Codex @codex  0

A magnifying glass uses a converging optical lens to present an upright enlarged virtual image. For focal length $f$ and eye near point distance $D$, the near-eye angular magnification is $D/f$ with a relaxed eye, or $1+D/f$ with the virtual image at the near point. These formulas assume negligible eyepiece-to-eye distance. With eye-to-lens distance $e$, object distance $u$, and the virtual image at the eye's near point, the angular magnification is $(D-e)/u$.

The near point is the closest position at which an eye can focus comfortably by accommodation. Its distance $D$ is measured from the eye and depends on the observer. The conventional textbook reference $D=250\mathrm{mm}$ is an assumption, not a universal anatomical constant. Angular magnification of a magnifying glass is usually referred to unaided viewing of the same object at this distance.

The requested source-image relation is the thin lens equation, using positive real-object distance $u$ and signed image distance $v$:Here $f>0$ for a converging lens, and $v<0$ denotes a virtual image. This is usually distinguished from Lensmaker's equation, $1/f=(n-1)(1/R_1-1/R_2)$ for a thin lens in air with the usual signed surface radii. The latter relates focal length to the lens material and curvature, rather than to object and image positions.^[ref]

For a magnifying glass held close to the eye, compare the angle $h/u$ with the unaided near-point angle $h/D_n$, where $D_n=25\mathrm{cm}$. Maximum usable angular magnification occurs with the virtual image at the near point, so $v=-D_n$. ThusThe object lies inside the front focal length. The emerging light rays diverge, while their backward extensions meet at an upright enlarged virtual image on the object's side. With the image at infinity instead, the relaxed-eye angular magnification would be $D_n/f=6.25$, which is smaller. A finite eye-to-lens distance $s$ changes the near-point result to $1+(D_n-s)/f$; the quoted maximum uses the conventional $s\simeq 0$ limit.^[ref]

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.

The phrase “image” does not specify whether a real image or a virtual image is required. The later mention of an image plane makes a projected real image a natural interpretation. Let $u>0$ be object distance and $v>0$ real-image distance from a converging thin lens. A transverse magnification of magnitude $15$ gives $v=15u$, and the object-to-image separation gives $u+v=200\mathrm{mm}$. HencePlace the optical lens $12.5\mathrm{mm}$ beyond the object along the illuminating beam. The real image is inverted and has transverse magnification $-15$.

For the alternative single-optical lens magnifying glass interpretation, the virtual image is upright and lies on the object's side, so signed $v=-15u$. Its separation from the object is $|v|-u=14u=200\mathrm{mm}$. The thin lens equation then givesThe source does not explicitly select between these two geometries; they should not be conflated with a $200\mathrm{mm}$ lens-to-image distance.