Mirrors — plane, concave, convex; mirror equation 1 - v + 1 - u = 1 - f: Worked examples

Before anything, let me re-state the tools in plain words so nothing is used unexplained.

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

Everything below is just those three lines, fed different numbers.

The scenario matrix

Think of a mirror problem as choosing three things: which mirror, where the object sits, and whether the setup is normal, degenerate, or extreme. The table below enumerates every class of outcome. Every cell is covered by a worked example (the "Ex" column). Recall $C$ is at signed distance $R = 2 f$ from the pole.

#	Mirror	Object position	Image: real/virtual, erect/inverted, size	Ex
A0	Concave	At infinity ( $u \to - \infty$ )	real point image at $F$ (limiting)	1b
A	Concave	Beyond $C$ ( $∥ u ∥ > ∥ R ∥$ )	real, inverted, diminished	1
B	Concave	At $C$ ( $u = R$ )	real, inverted, same size (degenerate)	2
C	Concave	Between $C$ and $F$	real, inverted, magnified	3
D	Concave	At $F$ ( $u = f$ )	image at infinity (limiting)	4
E	Concave	Inside $F$ ( $∥ u ∥ < ∥ f ∥$ )	virtual, erect, magnified	5
F	Convex	Anywhere real	always virtual, erect, diminished	6
G	Plane	Anywhere ( $f \to \infty$ )	virtual, erect, same size (limiting $R \to \infty$ )	7
H	Real-world word problem	(convex car mirror)	field-of-view / safety-distance	8
I	Exam twist	image = given multiple of object, find $f$	two-case sign trap	9
J	Degenerate	object at pole ( $u = 0$ ) & virtual object	image at pole $m = 1$ ; virtual object case	10

Notice the spine of the table is the concave mirror: as the object walks from far away (A) toward the pole (E), the image does a full life-cycle. That walk is the single most important picture in this whole topic — here it is (front of the mirror is to the left, the pole $P$ at $x = 0$ , and the object slides in from A to E):

Case A0 & A — Concave, object at infinity and beyond C

Worked example Ex 1b (cell A0) — the limiting case: object infinitely far

Concave $f = - 10 cm$ , object at infinity ( $u \to - \infty$ , e.g. a distant star). Where is the image?

Forecast: Rays from something infinitely far arrive parallel to the axis. By the very definition of $F$ , parallel rays converge at $F$ . So predict the image lands exactly at $F$ .

Take the limit: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ . As $u \to - \infty$ , $\frac{1}{u} \to 0$ , so $\frac{1}{v} \to \frac{1}{f} \Rightarrow v \to f = - 10 cm$ . Why this step? Sending $\frac{1}{u} \to 0$ is the algebra's way of encoding "object infinitely far, incoming rays parallel." What's left, $\frac{1}{v} = \frac{1}{f}$ , says the image sits at the focus — precisely the property that defines $F$ .
Magnification: $m = - \frac{v}{u} \to 0$ (a finite $v$ divided by an infinite $∣ u ∣$ ). Why this step? $m \to 0$ means the image shrinks to a point — an infinitely distant object cannot make a finite-height image. This is how a concave mirror forms the sharp point-image of a star in a telescope.

Verify: With a huge $∣ u ∣ = 1 0^{6}$ : $\frac{1}{v} = \frac{1}{- 10} - \frac{1}{- 1 0 ^{6}}$ , giving $v \approx - 10.0001 cm \approx f$ . ✔

Worked example Ex 1 (cell A) — the classic

Concave mirror, $R = 20 cm$ , object $30 cm$ in front. Find $v$ , $m$ , and describe the image.

Forecast: Object is beyond $C$ (since $∣ R ∣ = 20 < 30$ ). Guess the image type before computing — where does the s01 picture say A lands?

Sign the inputs. Use $f = \frac{R}{2}$ ; concave ⇒ $R < 0$ , so $R = - 20 cm$ and $f = - 10 cm$ . Real object in front ⇒ $u = - 30 cm$ . Why this step? The equation is only true with signed numbers, and $f = R /2$ turns the given radius into the focal length. Skip signs → nonsense.
Solve for $v$ : $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{- 10} - \frac{1}{- 30} = - \frac{3}{30} + \frac{1}{30} = - \frac{2}{30}$ $v = - 15 cm$ Why this step? Rearranging $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ isolates the one unknown.
Interpret sign: $v < 0$ ⇒ image is in front ⇒ real.
Magnification: $m = - \frac{v}{u} = - \frac{- 15}{- 30} = - 0.5$ . Negative ⇒ inverted; $∣ m ∣ < 1$ ⇒ diminished.

Verify: $\frac{1}{- 15} + \frac{1}{- 30} = - \frac{2}{30} - \frac{1}{30} = - \frac{3}{30} = - \frac{1}{10} = \frac{1}{f}$ . ✔ And $∣ v ∣ = 15$ sits between $F$ ( $10$ ) and $C$ ( $20$ ) — exactly the "real, inverted, diminished" cell A predicted.

Case B — Concave, object exactly at C (degenerate: same size)

Case C — Concave, object between C and F (real, magnified)

Case D — Concave, object at F (limiting: image at infinity)

Worked example Ex 4 (cell D) — the limit that makes a searchlight

Concave $f = - 10 cm$ , object placed exactly at the focus, $u = - 10 cm$ . Where is the image?

Forecast: This is the reverse of "parallel rays converge to $F$ ." So rays should leave parallel — image at infinity. Watch the algebra break gracefully.

Solve: $\frac{1}{v} = \frac{1}{- 10} - \frac{1}{- 10} = 0 \Rightarrow v = \infty$ . Why this step? $\frac{1}{v} = 0$ is the algebra's way of saying "no finite meeting point" — the reflected rays never converge, they go out parallel. This is the limiting case between real (C) and virtual (E): the exact frontier where the image flips sides.
Magnification: $m = - \frac{v}{u} \to \infty$ (infinitely stretched, undefined size). Physically the image is "at infinity." Why this step? An infinite $v$ over a finite $u$ blows the magnification up — the parallel outgoing rays would only "meet" infinitely far away, so no finite-size image exists. This is the exact reverse of case A0.

Verify: As $u \to f^{-}$ (object just beyond $F$ , e.g. $u = - 10 - ε$ ), $\frac{1}{v} = \frac{1}{- 10} - \frac{1}{- 10 - ε} \to 0^{-}$ , so $v \to - \infty$ (real, far in front); as $u \to f^{+}$ (object just inside $F$ , e.g. $u = - 10 + ε$ ), $\frac{1}{v} \to 0^{+}$ so $v \to + \infty$ (virtual, far behind). The image genuinely swings through infinity and switches sides. This is why a bulb at the focus of a torch mirror makes a parallel beam. ✔

Case E — Concave, object inside F (virtual, erect, magnified)

Case F — Convex mirror (always virtual, always shrunk)

Worked example Ex 6 (cell F) — prove it can never do otherwise

Convex $f = + 15 cm$ , object $u = - 20 cm$ . Find $v$ , $m$ , and argue the result is forced.

Forecast: For convex, $f > 0$ and a real object gives $u < 0$ . Can $v$ ever come out negative (real)? Try to prove no before computing.

Solve: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} - \frac{1}{- 20} = \frac{1}{15} + \frac{1}{20} = \frac{4 + 3}{60} = \frac{7}{60}$ . $v = \frac{60}{7} \approx + 8.57 cm$ Why this step? Both $\frac{1}{f}$ (positive, convex) and $- \frac{1}{u}$ (positive, since $u < 0$ ) add up, forcing $\frac{1}{v} > 0$ — this is the algebra showing us the image must be behind the mirror before we even interpret it.
Magnification: $m = - \frac{v}{u} = - \frac{60/7}{- 20} = \frac{60}{140} = \frac{3}{7} \approx + 0.43$ . Erect, diminished. Why this step? $0 < m < 1$ and $m > 0$ means every convex-mirror image is smaller and upright.
The forced argument. Let the real object be at distance $d = ∣ u ∣ > 0$ , so $u = - d$ and $f > 0$ . Then $\frac{1}{v} = \frac{1}{f} + \frac{1}{d} > 0 \Rightarrow v > 0 always (image virtual, behind).$ For the size, $\frac{1}{v} = \frac{1}{f} + \frac{1}{d} > \frac{1}{d}$ , so $v < d = ∣ u ∣$ , giving $∣ m ∣ = \frac{v}{∣ u ∣} < 1$ always (diminished). No claim about $∣ f ∣$ is needed — only $f > 0$ and $u < 0$ .

Verify: $\frac{7}{60} + \frac{1}{- 20} = \frac{7}{60} - \frac{3}{60} = \frac{4}{60} = \frac{1}{15} = \frac{1}{f}$ . ✔ This is exactly why a car side-mirror (convex) shows a wide, shrunken world: it can never produce a big scary real image.

Case G — Plane mirror as the $R \to \infty$ limit

Worked example Ex 7 (cell G) — a plane mirror is a mirror with

$f = \infty$ Treat a plane mirror as a spherical mirror with $R \to \infty$ , so $f = \frac{R}{2} \to \infty$ . Object $u = - 25 cm$ . Find $v$ and $m$ .

Forecast: Flat mirrors put your image "as far behind as you are in front," same size, upright. Does the equation reproduce that?

Take the limit: $\frac{1}{f} \to 0$ , so $\frac{1}{v} + \frac{1}{u} = 0 \Rightarrow \frac{1}{v} = - \frac{1}{u} \Rightarrow v = - u$ . Why this step? Setting $\frac{1}{f} = 0$ is the mathematical meaning of "no curvature" ( $R \to \infty$ ). It collapses the mirror equation to a mirror-image about the pole.
Plug: $v = - u = - (- 25) = + 25 cm$ . Positive ⇒ behind the mirror ⇒ virtual, and $∣ v ∣ = 25 = ∣ u ∣$ ⇒ equal distance. Why this step? Substituting the actual object distance turns the general relation $v = - u$ into the concrete "25 cm behind" fact, and the positive sign is what tells us it is a virtual image at equal distance — the defining property of a plane mirror.
Magnification: $m = - \frac{v}{u} = - \frac{25}{- 25} = + 1$ . Erect, same size. Why this step? $m = + 1$ is the numerical fingerprint of "upright and identical in size," confirming the plane-mirror behaviour falls straight out of the general equation.

Verify: $\frac{1}{25} + \frac{1}{- 25} = 0 = \frac{1}{f}$ . ✔ Virtual, erect, same size, equidistant — the textbook plane-mirror facts, recovered as a limit of the general equation.

Case H — Real-world word problem (convex car mirror)

Worked example Ex 8 (cell H) — "objects are closer than they appear"

A car's convex side mirror has $R = 2.0 m$ . A motorcycle is $8.0 m$ behind the car (i.e. in front of the mirror). Find the image position and how magnified/diminished it looks.

Forecast: Convex ⇒ image virtual, shrunk, and very close to the mirror. Will $∣ m ∣$ be tiny?

Sign the inputs. Convex ⇒ $R > 0$ , so $f = \frac{R}{2} = + 1.0 m$ . Object in front ⇒ $u = - 8.0 m$ . Why this step? Word problems bury the signs in words; extracting them (and using $f = R /2$ ) correctly is half the battle.
Solve: $\frac{1}{v} = \frac{1}{1.0} - \frac{1}{- 8.0} = 1 + \frac{1}{8} = \frac{9}{8} \Rightarrow v = \frac{8}{9} \approx + 0.889 m$ . Why this step? This is the mirror equation doing its one job — turning known $f$ and $u$ into the image position; the positive result already flags a virtual image close behind the mirror.
Magnification: $m = - \frac{v}{u} = - \frac{8/9}{- 8} = \frac{1}{9} \approx + 0.111$ . Why this step? We need the size to answer the real-world question. $m \approx 0.11$ means the bike looks $\sim 9 \times$ smaller — the eye reads "small" as "far away," which is why the etched warning says objects are closer than they appear.

Verify: $\frac{9}{8} + \frac{1}{- 8} = \frac{8}{8} = 1 = \frac{1}{f}$ . ✔ Virtual ( $v > 0$ ), heavily diminished — safe wide-angle view, cell H.

Case I — Exam twist: image is a given multiple of the object, find f

Worked example Ex 9 (cell I) — two answers, one sign trap

A concave mirror produces an image 3 times the size of a real object placed $20 cm$ from it. Find the focal length. (There are two valid setups — find both.)

Forecast: "3 times" doesn't say erect or inverted! Concave can give a magnified image two ways: real inverted ( $m = - 3$ ) or virtual erect ( $m = + 3$ ). Expect two focal lengths.

Set up magnification. $u = - 20 cm$ . From $m = - \frac{v}{u}$ we get $v = - m u$ . Why this step? We turn the size clue into a relation between $v$ and $u$ before touching the mirror equation, because the mirror equation needs a numeric $v$ .
Case I-a, real image ( $m = - 3$ ): $v = - (- 3) (- 20) = - 60 cm$ . Then $\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{- 60} + \frac{1}{- 20} = - \frac{1}{60} - \frac{3}{60} = - \frac{4}{60} = - \frac{1}{15}$ $f = - 15 cm$ Why this step? A real magnified image from a concave mirror is inverted, so $m$ is negative; plugging its $v$ into the equation gives the first focal length. $f < 0$ confirms concave — consistent with the problem.
Case I-b, virtual image ( $m = + 3$ ): $v = - (+ 3) (- 20) = + 60 cm$ . Then $\frac{1}{f} = \frac{1}{60} + \frac{1}{- 20} = \frac{1}{60} - \frac{3}{60} = - \frac{2}{60} = - \frac{1}{30}$ $f = - 30 cm$ Why this step? A virtual magnified image is erect, so $m$ is positive and $v > 0$ ; the equation then hands us the second focal length, again $f < 0$ (concave), so both setups are physically valid.

Verify: I-a: $\frac{1}{- 60} + \frac{1}{- 20} = - \frac{4}{60} = - \frac{1}{15} = \frac{1}{f}$ . ✔ I-b: $\frac{1}{60} + \frac{1}{- 20} = - \frac{2}{60} = - \frac{1}{30} = \frac{1}{f}$ . ✔ The trap is answering with only one; the sign of $m$ was never pinned down.

Case J — Object at the pole and the virtual-object edge case

Worked example Ex 10 (cell J) — the degenerate and the virtual-object cases

(J-a) Object at the pole, $u = 0$ . What does the mirror equation say?

Forecast: If the object is on the mirror surface, its image should be right there too — nothing to magnify.

Take the limit $u \to 0^{-}$ : $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ . As $u \to 0^{-}$ , $\frac{1}{u} \to - \infty$ , so $\frac{1}{v} \to + \infty$ , meaning $v \to 0$ . Why this step? The finite $\frac{1}{f}$ is swamped by the huge $\frac{1}{u}$ term, so the image collapses onto the pole with the object. This shows $u = 0$ is a genuine boundary, not a normal image-forming setup.
Magnification: since $\frac{1}{v} \approx - \frac{1}{u}$ near the limit, $v \approx u$ , so $m = - \frac{v}{u} \to + 1$ . Image at the pole, same size. Why this step? Working the ratio in the limit (rather than blindly writing $- 0/0$ ) shows the size ratio tends to $+ 1$ — object and image sit together on the mirror, upright and identical. This is the degenerate edge of the matrix.

(J-b) Virtual object ( $u > 0$ ). If converging light heading toward a point behind the mirror is intercepted first, that point acts as a virtual object with $u > 0$ . The mirror equation still holds — you just feed a positive $u$ . Example: concave $f = - 10$ , virtual object $u = + 30$ :

Solve: $\frac{1}{v} = \frac{1}{- 10} - \frac{1}{30} = - \frac{3}{30} - \frac{1}{30} = - \frac{4}{30} \Rightarrow v = - 7.5 cm$ (real, in front). Why this step? Feeding a positive $u$ is the only change; the equation itself never breaks, showing the same formula covers converging-beam inputs from multi-mirror systems.

Verify: J-b: $\frac{1}{- 7.5} + \frac{1}{30} = - \frac{4}{30} + \frac{1}{30} = - \frac{3}{30} = - \frac{1}{10} = \frac{1}{f}$ . ✔ The equation never breaks; only the interpretation of a positive $u$ changes. (Virtual objects arise in multi-mirror/lens systems — see Magnification & Image Formation.)

Recall

Recall Which cell is it? (forecast then check)

Concave, object at infinity → image where? ::: at the focus $F$ ( $v = f$ ), point-size ( $m \to 0$ ) Concave, object at $2 f$ (i.e. at $C$ ) → same-size real inverted (cell B) ::: $u = v = R$ , $m = - 1$ Concave, object at focus → image at infinity (cell D) ::: $\frac{1}{v} = 0$ Convex, object anywhere real → cell F ::: always virtual, erect, diminished Plane mirror as a limit → $f = ?$ ::: $f = R /2 \to \infty$ , so $v = - u$ , $m = + 1$ Object exactly at the pole ( $u = 0$ ) → image where, size? ::: at the pole, $m = + 1$ (degenerate) "Magnified 3×" with no orientation given → how many answers? ::: two ( $m = \pm 3$ )

See also: Magnification & Image Formation, Spherical Aberration, Reflection of Light — Laws & Normal, and for the transmission analogue Refraction & Lenses — Lens Maker's Equation.