Let $u$ and $v$ be the real object and image distances from the <thin lens>. The object-screen separation gives $u+v=L$, and the <thin lens> equation gives $uv=fL$. Thus the two lens positions solve
$$
u(L-u)=fL,\qquad u_\pm=\frac{L\pm\sqrt{L^2-4fL}}2.
$$
Their separation is $d=u_+-u_-=\sqrt{L^2-4fL}$, so
$$
\boxed{f=\frac{L^2-d^2}{4L}.}
$$
This is the <Bessel lens displacement method> for measuring <focal length>. It uses the screen-object separation and the displacement between two sharp-image settings, reducing the need to locate the lens's effective optical center. It assumes the thin-lens or corresponding principal-plane approximation; $L>4f$ is what provides two distinct real-image settings.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 338 / 3 / a / Solution

{c}

For fixed object-screen separation $L>4f$, a converging <thin lens> focuses at two positions separated by $d=\sqrt{L^2-4Lf}$. Thus $f=(L^2-d^2)/(4L)$. The method finds <focal length> without needing to locate a thin lens's optical center precisely.

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 Incoming links: Bessel lens displacement method