Far field intensity distribution

[gboff21]

Homework Statement

Describe the far-field intensity distribution if

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

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

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

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?

 

[member 553083]

The far field intensity should then be \frac{\textbf{I}_{0}}{\lambda^{2} z^{2}} |\frac{\pi D^{2}}{4} Jinc(\frac{k_{x} D}{2}) \otimes \delta ( k_{x}/2 ) + \delta ( - k_{x}/2 )|^{2} = \frac{\textbf{I}_{0}}{\lambda^{2} z^{2}} (\frac{\pi D^{2}}{2} Jinc(\frac{k_{x} D}{2}) )^{2} Is this right?