= Solution
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.
Back to article page
