= Solution
The requested source-image relation is the <thin lens> equation, using positive real-object distance $u$ and signed image distance $v$:
$$
\boxed{\frac1f=\frac1u+\frac1v.}
$$
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.<University Physics Volume 3 by Samuel J. Ling and others>[§2.4: thin-lens equation, sign conventions, and <lensmaker's equation>]{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 $v=-D_n$. Thus
$$
u=\frac{fD_n}{D_n+f}=\frac{100}{29}\,\mathrm{cm}\simeq3.45\,\mathrm{cm},\qquad
\boxed{M_{\max}=\frac{D_n}{u}=1+\frac{D_n}{f}=7.25.}
$$
The 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 $s\simeq0$ limit.<University Physics Volume 3 by Samuel J. Ling and others>[§2.7, equations 2.29–2.30: near-point <angular magnification> and lens-eye separation]{ref}
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2017-magnifier-near-point.png]
{title=Near-point magnification by a converging thin lens}
{description=Solid rays propagate towards the eye on the right. Dashed backward extensions locate the upright virtual image 25 cm to the left of the lens. The height scale is enlarged for readability.}
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