= Solution
For the normally incident <plane wave>, the incident <complex amplitude> and <phase> are uniform across the aperture. Treat the field using scalar <diffraction>, with the aperture represented by its prescribed <amplitude transmittance> $h(x,y)$. Let $R$ be the distance from the aperture centre to the observation point and let $\ell_x,\ell_y$ be the transverse direction cosines of that point.
Across an aperture of characteristic size $a$, the <Fraunhofer diffraction> assumptions are that $a/R\ll1$, so $1/r\simeq1/R$ and the <obliquity factor> is nearly constant, and that the quadratic contribution to the propagation <phase> is negligible:
$$
\frac{ka^2}{2R}\ll1,\qquad\text{or, in order of magnitude,}\quad \frac{a^2}{\lambda R}\ll1.
$$
The remaining path-length dependence is linear:
$$
r\simeq R-\ell_xx-\ell_yy.
$$
A coherent <monochromatic wave> and an aperture large enough relative to <wavelength> for the scalar aperture model complete the usual physical assumptions. The <complex amplitude> becomes
$$
\boxed{\psi_P=C\iint h(x,y)e^{-ik(\ell_xx+\ell_yy)}\,dx\,dy,}
$$
where $C$ includes the common propagation <phase>, distance factor and nearly constant <obliquity factor>. For the central part of a planar screen in the <paraxial approximation>, $\ell_x\simeq X/R$ and $\ell_y\simeq Y/R$. Small observation angles are useful for that screen-coordinate formula; a general far-field <Fourier transform> can instead use the exact direction cosines. A lens can also produce this pattern in its focal plane without requiring a physically very distant screen.
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