Past exam of the physics 2 paper of the University of Oxford / 2017 / 10

In Fraunhofer diffraction, the source and observation geometry make the incident and outgoing wavefronts effectively planar across the aperture. For direct propagation over distance $z$, the quadratic variation in propagation phase across a slit of width $w$ must be negligible, approximately $w^2/(\lambda z)\ll 1$. The far-field angular amplitude is then the Fourier transform of the aperture's complex illumination. A converging lens also produces this angular distribution in its rear focal plane for collimated illumination, without requiring a physically enormous observation distance. The approximation concerns the aperture-dependent phase, not the absence of diffraction.

Let the slit run over $-w/2<x<w/2$ and let its coherent illumination have constant complex amplitude $A_0$. In a direction making angle $\theta$ to the normal, the relative propagation phase from position $x$ is $-kx\sin \theta$, with $k=2\pi /\lambda$. Summing the secondary amplitudes givesThe intensity is proportional to the squared modulus, hence within the usual scalar Fraunhofer diffraction approximationThe central value is defined by the continuous limit $\sin \beta /\beta \to 1$. The nonzero minima satisfy $w\sin \theta =m\lambda$ for integer $m\ne 0$. This expression neglects the slowly varying obliquity factor and uses a uniform slit; it describes the narrow-angle diffraction pattern for $w\gg \lambda$.

The Rayleigh criterion declares two equal incoherent point images just resolved when the central maximum of one lies at the first minimum of the other's point spread function. For a single rectangular slit, the first minimum occurs at $w\sin {\theta }_1=\lambda$. ThusThis is a resolution in the direction across the slit width, and the last approximation requires $\lambda \ll w$. The central maximum's full width between its two first minima is approximately $2\lambda /w$; it is not the Rayleigh criterion's separation. Two coherent sources can interfere with one another, so a resolution statement for them also needs their relative phase.

A feature of transverse size $s=0.03\mathrm{m}$ at height $H=2.0\times 10^5\mathrm{m}$ subtendsChoose a representative visible wavelength $\lambda =550\mathrm{nm}$. For an unobstructed circular mirror, the first Airy disk minimum gives $\Delta {\theta }_R=1.22\lambda /D$, soUsing the single-slit scaling alone would give $D\sim \lambda H/s=3.67\mathrm{m}$; it gives the same metre-scale estimate but omits the circular-aperture factor. Choosing visible wavelengths between $400$ and $700\mathrm{nm}$ changes the circular-aperture estimate to roughly $3.3$–$5.7\mathrm{m}$. These are diffraction-only estimates, not a claim that resolving one stroke guarantees reading arbitrary text with limited contrast or noise.

At focal length $f=80\mathrm{m}$, the stroke maps to a focal-plane width $w_i=fs/H=12\mu \mathrm{m}$. A practical estimate with two detector samples across that stroke isThe required sensor extent in that direction is $f(500\mathrm{m})/H=0.20\mathrm{m}$. One $12\mu \mathrm{m}$ pixel per stroke would give about $1.67\times 10^4$ pixels across, but cannot reliably encode a stroke's position and contrast as the image moves relative to the pixel grid.

For a precise interpretation of “fully resolve” as sampling the entire diffraction-limited incoherent image, use the optical intensity-transfer cutoff ${\nu }_c=D/(\lambda f)$. The Nyquist–Shannon sampling theorem requires $1/(2p)\ge {\nu }_c$. With the circular mirror chosen in 10d,This stricter value retains the full circular-aperture spatial bandwidth; the $6\mu \mathrm{m}$ and $3.3\times 10^4$ estimates use two samples per specified feature instead. An engineering design would provide further sampling and contrast margin. The distinction states the sampling convention rather than treating the order-of-magnitude estimates as contradictory.