Past exam of the physics 2 paper of the University of Oxford / 2022 / 7

Take coordinate $y$ across the slit and observation angle $\theta$ from its normal vector. Under Fraunhofer diffraction, the relative optical path difference from $y$ is $y\sin \theta$. Uniform coherent illumination makes the complex amplitude proportional to the aperture integral:The light intensity is proportional to $|\mathcal{E}{|}^2$, soThe central value is the continuous limit of the sinc function. The first minima satisfy $\sin \theta =\pm \lambda /w$, if $\lambda \le w$. Infinite slit height removes spreading in the other transverse direction; this expression describes the slit-width direction.

Strictly, distance $D\gg w$ alone does not ensure Fraunhofer diffraction: one also needs $w^2/(\lambda D)\ll 1$, so that quadratic aperture-dependent wave phase is negligible, or observation in a suitable focal plane. For example, $w=1\mathrm{mm}$, $\lambda =500\mathrm{nm}$ and $D=1\mathrm{m}$ have $D/w=1000$ but $w^2/(\lambda D)=2$; the Fresnel diffraction correction is then substantial. The displayed result uses the intended far-field regime of the question and the usual scalar aperture approximation.

For viewing near the nadir, the required angular resolution isUsing the Rayleigh criterion with the first minimum of the single-slit diffraction pattern gives $\delta \theta \simeq \lambda /w$. Thus the minimum collecting width in that model isThe numerical coefficient depends on the aperture shape and the chosen resolution criterion. If the mirror has an unobstructed circular aperture, the Airy disk version of the Rayleigh criterion is $\delta \theta =1.22\lambda /D$, giving $D_{\min }=0.12298\mathrm{m}\simeq 12.3\mathrm{cm}$. The question does not specify mirror shape; $10.1\mathrm{cm}$ follows the slit-width model of part (i).