 Transverse magnification

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,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.^[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 $h_i=-h_{o}v/u$. This is the signed similar triangles argument. Combining it with the thin lens equation givesFor $f>0$, $u>2f$ 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.