Sign convention for mirrors and lenses: Exercises

Before anything else, meet the origin that every measurement starts from, and the four letters that appear in every problem on this page. The letters are not lengths — they are signed coordinates on one axis.

Now the shared picture. Every solution below points back to it, so read it carefully once.

Figure — Sign convention for mirrors and lenses

Trace it with your finger: the black dot at the centre is the pole (mirror) / optical centre (lens) — the origin all distances start from. The blue arrow is the incident light going left→right — that chosen direction is $+ x$ . The orange arrow is a real object, sitting upstream (left) — that is why $u < 0$ in every problem below. The red half-line is the negative (upstream) region; the green half-line is the positive (downstream) region. The two vertical black arrows at the right show the height rule: a point above the axis has $h > 0$ , below has $h < 0$ — that is the fact that decides the sign of $m$ in the analysis problems. Whenever a solution says "upstream ⇒ negative" or "above the axis ⇒ $h > 0$ ", it is reading straight off this figure.

Level 1 — Recognition

These test only: can you assign the correct sign to a quantity? No formula yet. We deliberately walk through all four elements plus a refracting surface and a height, so no sign is left untested.

Recall Solution — L1·Q1

Read every quantity off the shared figure above (all distances from the pole).

Object is on the left, upstream (red region) ⇒ $u = - 25 cm$ .
A concave mirror's focus lies in front of the mirror (left) ⇒ upstream ⇒ $f = - 12 cm$ .
The object points up, above the axis ⇒ $h = + 3 cm$ .

Answer: $u = - 25 cm$ , $f = - 12 cm$ , $h = + 3 cm$ .

Recall Solution — L1·Q2

(a) A convex lens is converging: light passes through and the rays meet at the second focus downstream (right, green region) ⇒ $f = + 20 cm$ .
(b) A concave lens is diverging: the rays spread out and only appear to come from a focus upstream (left, red region) ⇒ $f = - 20 cm$ .

Answer: convex $f = + 20$ cm; concave $f = - 20$ cm.

Recall Solution — L1·Q3

$R$ is just the position of the centre of curvature $C$ , read off the same axis.

(a) $C$ downstream (right, green region) ⇒ $R > 0$ (e.g. $+ 20$ cm).
(b) $C$ upstream (left, red region) ⇒ $R < 0$ (e.g. $- 20$ cm).

Answer: (a) $R > 0$ ; (b) $R < 0$ .

Level 2 — Application

Now plug the signs into $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ (mirror) or $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ (lens). See Mirror formula derivation and Lens formula and lensmaker's equation for where those come from.

Recall Solution — L2·Q1

Step 1 (signs): $u = - 30$ , $f = - 10$ , $h = + 2$ cm. Both $u$ and $f$ upstream (red region of the figure) ⇒ both negative. Step 2 (formula): $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{- 10} - \frac{1}{- 30} = - \frac{1}{10} + \frac{1}{30} = - \frac{3}{30} + \frac{1}{30} = - \frac{2}{30} .$ Step 3: $v = - 15 cm$ . Step 4 (nature from the sign): $v < 0$ ⇒ image is on the left (in front) ⇒ real. See Real and virtual images. Step 5 (orientation): $m = - \frac{v}{u} = - \frac{- 15}{- 30} = - 0.5$ . $m < 0$ ⇒ inverted; $∣ m ∣ < 1$ ⇒ diminished. Image height $h^{'} = m h = (- 0.5) (+ 2) = - 1$ cm (below axis, confirming inverted). Answer: $v = - 15 cm$ , real, inverted, half-size ( $h^{'} = - 1$ cm), in front of the mirror.

Recall Solution — L2·Q2

Step 1: $u = - 20$ , $f = + 10$ . Step 2 (lens = minus): $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} + \frac{1}{- 20} = \frac{2}{20} - \frac{1}{20} = \frac{1}{20} .$ Step 3: $v = + 20 cm$ . Nature: $v > 0$ ⇒ image on the right (green region) ⇒ real (for a lens, real images are downstream). Object at $2 f$ ⇒ image at $2 f$ . ✔

Level 3 — Analysis

Here the sign of the answer must be interpreted, and magnification $m$ (defined at the top as $h^{'} / h$ ) enters — see Magnification in optics. We deliberately include both lens edge cases here: a convex lens with the object inside its focus, and a concave lens — the two configurations that give lenses their characteristic virtual images.

Recall Solution — L3·Q1

Step 1: $u = - 20$ , $f = + 15$ (convex focus is behind the mirror, downstream, green region ⇒ positive), $h = + 4$ cm. Step 2: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} - \frac{1}{- 20} = \frac{1}{15} + \frac{1}{20} = \frac{4}{60} + \frac{3}{60} = \frac{7}{60} .$ Step 3: $v = \frac{60}{7} = + 8.571 cm$ . Positive ⇒ behind the mirror ⇒ virtual. Step 4 (magnification, mirror): $m = - \frac{v}{u} = - \frac{60/7}{- 20} = + \frac{60}{140} = + 0.4286.$ Step 5 (image height): $h^{'} = m h = (+ 0.4286) (+ 4) = + 1.71 cm$ . Positive ⇒ above the axis ⇒ erect, and shorter than 4 cm ⇒ diminished. Interpret: $m > 0$ ⇒ erect; $∣ m ∣ < 1$ ⇒ diminished — exactly what a convex mirror always produces. ✔ Answer: $v \approx + 8.57$ cm, virtual, erect, diminished, $h^{'} \approx + 1.71$ cm ( $m \approx + 0.43$ ).

Recall Solution — L3·Q2

Step 1: $u = - 8$ , $f = - 12$ (object inside the focus), $h = + 2$ cm. Step 2: $\frac{1}{v} = \frac{1}{- 12} - \frac{1}{- 8} = - \frac{1}{12} + \frac{1}{8} = - \frac{2}{24} + \frac{3}{24} = \frac{1}{24} .$ Step 3: $v = + 24 cm$ . Positive ⇒ behind the mirror ⇒ virtual (a concave mirror gives a virtual image when the object is inside $f$ — like a shaving mirror up close). Step 4: $m = - \frac{v}{u} = - \frac{24}{- 8} = + 3.$ Step 5 (image height): $h^{'} = m h = (+ 3) (+ 2) = + 6 cm$ . Positive ⇒ above the axis ⇒ erect, and taller ⇒ magnified. ✔ Answer: $v = + 24$ cm, virtual, erect, magnified $3 \times$ , $h^{'} = + 6$ cm.

Recall Solution — L3·Q3

Step 1: $u = - 6$ , $f = + 10$ , $h = + 2$ cm. Note $∣ u ∣ = 6 < f = 10$ : the object is inside the focal length. Step 2 (lens = minus): $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} + \frac{1}{- 6} = \frac{3}{30} - \frac{5}{30} = - \frac{2}{30} = - \frac{1}{15} .$ Step 3: $v = - 15 cm$ . Negative ⇒ image on the left, the same side as the object ⇒ virtual (for a lens, a virtual image is upstream). Step 4 (magnification, lens): $m = + \frac{v}{u} = \frac{- 15}{- 6} = + 2.5.$ Step 5 (image height): $h^{'} = m h = (+ 2.5) (+ 2) = + 5 cm$ . Positive ⇒ above the axis ⇒ erect, and taller ⇒ magnified. Interpret: this is exactly how a magnifying glass works — object inside $f$ gives a virtual, erect, magnified image on the object's side. ✔ Answer: $v = - 15$ cm, virtual, erect, magnified $2.5 \times$ , $h^{'} = + 5$ cm.

Recall Solution — L3·Q4

Step 1: $u = - 30$ , $f = - 20$ (concave lens ⇒ focus upstream ⇒ negative), $h = + 3$ cm. Step 2 (lens = minus): $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{- 20} + \frac{1}{- 30} = - \frac{3}{60} - \frac{2}{60} = - \frac{5}{60} = - \frac{1}{12} .$ Step 3: $v = - 12 cm$ . Negative ⇒ image on the left (same side as object) ⇒ virtual. Step 4 (magnification, lens): $m = + \frac{v}{u} = \frac{- 12}{- 30} = + 0.4.$ Step 5 (image height): $h^{'} = m h = (+ 0.4) (+ 3) = + 1.2 cm$ . Positive ⇒ erect, and smaller ⇒ diminished. Interpret: a diverging lens always gives a virtual, erect, diminished image, for any real object position — that is its signature. ✔ Answer: $v = - 12$ cm, virtual, erect, diminished, $h^{'} = + 1.2$ cm ( $m = + 0.4$ ).

Level 4 — Synthesis

Multi-step: chain two elements, or solve backwards from a magnification. See Combination of thin lenses. First, one idea that Level-1 did not cover:

The figure shows it: converging rays (blue) heading toward a point $P$ on the right; the second lens (green) sits before $P$ , so for that lens, $P$ is a virtual object with $u > 0$ .

Recall Solution — L4·Q1

Idea: the image made by lens 1 becomes the object for lens 2. Handle them one at a time, each with the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ .

Lens 1: $u_{1} = - 30$ , $f_{1} = + 10$ . $\frac{1}{v _{1}} = \frac{1}{10} + \frac{1}{- 30} = \frac{3}{30} - \frac{1}{30} = \frac{2}{30} \Rightarrow v_{1} = + 15 cm$ (15 cm to the right of lens 1).

Transfer: lens 2 is 5 cm right of lens 1, so this image would form $15 - 5 = 10$ cm to the right of lens 2 — beyond it. The rays are still converging toward it when they reach lens 2, so it is a virtual object (the definition above): $u_{2} = + 10 cm$ .

Lens 2: $f_{2} = - 15$ . $\frac{1}{v _{2}} = \frac{1}{f _{2}} + \frac{1}{u _{2}} = \frac{1}{- 15} + \frac{1}{10} = - \frac{2}{30} + \frac{3}{30} = \frac{1}{30} \Rightarrow v_{2} = + 30 cm .$ Answer: final image is 30 cm to the right of the concave lens, real.

Recall Solution — L4·Q2

Step 1 (read the words into signs): real inverted image ⇒ $m = - 0.5$ (inverted ⇒ $m < 0$ ; half size ⇒ $∣ m ∣ = 0.5$ ). For a mirror $m = - \frac{v}{u}$ , so $- \frac{v}{u} = - 0.5 \Rightarrow v = 0.5 u$ . Step 2 (formula): $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ with $f = - 20$ . Substitute $v = 0.5 u$ : $\frac{1}{0.5 u} + \frac{1}{u} = \frac{2}{u} + \frac{1}{u} = \frac{3}{u} = \frac{1}{- 20} .$ Step 3: $u = 3 \times (- 20) = - 60 cm$ . Answer: object is 60 cm in front of the mirror. (Check: $v = 0.5 u = - 30$ cm, real, in front. ✔)

Level 5 — Mastery

Recall Solution — L5·Q1

Step 1 (signs): the object sits at the focus, on the object side, so $u = - 18$ ; concave ⇒ $f = - 18$ . Here $u = f$ . Step 2: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{- 18} - \frac{1}{- 18} = 0.$ Step 3: $\frac{1}{v} = 0 \Rightarrow v \to \infty$ . Interpret: when the object is at the focus, the reflected rays leave parallel, so they never meet — the image is at infinity (this is the searchlight/torch principle run in reverse). A degenerate but perfectly consistent case: whenever $u = f$ , the formula forces $v \to \infty$ . ✔ Answer: $v \to \infty$ (reflected rays parallel, no finite image).

Recall Solution — L5·Q2

Step 1 (signs): $u = - 40$ , $R = + 20$ (centre of curvature downstream, right, green region — exactly L1·Q3 case a), $n_{1} = 1, n_{2} = 1.5$ . Step 2: $\frac{n _{2}}{v} = \frac{n _{2} - n _{1}}{R} + \frac{n _{1}}{u} = \frac{1.5 - 1}{20} + \frac{1}{- 40} = \frac{0.5}{20} - \frac{1}{40} = \frac{1}{40} - \frac{1}{40} = 0.$ Step 3: $\frac{1.5}{v} = 0 \Rightarrow v = \infty$ . Interpret: the object sits exactly at the surface's first focal point, so the refracted rays emerge parallel — the image is at infinity. A degenerate but perfectly consistent case, mirroring L5·Q1. ✔ Answer: $v \to \infty$ (rays exit parallel).

Recall Solution — L5·Q3

Step 1: $u = - 30$ , $R = + 20$ , $n_{1} = 1$ , $n_{2} = 1.5$ . Step 2: $\frac{1.5}{v} = \frac{0.5}{20} + \frac{1}{- 30} = \frac{1}{40} - \frac{1}{30} .$ Common denominator 120: $\frac{3}{120} - \frac{4}{120} = - \frac{1}{120} .$ Step 3: $\frac{1.5}{v} = - \frac{1}{120} \Rightarrow v = 1.5 \times (- 120) = - 180 cm .$ Interpret: $v < 0$ ⇒ image on the same side as the object (left, in the incoming air) ⇒ virtual. Because the object is now inside the first focal length, the refracted rays diverge and only appear to come from a point 180 cm behind. ✔ Answer: $v = - 180$ cm, virtual, on the object's side.

Recall check

Recall Where is the origin (from which

$u, v, f$ are measured)? The pole for a mirror (centre of the reflecting surface) or the optical centre for a lens (the middle point a ray passes through undeviated). Both are placed at $x = 0$ .

Recall Object distance

$u$ — sign for a normal real object? $u < 0$ . A real object sits upstream (left, red region of the figure), against the incident-light direction, so its coordinate is negative.

Recall Focal length

$f$ — signs for the four elements? Concave mirror $f < 0$ (focus in front, left). Convex mirror $f > 0$ (focus behind, right). Convex lens $f > 0$ (rays converge downstream, right). Concave lens $f < 0$ (rays seem to come from upstream, left).

Recall Convex lens, object inside the focus (

$∣ u ∣ < f$ ) — what image? Virtual, erect, magnified, on the object's side ( $v < 0$ , $m > 0$ , $∣ m ∣ > 1$ ) — the magnifying-glass case.

Recall Concave (diverging) lens — what image for any real object?

Always virtual, erect, diminished, on the object's side ( $v < 0$ , $0 < m < 1$ ).

Recall Image distance

$v$ — how does its sign reveal the image? Mirror: $v < 0$ ⇒ real (in front, left); $v > 0$ ⇒ virtual (behind). Lens/surface: $v > 0$ ⇒ real (downstream, right); $v < 0$ ⇒ virtual (same side as object).

Recall Magnification

$m$ and image height $h^{'}$ — what do they tell you? $m = \frac{h ^{'}}{h}$ ; mirror $m = - \frac{v}{u}$ , lens $m = + \frac{v}{u}$ . Then $h^{'} = m h$ . $m > 0$ ⇒ erect ( $h^{'}$ above axis); $m < 0$ ⇒ inverted ( $h^{'}$ below axis); $∣ m ∣ > 1$ ⇒ magnified.

Recall What happens when

$u = f$ (mirror) or object at the surface's focus? $\frac{1}{v} \to 0$ , so $v \to \infty$ : the outgoing rays are parallel and no finite image forms.

Recall Radius

$R$ of a refracting surface — sign rule? $R > 0$ if the centre of curvature lies downstream (right); $R < 0$ if upstream (left). Same rule as $u, v, f$ .

Recall Virtual object — what is it and what sign?

A downstream point ( $u > 0$ ) toward which incoming rays are already converging when the element intercepts them. Occurs at the 2nd lens of a chain when the 1st image would form beyond it.

Parent: Sign convention for mirrors and lenses.