= Solution
Use a converging <thin lens> as a <magnifying glass>, with the screen at or just inside its front <focal plane>. It presents an upright <virtual image> at infinity or at the eye's <near point>, respectively. Let the <near point> distance be $D$; the conventional numerical assumption is $D=250\,\mathrm{mm}$, which is not specified in the paper.
Under the usual approximation that the eye is close to the eyepiece, interpreting $50\,\mathrm{mm}$ as the <focal length> $f$ gives the familiar <angular magnifications>
$$
M_\infty=\frac Df=5,\qquad
\boxed{M_{\rm near}=1+\frac Df=6\quad(f=50\,\mathrm{mm},\ D=250\,\mathrm{mm}).}
$$
The first is for a relaxed eye with the screen at the front <focal plane>. For the second, the <virtual image> is at signed distance $v=-D$ and the <thin lens> equation gives screen distance $u=fD/(D+f)=41.7\,\mathrm{mm}$. Thus the often intended maximum $6$ requires a screen-to-eye distance of approximately $41.7\,\mathrm{mm}$, rather than exactly $50\,\mathrm{mm}$.
If $50\,\mathrm{mm}$ instead fixes the physical screen-to-eye separation $s$, an eye adjacent to the <optical lens> sees the <angular magnification>
$$
\boxed{M=\frac Ds=5\quad(s=50\,\mathrm{mm},\ D=250\,\mathrm{mm}).}
$$
This is achieved by $f=s$ for relaxed viewing, or by $f=sD/(D-s)=62.5\,\mathrm{mm}$ for a <virtual image> at the <near point>. The latter follows from $1/f=1/s-1/D$.
More generally, the source does not specify the eyepiece-to-eye distance either. If that distance is $e$, then $u=s-e$ and a <virtual image> at the <near point> has <transverse magnification> $(D-e)/u$. Its <angular magnification> relative to unaided viewing at $D$ is
$$
M=\frac{D-e}{s-e}.
$$
For example, $s=50\,\mathrm{mm}$ and $e=25\,\mathrm{mm}$ give $M=9$, with $u=25\,\mathrm{mm}$ and $f=28.125\,\mathrm{mm}$. This is a concrete counterexample to a universal maximum of $5$ at fixed screen-to-eye separation. In ideal <geometrical optics>, $e\to s^-$ allows arbitrarily large $M$ with an arbitrarily short <focal length>; practical <aperture> and <optical aberrations> constrain this. A unique numerical maximum therefore needs the conventional near-eye eyepiece geometry and <near point> assumption. The conventional $f=50\,\mathrm{mm}$ answer is $6$, with those assumptions stated explicitly.
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