= Angular magnification
{title2=$M_\theta=\theta_i/\theta_o$}
<Angular magnification> is the ratio of the signed viewing <angle> with an <optical instrument> to the signed viewing <angle> in a stated reference arrangement. Measure both as apparent directions from the eye, positive toward the same side of the <optical axis>; then an inverted view has negative $M_\theta$. If only positive angular sizes are compared, the result is $|M_\theta|$ and carries no orientation information. https://openstax.org/books/college-physics-2e/pages/26-5-telescopes[Urone and Hinrichs, College Physics 2e, §26.5: Telescope angular magnification and inversion]{ref}.
For the upright <virtual image> of a converging <thin lens>, let the eye be at $x=e\geq0$ and take the unaided reference to be the same small object at a distance $D$ from the eye. Within the <small-angle approximation>,
$$
\theta_i\simeq\frac{h_i}{e-v},\qquad \theta_o\simeq\frac{h_o}{D},\qquad M_\theta\simeq m\frac{D}{e-v}.
$$
Specify $D$: keeping the object in its original position means $D=e+u$, whereas the customary magnifier rating uses a near-point reference, often $D=25\,\mathrm{cm}$. A reference distance is a convention, not a claim that every eye has that near point. https://openstax.org/books/university-physics-volume-3/pages/2-7-the-simple-magnifier[Ling, Sanny and Moebs, University Physics Volume 3, §2.7: Angle ratio and viewing distances]{ref}.
For the <virtual image> diagram, $m=4$, $v=-30\,\mathrm{cm}$ and $e=5\,\mathrm{cm}$. With $D=25\,\mathrm{cm}$, $M_\theta\simeq20/7$, whereas keeping the object at $D=e+u=12.5\,\mathrm{cm}$ gives $M_\theta\simeq10/7$. Thus neither viewing comparison gives the <transverse magnification> four. A person may also be unable to focus on the unaided object at the shorter reference distance.
At $u=f$, each outgoing bundle is parallel with slope $-h_o/f$; its apparent viewing direction has <angle> $\theta_i\simeq h_o/f$. Consequently $M_\theta\simeq D/f$ remains finite even though $m$ is undefined at a finite image plane. Equivalently, in the limit $u\to f^-$, the factors $m$ and $e-v$ both grow without bound but their ratio tends to $1/f$. For $f=10\,\mathrm{cm}$ and $D=25\,\mathrm{cm}$, the relaxed-view prediction is $M_\theta\simeq2.5$. https://openstax.org/books/university-physics-volume-3/pages/2-7-the-simple-magnifier[Ling, Sanny and Moebs, §2.7: Image at infinity]{ref}.
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