Plane and Spherical Mirrors

Lean toward a make-up or shaving mirror and your face looms back at you, upright and enlarged. Glance at a car's passenger wing-mirror and a whole street compresses into a small, upright picture stamped with the warning objects in mirror are closer than they appear. A headlight or a telescope takes a curved sheet of metal-coated glass and throws out — or gathers in — a tight parallel beam. All three are the same device: a mirror whose surface is a piece of a sphere, forming images by nothing more than the law of reflection applied point by point across the curve. Because reflection sends every wavelength the same way, a mirror focuses red and blue light identically — no chromatic aberration, the single reason every large telescope since the nineteenth century is a reflector rather than a lens. And because the geometry is local, the whole imaging behaviour collapses, in the paraxial (small-angle) regime, into one compact relation: the mirror equation $1/s_o+1/s_i=1/f$ with focal length $f=R/2$ , half the radius of curvature. A flat mirror is just the $R\to\infty$ limit; a convex mirror is the same equation with $f<0$ . The subtlety — and the whole discipline of this article — is sign. Object distance, image distance, and the magnification $M=-s_i/s_o$ are all signed, and it is their signs, not their magnitudes, that say whether an image is real or virtual, erect or inverted, enlarged or shrunk. Get the convention right and one equation classifies every mirror image you will ever meet; get a single sign wrong and you will confidently call an upright image upside-down.

Key takeaways

A curved mirror images light by the law of reflection applied across its surface; in the paraxial regime this reduces to one relation, the mirror equation $1/s_o+1/s_i=1/f$ , with $f=R/2$ .
Hecht sign convention: a real object has $s_o>0$ ; a real image (in front, on the object's side) has $s_i>0$ ; a virtual image (behind the mirror) has $s_i<0$ ; $f>0$ for a concave mirror, $f<0$ for a convex one.
Transverse magnification $M=-s_i/s_o$ : its sign gives orientation ( $M<0$ inverted, $M>0$ erect) and its magnitude gives size ( $|M|>1$ magnified, $|M|<1$ reduced).
A concave mirror ( $f>0$ ) makes a real inverted image when $s_o>f$ and a virtual erect magnified image when $s_o<f$ — the make-up-mirror effect; a convex mirror ( $f<0$ ) makes a virtual, erect, minified image for every real object.
A plane mirror is the $R\to\infty$ limit: $s_i=-s_o$ and $M=+1$ , an equal-size erect virtual image as far behind as the object is in front; and because reflection is achromatic, telescopes are reflectors.
Key idea: one signed equation $1/s_o+1/s_i=1/f$ with $f=R/2$ and $M=-s_i/s_o$ governs every plane, concave, and convex mirror — and it is the signs, not the magnitudes, that decide real-vs-virtual and erect-vs-inverted.

The core insight

A curved mirror forms images by the law of reflection alone, and in the paraxial limit its conjugate distances obey $1/s_o+1/s_i=1/f$ with $f=R/2$ . The transverse magnification $M=-s_i/s_o$ carries all the qualitative information in its sign: real or virtual, erect or inverted, enlarged or reduced. Reflection is wavelength-independent, so the image never suffers chromatic aberration.

Where this lives today

The most spectacular spherical-mirror descendants fly. The James Webb Space Telescope collects infrared starlight on eighteen hexagonal beryllium segments, gold-coated and figured into a single 6.5-metre reflector that departs from a sphere by just enough to be an aberration-corrected conic; actuators nudge each segment until the eighteen wavefronts stack to a fraction of a wavelength. The same curvature-focus physics runs the mundane world: a headlight places a bulb at the focus of a concave reflector to throw a parallel beam, a dentist's concave mirror gives a magnified erect view of a tooth ( $s_o<f$ ), and the convex mirror keeps its everyday jobs — the wide-angle security dome and the passenger wing-mirror whose shrunken erect image warns that things are closer than they appear. Even a laser lives between two spherical mirrors, whose radii set the cavity's stability and beam size.

The physical picture

A plane mirror does the simplest thing: it refolds the diverging cone of rays from each object point into an identical cone that appears to stream from a point an equal distance behind the surface. No ray reaches that point, so the image is virtual — you cannot catch it on a screen — yet it is erect, life-size, and shows the familiar front-to-back handedness flip that turns your right hand into the reflection's left. Curving the surface breaks this symmetry in a controllable way. A concave mirror bows away from the light, so its normal tilts more steeply toward the axis as you move out from the vertex; it deflects the outer rays inward more strongly than the inner ones and brings a parallel beam to a focus. A convex mirror bows the other way, spreading a parallel beam so the reflected rays only appear to come from a point behind the mirror — which is why a convex wing-mirror packs a wide scene into a small, upright, shrunken image. To locate the image of an extended object, trace three principal rays from the object tip, each reflecting by a rule you can draw without calculation.

Figure

Concave mirror, object beyond C — three principal rays cross to form a real, inverted, reduced image

Notice what the diagram proves: all three reflected rays meet at one point, so every ray from the object tip is sent to the same image tip — the mirror images stigmatically. With the object beyond the centre of curvature $C$ , that point lies between $F$ and $C$ , inverted and smaller than the object; slide the object inward and the image marches outward and grows. The factor $f=R/2$ that fixes $F$ is pure geometry: a parallel ray strikes at height $h$ , where the normal points toward $C$ at angle $\theta\approx h/R$ to the axis; reflection doubles that to $2\theta$ , so the ray crosses the axis at $h/(2\theta)=R/2$ , independent of $h$ for small angles. Everything qualitative about a concave mirror — real or virtual, upright or flipped, bigger or smaller — is encoded in a single curve: the image distance plotted against the object distance. That curve is a hyperbola, and its most important feature is the place where it blows up.

Figure

Image distance vs object distance for a concave mirror f = 20 cm — the crossover at s_o = f where the image flips real ↔ virtual and M changes sign

Read the punchline from the crossover outward. For $s_o>f$ the image distance is positive — the rays truly converge and the image is real and inverted ( $M<0$ ). As the object approaches the focus from outside, $s_i$ races to $+\infty$ : the reflected rays come out parallel (the searchlight, run backwards). Cross to $s_o<f$ and $s_i$ reappears from $-\infty$ as a negative number — the rays only seem to diverge from a point behind the mirror, so the image is virtual, erect ( $M>0$ ), and enlarged. The whole real-to-virtual transition, and the sign flip of $M$ with it, happens at the one place $s_o=f$ . That single sign change is what this article asks you to master.

The core equations

Every result below assumes the Hecht sign convention, stated once and used everywhere. Light travels toward the mirror; distances are measured from the vertex. The object distance $s_o$ is positive for a real object in front of the mirror. The image distance $s_i$ is positive for a real image in front of the mirror (the object's side, where the light goes) and negative for a virtual image behind it. Radius and focal length are positive for a concave mirror and negative for a convex one, with the focal length exactly half the radius. Three equations then carry the whole subject. The central relation ties the two conjugate distances to the focal length — the paraxial statement that every ray from one axial point returns to a single image point, derived below from reflection alone.

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

where: $s_o$ — object distance from the vertex (positive for a real object); $s_i$ — image distance ( $>0$ real/in front, $<0$ virtual/behind); $f$ — focal length. Read it as a see-saw: as $s_o$ falls toward $f$ , $1/s_i\to 0$ and the image recedes to infinity; push $s_o$ below $f$ and $1/s_i$ turns negative — the image goes virtual. The focal length is not an independent input; it is fixed by the mirror's shape. A concave surface of radius $R$ folds a parallel beam to a point midway between the surface and its centre of curvature.

f=\frac{R}{2} \qquad (2)

where: $f$ — focal length (concave $f>0$ , convex $f<0$ ); $R$ — radius of curvature (concave $R>0$ , convex $R<0$ ). The factor of two is the reflection doubling the incidence angle; there is no refractive index and no wavelength here, which is precisely why a mirror's focus is the same for every colour. Finally, the size and orientation of the image follow from a single signed ratio, the transverse magnification.

M=-\frac{s_i}{s_o} \qquad (3)

where: $M$ — ratio of image height to object height; $s_i,\,s_o$ — the conjugate distances. The minus sign is doing real work: a real image ( $s_i>0$ ) of a real object ( $s_o>0$ ) gives $M<0$ , i.e. inverted, while a virtual image ( $s_i<0$ ) gives $M>0$ , erect. The magnitude $|M|$ sets the size — larger than one is magnified, smaller than one is reduced.

Derivation AMirror equation and f = R/2 from paraxial reflection

We want to establish, from the law of reflection and nothing else, that all paraxial rays leaving an axial object point $O$ reconverge to a single image point $I$ , and that the two distances obey $1/s_o+1/s_i=2/R=1/f$ . We adopt the Hecht convention up front — object and real image on the reflecting side counted positive, concave $R>0$ — and work to first order in the small angles of the paraxial regime, where a ray stays near the axis and strikes near the vertex. The whole proof is the geometry of one ray plus the equality of two angles where it reflects.

Step 1. Place the vertex $V$ at the origin, the centre of curvature $C$ a distance $R$ along the axis, the object $O$ at $s_o$ and the image $I$ at $s_i$ . A ray from $O$ meets the mirror at a point $P$ a small height $h$ above the axis. Define the small angles that $OP$ , $CP$ and $IP$ make with the axis: $\alpha\approx h/s_o$ , $\beta\approx h/R$ , $\gamma\approx h/s_i$ . Why: in the paraxial limit each angle is small enough that $\tan\theta\approx\theta\approx(\text{height})/(\text{distance})$ , linearising the geometry so the single height $h$ appears in all three.

Step 2. The line $CP$ is the surface normal at $P$ , because a sphere's normal always passes through its centre. In triangle $OPC$ the angle $\beta$ at $C$ is an exterior angle, so by the exterior-angle theorem it equals the sum of the two remote interior angles, $\beta=\alpha+\theta_i$ , giving the angle of incidence $\theta_i=\beta-\alpha$ . Why: measuring the incidence angle from the normal $CP$ rather than from the axis is what lets the law of reflection enter cleanly in the next step.

Step 3. By the law of reflection the reflected ray leaves at the equal angle $\theta_r=\theta_i$ on the far side of the normal, heading toward $I$ . In triangle $CPI$ the angle $\gamma$ at $I$ is again an exterior angle, so $\gamma=\beta+\theta_r$ , i.e. $\theta_r=\gamma-\beta$ . Setting $\theta_i=\theta_r$ gives $\beta-\alpha=\gamma-\beta$ , that is $2\beta=\alpha+\gamma$ . Why: reflection is the only physical law used; equal angles about the normal collapse the two triangles into one linear relation among the three axis angles.

Step 4. Substitute the paraxial angles: $2(h/R)=h/s_o+h/s_i$ , and the common height $h$ cancels to leave $1/s_o+1/s_i=2/R$ . Because $h$ dropped out, every paraxial ray from $O$ lands on the same $I$ — the image is stigmatic. Sending the object to infinity ( $\alpha\to0$ ) gives $1/s_i=2/R$ , so a distant source focuses at $s_i=R/2$ ; we name that distance the focal length, $f=R/2$ , and the relation becomes $1/s_o+1/s_i=1/f$ . Why: the cancellation of $h$ is the content of "imaging"; the object-at-infinity case is what physically defines $f$ and ties it to the radius. *Verified symbolically in verify_plane-spherical-mirrors.py: $2\beta=\alpha+\gamma$ from $\theta_i=\theta_r$ , the reduction to $1/s_o+1/s_i=2/R$ , and the object-at-infinity result $s_i=R/2=f$ .*

Go deeper · Derivation B · 2 minTransverse magnification M = -s_i/s_o from similar triangles

Derivation BTransverse magnification M = -s_i/s_o from similar triangles

Having located the image, we want its height and orientation, and to show a single signed number $M=-s_i/s_o$ reports both. The tool is one more principal ray — the chief ray from the object tip straight to the vertex — plus the fact that at the vertex the surface is perpendicular to the axis, so the axis is the local normal. Two similar right triangles, one each side of the axis, fix the height ratio, and the sign of that ratio encodes erect versus inverted.

Step 1. Send a ray from the object tip, a height $y_o$ above the axis at distance $s_o$ , down to the vertex $V$ . Since the axis is the normal at $V$ , this ray reflects with equal angles above and below the axis. Why: choosing the vertex as the reflection point makes the axis the mirror line, so the incident and reflected rays are simple mirror images of each other — no curvature bookkeeping needed.

Step 2. The incident ray makes an angle $u$ with the axis where $\tan u=y_o/s_o$ . The reflected ray makes the same angle $u$ on the opposite side, and reaches the image tip at distance $s_i$ and height $y_i$ , so $\tan u=-y_i/s_i$ : the minus sign records that for a real image the tip has dropped below the axis while the object tip was above it. Why: equal reflection angles force equal tangents, and the opposite side of the axis is exactly what inverts a real image.

Step 3. Equate the two expressions for $\tan u$ : $y_o/s_o=-y_i/s_i$ , hence $M\equiv y_i/y_o=-s_i/s_o$ . Why: the height ratio is the magnification by definition, and eliminating the shared angle $u$ leaves it in terms of the conjugate distances alone.

Step 4. Now read the signs. A real image ( $s_i>0$ ) gives $M<0$ — inverted; a virtual image ( $s_i<0$ ) gives $M>0$ — erect. The magnitude decides size: $|M|>1$ magnified, $|M|<1$ reduced, $|M|=1$ same size. With the mirror equation this classifies every image a mirror can form (table below). Why: packing orientation and size into one signed quantity is what makes the sign convention non-negotiable — drop the sign and you lose half the answer. *Verified in verify_plane-spherical-mirrors.py: $M=-s_i/s_o$ from the vertex chief ray, and the sign/magnitude classification across the concave regimes and the convex case.*

Together the two derivations sort a concave mirror's images into five regimes, set by where the object sits relative to $F$ and $C$ .

Object position	Image distance $s_i$	Real / virtual	Orientation	Size
Beyond $C$ ( $s_o>2f$ )	$f<s_i<2f$	real	inverted ( $M<0$ )	reduced ( $\|M\|<1$ )
At $C$ ( $s_o=2f$ )	$s_i=2f$	real	inverted	same size ( $M=-1$ )
Between $C$ and $F$ ( $f<s_o<2f$ )	$s_i>2f$	real	inverted	magnified ( $\|M\|>1$ )
At $F$ ( $s_o=f$ )	$s_i\to\infty$	none	rays emerge parallel	—
Inside $F$ ( $s_o<f$ )	$s_i<0$	virtual	erect ( $M>0$ )	magnified ( $\|M\|>1$ )

Limiting cases

The mirror equation is paraxial, and its edges recover every special mirror you already know.

Case	Result	Meaning
Object at infinity	$s_o\to\infty\ \Rightarrow\ s_i=f=R/2$	parallel starlight focuses at $F$ — the telescope primary and the headlight run in reverse
Object at the focus	$s_o=f\ \Rightarrow\ s_i\to\infty$	reflected rays emerge parallel — the searchlight and collimator configuration
Plane mirror	$R\to\infty,\ f\to\infty\ \Rightarrow\ s_i=-s_o,\ M=+1$	virtual, erect, same-size image as far behind as the object is in front
Convex mirror	$f<0\ \Rightarrow\ s_i<0,\ 0<M<1$ for every $s_o>0$	always virtual, erect and minified — the wide-angle wing-mirror
Marginal (non-paraxial) rays	$s_i^{\text{marginal}}<s_i^{\text{paraxial}}$	outer rays cross the axis nearer the vertex: spherical aberration, cured on-axis by a paraboloid

Common trap"Distances are always positive" — the sign-convention trap

The commonest mirror mistake is to treat $s_i$ and $M$ as magnitudes and paste the "real/inverted" or "virtual/erect" label on by hand. The convention already contains that information — but only if you keep the sign. Solve the mirror equation for $1/s_i$ and carry the sign it comes out with: a positive $s_i$ is a real image in front of the mirror, a negative $s_i$ a virtual image behind it. Then $M=-s_i/s_o$ gives the orientation automatically — a virtual image ( $s_i<0$ ) gives $M>0$ , correctly flagging an erect image. Drop the minus sign hiding in a negative $s_i$ and $M$ flips with it, so you call an upright image inverted. A second myth is worth killing outright: "a concave mirror always magnifies" is false. Beyond the centre of curvature ( $s_o>2f$ ) it forms a reduced image — at $s_o=2f$ it is exactly life-size ( $M=-1$ ) — and it magnifies ( $|M|>1$ ) only when the object is inside $2f$ .

Worked examples

1 · One concave mirror, both a real and a virtual image

A concave shaving mirror has radius of curvature $R=40\ \mathrm{cm}$ , so $f=R/2=20\ \mathrm{cm}$ . Find the image (a) for an object at $s_o=30\ \mathrm{cm}$ and (b) for an object at $s_o=10\ \mathrm{cm}$ , and describe each.

Step 1. Case (a), object outside the focus. From the mirror equation, $1/s_i=1/f-1/s_o=1/20-1/30=(3-2)/60=1/60$ , so $s_i=+60\ \mathrm{cm}$ . The positive sign means a real image, 60 cm in front of the mirror.

Step 2. Its magnification is $M=-s_i/s_o=-60/30=\mathbf{-2}$ . The negative sign says inverted, the magnitude says twice life-size: real, inverted, magnified 2×. This is the sharp, upside-down image you see when you hold a concave mirror at arm's length.

Step 3. Case (b), object inside the focus. Now $1/s_i=1/20-1/10=(1-2)/20=-1/20$ , so $s_i=\mathbf{-20\ cm}$ . The image distance is negative — a virtual image, 20 cm behind the mirror, which no screen can catch.

Step 4. Here $M=-s_i/s_o=-(-20)/10=\mathbf{+2}$ : positive, so erect, twice as large — virtual, erect, magnified 2×, the make-up-mirror view of your face. The same mirror gave $M=-2$ and $M=+2$ ; carrying the object across the focal point at $s_o=f=20\ \mathrm{cm}$ is exactly what flipped $s_i$ from $+60$ to $-20$ and $M$ from inverted to erect.

2 · The convex wing-mirror and its wide field of view

A car's passenger-side mirror is convex with $f=-15\ \mathrm{cm}$ . A car sits $s_o=60\ \mathrm{cm}$ behind it. Where is the image, how large is it, and why does the mirror widen your view?

Step 1. Convex means $f<0$ . From the mirror equation, $1/s_i=1/f-1/s_o=1/(-15)-1/60=-4/60-1/60=-5/60=-1/12$ , so $s_i=\mathbf{-12\ cm}$ : a virtual image, 12 cm behind the mirror, as expected for a convex surface.

Step 2. The magnification is $M=-s_i/s_o=-(-12)/60=\mathbf{+0.2}$ : positive and less than one, so the image is erect and one-fifth life-size. Solving $1/s_i=1/f-1/s_o$ with $f<0$ and $s_o>0$ always gives $s_i<0$ and $0<M<1$ — a convex mirror does this for every real object.

Step 3. That shrinkage is the point. By compressing objects, the mirror squeezes a much wider slice of the road onto the same glass, giving a larger field of view than a flat mirror of equal size — invaluable for the blind spot.

Step 4. The trade-off is written on the mirror: because everything looks five times smaller, your brain reads it as five times farther, so the warning objects in mirror are closer than they appear. The reduced $|M|$ buys field of view at the cost of an honest sense of distance.

Physics of the portrait

The spherical mirror is not a person but an idea with a long biography — the oldest imaging instrument humans made, and among the most advanced in its refined descendants. Its single defect and the cure for it shaped the whole history of the telescope.

Origins · 1 minspherical mirror and how this was discovered

spherical mirror

reflecting optics · antiquity to today · from polished bronze to the telescope

Long before anyone could say why a curved mirror magnifies, people ground obsidian and cast bronze into mirrors — the reflecting surface predates the lens by millennia. The first deep theory came from Ibn al-Haytham (Alhazen) around 1021, who posed the problem that still bears his name: given a source and an eye, find the point on a spherical mirror where a single ray reflects from one to the other — a quartic geometry problem he solved with conic sections. The curved mirror's flaw and its remedy define the modern story. A sphere does not bring a wide parallel beam to a point — the marginal rays cross the axis short of the paraxial focus, smearing a star into a bright cusp-shaped surface, the caustic: spherical aberration. Replacing the sphere with a paraboloid focuses every axial parallel ray to one exact point, which is why Gregory, Newton, and Hadley turned to reflectors, and why a paraboloidal or hyperboloidal mirror is the heart of every large telescope from Newton's palm-sized speculum disc to the segmented, gold-coated primary of the James Webb Space Telescope.

History

6000

BCE — obsidian is ground and polished into mirrors at Çatalhöyük in Anatolia, among the oldest known mirrors.

1021

— Ibn al-Haytham poses and studies the reflection-point problem on a spherical mirror (Alhazen's problem) in the Kitāb al-Manāẓir.

1663

— James Gregory designs a reflecting telescope with a paraboloidal primary in Optica Promota.

1668

— Isaac Newton builds the first working reflecting telescope, sidestepping the chromatic aberration of lenses.

1721

— John Hadley demonstrates a practical paraboloidal reflecting telescope to the Royal Society.

Going further

Texts. Hecht, Optics, 5th ed., ch. 5 (Geometrical Optics) · Saleh & Teich, Fundamentals of Photonics, 3rd ed., ch. 1 (Ray Optics) · Born & Wolf, Principles of Optics, ch. 4 (image-forming instruments and aberrations) · Demtröder, Laser Spectroscopy, Vol. 1 (mirror resonators and collimators).

Provenance. All results are standard geometrical optics (Hecht ch. 5; Saleh & Teich ch. 1). A verifier should reproduce, from the stated Hecht convention: $f=R/2$ by the paraxial doubling of the incidence angle; the mirror equation $1/s_o+1/s_i=1/f=2/R$ from reflection at a spherical surface; $M=-s_i/s_o$ from the vertex chief ray; the real, virtual, convex, and plane cases tabulated on the cheat sheet. The Newtonian form of the same law is $(s_o-f)(s_i-f)=f^2$ , stated here in its correct closed form. No fabricated constants are used, and every sign is fixed by the convention (concave $R>0$ , real image $s_i>0$ ).

Prerequisites. Law of Reflection · Ray Model of Light · Fermat's Principle of Least Time · Refraction at a Spherical Surface. Leads to. Image Formation and Magnification · Thin Lens Equation and Lensmaker's Equation · Thick Lenses and Principal Planes · Stops, Pupils, and Field of View.

Why is the focal length exactly half the radius, f = R/2, and not R?

Because reflection doubles the angle. A parallel ray strikes at height $h$ , where the normal points to the centre of curvature $C$ at angle $\theta\approx h/R$ ; reflection turns the ray through $2\theta$ , so it crosses the axis at $h/(2\theta)=R/2$ — halfway to $C$ . The centre $C$ is where a ray aimed straight at the mirror returns on itself ( $s_o=R$ maps to $s_i=R$ ), while the focus $F$ at $R/2$ is where parallel rays converge — different points precisely because of that factor of two.

What is the real physical difference between a real and a virtual image?

A real image is a place where light rays actually cross and pile up; put a screen there and you record it, and in the mirror convention it has $s_i>0$ (in front of the mirror). A virtual image is a place the rays only appear to come from — their backward extensions meet there but no light reaches it — so a screen catches nothing and $s_i<0$ (behind the mirror). Your eye still sees a virtual image because its lens converges the diverging rays onto your retina, forming a real image of the virtual one. That is why the face in a make-up mirror looks solid even though it lives behind the glass.

Isn't magnification always positive? Why does M come out with a sign?

Everyday usage treats "magnification" as a positive zoom factor, but in optics $M=-s_i/s_o$ is deliberately signed so one number reports both size and orientation. The magnitude $|M|$ is the zoom factor; the sign tells you which way up ( $M<0$ inverted, $M>0$ erect). So $M=-2$ is an upside-down image twice life-size, while $M=+2$ from the same mirror at a closer object is upright. If a problem only asks "how many times bigger," quote $|M|$ — but never throw the sign away while solving, because it distinguishes the two cases.

My make-up mirror magnifies my face, yet reflects a tiny upside-down room across the hall. How can one mirror do both?

It is the mirror equation at two object distances. Your face is inside the focal length ( $s_o<f$ ), giving a virtual, erect, magnified image. The far wall is many focal lengths away ( $s_o\gg f$ ), beyond the centre of curvature, where the image is real, inverted, and reduced — the little flipped room, floating as a real image in front of the mirror. The switch happens as an object crosses the focal point — the crossover in the punchline figure.

Why are large telescopes built with mirrors instead of lenses?

Two structural reasons. First, reflection is achromatic: $f=R/2$ contains no refractive index and no dispersion, so a mirror sends red and blue to the same focus, whereas a lens focuses colours at different points and needs extra glass to correct it. Second, a mirror is supported across its whole back and only one face is figured, so it can be made enormous, while a large lens sags under its own weight and can only be held at its rim. Hence every observatory-class instrument since the nineteenth century is a reflector.

What is spherical aberration, and why does a parabola fix it?

A spherical surface only obeys $f=R/2$ for paraxial rays. Rays striking far from the vertex reflect through the axis a little closer to the mirror than the paraxial focus, so a wide parallel beam converges not to a point but to a bright cusped surface, the caustic — this blur is spherical aberration. A paraboloid is defined so that every ray parallel to its axis travels an equal optical path to one focus (Fermat's principle), removing the on-axis defect exactly. The price is that a paraboloid is perfect only on-axis; off-axis it introduces coma, which is why the best telescopes use paired conic surfaces (the Ritchey–Chrétien design) to tame both.

Does the mirror equation really contain the plane mirror as a special case?

Yes — a flat mirror is just the limit of vanishing curvature. Let $R\to\infty$ , so $f=R/2\to\infty$ and $1/f\to0$ ; the mirror equation becomes $1/s_o+1/s_i=0$ , i.e. $s_i=-s_o$ — the image as far behind the mirror as the object is in front, and virtual (negative $s_i$ ). The magnification is $M=-s_i/s_o=-(-s_o)/s_o=+1$ : erect and life-size. So the bathroom mirror falls straight out of the same equation that governs concave and convex mirrors, with no extra rules.