Magnification

Magnification compares an image's scale with a specified reference. Transverse magnification compares signed lengths in object and image planes; angular magnification compares viewing angles at an observer. Both are dimensionless quantities, but they answer different questions. A projected real image twice as tall has $|m|=2$; an optical telescope's advertised viewing enlargement normally concerns angular magnification. ^[ref].

Always specify the sign and the reference arrangement. For signed transverse magnification, $m<0$ means inverted and $m>0$ means upright; $|m|>1$ means enlarged and $0<|m|<1$ means reduced. A larger virtual image can subtend a modest viewing angle if it is farther away. At $u=f$, an ideal converging thin lens has no finite image height and hence no finite transverse magnification, while angular magnification can remain finite. Enlargement also does not guarantee extra detail: diffraction-limited resolution and optical aberrations limit the information available to enlarge. ^[ref].

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.

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. ^[ref].

For the upright virtual image of a converging thin lens, let the eye be at $x=e\ge 0$ and take the unaided reference to be the same small object at a distance $D$ from the eye. Within the small-angle approximation,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. ^[ref].

For the virtual image diagram, $m=4$, $v=-30\mathrm{cm}$ and $e=5\mathrm{cm}$. With $D=25\mathrm{cm}$, $M_{\theta }\simeq 20/7$, whereas keeping the object at $D=e+u=12.5\mathrm{cm}$ gives $M_{\theta }\simeq 10/7$. 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 $M_{\theta }\simeq 2.5$. ^[ref].