= Solution
The projection lens maps the mask's angular spectrum into its back focal plane: within scalar Fourier optics this is the \b[<Fraunhofer diffraction> or spatial-Fourier plane]. A periodic mask produces discrete diffraction orders, separated by $1/P$ in object-side spatial frequency after accounting for projection magnification. The finite <numerical aperture> bounds the transverse wave vectors admitted by the lens, so the pupil is a spatial-frequency filter.
For an ideal on-axis coherent system,
$$
\widetilde U_{\rm image}(\nu_x,\nu_y)=P(\nu_x,\nu_y)\widetilde U_{\rm mask}(\nu_x,\nu_y),
\qquad |\boldsymbol\nu|<\frac{\mathrm{NA}}{\lambda}.
$$
A second Fourier transformation forms the image, and the recorded intensity is $|U_{\rm image}|^2$. If only the zero order passes, the field is uniform and the line pattern disappears. Passing zero and first orders restores a periodic modulation; passing more orders sharpens the edges. Thus the loss of fine structure is explained by rejected diffraction orders, not by a geometrical inability to draw narrow lines. The field-transfer cutoff differs from the twice-as-large incoherent intensity-transfer cutoff.<Advanced Optical Imaging: Incoherent imaging by Delft University of Technology>[coherent and incoherent transfer: pupil filtering and intensity formation]{ref}
\Image[/past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2022/paper-1-grating-pupil.png]
{title=Coherent grating orders admitted or rejected by a finite pupil}
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