. Relevant equations
The fourier transform of the circ function was given to us earlier:
$\textbf{F} [ circ(\frac{\rho}{D}) ] = \frac{\pi D^{2}}{4} Jinc(\frac{k_{\rho} D}{2})$
where Jinc is a bessel function

I know that the fourier transform of a product of two functions is the convolution of the two fourier transforms. (Convolution theorem)

The far field intensity is given by $\textbf{I}^{(z)} = \frac{\textbf{I}_{0}}{\lambda^{2} z^{2}} |\textbf{F}[f(x',y')]|^{2}$

Also $k_{x} = \frac{k x}{z} = k \theta$

3. The attempt at a solution

so the fourier transform reads: $\frac{\pi D^{2}}{4} Jinc(\frac{k_{\rho} D}{2}) \otimes \delta ( k_{x} - \frac{k \theta}{2} ) + \delta ( k_{x} + \frac{k \theta}{2} )$

then the $\delta ( k_{x} - \frac{k \theta}{2} )$ simplifies to $\delta (k_{x}/2 )$ and similar for the other one.
So there would be two bessel functions on the far field plane.
Is this right?

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