= Thin lens
{title2=$1/u+1/v=1/f$}
{wiki}
A <thin lens> models an <optical lens> by collapsing its refracting surfaces into one plane, neglecting thickness relative to the relevant <radius of curvature> values and object and image distances. Here use an ideal converging <thin lens> in air and the <paraxial approximation>: <light rays> make small <angles> with the <optical axis>, so <Snell's law> can be linearized using the <small-angle approximation>. Equal surrounding <refractive indices> on both sides give the familiar symmetric equation.<MIT 2.71 and 2.710 Optics, Spring 2014 by Nicholas X. Fang/Lecture Notes on Geometrical Optics>[§§E–G]{ref}
Put the <thin lens> at $x=0$, with incident <light> travelling to the right. An object at $x=-u$ has $u>0$. The image coordinate is $x=v$: $v>0$ for a <real image> and $v<0$ for a <virtual image>. The converging <thin lens> has <focal length> $f>0$, and heights $h_o,h_i$ are signed relative to the <optical axis>. The <thin lens> equation and <transverse magnification> are
$$
\frac1f=\frac1u+\frac1v,\qquad v=\frac{fu}{u-f},\qquad m=\frac{h_i}{h_o}=-\frac vu.
$$
The expression for $v$ requires $u\ne f$. Negative <transverse magnification> means inverted; positive <transverse magnification> means upright. Remember that the positive distance $u$ differs in sign from the object's coordinate $-u$. https://openstax.org/books/university-physics-volume-3/pages/2-4-thin-lenses[Ling, Sanny and Moebs, OpenStax University Physics Volume 3, §2.4: Sign conventions]{ref}.
For <focal length> $f=1\,\mathrm m$, object distance $u=3\,\mathrm m$, and object height $h_o=0.1\,\mathrm m$, calculation gives
$$
\frac1v=\left(1-\frac13\right)\,\mathrm{m^{-1}}=\frac23\,\mathrm{m^{-1}},\qquad v=1.5\,\mathrm m,\qquad m=-0.5,\qquad h_i=-0.05\,\mathrm m.
$$
The <real image> is inverted, half as tall, and located $1.5\,\mathrm m$ to the right of the <thin lens>. A screen at that position receives the image because the outgoing <light rays> actually meet there.
\Image[/thin-lens-real-image.png]
{title=A converging <thin lens> forms a <real image>: $f=1\,\mathrm m$, $u=3\,\mathrm m$, $v=1.5\,\mathrm m$, $m=-0.5$}
{description=The object height is exactly $0.1\,\mathrm m$ and the image height is $-0.05\,\mathrm m$ in the ideal <thin lens> model. Both outgoing <light rays> travel to the right and intersect at the image tip. The vertical scale is enlarged for visibility; the drawing is a schematic of <geometrical optics>.}
For the same <focal length> $f=1\,\mathrm m$ and object height $h_o=0.1\,\mathrm m$, move the object inside the front <optical focus>, to $u=0.5\,\mathrm m$. Then
$$
\frac1v=(1-2)\,\mathrm{m^{-1}}=-1\,\mathrm{m^{-1}},\qquad v=-1\,\mathrm m,\qquad m=2,\qquad h_i=0.2\,\mathrm m.
$$
The <virtual image> is upright, twice as tall, and apparently $1\,\mathrm m$ to the left of the <thin lens>. Outgoing <light rays> still travel to the right; only their backward extensions intersect at the <virtual image>. A screen there cannot collect that image from the outgoing bundle. This geometry explains why a converging <optical lens> can act as a magnifier when held close to an object. https://openstax.org/books/college-physics-2e/pages/25-6-image-formation-by-lenses[Urone and Hinrichs, OpenStax College Physics 2e, §25.6: Virtual images inside the focus]{ref}.
\Image[/thin-lens-virtual-image.png]
{title=A converging <thin lens> forms a <virtual image>: $f=1\,\mathrm m$, $u=0.5\,\mathrm m$, $v=-1\,\mathrm m$, $m=2$}
{description=The object height is exactly $0.1\,\mathrm m$ and the image height is $0.2\,\mathrm m$ in the ideal <thin lens> model. Solid outgoing <light rays> have arrows to the right. Dashed backward extensions meet at the image tip; they are not transmitted <light rays>. The vertical scale is enlarged for visibility.}
Both diagrams use two rules for <ray tracing>. A <light ray> from the object's tip parallel to the <optical axis> reaches the <thin lens> at $(0,h_o)$ and exits along the <straight line> through the rear <optical focus> $(f,0)$. Another <light ray> through the centre $(0,0)$ continues undeviated in the ideal <thin lens> model. Their intersection, or the intersection of their backward extensions, locates the image tip. https://openstax.org/books/university-physics-volume-3/pages/2-4-thin-lenses[Ling, Sanny and Moebs, OpenStax University Physics Volume 3, §2.4: Ray tracing]{ref}.
A short geometric proof outline makes the signs transparent. The two outgoing <straight lines> are
$$
y_{\rm parallel}(x)=h_o\left(1-\frac xf\right),\qquad y_{\rm centre}(x)=-\frac{h_o}{u}x.
$$
For nonzero $h_o$, equating their heights at $x=v$ gives $1-v/f=-v/u$, hence the <thin lens> equation; the centre <light ray> gives $h_i/h_o=-v/u$, hence the <transverse magnification>. This derives the image equations from the ideal <ray tracing> rules, whose physical justification comes from <Snell's law> in the <paraxial approximation>.
At $u=f$, the two outgoing <straight lines> are parallel: there is no finite image position. For $u>f$ the <real image> has $m<0$; for $0<u<f$ the <virtual image> has $m>0$. These cases let an <optical camera> designer choose a receiving position and let a user understand a magnifier. An actual <optical lens> has finite <apertures>, which cause <diffraction> and impose <diffraction-limited resolution>; <optical aberrations> can further blur the image. The diagrams are ideal <geometrical optics>, with exact coordinates within that model, rather than exact predictions of <Maxwell's equations>. https://www.feynmanlectures.caltech.edu/I_27.html[Feynman, Leighton and Sands, The Feynman Lectures on Physics I, §§27-6–27-7: Aberrations and resolving power]{ref}.
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