Past exam of the physics 2 paper of the University of Oxford / 2021 resit / 6 / b

Use the principle of dark-field microscopy: block the unscattered beam where it is concentrated, while transmitting most light scattering at nonzero angles. All direct light rays of the collimated illumination are parallel, so the thin lens focuses them to one point in its back focal plane. Put a small opaque circular stop centred on that point, one focal length behind the optical lens. In the real image geometry above this is $11.719\mathrm{mm}$ behind the optical lens, well before the real image at $187.5\mathrm{mm}$.

A light ray leaving object coordinate $y$ with a small initial direction $\theta$ reaches the optical lens at $y+u\theta$. The thin lens changes its direction to $\theta -(y+u\theta )/f$. At the back focal plane its coordinate is thereforeThus direct illumination with $\theta =0$ is concentrated at the centre, whereas scattered light at different directions is distributed away from it, independently of $y$. A stop of radius $r_s$ removes approximately $|\theta |\lesssim r_s/f$; choose it just large enough to cover the direct beam's finite diffraction spot, illumination divergence and alignment errors. Scattered light inside that small angular cone is also lost, so negligible loss requires most useful scattered light to lie outside it. A stop in the object's real image plane instead removes object detail and does not generally separate scattered and direct light. For the virtual image interpretation, the same physical back focal plane filtering requires a downstream observation arrangement that actually passes through this plane.