Past exam of the physics A course of the University of Cambridge / 2017 / Paper 2 / Section B / B8 / Why the given slit width removes the chosen order

For a slit of width $a$, the single-slit diffraction amplitude is proportional to $\mathrm{sinc}(\pi a\sin \theta /\lambda )$. At a grating maximum, $D\sin \theta =m\lambda$, so the envelope factor isThe specified $a=1\mu \mathrm{m}=D/2$ gives $ma/D=1$ at $m=2$. ThusThe grating maximum coincides exactly with the first single-slit minimum for each wavelength. The preceding width and detector calculations describe the ideal array factor, but this particular rectangular-aperture grating cannot deliver the assumed second-order signal. Pixel refinement cannot repair the missing optical peak. One must change the aperture duty cycle or grating profile, or use an available order and recalculate its resolving power.