 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 magnificationsThe 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 magnificationThis 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 $D$ isFor 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.