= Solution
In <Fraunhofer diffraction>, the relative propagation <phase> across the <aperture> is linear in <aperture> coordinates. If $b$ is its largest transverse size, the neglected quadratic path term must be small: for an observer at distance $R$,
$$
\boxed{\frac{b^2}{\lambda R}\ll1}.
$$
For a finite-distance point source a distance $s$ away, also require $b^2/(\lambda s)\ll1$ if the illumination is to be treated as a <plane wave>. Numerical factors depend on whether $b$ is a radius or full width and on the permitted <phase> error; the condition expresses a small <Fresnel number>.
One can place both source and screen sufficiently far away, or collimate a point source with a <lens (optics)> and observe the <aperture> in the back focal plane of a second <lens (optics)>. In the latter arrangement rays of each outgoing direction focus at one point, with $X\simeq f\theta_x$; the lens removes the need for a physically distant screen. A laser beam approximately uniform across the <aperture> can supply the incident <plane wave>. The sketches distinguish an actual far field from its focal-plane equivalent.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2018/paper-2-b7-fraunhofer-geometry.png]
{title=Far-field diffraction geometry and the lens arrangement that produces the same angular pattern}
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