= Transverse magnification
{title2=$m=h_i/h_o$}
<Transverse magnification> is the ratio of corresponding signed lengths perpendicular to the <optical axis>. For an ideal image of a small arrow of nonzero height,
$$
m=\frac{h_i}{h_o}.
$$
Use the same upward-positive height convention in both planes. Thus an upright $0.50\,\mathrm{cm}$ arrow imaged as $-0.25\,\mathrm{cm}$ has $m=-1/2$: inverted and half as tall. The sign describes orientation; <real image> versus <virtual image> describes ray intersections. They are separate properties, even when a particular <thin lens> setup correlates them.<MIT 2.71 and 2.710 Optics, Spring 2014 by Nicholas X. Fang/Lecture Notes on Geometrical Optics>[§H: Magnification]{ref}
For a <thin lens> with object coordinate $-u$ and image coordinate $v$, the undeviated centre <light ray> is the <straight line> through $(-u,h_o)$ and $(0,0)$, so its height at $x=v$ is $h_i=-h_ov/u$. This is the signed <similar triangles> argument. Combining it with the <thin lens> equation gives
$$
m=-\frac vu=-\frac{f}{u-f}\qquad (u\ne f).
$$
For $f>0$, $u>2f$ gives $-1<m<0$; $u=2f$ gives $m=-1$; $f<u<2f$ gives $m<-1$; and $0<u<f$ gives $m>1$. At $u=f$, the image is conventionally “at infinity”: the divergent expressions for $v$ and $m$ describe the limiting behaviour, not a finite image of infinite height. Different object points produce parallel bundles in different directions, which is why <angular magnification> remains meaningful.
The <real image> experiment tests $m$ in two independent ways: signed ratios of measured corresponding intervals, and $-v/u$ from measured longitudinal distances. As a further prediction, moving from $u=30\,\mathrm{cm}$ to $u=15\,\mathrm{cm}$ for $f=10\,\mathrm{cm}$ swaps $u,v$, changing $m=-1/2$ to $m=-2$. The two image sizes differ by a factor of four, while both images remain inverted. This reversal follows from the symmetric <thin lens> equation and connects to the <Bessel lens displacement method>.
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