Visual walkthrough — Mirrors — plane, concave, convex; mirror equation 1 - v + 1 - u = 1 - f

We only ever use two facts: light bounces so that ==angle in = angle out (see Reflection of Light — Laws & Normal), and two triangles with the same angles have proportional sides== (similar triangles). Everything below is those two facts, repeated.

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Step 0 — The stage we draw on

WHAT. Before any ray, we lay down the fixed furniture of a concave mirror.

WHY. Every distance in the final formula is measured from one special point (the pole). If we don't pin down that origin and a direction for "positive," the signs will be nonsense.

PICTURE. Look at the figure. The curved arc is the mirror. The straight line through it is the principal axis — our number line. The point where the axis meets the mirror is the pole $P$ (our origin, the "zero"). The dot $C$ is the centre of curvature — the centre of the sphere this mirror is a slice of. Halfway between them sits $F$, the focus.

Incident light travels left → right. Anything to the left of $P$ (against the light) counts as negative; to the right, positive. Heights up are positive. For a concave mirror $F$ and $C$ sit to the left of $P$, so in signed coordinates $f<0$ and $R<0$ — exactly the parent note's convention. Their ruler lengths $|f|,R$ are of course positive; keep the two apart.

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Step 1 — One parallel ray, and the normal that tames it

WHAT. Send a single ray straight in, parallel to the axis, striking the mirror at a point $M$ a little above $P$.

WHY. A ray parallel to the axis is the definition of what makes the focus. To bounce it we need the normal (the "which way is straight-out-of-the-surface" line) at $M$. For a sphere that normal is easy: it points at the centre $C$, because a radius always meets a sphere at a right angle.

PICTURE. The blue ray comes in flat. At $M$ the orange line is the normal $MC$. The reflected green ray leaves $M$ and crosses the axis at $F$.

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Step 2 — Why the focus sits (almost) exactly halfway ($f=R/2$)

WHAT. Track the three angles $\theta$ around triangle $MFC$ and prove $F$ is the midpoint of $PC$ — in the near-axis limit.

WHY. We need to know where $F$ is in terms of $R$ before we can talk about $f$ at all. This is the first number the mirror hands us for free.

PICTURE. The incoming blue ray is parallel to the axis, and $MC$ is a transversal cutting two parallel lines (the ray and the axis). So the angle at $C$, $\angle MCF$, equals $\theta$ (alternate angles). We already have $\angle FMC=\theta$ from the bounce. A triangle with two equal base angles is isosceles, so the two sides opposite them are equal: $|FM|=|FC|$.

Now the one and only approximation on this page — the paraxial (near-axis, small-angle) limit:

Taking that limit, so that $|FM|\to |FP|$, and combining with $|FM|=|FC|$:

$\underset{=|f|}{\underbrace{|FP|}}=\underset{=R-|f|}{\underbrace{|FC|}}⟹|f|=R-|f|⟹\boxed{|f|=\frac{R}{2}}$

Here $|FP|$ is the ruler focal length $|f|$; $|FC|$ is what's left of the radius after $|f|$, i.e. $R-|f|$ (since $F$ lies between $P$ and $C$). Solving gives the halfway result. Restoring Step 0 signs (both $f$ and $R$ negative for concave) leaves the tidy $f=R/2$.

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Step 3 — Place a real object; find the two rays that build its image

WHAT. Stand an arrow of height $h$ on the axis, beyond $C$. Draw the two rays whose meeting point locates the image tip.

WHY. One ray tells you a direction, but a point needs two lines crossing. We pick the two rays whose geometry we already understand: the parallel ray (Step 1) and the ray straight to the pole.

PICTURE. Red arrow = object, height $h$, at distance $|u|$ left of $P$. Green ray from its tip goes parallel, bounces through $F$. Blue ray from its tip hits the pole $P$ and bounces off the axis. Where they cross, the image tip forms: an inverted arrow of height $h^{\prime }$ at distance $|v|$.

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Step 4 — The pole ray gives magnification $m=-v/u$

WHAT. Zoom in on just the blue pole ray and the two right triangles it makes with the axis.

WHY. The pole ray reflects with the axis itself as its normal, so it makes equal angles above and below the axis. Equal angles ⇒ the object triangle and image triangle are similar ⇒ their sides are in the same ratio. That ratio is what "magnification" means.

PICTURE. Left triangle: object height $|h|$ over base $|u|$. Right triangle: image height $|h^{\prime }|$ over base $|v|$. The shared vertex angle at $P$ is equal on both sides (law of reflection), so the triangles are similar.

Now translate to signed coordinates. In Step 0's convention the image points down ($h^{\prime }<0$) while the object points up ($h>0$), and both $u,v<0$. Substituting $|h^{\prime }|=-h^{\prime }$, $|v|=-v$, $|u|=-u$:

$\frac{-h^{\prime }}{h}=\frac{-v}{-u}⟹\frac{h^{\prime }}{h}=-\frac{v}{u}$

$\boxed{m=\frac{h^{\prime }}{h}=-\frac{v}{u}}$

The minus sign is not decoration — it is the geometry telling you a real image flips over.

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Step 5 — The parallel ray gives a second ratio

WHAT. Now use the green parallel ray. Two more similar triangles appear, both sharing the focus $F$.

WHY. We have one equation linking $h^{\prime }/h$ to distances. To eliminate the heights and get a pure $u,v,f$ relation we need a second, independent equation. The parallel-through-$F$ ray supplies it.

PICTURE — why the mirror-height equals $h$. The green ray leaves the object tip travelling exactly horizontally (parallel to the axis). A horizontal line keeps the same height at every $x$. So when it reaches the mirror right next to $P$ (which sits essentially on the axis in the paraxial limit), its height there is still the object's height $|h|$. In the figure, notice the green segment from object tip to mirror is perfectly flat — equal heights at both ends is just "a horizontal line is level." That level segment is one leg; the drop from mirror down through $F$ to the image tip is the other.

So near $P$ the mirror carries height $|h|$. From there the ray dives straight through $F$ to the image tip at height $|h^{\prime }|$. This makes two right triangles that share the vertex angle at $F$:

• Triangle A — base $|f|$ (from $P$ to $F$ along the axis), upright side $|h|$ (the mirror height).
• Triangle B — base $|v|-|f|$ (from $F$ out to the image foot), upright side $|h^{\prime }|$ (the image height).

Same apex angle at $F$ ⇒ similar ⇒ matching sides in proportion.

Now translate to signed coordinates using Step 0 (concave: $u,v,f$ all negative; $h^{\prime }<0$, $h>0$). Substitute $|h^{\prime }|=-h^{\prime }$, $|h|=h$, $|v|=-v$, $|f|=-f$:

$\frac{-h^{\prime }}{h}=\frac{(-v)-(-f)}{-f}=\frac{-v+f}{-f}=\frac{v-f}{f}$

so the signed second ratio is

$\frac{h^{\prime }}{h}=-\frac{v-f}{f}.$

$v-f$ = signed gap between image and focus; $f$ = signed focal length. This is our second expression for the same height ratio — now written in the same signed language as Step 4.

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Step 6 — Set the two ratios equal and solve

WHAT. Both Step 4 and Step 5 describe $h^{\prime }/h$. Equate them; the heights vanish; algebra alone finishes the job.

WHY. No new physics remains — reflection did its work in Steps 1–5. What's left is bookkeeping in one consistent sign language.

PICTURE. The figure overlays both triangle-pairs so you can literally see the two $h^{\prime }/h$ ratios meeting at the same image tip.

Both signed expressions equal $\frac{h^{\prime }}{h}$, so set them equal:

$\underset{\text{Step 4}}{\underbrace{-\frac{v}{u}}}=\underset{\text{Step 5}}{\underbrace{-\frac{v-f}{f}}}$

The two minus signs cancel, giving a clean start with no ambiguity:

$\frac{v}{u}=\frac{v-f}{f}$

• Cross-multiply (multiply both sides by $uf$): $vf=u(v-f)=uv-uf$.
• Collect the $uv$ term: $vf+uf=uv$.
• Divide every term by $uvf$: $\frac{vf}{uvf}+\frac{uf}{uvf}=\frac{uv}{uvf}⟹\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$

Every move above is ordinary algebra on signed quantities — no sign was flipped by hand; the signs simply rode along from Step 0. That is why the same equation will work for convex and virtual cases without amendment.

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Step 7 — Edge & degenerate cases (walk every one)

WHAT. Check that the picture still behaves at the boundaries: object at $F$, object at $C$, and a convex mirror.

WHY. A derivation you trust must survive its extremes. If any case breaks, the formula is wrong.

PICTURE. Three mini-scenes — parallel-out (object at $F$), self-return (object at $C$), and the always-behind image of a convex mirror.

• Object at focus ($u=f$): $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=0\Rightarrow v\to \infty$. The rays leave parallel — no image, image "at infinity." The two triangles of Step 5 become degenerate (zero base at $F$), exactly matching.
• Object at centre ($u=2f$, since $R=2f$): algebra gives $v=2f$ too. Image sits back on $C$, same size, inverted — the mirror maps $C$ to itself. $m=-v/u=-1$. ✔
• Convex mirror: here $f>0$. For any real object ($u<0$), $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$ is a sum of two positives ⇒ $v>0$ always ⇒ image always behind, virtual, erect, shrunk. The parallel ray now appears to come from $F$ behind the mirror; the same similar-triangle skeleton runs with dashed virtual rays. See the always-virtual image in the figure.

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The one-picture summary

Everything above collapses into a single diagram: two rays (pole ray blue, parallel ray green) from one object tip, four labelled distances, two pairs of similar triangles, and the boxed equation they force.

Recall Feynman retelling (say it out loud, no symbols)

I put an arrow in front of a curved mirror. I trace two rays from its tip: one straight to the middle of the mirror that bounces off at an equal angle, and one running flat that bounces through the focus. Where they cross is where the image tip lands. The flat ray's bounce, using only "a radius hits a sphere at a right angle" and "angle in equals angle out," proves the focus sits halfway to the sphere's centre — as long as I stay near the axis. The middle ray makes two look-alike triangles, telling me the image is stretched by the ratio of the two distances (with a flip). The flat ray makes two more look-alike triangles giving that same stretch a second way. I build both ratios with plain ruler lengths, then translate each into the signed number line, so both speak the same language. Setting them equal, the heights cancel, and after dividing through by the product of the distances I'm left with: one-over-image plus one-over-object equals one-over-focus. I never memorised it — I drew it.

Recall Forecast-then-verify

Concave, $f=-10$ cm, object at $C$ ($u=-20$): predict then compute $v,m$. ::: $\frac{1}{v}=\frac{1}{-10}-\frac{1}{-20}=-\frac{1}{20}\Rightarrow v=-20$ cm, $m=-1$: same size, inverted, real — object mapped onto itself. ✔ Object exactly at focus ($u=f=-10$): what image? ::: $\frac{1}{v}=\frac{1}{-10}-\frac{1}{-10}=0\Rightarrow v=\infty$: reflected rays parallel, no finite image. Convex $f=+15$, $u=-30$: sign of $v$ before computing? ::: Positive — sum of two positive reciprocals — so virtual, behind the mirror. ($v=+10$, $m=+\frac{1}{3}$.)

Related builds: Magnification & Image Formation uses the $m=-v/u$ from Step 4; Refraction & Lenses — Lens Maker's Equation runs the same similar-triangle machinery for lenses; the paraxial caveat of Step 2 is the root of Spherical Aberration.