Image Formation and Magnification

Point a lens at a candle and slide a card behind it until the flame snaps into focus: you have just built an image. Every object is a field of point sources, each spraying a cone of rays, and an imaging element — a lens, a mirror, a single curved surface — is a machine for gathering each cone back toward a single point. Where the cones physically reconverge, a real image hangs in space and can be caught on a screen, a sensor, or the retina. Where the rays only appear to diverge from a point behind the element, a virtual image floats, seen by the eye but impossible to project. The bookkeeping of this reconvergence is astonishingly compact. One conjugate equation, $1/s_o+1/s_i=1/f$ , says where the image lands; one magnification, $M_T=-s_i/s_o$ , says how big and which way up. But an object is three-dimensional, and the third dimension does not scale the way the first two do. Depth is stretched by the square of the transverse scale and its order is reversed — the longitudinal magnification is $M_L=-M_T^2$ — which is exactly why a magnifier makes a beetle look grotesquely deep, and why a solid cube never images to a similar cube. A third quantity, the angular magnification, ties the bundle together: gain size and you lose ray-angle, in the strict ratio $M_T M_\alpha=1$ . Those three magnifications are not independent decorations on the conjugate equation. They descend from a single conserved quantity of paraxial optics — the Lagrange invariant $n y u$ — so that image formation is, at bottom, one linear map with one conservation law. Fix a sign convention and hold it, and that map builds the camera, the eye, the microscope, and the telescope from the same short line of algebra.

Key takeaways

An image forms where the rays from an object point are made to reconverge (real, projectable) or where they only appear to diverge from (virtual, not projectable); ray tracing tells the two apart.
The conjugate equation $1/s_o+1/s_i=1/f$ locates the image; the transverse magnification $M_T=-s_i/s_o$ fixes its height and orientation (negative = inverted).
Newton's form $x_o x_i=f^2$ (distances measured from the focal points) gives the same imaging with $M_T=-f/x_o=-x_i/f$ .
Longitudinal magnification $M_L=-M_T^2$ : depth scales as the square of width and reverses order — 3-D objects always image distorted in depth.
Angular magnification $M_\alpha=1/M_T$ , so $M_T M_\alpha=1$ ; all three follow from the conserved Lagrange invariant $n y u$ .
Key idea: every object point maps to one conjugate image point by the linear paraxial relation $1/s_o+1/s_i=1/f$ ; the single transverse magnification $M_T=-s_i/s_o$ then fixes lateral size and orientation, while depth ( $M_L=-M_T^2$ ) and ray-angle ( $M_\alpha=1/M_T$ ) follow from the invariant $nyu$ .

The core insight

Every object point maps to a unique conjugate image point through the linear paraxial relation $1/s_o+1/s_i=1/f$ . The transverse magnification $M_T=-s_i/s_o$ fixes the image's height and orientation; the longitudinal magnification $M_L=-M_T^2$ scales depth (and reverses it); the angular magnification $M_\alpha=1/M_T$ scales ray angles. All three are consequences of the Lagrange (Smith–Helmholtz) invariant $nyu=\text{const}$ .

Where this lives today

The conjugate equation and its three magnifications are quietly inside every imaging device you own. A camera autofocus motor solves $1/s_o+1/s_i=1/f$ in real time, moving the lens so a distant scene ( $s_i\to f$ ) or a close macro subject lands sharply on the sensor; the "focus breathing" videographers dislike is the focal length shifting as the elements move. The human eye does the same with a deformable lens, forming the real inverted retinal image Kepler first described, at a fixed image distance while the power changes to accommodate. Longitudinal magnification is a working nuisance in microscopy and 3-D metrology: confocal and light-sheet microscopes must calibrate the $M_L=-M_T^2$ depth stretch to reconstruct true specimen shape, and stereo-photogrammetry software carries an explicit vertical-exaggeration factor for exactly this reason. And the reciprocity $M_T M_\alpha=1$ — étendue conservation in disguise — sets a hard limit that no lens can beat: you cannot simultaneously image a wide field and a wide aperture without paying in one to gain in the other, the constraint that shapes every telescope and endoscope design.

The physical picture

We use the Hecht sign convention throughout: light travels left to right; the object distance $s_o>0$ for a real object left of the element, the image distance $s_i>0$ for a real image on the outgoing side, and $f>0$ for a converging element. A negative $s_i$ signals a virtual image, a negative $M_T$ an inverted one. Picture a point on the object tip spraying a fan of rays. The lens adds a fixed dose of convergence, and the fan re-crosses at the image tip. You never need the whole fan: three rays whose paths are known by definition pin the image down — the ray that arrives parallel to the axis and leaves through the back focus $F'$ , the ray straight through the lens centre, and the ray through the front focus $F$ that leaves parallel. Their crossing is the image.

Figure

A converging lens (f = 10 cm) images an object at s_o = 15 cm to a real inverted image at s_i = 30 cm; the three construction rays cross at the image tip and M_T = −s_i/s_o = −2

Magnification is really a statement about how the ray bundle is squeezed. If the image cone reconverges over a shorter run than the object cone diverged, its rays are steeper and its height smaller — angular gain bought with transverse loss. Depth is the odd one out: because $M_L=-M_T^2$ , a lens that doubles heights quadruples depths and swaps near for far. Slide the object toward the front focus and the story runs to its two limits — image size runs off to infinity as $s_o\to f$ , and flips from inverted-real to erect-virtual as the object crosses inside the focus.

Figure

Punchline — transverse magnification M_T = −f/(s_o−f) for f = 10 cm: the real branch (s_o > f, inverted, M_T < 0) and the virtual branch (s_o < f, erect, M_T > 0) diverge at s_o = f where the image runs to infinity; a real M_T = −2 at 15 cm and a virtual M_T = +2 at 5 cm are marked

The curve is the whole story of a single converging lens. To the right of the asymptote at $s_o=f$ lies the real, inverted world of cameras and projectors — the magnification is negative, its size set by how far past the focus the object sits. To the left lies the virtual, erect world of the magnifying glass, magnification positive and unbounded near the focus. At $s_o=2f$ the two conjugates are symmetric and $M_T=-1$ ; push to infinity and the image shrinks into the focal plane.

The core equations

Everything begins with the conjugate relation for a thin lens (or, with the right power, any paraxial element) in a single medium. Two or three sentences of context: the reciprocal distances of object and image add to the reciprocal focal length, so as the object approaches the focus the image recedes, and the sign of $s_i$ decides real versus virtual.

\frac{1}{s_o}+\frac{1}{s_i}=\frac{1}{f} \qquad (1)

where: $s_o>0$ — object distance (real object, left of the element); $s_i$ — image distance, $s_i>0$ real (outgoing side), $s_i<0$ virtual; $f>0$ — converging focal length. Solving, $s_i=\dfrac{f s_o}{s_o-f}$ : for $s_o>f$ the image is real, for $s_o<f$ virtual. The transverse magnification compares image height to object height and follows from the undeviated centre ray:

M_T=\frac{y_i}{y_o}=-\frac{s_i}{s_o} \qquad (2)

where: $y_o,y_i$ — object and image heights; $M_T<0$ inverted, $M_T>0$ erect; $|M_T|>1$ enlarged. Substituting $s_i=fs_o/(s_o-f)$ gives the compact form $M_T=-\dfrac{f}{s_o-f}$ plotted above. Measuring distances not from the lens but from the two focal points gives Newton's equations, often the fastest route to a number:

x_o\,x_i=f^2,\qquad M_T=-\frac{f}{x_o}=-\frac{x_i}{f} \qquad (3)

where: $x_o=s_o-f$ — object distance from the front focus $F$ ; $x_i=s_i-f$ — image distance from the back focus $F'$ . The product $x_o x_i=f^2$ is exactly equivalent to $(1)$ ; both signs of $x_o$ (object outside or inside the focus) are allowed. Depth does not scale like height. Differentiating $(1)$ at fixed $f$ gives the longitudinal magnification of a small on-axis element:

M_L=\frac{\mathrm{d}s_i}{\mathrm{d}s_o}=-M_T^2 \qquad (4)

where: $M_L$ — axial magnification (single medium). The minus sign is depth reversal: the near face of a solid object becomes the far face of its image. Because $M_L=-M_T^2$ , a transverse gain of $10$ is an axial gain of $100$ . Finally the angular magnification and the invariant that ties the three together:

M_\alpha=\frac{u_i}{u_o}=\frac{1}{M_T},\qquad n\,y\,u=n'\,y'\,u' \qquad (5)

where: $u_o,u_i$ — marginal ray angles in object and image space; $nyu$ — the Lagrange (Smith–Helmholtz) invariant, conserved through any paraxial system. In one medium $M_T M_\alpha=1$ : transverse and angular magnification are strict reciprocals, so no imaging system can enlarge both field and aperture without limit — the paraxial ancestor of étendue conservation.

Derivation ATransverse and longitudinal magnification from the conjugate relation

This card earns the two magnifications that do the day-to-day work of imaging, and exposes why depth behaves so differently from width. The strategy is geometric for the transverse case — the undeviated centre ray makes two similar triangles whose height ratio is the magnification — and analytic for the longitudinal case, where differentiating the conjugate equation at fixed focal length turns a ratio of small displacements into $-M_T^2$ . The sign that survives each step is the whole point: it labels inverted versus erect, and near-face versus far-face.

Step 1. Send the ray from the object tip (height $y_o$ , distance $s_o$ ) straight through the lens centre; a thin lens does not deviate it. On the object side it drops from $y_o$ to the axis over a run $s_o$ ; on the image side it climbs from the axis to $y_i$ over a run $s_i$ . Why the centre ray: the two lens faces are parallel and close at the axis, so this ray emerges undeflected — its slope is a pure geometric fact.

Step 2. The two right triangles share the centre-ray slope, so $y_o/s_o=|y_i|/s_i$ . Restoring the convention (a real image sits below the axis for an erect object), $M_T=y_i/y_o=-s_i/s_o$ . Why the minus sign: on the far side the ray has crossed the axis, so the image tip lies on the opposite side — a real image is inverted.

Step 3. For depth, treat a short axial object of thickness $\mathrm{d}s_o$ imaged to $\mathrm{d}s_i$ . Differentiate $1/s_o+1/s_i=1/f$ with $f$ fixed: $-\mathrm{d}s_o/s_o^2-\mathrm{d}s_i/s_i^2=0$ , so $\mathrm{d}s_i/\mathrm{d}s_o=-s_i^2/s_o^2$ . Why differentiate: longitudinal magnification is by definition the ratio of conjugate axial displacements, which is a derivative of the imaging map.

Step 4. But $s_i^2/s_o^2=M_T^2$ , so $M_L=-M_T^2$ . The magnitude is the square of the transverse scale; the sign is negative, meaning axial order reverses. For $M_T=-2$ this is $M_L=-4$ : a $1\,\mathrm{mm}$ deep object images $4\,\mathrm{mm}$ deep, front-to-back flipped. Why it matters: solid objects can never image to similar solids — microscopy and photogrammetry must correct this built-in depth distortion. *Verified in verify_image-formation-magnification.py: $M_T=-s_i/s_o$ and $M_L=-M_T^2$ as sympy identities; the real case ( $f=10$ , $s_o=15\Rightarrow s_i=30$ , $M_T=-2$ , $M_L=-4$ ) and the virtual case ( $s_o=5\Rightarrow s_i=-10$ , $M_T=+2$ ).*

Go deeper · Derivation B · 1 minNewton's form and the equivalence x_o x_i = f²

Derivation BNewton's form and the equivalence x_o x_i = f²

Newton's equations are the same imaging law wearing different coordinates, and they are often the quickest path to an answer because the product $x_o x_i=f^2$ is symmetric and sign-transparent. This card derives them from the Gaussian relation and shows the two magnification formulas $M_T=-f/x_o=-x_i/f$ fall straight out, then confirms they reproduce both a real and a virtual conjugate without special-casing the algebra. It is also the cleanest way to see longitudinal magnification: with $x_i=f^2/x_o$ , depth scaling is one derivative away.

Step 1. Measure the object from the front focus, $x_o=s_o-f$ , and the image from the back focus, $x_i=s_i-f$ . Substitute $s_o=x_o+f$ , $s_i=x_i+f$ into $1/s_o+1/s_i=1/f$ . Why shift the origin: the focal points are the natural landmarks of a lens, and referencing them symmetrises the equation.

Step 2. Clearing denominators, $(x_i+f)f+(x_o+f)f=(x_o+f)(x_i+f)$ . Expanding and cancelling the $f^2$ and cross terms leaves $x_o x_i=f^2$ . Why so clean: the Gaussian relation is a rectangular hyperbola in $(s_o,s_i)$ ; shifting to the focal points centres it, and a centred rectangular hyperbola is simply $x_o x_i=\text{const}$ .

Step 3. From $M_T=-s_i/s_o=-(x_i+f)/(x_o+f)$ and $x_o x_i=f^2$ , algebra collapses this to $M_T=-f/x_o=-x_i/f$ . Why two equal forms: $x_o x_i=f^2$ means $x_i/f=f/x_o$ , so the image-side and object-side magnification expressions are identically equal.

Step 4. Test both conjugates for $f=10\,\mathrm{cm}$ . Real: $s_o=15\Rightarrow x_o=5$ , $x_i=f^2/x_o=20$ , $s_i=30$ , $M_T=-f/x_o=-2$ . Virtual: $s_o=5\Rightarrow x_o=-5$ , $x_i=-20$ , $s_i=-10$ , $M_T=+2$ . Why this matters: the same formula, with no case split, delivers the real inverted image and the virtual erect magnifier — a sign-safe route students can trust. *Verified in verify_image-formation-magnification.py: $x_o x_i-f^2=0$ symbolically; $M_T=-f/x_o=-x_i/f$ ; both conjugates ( $100=5\cdot20$ and $100=(-5)(-20)$ ) reproduce $s_i$ and $M_T$ .*

Limiting cases

Case	Result	Meaning
Object at the focus, $s_o=f$	$s_i\to\infty$ , $M_T\to\infty$	rays emerge parallel; the image recedes to infinity — the collimator
Distant object, $s_o\to\infty$	$s_i\to f$ , $M_T\to 0$	image forms in the focal plane, shrunk — a camera focused at infinity
Symmetric conjugates, $s_o=s_i=2f$	$M_T=-1$	same-size inverted real image; minimum object-to-image separation, $4f$
Inside the focus, $s_o<f$	$s_i<0$ , $M_T>+1$	virtual, erect, enlarged — the simple magnifier
Plane mirror / plane interface, $R\to\infty$	$M_T=+1$ , $M_L=-1$	erect same-size virtual image, front–back depth reversed
Across media, $n\neq n'$	$M_T M_\alpha=n/n'$ , $M_L=(n'/n)M_T^2$	the invariant $nyu$ sets the reciprocity; equal only for one medium

Common trap“The magnification is one number, so the image is just a scaled copy of the object”

It is not — depth betrays the copy. Transverse magnification $M_T$ scales the two dimensions across the axis, but the dimension along the axis scales by $M_L=-M_T^2$ , a different number and a different sign. A lens that renders a face at $M_T=-3$ (three times tall, inverted) renders its nose-to-ear depth at $M_L=-9$ : nine times deep and turned inside out, near becoming far. This is why high-magnification microscope images of thick specimens look impossibly deep, why stereo photogrammetry carries a "vertical exaggeration" factor, and why a solid object never images to a similar solid. Say "the transverse magnification is $M_T$ , and the depth magnification is $-M_T^2$ ," and you keep the whole object honest.

Worked examples

1 · A real, inverted image (the projector case)

A thin converging lens of focal length $f=10\ \mathrm{cm}$ has an object $s_o=15\ \mathrm{cm}$ in front of it. Locate the image and give its size, orientation, and depth scaling.

Step 1. Conjugate equation: $\dfrac{1}{s_i}=\dfrac{1}{f}-\dfrac{1}{s_o}=\dfrac{1}{10}-\dfrac{1}{15}=\dfrac{1}{30}$ , so $\boxed{s_i=+30\ \mathrm{cm}}$ — positive, hence a real image on the far side.

Step 2. Transverse magnification: $M_T=-\dfrac{s_i}{s_o}=-\dfrac{30}{15}=\boxed{-2}$ — twice the height and inverted (the negative sign). Newton's form agrees: $x_o=5$ , $x_i=f^2/x_o=20$ , $M_T=-f/x_o=-2$ .

Step 3. Depth: $M_L=-M_T^2=\boxed{-4}$ . A specimen $1\ \mathrm{mm}$ thick along the axis images $4\ \mathrm{mm}$ deep with its near and far faces swapped.

Step 4. Angles: $M_\alpha=1/M_T=-0.5$ , so the ray cone is steeper by two in the image space — and $M_T M_\alpha=(-2)(-0.5)=1$ , the Lagrange invariant honoured. This is exactly the geometry of a slide projector throwing an enlarged inverted image on a wall.

2 · A virtual, erect image (the magnifying glass)

Same lens, but now the object sits inside the focus at $s_o=5\ \mathrm{cm}$ . Where is the image, and what does the eye see?

Step 1. $\dfrac{1}{s_i}=\dfrac{1}{10}-\dfrac{1}{5}=-\dfrac{1}{10}$ , so $\boxed{s_i=-10\ \mathrm{cm}}$ — negative, a virtual image $10\ \mathrm{cm}$ in front of the lens, on the same side as the object.

Step 2. $M_T=-\dfrac{s_i}{s_o}=-\dfrac{-10}{5}=\boxed{+2}$ — erect and doubled. This is the working regime of a magnifier: the object must sit inside the focal length to give an enlarged, upright virtual image the eye can view.

Step 3. It cannot be caught on a screen. The real rays leaving the lens still diverge; only their backward extensions meet at $s_i=-10\ \mathrm{cm}$ , so no energy concentrates there. An eye placed beyond the lens refocuses those diverging rays onto its retina and perceives the enlarged image.

Step 4. Depth again squares and flips: $M_L=-M_T^2=-4$ . Even the virtual, erect image distorts depth by four and reverses it — a reminder that $M_L=-M_T^2$ holds regardless of whether the image is real or virtual.

Physics of the portrait

Before Kepler, no one could say where an image was. The ancients and the medieval perspectivists argued about how the eye "sees," but it was Kepler who first treated the eye as an optical instrument that forms a real, inverted picture on a screen at the back — the retina — and who worked out how lenses build images point by point.

Origins · 1 minJohannes Kepler and how this was discovered

Johannes Kepler

1571–1630 · Graz, Prague & Linz

Kepler is remembered for planetary orbits, but he founded modern geometrical optics almost as a side project. In Ad Vitellionem Paralipomena (1604), written while he was imperial mathematician in Prague, he correctly explained the pinhole image and — decisively — argued that the eye forms a real, inverted image on the retina, the picture the brain later re-inverts; he cheerfully accepted the paradox that we see the world upside-down on our retinas. In Dioptrice (1611) he gave the first systematic theory of refraction by lenses, worked out real and virtual images, and designed the "astronomical" (Keplerian) telescope of two convex lenses, which inverts the image but gives a wider field than Galileo's. He never had an accurate law of refraction — Snell's sine law came a decade later — yet by reasoning paraxially he got the geometry of image formation essentially right, and every conjugate equation in this article is the harvest of the program he began.

History

1604

— Kepler, in Ad Vitellionem Paralipomena, explains the pinhole image and the real inverted retinal image

1611

— Kepler's Dioptrice gives the first theory of image formation by lenses and the Keplerian telescope

1637

— Descartes, in La Dioptrique, analyses image formation through refracting surfaces with the sine law

1841

— Gauss, in Dioptrische Untersuchungen, founds first-order (paraxial) imaging on the cardinal points

Going further

Texts. Hecht, Optics, 5th ed., §5.2–5.3 (imaging, conjugate and magnification relations) · Saleh & Teich, Fundamentals of Photonics, 3rd ed., Ch. 1 (Ray Optics) · Born & Wolf, Principles of Optics, §4.4 (geometrical theory of imaging, the Lagrange invariant) · Jenkins & White, Fundamentals of Optics, Ch. 4–5.

Provenance. Definitions and derivations follow Hecht §5.2–5.3 (Eq. 5.26 conjugates, 5.36 longitudinal magnification) and Born & Wolf §4.4 (Smith–Helmholtz invariant). The base article stated the conjugate relation in the Cartesian convention ( $s_o<0$ for a real object); it is recast here into the Hecht real-positive convention ( $s_o>0$ , $1/s_o+1/s_i=1/f$ , $M_T=-s_i/s_o$ ) for consistency with the rest of this track, with no change to any physical result. verify_image-formation-magnification.py reproduces: $M_T=-s_i/s_o$ , Newton $x_o x_i=f^2$ , $M_L=-M_T^2$ (sympy); the real conjugate ( $f=10$ , $s_o=15\Rightarrow s_i=30$ , $M_T=-2$ , $M_L=-4$ ) and the virtual conjugate ( $s_o=5\Rightarrow s_i=-10$ , $M_T=+2$ ); the image-type table; $M_T M_\alpha=1$ ; the plane-mirror limit ( $M_T=+1$ , $M_L=-1$ ).

Builds on: Ray Model of Light · Snell's Law of Refraction · Fermat's Principle of Least Time · Refraction at a Spherical Surface. Leads to: Thin Lens Equation and Lensmaker's Equation · Thick Lenses and Principal Planes · Stops, Pupils, and Field of View · Plane and Spherical Mirrors.

How do I decide from the numbers alone whether an image is real or virtual?

Read the sign of $s_i$ . Solve $1/s_i=1/f-1/s_o$ : if $s_i>0$ the image is real and sits on the outgoing side (projectable); if $s_i<0$ it is virtual and sits on the object side (only viewable). For a single converging lens the switch happens at $s_o=f$ : objects beyond the focus give real images, objects inside give virtual ones. The magnification sign then tells you orientation — $M_T=-s_i/s_o$ is negative (inverted) for a real image and positive (erect) for the virtual case. You never need to trace rays to classify the image; the two signs do it.

Why is a real image inverted but a virtual image erect, for the same lens?

Because a real image forms after the rays have crossed the axis, and a virtual image forms before they would. When the object is beyond the focus, the converging rays meet and cross on the far side, so the tip that started above the axis ends below it — inverted, $M_T<0$ . When the object is inside the focus, the lens cannot bend the rays enough to cross; they still diverge, and their backward extensions meet on the same side, above the axis — erect, $M_T>0$ . The sign of $s_i$ (which flips at $s_o=f$ ) carries this directly into $M_T=-s_i/s_o$ .

Where does M_L = −M_T² actually come from — it looks arbitrary?

It is a derivative. Longitudinal magnification is the ratio of a small image-depth to the small object-depth that produced it, $M_L=\mathrm{d}s_i/\mathrm{d}s_o$ . Differentiate the conjugate equation $1/s_o+1/s_i=1/f$ at fixed $f$ and you get $\mathrm{d}s_i/\mathrm{d}s_o=-s_i^2/s_o^2$ , and since $M_T=-s_i/s_o$ , that ratio is exactly $-M_T^2$ . The square is the two powers of distance in $1/s^2$ ; the minus sign is the depth reversal built into the imaging map. It holds for real and virtual images alike, in a single medium.

What is the Lagrange invariant and why should I care?

The Lagrange (or Smith–Helmholtz) invariant is the quantity $n y u$ — index times ray height times ray angle — evaluated for a chief and a marginal ray; it is conserved through every paraxial surface of a system. Its payoff is the reciprocity $M_T M_\alpha=n/n'$ (equal to $1$ in one medium): whatever transverse magnification you gain, you lose in angular magnification, and vice versa. Physically it is conservation of étendue (optical throughput) — the reason you cannot build a lens that magnifies a wide field at a wide aperture for free. It is the paraxial seed of the Abbe sine condition and of the brightness theorem in radiometry.

The camera focuses a far scene near the focal plane — how does the conjugate equation say that?

Set $s_o\to\infty$ in $1/s_o+1/s_i=1/f$ : the $1/s_o$ term vanishes and $s_i\to f$ . So a very distant object images almost exactly one focal length behind the lens, in the focal plane — which is why "focused at infinity" means the sensor sits at $f$ . As the subject comes closer, $1/s_o$ grows and the image distance $s_i=fs_o/(s_o-f)$ increases past $f$ , so the lens must move away from the sensor to keep focus. A macro shot at $s_o=2f$ needs $s_i=2f$ , doubling the lens-to-sensor gap — the reason macro lenses extend so far.

Does the same magnification framework work for mirrors and single surfaces, or only thin lenses?

It is universal within the paraxial regime. Plane and spherical mirrors obey $1/s_o+1/s_i=1/f$ with $f=R/2$ , and refraction at a single spherical surface obeys $n_1/s_o+n_2/s_i=(n_2-n_1)/R$ ; both share the magnification structure, with $M_T=-s_i/s_o$ for the mirror and $M_T=-(n_1 s_i)/(n_2 s_o)$ across a surface. The three magnifications and the Lagrange invariant carry over unchanged. That universality is the point: fix one sign convention and the same short algebra images through a mirror, a surface, a thin lens, or a whole instrument built by cascading them.