 Solution

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]