= Bessel two focus positions
{c}
{title2=$x_\pm=(L\pm\sqrt{L^2-4Lf})/2$}
Place a converging <thin lens> at coordinate $x$ between the object at $0$ and the screen at $L$. In air and the <paraxial approximation>, the required object and image distances are $u=x$ and $v=L-x$. The <thin lens> equation gives a <quadratic equation>:
$$
\frac1f=\frac1x+\frac1{L-x}=\frac{L}{x(L-x)},\qquad x^2-Lx+Lf=0.
$$
Its <discriminant> is $L^2-4Lf$. For $L>4f>0$, the two roots and their separation are
$$
x_\pm=\frac{L\pm d}{2},\qquad d=x_+-x_-=\sqrt{L^2-4Lf},\qquad x_-+x_+=L.
$$
Both roots lie inside $(0,L)$, and both object and image distances exceed $f$. The two roots therefore describe finite <real images> on the screen. Exchanging the roots exchanges $u,v$; from <transverse magnification> $m=-v/u$,
$$
m_-=-\frac{L+d}{L-d},\qquad m_+=-\frac{L-d}{L+d},\qquad m_-m_+=1.
$$
The image-height ratio between the enlarged and reduced settings is $|m_-|/|m_+|=|m_-|^2$, rather than $|m_-|$.
At $L=4f$, the roots coincide: $d=0$, $x=L/2=2f$, $u=v=2f$ and $m=-1$. There is one setting, not two separately measurable settings. For $0<L<4f$ the <discriminant> is negative, so no position between the fixed object and screen gives a sharp <real image> there. This does not forbid a <real image> elsewhere if the screen is moved. Another short proof observes that $x(L-x)$ has maximum $L^2/4$ at $x=L/2$; the focusing requirement $x(L-x)=Lf$ is possible exactly when $L\ge4f$.
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