= Solution
Let $x_i=L/f_i$, with positive $f$ focusing in the first transverse plane. The doublet matrix is $M=Q_2D(L)Q_1$, with $Q_i=\bigl(\begin{smallmatrix}1&0\\-1/f_i&1\end{smallmatrix}\bigr)$. Its lower-left entry is $C_x=-(x_1+x_2-x_1x_2)/L$. In the other plane both quadrupole signs reverse, giving $C_y=-(-x_1-x_2-x_1x_2)/L$. Thus positive effective focusing power in both planes requires
$$
\boxed{x_1x_2<-|x_1+x_2|.}
$$
For a parallel incident ray bundle, actual downstream focal distances are $\ell_x=L(1-x_1)/(x_1+x_2-x_1x_2)$ and $\ell_y=L(1+x_1)/(-x_1-x_2-x_1x_2)$. Requiring \i[both] foci downstream adds $|x_1|<1$. Positive effective power alone need not imply a <real image> after the final element. The original schematic below distinguishes these conditions.
\Image[/past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2016-doublet-and-brightness.png]
{title=Doublet focusing region and wavelength-dependent emittance benefit under fixed source optics}
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