Sign convention for mirrors and lenses: Worked examples

Before anything, let us re-earn the two tools we will use over and over, so no symbol appears unexplained.

The scenario matrix

Every problem on this topic is one cell of this table. The last column names the example that clears it. (Recall from the definition above: $C$ = the point at distance $2∣ f ∣$ , with the same sign as $f$ .)

#	Element	Case class (the "cell")	What is being stress-tested	Example
A	Concave mirror	Object beyond $C$ (i.e. $	u	>2
B	Concave mirror	Object between $F$ and pole ($	u	<
C	Concave mirror	Object exactly at $F$ ( $u = f$ )	degenerate: image at infinity	Ex 3
D	Convex mirror	Any real object	always virtual, erect, diminished; $v > 0$	Ex 4
E	Plane mirror	Limiting case $f \to \infty$	$v = - u$ , virtual, same size	Ex 5
F	Convex lens	Object beyond $2 f$	real, inverted, diminished	Ex 6
G	Convex lens	Object inside $f$ (magnifier)	virtual, erect, magnified; $v < 0$	Ex 7
H	Convex lens	Object exactly at $F$ ( $u = f$ )	degenerate: image at infinity	Ex 8
I	Concave lens	Any real object	always virtual, erect, diminished	Ex 9
J	Word problem	Real-world (rear-view mirror)	translate words → signs	Ex 10
K	Exam twist	Two-lens combination + magnification	chaining $v$ → next $u$	Ex 11

Cells A–K together cover every quadrant of sign, both mirror and lens degenerate/limiting inputs (C, E, H), a word problem (J) and an exam twist (K).

Mirror examples

Figure — Sign convention for mirrors and lenses

Reading Figure 1. The horizontal chalk line is the principal axis (our $+ x$ , pointing right). The tall curved chalk stroke on the right is the concave mirror, with its pole P (white dot) where it meets the axis. To the left of P sit two marked points: the yellow dot is the focus $F$ and the blue dot is the centre of curvature $C$ (the $2 f$ point). The pink arrow along the top shows the incident light travelling left→right (+x). Three upright object arrows stand left of the pole — this is why every $u$ below is negative:

yellow arrow, beyond $C$ = Ex 1 (Cell A),

blue arrow, between $F$ and P = Ex 2 (Cell B),

pink arrow, planted on $F$ = Ex 3 (Cell C), whose reflected rays leave parallel (image at infinity).

The bottom labels remind you: everything left of P is negative, everything behind P (right) is positive.

Worked example Ex 1 — Cell A: concave mirror, object beyond C

A candle is 40 cm in front of a concave mirror of focal length 15 cm. Locate the image and describe it.

Forecast: In the figure this is the yellow arrow, sitting to the left of $C$ (the $2 f = 30$ cm point). Guess: real, inverted, smaller than the object. Where does $v$ land?

Assign signs. $u = - 40$ cm, $f = - 15$ cm. Why this step? In the figure the yellow object arrow is left of the pole → against incident light → negative; the concave focus $F$ is also drawn left of the pole → negative.
Substitute into the mirror formula. $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{- 15} - \frac{1}{- 40} = - \frac{1}{15} + \frac{1}{40}$ . Why this step? We use the solved-for- $v$ mirror form $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ from the summary above.
Compute. Common denominator 120: $- \frac{8}{120} + \frac{3}{120} = - \frac{5}{120} = - \frac{1}{24}$ , so $v = - 24$ cm. Why this step? Combine the two fractions into one, then invert to get $v$ itself.
Interpret the sign. $v < 0$ ⇒ (mirror rule) image lies in $- x$ , in front of the mirror = real — the same left side as the yellow object.
Magnification. $m = - \frac{v}{u} = - \frac{- 24}{- 40} = - 0.6$ ⇒ inverted, $0.6 \times$ size. Why this step? Apply the mirror magnification rule; its sign reveals orientation, its size the scaling.

Verify: $- 24$ cm lies between $F$ (15) and $C$ (30) as it must for an object beyond $C$ — consistent. Plug back: $\frac{1}{- 24} + \frac{1}{- 40} = - \frac{5}{120} - \frac{3}{120} = - \frac{8}{120} = - \frac{1}{15} = \frac{1}{f}$ . ✔

Worked example Ex 2 — Cell B: concave mirror, object inside F

A tooth is 6 cm in front of a dentist's concave mirror of focal length 10 cm. Find the image.

Forecast: This is the blue arrow in the figure, tucked between $F$ and the pole. Concave mirror + object inside $F$ is the "magnifying mirror" regime → guess virtual, erect, enlarged, so $v > 0$ .

Signs. $u = - 6$ cm, $f = - 10$ cm. Why this step? The blue object arrow and the focus both lie left of the pole in the figure → both negative.
Substitute. $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{- 10} - \frac{1}{- 6} = - \frac{1}{10} + \frac{1}{6}$ . Why this step? Same solved-for- $v$ mirror form $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ .
Compute. Denominator 30: $- \frac{3}{30} + \frac{5}{30} = \frac{2}{30} = \frac{1}{15}$ , so $v = + 15$ cm. Why this step? Merge to one fraction and invert; the positive result already signals a switch of side.
Interpret. $v > 0$ ⇒ (mirror rule) image lies in $+ x$ , behind the mirror = virtual (the positive/right zone in the figure).
Magnification. $m = - \frac{v}{u} = - \frac{15}{- 6} = + 2.5$ ⇒ erect, $2.5 \times$ magnified. Why this step? The sign of $m$ confirms "erect", matching the dentist-mirror expectation.

Verify: Positive $m > 1$ and erect matches the dentist-mirror purpose. Plug back: $\frac{1}{15} + \frac{1}{- 6} = \frac{2}{30} - \frac{5}{30} = - \frac{3}{30} = - \frac{1}{10} = \frac{1}{f}$ . ✔

Worked example Ex 3 — Cell C: concave mirror, object AT the focus (degenerate)

Object placed exactly at the focus of a concave mirror, $f = - 12$ cm, so $u = - 12$ cm.

Forecast: In the figure this is the pink arrow, planted right on $F$ . It is the boundary between the "real, beyond" and "virtual, inside" regimes. Guess: image nowhere finite — rays come out parallel.

Signs. $u = - 12$ cm, $f = - 12$ cm. Why this step? Object and focus coincide, both left of the pole → both take the same negative value.
Substitute. $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{- 12} - \frac{1}{- 12} = 0$ . Why this step? When $u = f$ the two identical fractions cancel exactly, driving $1/ v$ to zero.
Interpret. $\frac{1}{v} = 0 \Rightarrow v = \pm \infty$ : the reflected rays are parallel, the image forms at infinity. Why this step? $1/ v \to 0$ is the mathematical face of "no crossing point" — matching the pink object's parallel outgoing rays.

Verify: This is the reverse of "parallel rays converge to $F$ ": an object at $F$ must produce parallel rays (reversibility of light). $\frac{1}{u} = \frac{1}{f}$ ⇒ $\frac{1}{v} = 0$ exactly. ✔ Degenerate case handled.

Worked example Ex 4 — Cell D: convex mirror (always virtual)

A convex security mirror has $f = + 20$ cm. A person stands 60 cm away.

Forecast: Convex mirrors always give a virtual, erect, diminished image behind the mirror → $v > 0$ , $0 < m < 1$ .

Signs. $u = - 60$ cm, $f = + 20$ cm. Why this step? The object is still on the left → $u < 0$ ; but a convex focus lies behind the mirror = downstream of light → $f > 0$ .
Substitute. $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{20} - \frac{1}{- 60} = \frac{1}{20} + \frac{1}{60}$ . Why this step? Solved-for- $v$ mirror form; subtracting a negative $u$ turns into an addition.
Compute. Denominator 60: $\frac{3}{60} + \frac{1}{60} = \frac{4}{60} = \frac{1}{15}$ , so $v = + 15$ cm. Why this step? One fraction, then invert; positive $v$ confirms the image is behind the mirror.
Interpret. $v > 0$ ⇒ (mirror rule) image in $+ x$ , behind mirror = virtual.
Magnification. $m = - \frac{v}{u} = - \frac{15}{- 60} = + 0.25$ ⇒ erect, quarter size. Why this step? The magnification rule pins down orientation (erect) and shrinkage ( $0.25 \times$ ).

Verify: $0 < v < f$ and $0 < m < 1$ — the fingerprint of every convex mirror. Plug back: $\frac{1}{15} + \frac{1}{- 60} = \frac{4}{60} - \frac{1}{60} = \frac{3}{60} = \frac{1}{20} = \frac{1}{f}$ . ✔

Worked example Ex 5 — Cell E: plane mirror as the limit

$f \to \infty$ A plane mirror is a mirror of infinite radius, so $f \to \infty$ and $\frac{1}{f} \to 0$ . Object at $u = - 25$ cm.

Forecast: Everyone knows a plane mirror gives an image "as far behind as the object is in front," same size, erect. Let the formula prove it.

Take the limit. $\frac{1}{f} = 0$ , so the mirror formula becomes $\frac{1}{v} + \frac{1}{u} = 0$ . Why this step? A flat mirror is the $R \to \infty$ end of the spherical family, so its focal term vanishes.
Solve. $\frac{1}{v} = - \frac{1}{u} \Rightarrow v = - u = - (- 25) = + 25$ cm. Why this step? With the focal term gone, $v$ is forced to be the exact mirror-partner of $u$ .
Interpret. $v > 0$ ⇒ (mirror rule) image 25 cm behind the mirror = virtual.
Magnification. $m = - \frac{v}{u} = - \frac{25}{- 25} = + 1$ ⇒ erect, same size. Why this step? $m = + 1$ is the algebraic statement of "identical, upright mirror image."

Verify: $v = - u$ and $m = + 1$ is exactly the everyday "mirror image" rule. Limiting case confirmed. ✔

Lens examples

Reading Figure 2. The horizontal chalk line is again the principal axis (+x). The double-curved chalk shape at the centre is a convex lens, with its optical centre O on the axis. The two yellow dots mark the two foci $F$ (left) and $F^{'}$ (right). The pink top arrow is the incident light (+x). Crucially, the page is now split into two shaded bands, drawn straight from the "meaning of $v$ " rule above:

the blue band on the left is the $v < 0$ zone — for a lens that is the virtual side (same side as the object). Ex 7's answer lands here.

the yellow band on the right is the $v > 0$ zone — for a lens that is the real (transmitted) side. Ex 6's answer lands here.

The blue object arrow stands on the left ( $u < 0$ ). Because light passes through the lens, the real side is the right — the mirror image of the mirror case, and exactly why the lens formula carries a minus. In Ex 8 the object is pushed onto $F$ and the image runs off the right edge to infinity.

Worked example Ex 6 — Cell F: convex lens, object beyond 2f

A slide is 30 cm to the left of a projector lens with $f = + 10$ cm. Find the image.

Forecast: Object beyond $2 f$ (which is 20 cm) → real, inverted, diminished, image landing in the yellow right-hand zone of the figure between $f$ and $2 f$ .

Signs. $u = - 30$ cm, $f = + 10$ cm. Why this step? The blue object arrow is left of the lens → $u < 0$ ; a convex lens converges rays into the yellow zone on the right → focus positive.
Lens formula. $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} + \frac{1}{- 30}$ . Why this step? We use the solved-for- $v$ lens form $\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$ from the summary (the minus in $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ became a plus when we moved $\frac{1}{u}$ across).
Compute. Denominator 30: $\frac{3}{30} - \frac{1}{30} = \frac{2}{30} = \frac{1}{15}$ , so $v = + 15$ cm. Why this step? Combine to one fraction and invert; positive $v$ places the image in the yellow (real) zone.
Interpret. $v > 0$ ⇒ (lens rule) image in $+ x$ on the transmitted right side = real.
Magnification. $m = + \frac{v}{u} = \frac{15}{- 30} = - 0.5$ ⇒ inverted, half size. Why this step? The lens magnification rule fixes orientation (inverted) and scale (half).

Verify: $15$ cm lies between $f = 10$ and $2 f = 20$ as required for an object beyond $2 f$ . Plug back: $\frac{1}{15} - \frac{1}{- 30} = \frac{2}{30} + \frac{1}{30} = \frac{3}{30} = \frac{1}{10} = \frac{1}{f}$ . ✔

Worked example Ex 7 — Cell G: convex lens as a magnifier (object inside f)

A stamp sits 6 cm to the left of a magnifying lens with $f = + 10$ cm.

Forecast: Object inside the focus of a converging lens → the "magnifying glass" mode: virtual, erect, enlarged, image on the same side as the object, i.e. back in the blue $v < 0$ zone of the figure.

Signs. $u = - 6$ cm, $f = + 10$ cm. Why this step? Object on the left → $u < 0$ ; the lens is convex → $f > 0$ . The object now sits between $F$ and the lens.
Substitute. $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} + \frac{1}{- 6} = \frac{1}{10} - \frac{1}{6}$ . Why this step? Same solved-for- $v$ lens form $\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$ ; because $∣ u ∣ < f$ the second term now dominates, so watch for a sign flip.
Compute. Denominator 30: $\frac{3}{30} - \frac{5}{30} = - \frac{2}{30} = - \frac{1}{15}$ , so $v = - 15$ cm. Why this step? Merge and invert; the negative result tells us the image jumped to the left (blue) zone — a virtual image.
Interpret. $v < 0$ ⇒ (lens rule) image in $- x$ on the object's side (left) = virtual.
Magnification. $m = + \frac{v}{u} = \frac{- 15}{- 6} = + 2.5$ ⇒ erect, $2.5 \times$ enlarged. Why this step? The lens magnification rule confirms erect and enlarged — exactly what a magnifying glass should do.

Verify: Virtual + erect + enlarged is precisely what a magnifying glass does. Plug back: $\frac{1}{- 15} - \frac{1}{- 6} = - \frac{2}{30} + \frac{5}{30} = \frac{3}{30} = \frac{1}{10} = \frac{1}{f}$ . ✔

Worked example Ex 8 — Cell H: convex lens, object AT the focus (degenerate)

A collimator uses a convex lens with $f = + 10$ cm and an object placed exactly at its focus, so $u = - 10$ cm.

Forecast: This is the lens twin of Ex 3. In the figure the object arrow slides onto $F$ itself; guess the emerging rays are parallel and the image runs off to infinity — the boundary between Ex 6 (real, right) and Ex 7 (virtual, left).

Signs. $u = - 10$ cm, $f = + 10$ cm. Why this step? Object on the left → $u < 0$ ; convex lens → $f > 0$ ; here their magnitudes are equal because the object sits on $F$ .
Substitute. $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} + \frac{1}{- 10} = 0$ . Why this step? Solved-for- $v$ lens form; the two equal-and-opposite fractions cancel exactly, forcing $1/ v$ to zero.
Interpret. $\frac{1}{v} = 0 \Rightarrow v = \pm \infty$ : the transmitted rays leave parallel, the image forms at infinity (off the right edge of the figure). Why this step? $1/ v \to 0$ is again the "no crossing point" signature — this is exactly how a collimator turns a point source into a parallel beam.

Verify: Reversibility: a source at $F$ of a converging lens produces a parallel beam, the inverse of "parallel rays focus at $F^{'}$ ." $\frac{1}{u} = - \frac{1}{f}$ ⇒ $\frac{1}{f} + \frac{1}{u} = 0$ exactly. ✔ Lens degenerate case handled.

Worked example Ex 9 — Cell I: concave lens (always virtual)

A concave lens has $f = - 20$ cm. An object is 30 cm to its left.

Forecast: Diverging lens → always virtual, erect, diminished, on the object side → $v < 0$ , $0 < m < 1$ .

Signs. $u = - 30$ cm, $f = - 20$ cm. Why this step? Object on the left → $u < 0$ ; a concave-lens focus lies upstream (left of the lens) → $f < 0$ .
Substitute. $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{- 20} + \frac{1}{- 30}$ . Why this step? Solved-for- $v$ lens form; both terms are now negative, so the image is guaranteed to land on the left.
Compute. Denominator 60: $- \frac{3}{60} - \frac{2}{60} = - \frac{5}{60} = - \frac{1}{12}$ , so $v = - 12$ cm. Why this step? Combine and invert; the negative $v$ confirms a virtual image on the object's side.
Interpret. $v < 0$ ⇒ (lens rule) image in $- x$ , same side as object = virtual.
Magnification. $m = + \frac{v}{u} = \frac{- 12}{- 30} = + 0.4$ ⇒ erect, diminished. Why this step? The lens rule fixes orientation (erect) and shrinkage ( $0.4 \times$ ).

Verify: $∣ v ∣ < ∣ u ∣$ and $0 < m < 1$ — the signature of every diverging lens. Plug back: $\frac{1}{- 12} - \frac{1}{- 30} = - \frac{5}{60} + \frac{2}{60} = - \frac{3}{60} = - \frac{1}{20} = \frac{1}{f}$ . ✔

The word problem and the exam twist

Worked example Ex 10 — Cell J: real-world (car rear-view mirror)

A car's convex rear-view mirror has $f = + 2.0$ m. A vehicle 10 m behind (i.e. 10 m in front of the mirror) is imaged. If the real vehicle is 1.5 m tall, how tall is its image and is it upright?

Forecast: Convex mirror → tiny, upright virtual image ("objects are closer than they appear").

Translate to signs. "10 m in front" ⇒ $u = - 10$ m; convex ⇒ $f = + 2$ m; height $h = + 1.5$ m. Why this step? Front of the mirror is the object/left side → negative distance; convex focus is behind → positive.
Find $v$ . $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{2} - \frac{1}{- 10} = \frac{1}{2} + \frac{1}{10} = \frac{5}{10} + \frac{1}{10} = \frac{6}{10}$ , so $v = + \frac{10}{6} \approx 1.667$ m. Why this step? Solved-for- $v$ mirror form; positive $v$ = image behind the mirror (virtual).
Magnification. $m = - \frac{v}{u} = - \frac{1.667}{- 10} = + 0.1667$ . Why this step? The magnification rule turns the two distances into orientation + scale.
Image height. $h^{'} = m h = 0.1667 \times 1.5 = + 0.25$ m, i.e. 25 cm, positive ⇒ upright. Why this step? $h^{'} = m h$ converts the abstract magnification into a real measurable height.

Verify: Positive small $m$ , upright, diminished ⇒ matches the real mirror warning label. Units: metres throughout, $h^{'}$ in metres. ✔

Worked example Ex 11 — Cell K: exam twist, two thin lenses in a row

Two thin convex lenses $L_{1}$ ( $f_{1} = + 10$ cm) and $L_{2}$ ( $f_{2} = + 20$ cm) are placed 40 cm apart. An object is 15 cm to the left of $L_{1}$ . Find the final image. (See Combination of thin lenses.)

Forecast: $L_{1}$ makes an intermediate image; that image becomes the object for $L_{2}$ . Chase the signs carefully — the $u$ for $L_{2}$ is measured from $L_{2}$ 's optical centre.

Lens 1. $u_{1} = - 15$ , $f_{1} = + 10$ : $\frac{1}{v _{1}} = \frac{1}{f _{1}} + \frac{1}{u _{1}} = \frac{1}{10} + \frac{1}{- 15} = \frac{3}{30} - \frac{2}{30} = \frac{1}{30}$ , so $v_{1} = + 30$ cm (real, 30 cm right of $L_{1}$ ). Why this step? Solve $L_{1}$ alone first with the solved-for- $v$ lens form to locate the intermediate image.
Object for lens 2. The intermediate image is 30 cm right of $L_{1}$ ; the lenses are 40 cm apart, so it is $40 - 30 = 10$ cm to the left of $L_{2}$ ⇒ $u_{2} = - 10$ cm. Why this step? Every distance for $L_{2}$ is measured from its own centre; still on $L_{2}$ 's left, it acts as a normal real object → negative.
Lens 2. $f_{2} = + 20$ : $\frac{1}{v _{2}} = \frac{1}{f _{2}} + \frac{1}{u _{2}} = \frac{1}{20} + \frac{1}{- 10} = \frac{1}{20} - \frac{2}{20} = - \frac{1}{20}$ , so $v_{2} = - 20$ cm. Why this step? Apply the lens formula again to get the final image position from $L_{2}$ .
Interpret. $v_{2} < 0$ ⇒ (lens rule) final image 20 cm to the left of $L_{2}$ = virtual (for $L_{2}$ ).
Total magnification. $m = m_{1} m_{2} = \frac{v _{1}}{u _{1}} \cdot \frac{v _{2}}{u _{2}} = \frac{30}{- 15} \cdot \frac{- 20}{- 10} = (- 2) \times (2) = - 4$ ⇒ inverted, $4 \times$ . Why this step? Magnifications of a lens chain multiply; the product's sign gives the final orientation.

Verify: $m_{1} = - 2$ (inverted), $m_{2} = + 2$ , product $- 4$ . Each stage reproduces $1/ f$ : stage 1 $\frac{1}{30} - \frac{1}{- 15} = \frac{1}{10}$ ; stage 2 $\frac{1}{- 20} - \frac{1}{- 10} = \frac{1}{20}$ . ✔

Recall Fast self-test: which cell?

An erect, magnified, virtual image from a single curved mirror — which element and where is the object? ::: Concave mirror, object inside the focus (Cell B). A real, diminished, inverted image from a single lens — which lens and where? ::: Convex lens, object beyond $2 f$ (Cell F). $\frac{1}{v} = 0$ came out of your algebra for a mirror — what happened physically? ::: Object sat at the focus; image at infinity, rays parallel (Cell C). $\frac{1}{v} = 0$ came out for a convex lens — what device is this? ::: A collimator: object at $F$ , parallel beam out, image at infinity (Cell H). $v > 0$ for a lens vs $v > 0$ for a mirror — same physical meaning? ::: No: $v > 0$ always means "image in $+ x$ ", but for a lens that is real (right, transmitted) while for a mirror it is virtual (behind).