Far field intensity distribution

AI Thread Summary
The discussion focuses on determining the far-field intensity distribution for a given function involving a circular aperture and a cosine modulation. The Fourier transform of the circular function is identified as a product involving the Jinc function, and the convolution theorem is applied to combine it with delta functions. The resulting far-field intensity expression incorporates the intensity scaling factor and the squared magnitude of the Fourier transform. The final expression suggests that the far-field intensity will feature two Jinc functions, indicating a specific distribution pattern. The correctness of these calculations and interpretations is questioned, prompting further validation.
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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?
 
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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?
 
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