The Mirror Formula & Sign Convention

Ray diagrams are beautiful, and slow. Draw one carefully and you can see that a concave mirror throws a small, upside-down image of a distant tree onto a wall — but pin down exactly how far and exactly how tall, from a diagram, and you are fighting your ruler. Exams do not have time for rulers. They have time for one equation, applied in fifteen seconds, that turns "the object is 20 cm from a mirror of focal length 15 cm" into "the image is a real, inverted, three-times-magnified picture 60 cm in front". That equation is the mirror formula, and this article both proves it and drills it.

\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

Three symbols — the object distance $u$ , the image distance $v$ , the focal length $f$ — and a companion for size, the magnification $m = -v/u$ . Feed in any two and it hands you the third, together with the image's nature: real or virtual, upright or inverted, bigger or smaller. It is the single most examined relation in ray optics, and it is worth every minute you spend on it. But the formula is only as trustworthy as your signs. The mirror formula is written in the New Cartesian sign convention, and every quantity you put in carries a plus or a minus that is not decoration — it is the physics. A real object is always $u < 0$ . A concave mirror has $f < 0$ ; a convex one has $f > 0$ . Drop a single minus and the image lands on the wrong side of the mirror, real turns virtual, and a correct method earns zero. This page derives $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$ honestly, from similar triangles on a paraxial ray diagram; derives the magnification the same way; then works the exact cases every board reuses — concave and convex, object outside and inside the focus — with the signs shown at every step.

Key takeaways

The mirror formula $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$ relates object distance $u$ , image distance $v$ and focal length $f$ for a spherical mirror; give it any two and it returns the third.
Magnification $m = -\dfrac{v}{u} = \dfrac{h'}{h}$ gives the image's size and orientation: $m < 0$ is real and inverted, $m > 0$ is virtual and erect, $|m| > 1$ is magnified.
It is written in the New Cartesian convention: real object $u < 0$ ; concave $f < 0$ ; convex $f > 0$ — the signs are the physics, not bookkeeping.
Worked anchors (concave $f = -15$ cm): object at $u = -20 \to v = -60$ cm, $m = -3$ (real, inverted); at $C$ , $u = -30 \to v = -30$ , $m = -1$ ; inside $F$ , $u = -10 \to v = +30$ , $m = +3$ (virtual, erect).
A convex mirror ( $f = +15$ , $u = -20$ ) always gives $v = +8.571$ cm, $m = +0.4286$ — virtual, erect, diminished, every time.
Key idea: one equation, $1/v + 1/u = 1/f$ with $m = -v/u$ , locates and classifies every mirror image — but only if each of $u$ , $v$ and $f$ is entered with its New Cartesian sign.

The core insight

The mirror formula $1/v + 1/u = 1/f$ , with magnification $m = -v/u = h'/h$ , is derived from similar triangles on a paraxial ray diagram and locates and classifies any spherical-mirror image — provided every distance carries its New Cartesian sign, so that a real object is negative and a concave focal length is negative.

Where this lives today

The mirror formula is the design equation for anything that focuses light with a curved reflector. An astronomer choosing a telescope mirror uses $f = R/2$ and equation (1) to fix where the eyepiece or camera must sit; a headlamp or searchlight engineer places the filament at $u = f$ so the reflected beam comes out parallel ( $v \to \infty$ ); a solar cooker designer puts the pot at the focus of a concave dish so distant sunlight ( $u \to -\infty$ ) concentrates at $v = f$ . The wide-angle convex mirror on a car or at a blind junction is equation (1) with $f > 0$ , guaranteeing the shrunken, upright, always-virtual image that fits a whole road into a glance. In a dentist's or make-up mirror the object sits inside the focus, and the formula's jump to $v > 0$ , $m > 1$ is exactly the enlarged, upright view that makes the tool useful. Even the autofocus and image-stabilisation in a phone camera rest on the same paraxial imaging relations, scaled and computerised. One equation, entered with its signs, quietly sizes and places every reflected image in the built world.

The physical picture

A ray diagram and the mirror formula are the same statement told two ways. The diagram shows two rays from a point on the object crossing at the image; the formula is the algebra of exactly those crossings, distilled so you never have to draw them again. What makes the distillation possible is the paraxial approximation from the last article — that all the rays we use stay close to the axis — because only then do the reflection points sit effectively at the pole and the geometry becomes clean similar triangles. Set up the standard picture for a concave mirror: an upright object of height $h$ standing a distance $u$ in front of the mirror, and the image it forms, of height $h'$ , at distance $v$ . Trace two rays from the top of the object. One runs parallel to the axis and reflects through the focus $F$ . The other runs straight to the pole $P$ and reflects off it, obeying the law of reflection about the axis. Where the two reflected rays cross is the top of the image, and dropping a perpendicular to the axis completes it. The formula falls out of comparing the triangles those rays make — which is exactly what the derivation below does.

Figure

The similar-triangles construction behind the mirror formula: from the top of the object one ray runs parallel to the axis and reflects through F, another runs to the pole P and reflects about the axis, and the two meet at the top of a real inverted image; triangles at the pole and at the focus are similar

The two working equations

Everything numerical about a spherical mirror is in two lines. The first locates the image:

\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \qquad (1)

where: $u$ is the object distance, $v$ the image distance and $f$ the focal length, all measured from the pole with New Cartesian signs. Read it as a rule for combining "closeness": the reciprocal distances of object and image add up to the reciprocal focal length. Rearranged for the usual unknown, $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$ , so once you know $u$ and $f$ , the image distance drops straight out — and its sign tells you the side: $v < 0$ means a real image in front of the mirror, $v > 0$ means a virtual image behind it. The second equation gives the size and orientation:

m = \frac{h'}{h} = -\frac{v}{u} \qquad (2)

where: $h$ is the object height, $h'$ the image height, and $m$ the magnification. The minus sign is part of the definition in this convention, and it does real work: it makes $m$ come out negative for a real, inverted image (where $u$ and $v$ have the same sign) and positive for a virtual, erect image (where they differ). So the sign of $m$ reports the orientation for free — negative is inverted, positive is erect — and its size reports the scaling: $|m| > 1$ magnified, $|m| < 1$ diminished, $|m| = 1$ same size. The next article is devoted to $m$ ; here it is the tool that lets each worked example end with the image's full nature, which is what boards actually mark.

The sign convention, in one picture

Before a single number goes into equation (1), it must be dressed in its New Cartesian sign. The convention is the one stated in full in the previous article; the diagram below is the whole of it at a glance. The pole is the origin. The incident light sets the positive direction, so everything in front of the mirror — where a real object sits and where the light actually is — counts as negative, and everything behind the mirror counts as positive. Heights above the axis are positive, below it negative.

Figure

The New Cartesian sign convention on a number line: the pole is the origin, incident light travels in the positive direction, distances in front of the mirror are negative and behind it positive, so a real object has u negative, a concave mirror has f and R negative, and a convex mirror has f and R positive

With the convention fixed, the mirror formula and magnification become universal: the same two equations handle concave and convex, real and virtual, without ever being rewritten. All that changes between cases is the signs you feed in — a real object at $u < 0$ , a concave mirror at $f < 0$ , a convex mirror at $f > 0$ .

Derivation AThe mirror formula from similar triangles

The mirror formula is not a fact to memorise but a two-triangle consequence of the law of reflection, and doing the derivation once fixes both the formula and the reason the signs behave. The goal is to prove $1/v + 1/u = 1/f$ for a concave mirror forming a real image, using only similar triangles on a paraxial ray diagram, and then to show the New Cartesian signs make the same line hold for every other case. We work first with plain positive lengths and attach signs at the end.

Step 1. Set up the diagram. An object $AB$ of height $h$ stands with its foot $A$ on the axis a distance $u$ from the pole $P$ ; its image $A'B'$ , height $h'$ , forms a distance $v$ from $P$ . From the top $B$ send two rays: one parallel to the axis, striking the mirror near the pole at a point $M$ (at height $h$ , since paraxial) and reflecting through the focus $F$ ; and one straight to the pole $P$ , reflecting about the axis. They cross at $B'$ , the top of the image.

Step 2. Use the pole ray. The incident ray $BP$ and reflected ray $PB'$ make equal angles with the axis, so right triangles $ABP$ and $A'B'P$ are similar (equal angles at $A$ and $A'$ , equal vertical angles at $P$ ). Corresponding sides are therefore in proportion: $\dfrac{A'B'}{AB} = \dfrac{PA'}{PA}$ , that is $\dfrac{h'}{h} = \dfrac{v}{u}$ .

Step 3. Use the parallel ray. Its reflection $MF B'$ passes through $F$ . The little triangle $MPF$ (with $MP = h$ , the object height, since $M$ sits at the object's height near the pole) is similar to triangle $A'B'F$ (equal vertical angles at $F$ , both right-angled). So $\dfrac{A'B'}{MP} = \dfrac{FA'}{FP}$ , and since $MP = AB = h$ , this is $\dfrac{h'}{h} = \dfrac{FA'}{FP} = \dfrac{v - f}{f}$ , because $FA' = PA' - PF = v - f$ .

Step 4. Equate and simplify. Steps 2 and 3 both equal $h'/h$ , so $\dfrac{v}{u} = \dfrac{v - f}{f}$ . Cross-multiplying, $vf = u(v - f) = uv - uf$ , hence $vf + uf = uv$ . Divide every term by $uvf$ : $\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$ . Finally attach New Cartesian signs — a real object and a concave mirror's real image and focus are all on the negative side — and the identical relation $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$ holds with every symbol signed, now valid for convex mirrors and virtual images too. A quick test: concave $f = -15$ , $u = -20$ gives $\dfrac{1}{v} = -\dfrac{1}{15} + \dfrac{1}{20} = -\dfrac{1}{60}$ , so $v = -60$ cm — real, in front, as the diagram promised. Verified: the anchor $v = -60$ cm from $f = -15$ , $u = -20$ , and the full formula on every worked case, are recomputed in verify_hs2-mirror-formula.py.

Go deeper · Derivation B · 1 minThe magnification is minus v over u

Derivation BThe magnification is minus v over u

The magnification formula $m = -v/u$ comes from the very same diagram, and the minus sign — the part students most often drop — is exactly what encodes whether the image is upright or upside-down. The goal is to derive $m = h'/h = -v/u$ from the pole-ray triangles and to see why the sign is not optional. This is the companion the next article builds a whole discussion on.

Step 1. Return to the pole-ray triangles $ABP$ and $A'B'P$ from the formula derivation. Their similarity already gave $\dfrac{|h'|}{|h|} = \dfrac{|v|}{|u|}$ in plain lengths — the image is scaled from the object by the ratio of the distances.

Step 2. Now put the signs back with the New Cartesian convention. For a real image the object is erect ( $h > 0$ ) but the image is inverted, hanging below the axis, so $h' < 0$ ; meanwhile both $u < 0$ and $v < 0$ . The magnitude relation $|h'|/|h| = |v|/|u|$ must be written so that a negative $h'$ results, which forces a minus sign: $\dfrac{h'}{h} = -\dfrac{v}{u}$ .

Step 3. Check the sign does the right job in every case. Real image ( $u, v$ same sign): $-v/u < 0$ , so $m < 0$ — inverted, correct. Virtual image ( $u < 0$ , $v > 0$ ): $-v/u > 0$ , so $m > 0$ — erect, correct. So the one formula $m = -v/u$ reports orientation automatically, and dropping its minus sign silently flips every image you classify.

Step 4. Read off the meaning of the number. With $m = -v/u = h'/h$ : the sign gives orientation (negative inverted, positive erect), and the magnitude gives scale ( $|m|>1$ magnified, $|m|<1$ diminished, $|m|=1$ same size). For concave $f = -15$ , $u = -20$ : $m = -(-60)/(-20) = -3$ , a real image three times the object's height and inverted — the same case the formula derivation located at $v = -60$ cm. Verified: $m = -3$ for that case and the sign behaviour across real and virtual images are recomputed in verify_hs2-mirror-formula.py.

Limiting cases

Case	Result	Meaning
$u \to -\infty$	$v \to f$	a very distant object images at the focus (the Sun on a screen)
$u \to f$ (object at focus)	$v \to \infty$	reflected rays leave parallel — no image forms
concave, object inside $F$	$v > 0$ , $m > 0$	image jumps behind the mirror: virtual, erect, magnified
convex, any real object	$v > 0$ , $0 < m < 1$	always virtual, erect, diminished
$u = v$	object and image at $C$	$m = -1$ , same size, inverted, at the centre of curvature
$f \to \infty$ (flat mirror)	$v = -u$ , $m = +1$	recovers the plane-mirror result of two articles ago

Common trapThe minus signs are the physics — dropping one changes the answer, not just its label

The single most expensive habit on the mirror formula is treating the signs as afterthoughts. They are not: a real object is $u < 0$ , a concave mirror is $f < 0$ , and if you feed in $u = +20$ or $f = +15$ for a real object in front of a concave mirror, the formula faithfully returns an image on the wrong side of the glass, with the wrong reality and the wrong orientation — and every mark that depended on it is gone. Three rules prevent it. First, a real object always has $u < 0$ — write the minus before you write the number. Second, put in $f < 0$ for concave and $f > 0$ for convex, from the mirror type, every time. Third, never mix New Cartesian with the older "real-is-positive" convention within one problem; if your textbook uses the latter, declare it and stay there. And finish every answer by reading the signs back: $v < 0$ is real and in front, $v > 0$ is virtual and behind, $m < 0$ is inverted, $m > 0$ is erect. Boards award marks for stating that nature — the arithmetic is only half the question.

Worked examples

Board level — one concave mirror, three object positions (CBSE and NEET pattern)

A concave mirror has focal length 15 cm. Find the image distance, the magnification and the full nature of the image when a 4 cm tall object is placed (a) 20 cm, (b) 30 cm, and (c) 10 cm in front of the mirror.

Step 1. Fix the signs once: concave, so $f = -15\ \mathrm{cm}$ ; real object throughout, so $u$ is negative in every part. Use $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$ and $m = -v/u$ .

Step 2. Part (a), $u = -20\ \mathrm{cm}$ : $\dfrac{1}{v} = -\dfrac{1}{15} + \dfrac{1}{20} = \dfrac{-4 + 3}{60} = -\dfrac{1}{60}$ , so $v = \boxed{-60\ \mathrm{cm}}$ . Then $m = -\dfrac{v}{u} = -\dfrac{-60}{-20} = \boxed{-3}$ . Nature: $v < 0$ real and in front, $m < 0$ inverted, $|m| > 1$ magnified — a real, inverted, magnified image, 12 cm tall, 60 cm from the mirror.

Step 3. Part (b), $u = -30\ \mathrm{cm}$ (the object is now at the centre of curvature, since $R = 2f = -30\ \mathrm{cm}$ ): $\dfrac{1}{v} = -\dfrac{1}{15} + \dfrac{1}{30} = \dfrac{-2 + 1}{30} = -\dfrac{1}{30}$ , so $v = \boxed{-30\ \mathrm{cm}}$ and $m = -\dfrac{-30}{-30} = \boxed{-1}$ : real, inverted, same size, sitting on the object at $C$ .

Step 4. Part (c), $u = -10\ \mathrm{cm}$ (now inside the focus): $\dfrac{1}{v} = -\dfrac{1}{15} + \dfrac{1}{10} = \dfrac{-2 + 3}{30} = \dfrac{1}{30}$ , so $v = \boxed{+30\ \mathrm{cm}}$ and $m = -\dfrac{+30}{-10} = \boxed{+3}$ : the image has jumped behind the mirror — $v > 0$ virtual, $m > 0$ erect, magnified. This is the make-up mirror. Notice how the single sign of $v$ flipped the whole story as the object crossed the focus.

Board level — the convex mirror always behaves (CBSE and ICSE pattern)

An object is placed 20 cm in front of a convex mirror of focal length 15 cm. Find the image distance, magnification and nature.

Step 1. Signs: convex, so $f = +15\ \mathrm{cm}$ ; real object, $u = -20\ \mathrm{cm}$ .

Step 2. $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{15} + \dfrac{1}{20} = \dfrac{4 + 3}{60} = \dfrac{7}{60}$ , so $v = \dfrac{60}{7} = \boxed{+8.571\ \mathrm{cm}}$ .

Step 3. $m = -\dfrac{v}{u} = -\dfrac{8.571}{-20} = \boxed{+0.4286}$ . Nature: $v > 0$ virtual and behind, $m > 0$ erect, $|m| < 1$ diminished — virtual, erect, diminished, as a convex mirror gives for every real object. The image is only about $43\%$ of the object's height and tucked $8.571\ \mathrm{cm}$ behind the glass, which is why a convex wing mirror packs a wide scene into a small pane.

Competitive level — invert the problem: find the mirror from the image (JEE pattern)

An object placed 20 cm in front of a concave mirror produces a real image magnified three times. Find the mirror's focal length and radius of curvature.

Step 1. Translate the words into signed quantities. Real object: $u = -20\ \mathrm{cm}$ . "Real image, magnified three times" means the image is inverted, so $m = -3$ (real is inverted, hence the minus), not $+3$ . This single sign choice is the whole question.

Step 2. Get $v$ from the magnification: $m = -\dfrac{v}{u} \Rightarrow -3 = -\dfrac{v}{-20} = -\dfrac{v}{20}$ ... rearranged, $v = -3 \times 20 = -60$ , so $v = -60\ \mathrm{cm}$ — a real image in front, consistent with the sign.

Step 3. Now use the mirror formula: $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} = -\dfrac{1}{60} - \dfrac{1}{20} = \dfrac{-1 - 3}{60} = -\dfrac{4}{60} = -\dfrac{1}{15}$ , so $f = \boxed{-15\ \mathrm{cm}}$ — concave, as the negative sign confirms. The radius is $R = 2f = \boxed{-30\ \mathrm{cm}}$ .

Step 4. Sanity check by running it forward: $f = -15$ , $u = -20$ gives $v = -60$ and $m = -3$ , exactly the data we started from. The loop closes, so the answer is secure. Timing note: about 75 seconds once you decide the sign of $m$ . The trap that eats the time — and the marks — is writing $m = +3$ for a real image; a real image is inverted, so its magnification is negative. Decide the sign of $m$ from the word "real" before you compute anything.

Exam traps

Dropping the minus signs. The number-one lost mark in mirror optics. A real object is $u < 0$ ; a concave mirror is $f < 0$ . Feeding in positive values for a real object in front of a concave mirror throws the image onto the wrong side and inverts its reality. Write the sign before the digit.

Mixing sign conventions. New Cartesian (concave $f < 0$ ) and the older real-is-positive convention (concave $f > 0$ ) give different-looking working, and blending them mid-problem is fatal. CBSE, ICSE, JEE, NEET and AP Physics 2 mark to New Cartesian; use it, declare it, and never switch mid-solution.

Forgetting $f = R/2$ . When a question gives only the radius of curvature, you must halve it (with its sign) before using the formula: $R = -30\ \mathrm{cm}$ means $f = -15\ \mathrm{cm}$ , not $-30$ . Putting $R$ where $f$ belongs doubles your focal length and wrecks the answer.

Not stating the nature. Boards award explicit marks for "real or virtual, erect or inverted, magnified or diminished". Reading it off the signs ( $v$ for side, $m$ for orientation and size) costs seconds and earns those marks; leaving the answer as a bare number throws them away.

Board scope. The mirror formula is examined by CBSE, ICSE, JEE Main, JEE Advanced, NEET, AP Physics 2 (Unit 13) and A-Level, and appears in IGCSE 0625 at the qualitative-to-basic level. IB has no geometric optics at SL or HL, so if you are an IB candidate the mirror equation is not on your paper — treat this as enrichment. Everyone else needs it cold, with signs.

Origins · 1 minIsaac Barrow and how this was discovered

Isaac Barrow

1630–1677 · Cambridge

Barrow was the first Lucasian Professor of Mathematics at Cambridge — the chair he would hand to his brilliant student Isaac Newton in 1669 — and a formidable geometer of optics. His Lectiones Opticae (Optical Lectures, 1669), which the young Newton helped prepare for the press, gave a purely geometric method for finding where a mirror or lens places the image of a point: follow the rays, intersect them, and read off the image. That construction is the ancestor of the mirror and lens equations, turning the ray diagram into something you can calculate. Barrow worked before the modern algebraic formula and its signs existed, but the idea that an image has a definite, constructible location on the axis — the idea this whole article computes — is his. He later left mathematics for theology, satisfied to have launched Newton.

History

1021

— Ibn al-Haytham gives a geometric account of where mirrors and lenses form images.

1611

— Johannes Kepler, in Dioptrice, distinguishes real from virtual images, the classification the formula's signs encode.

1637

— René Descartes, in La Dioptrique, treats image formation with coordinate geometry.

1669

— Isaac Barrow, in Lectiones Opticae, constructs the image location by rays — the ancestor of the mirror equation.

1841

— Carl Friedrich Gauss, in Dioptrische Untersuchungen, casts paraxial imaging in the modern form that underlies the mirror and lens equations.

Going further

Texts. NCERT Science Class 10, Ch. 10 ("Light — Reflection and Refraction"), the mirror formula and magnification · NCERT Physics Class 12, Ch. 9 ("Ray Optics"), reflection by spherical mirrors with sign convention · H. C. Verma, Concepts of Physics Vol. 1, "Geometrical Optics" · Cambridge A-Level 9702 curved-mirror imaging.

Provenance. Standard geometric optics. Recompute (concave $f = -15$ cm): $u = -20 \to v = -60$ , $m = -3$ ; $u = -30 \to v = -30$ , $m = -1$ ; $u = -10 \to v = +30$ , $m = +3$ ; convex $f = +15$ , $u = -20 \to v = +8.571$ , $m = +0.4286$ ; the inverse problem $u = -20$ , $m = -3 \to v = -60 \to f = -15$ , $R = -30$ ; and the image height $h' = m h = -3 \times 4 = -12$ cm. Limits: $u \to -\infty \Rightarrow v \to f$ ; $u = f \Rightarrow 1/v = 0$ .

Prerequisites. Spherical Mirrors: Concave and Convex — the anatomy, $f = R/2$ and the sign convention this article computes with. Leads to. Magnification by Mirrors, which builds the full nature table on $m = -v/u$ ; and the mirror-numericals drill that follows.

Go deeper (UG). The mirror equation is one line of the general paraxial (Gaussian) imaging theory, extended to lenses, mirror-lens systems and aberrations, in Plane and Spherical Mirrors in the university optics track. It is enrichment only; every result on this page follows from similar triangles and the sign convention, and no exam here will ask for the matrix or aberration methods.

Why is there a minus sign in m = -v/u?

Because the magnification must report orientation, not just scale, and the minus sign is what makes it do so. In the New Cartesian convention a real, inverted image has object and image distances of the same sign (both negative), so $-v/u$ comes out negative — correctly flagging the image as inverted. A virtual, erect image has them of opposite sign, so $-v/u$ is positive — correctly flagging erect. If you drop the minus, every real image would come out with a positive magnification and you would call an upside-down image upright. The sign is not decoration; it is the part of the formula that knows which way up the image is.

How do I know if the image is real or virtual without drawing?

Read the sign of $v$ after you compute it. In the New Cartesian convention a negative image distance ( $v < 0$ ) means the image is in front of the mirror, where the reflected rays actually cross — it is real and could be caught on a screen. A positive image distance ( $v > 0$ ) means the image is behind the mirror, where the rays only appear to meet — it is virtual. So you never need the diagram to classify: solve for $v$ , look at its sign. A concave mirror gives $v < 0$ (real) for objects beyond the focus and $v > 0$ (virtual) for objects inside it; a convex mirror gives $v > 0$ (virtual) always.

The question gives me the radius of curvature, not the focal length. What do I do?

Halve it, keeping the sign: $f = R/2$ . If the mirror is concave and $R = -30\ \mathrm{cm}$ , then $f = -15\ \mathrm{cm}$ ; if it is convex and $R = +30\ \mathrm{cm}$ , then $f = +15\ \mathrm{cm}$ . Only then put $f$ into the mirror formula. A very common slip is to use the radius directly in place of the focal length, which doubles $f$ and moves the image to the wrong place. Whenever a problem quotes $R$ , your first written line should be " $f = R/2 = \dots$ " with the sign attached, so the halving never gets forgotten under time pressure.

Can I use the same formula for both concave and convex mirrors?

Yes — that is the whole point of a sign convention. The single equation $1/v + 1/u = 1/f$ , with $m = -v/u$ , handles concave mirrors, convex mirrors, real images and virtual images without ever being rewritten. What changes between cases is only the signs of the numbers you substitute: concave takes $f < 0$ , convex takes $f > 0$ , a real object takes $u < 0$ , and the formula then returns a $v$ whose sign tells you the image's side and an $m$ whose sign tells you its orientation. One formula, many cases, all distinguished by signs — which is exactly why getting the signs right matters so much.

What happens when the object is exactly at the focus?

The formula gives $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{f} - \dfrac{1}{f} = 0$ , so $v \to \infty$ : no image forms at any finite distance. Physically, a point source placed at the focus of a concave mirror reflects into a beam of rays all parallel to each other — they never reconverge, so there is no image to catch. This is the exact reverse of a distant object imaging at the focus, and it is the principle of a searchlight or a car headlamp, where the bulb sits at the focus precisely so the light leaves as a parallel beam. Mathematically it is the boundary between the real-image regime (object beyond $F$ ) and the virtual-image regime (object inside $F$ ).

Why do real objects always have a negative distance?

Because of where they sit relative to the incoming light. The New Cartesian convention measures distances from the pole and takes the direction of the incident light as positive. A real object is always in front of the mirror, on the same side as the light travelling towards it — which is the direction opposite to how we count positive for the reflected side, so its distance comes out negative. It is not that objects are "special"; it is simply that "in front of the mirror" is the negative side in this convention. The only way to get a positive object distance is a virtual object — converging light headed towards a point behind the mirror — which you will not meet until multi-mirror or lens-mirror systems well beyond this chapter.