Past exam of the physics 2 paper of the University of Oxford / 2021 / 7

For a uniformly illuminated single slit of width $b$, summing the wave amplitudes across the slit gives the Fraunhofer diffraction profileThe first minimum occurs at $\sin \theta =\lambda /b$. The Rayleigh criterion places one source's central maximum at the other source's first minimum; hence the slit gives $\Delta \theta \simeq \lambda /b$. A circular aperture instead produces an Airy disk, whose first minimum is at $1.22\lambda /D$. For the camera's circular lens,The factor $1.22$ belongs to circular-aperture diffraction, rather than the single-slit profile.

The Rayleigh criterion gives a separation in the sensor plane $\delta =f\Delta \theta =2.44\mu \mathrm{m}$, using the $5\mathrm{mm}$ image distance. Counting one square pixel per such spatial resolution element givesThis is an estimate of the number of optically distinguishable elements. To sample all the spatial information, specify a signal sampling criterion as well. For incoherent imaging through an ideal circular aperture, the optical transfer function has cutoff spatial frequency ${\nu }_c=D/(\lambda f)$. The Nyquist–Shannon sampling theorem requires a pixel pitchA common intermediate rule, two pixels per Rayleigh separation, gives about $8.1$ million pixels; it is a heuristic and does not quite sample the full ideal optical transfer function bandwidth. Thus about $2$ million is the elementary spatial resolution element count, while $12$ million is the ideal monochrome Nyquist–Shannon sampling theorem answer to “full use.” Color-filter layout and finite pixel area would impose further practical conditions.