= Solution
Assume a converging <thin lens> with $f>0$, a real object, and a real image on the detector. For positive object and image distances $u,v$, the <thin lens> equation gives $uv=f(u+v)$. Put $S=u+v$. Then
$$
(u-v)^2=S^2-4uv=S(S-4f)\ge0.
$$
Since $S>0$, $S\ge4f$, with equality at $u=v=2f$. Therefore
$$
\boxed{S_{\min}=4f,\qquad u=v=2f.}
$$
Below this separation no real conjugate pair exists for one ideal thin lens; a diverging lens alone cannot form a real image of a real object on the detector.
Back to article page
