= Solution
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 therefore
$$
y_f=y+u\theta+f\left(\theta-\frac{y+u\theta}{f}\right)=f\theta.
$$
Thus 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.
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