= Image distance for two separated thin lenses
{title2=$d_2=\frac{f_2[(L-f_1)d_1-Lf_1]}{(L-f_1-f_2)d_1+f_1f_2-Lf_1}$}
Use positive distances for a real incident object and a real outgoing image. The first <thin lens> makes an image at $v_1=f_1d_1/(d_1-f_1)$. The second has object distance $L-v_1$, possibly negative, and the <thin lens> equation gives the displayed expression. At $L=0$ it becomes $d_2=Fd_1/(d_1-F)$ with $F^{-1}=f_1^{-1}+f_2^{-1}$. For $d_1>0$, a real final image requires both $F>0$ and $d_1>F$; the inequality alone is insufficient when $F<0$.
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