ARCHETYPE: Spherical-Mirror Numericals

Concept articles teach you what a concave mirror does. Exams ask you to compute it, in under two minutes, with the signs correct and the nature stated. That is a different skill, and it is the one this page drills. You already met the mirror formula $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$ and the magnification $m = -\dfrac{v}{u}$ ; here they stop being facts to admire and become a machine you crank. Feed in "concave, focal length 15 cm, object 20 cm away" and out drops "real, inverted, three times as tall, 60 cm in front" — every part of it a marked answer. The whole gym runs on one four-move recipe. Write the signs first (concave $f < 0$ , convex $f > 0$ , real object $u < 0$ ; halve $R$ with its sign if that is what you were handed). Solve $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$ for $v$ . Compute $m = -v/u$ . Read the signs back into real-or-virtual, erect-or-inverted, magnified-or-diminished. Master that loop and every spherical-mirror numerical — forward or backward, concave or convex — is the same fifteen-second exercise. This ARC uses the New Cartesian convention, exactly as the live mirror-formula and mirror-magnification articles shipped it, so a real object is negative and a concave focal length is negative. The numbers here are locked to those articles on purpose: a concave $f = -15$ cm mirror with an object at $u = -20$ cm gives $v = -60$ cm and $m = -3$ on this page and on every other page in the chapter. Get the recipe into your hands, and mirror numericals become free marks.

Key takeaways

The recipe: signs first (concave $f<0$ , convex $f>0$ , real object $u<0$ ) → $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$ → $m = -\dfrac{v}{u}$ → read the signs back into the image's nature.
Concave anchor ( $f = -15$ cm): $u = -20 \to v = -60$ , $m = -3$ (real, inverted, ×3); at $C$ , $u = -30 \to v = -30$ , $m = -1$ ; inside $F$ , $u = -10 \to v = +30$ , $m = +3$ (virtual, erect).
A 6 cm object at $u=-20$ therefore gives an image $h' = m h = -3 \times 6 = -18$ cm tall.
Convex ( $f = +15$ , $u = -20$ ): $v = +8.571$ cm, $m = +0.4286$ — virtual, erect, diminished, for every real object.
Run it backwards: a real image magnified ×3 at $u = -20$ means $m = -3$ (real is inverted), so $v = -60$ , $f = -15$ cm, $R = -30$ cm.
Key idea: one signed formula, cranked four ways, both locates and classifies every spherical-mirror image — and the marks are in the signs and the stated nature, not the bare number.

The core insight

Every spherical-mirror numerical is the same four-move drill — sign the inputs (concave $f<0$ , convex $f>0$ , real object $u<0$ ), solve $1/v = 1/f - 1/u$ , take $m = -v/u$ , and read the signs back into real/virtual, erect/inverted, magnified/diminished — so the arithmetic is trivial and the exam marks live entirely in the signs and the stated nature.

Where this lives today

Every one of these numerical cases is a design calculation someone actually does. A telescope maker uses $f = R/2$ and equation (1) to fix where the camera or eyepiece sits behind a concave primary; the distant-object limit $v \to f$ is why the sensor lives at the focus. A solar cooker or concentrating-solar dish puts the pot at the focus so that sunlight ( $u \to -\infty$ ) piles up at $v = f$ . A headlamp engineer inverts that: place the filament at $u=f$ so the reflected beam leaves parallel. The shrinking, always-upright image of a convex security or wing mirror is equation (1) with $f>0$ guaranteeing $0<m<1$ , packing a wide scene into a small pane. The magnified, erect image of a dentist's or make-up mirror is the object-inside- $F$ case where $v$ flips positive and $m>1$ . Even a phone camera's autofocus solves a paraxial imaging relation like these, thousands of times a second. One signed formula, cranked, quietly sizes and places reflected images across the built world — and the exam simply asks you to crank it by hand.

The setup

A spherical-mirror numerical is not a physics problem so much as a bookkeeping problem: the physics was settled when you derived $1/v + 1/u = 1/f$ from similar triangles, and all that remains is to carry the New Cartesian signs cleanly from the question to the answer. The pole $P$ is the origin, the incident light fixes the positive direction, and everything in front of the mirror — where a real object and a real image live — is negative. That single fact drives every sign in the toolkit. Three quantities describe the mirror and where it throws its image. The object distance $u$ (always negative for the real objects you are given), the image distance $v$ (its sign tells you which side), and the focal length $f$ (negative for concave, positive for convex, and equal to $R/2$ with $R$ 's sign). Two equations tie them together: the mirror formula for where, and the magnification for how big and which way up. The art is entirely in the signs — which is why the recipe insists you write them down before you touch the arithmetic.

Figure

A concave mirror of focal length 15 cm with a real object at u equals minus 20 cm forming a real inverted image three times as tall at v equals minus 60 cm, with F at minus 15 and C at minus 30 marked and every distance carrying its New Cartesian sign

The two working equations

Everything on this page is one of these two lines, cranked. 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 from the pole in New Cartesian signs. In practice you always rearrange it to $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$ , because $u$ and $f$ are what the question hands you. The sign of the answer is half the mark: $v < 0$ is a real image in front of the mirror; $v > 0$ is a virtual image behind it. The second gives 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 doing physics, not decoration: it makes $m$ come out negative for a real, inverted image (where $u,v$ share a sign) and positive for a virtual, erect image (where they differ). So $m$ 's sign reports orientation for free, and $|m|$ reports the scaling — $|m|>1$ magnified, $|m|<1$ diminished, $|m|=1$ same size. Finally, when a question quotes only the radius of curvature, convert first with $f = R/2$ , sign and all. This ARC does not re-derive these; the mirror-formula article proves them from similar triangles. Here they are tools. What follows is the recipe made concrete, over and over, until the signs are automatic.

The four regimes at a glance

Before the worked cases, fix the map. For a concave mirror ( $f<0$ ) the image nature depends entirely on where the object sits relative to $F$ and $C$ ; a convex mirror ( $f>0$ ) has only one behaviour, no matter where the object is.

Object position (concave $f=-15$ )	$v$	$m$	nature
beyond $C$ ( $u < -30$ )	small, negative	$-1<m<0$	real, inverted, diminished
at $C$ ( $u = -30$ )	$-30$	$-1$	real, inverted, same size
between $C$ and $F$ ( $-30<u<-15$ )	large, negative	$m<-1$	real, inverted, magnified
at $F$ ( $u=-15$ )	$\to \infty$	—	no image (reflected rays parallel)
inside $F$ ( $-15<u<0$ )	positive	$m>+1$	virtual, erect, magnified
convex, any real object	positive	$0<m<1$	virtual, erect, diminished

Figure

A strip of four object positions in front of a concave mirror — beyond the centre of curvature, at the centre, between centre and focus, and inside the focus — with the image nature written under each, and the convex case at the end always virtual erect and diminished

Derivation AThe master recipe on the concave f equals minus 15 set

The point of this ARC is not a new theorem but a habit: apply the same four moves so mechanically that the signs never slip under time pressure. The goal here is to run the recipe end to end on the standard concave mirror of focal length 15 cm for three object positions — 20 cm, 30 cm and 10 cm in front — showing every sign at every line, so that the pattern (a real, inverted image swinging to virtual and erect as the object crosses the focus) is something you can reproduce blind. We quote $1/v + 1/u = 1/f$ and $m = -v/u$ from the mirror-formula article and only crank them.

Step 1. Sign the fixed data once. The mirror is concave, so $f = -15\ \mathrm{cm}$ ; the object is real in every part, so $u$ is negative. The centre of curvature sits at $R = 2f = -30\ \mathrm{cm}$ , and the focus at $-15\ \mathrm{cm}$ , which is the map for reading the answers.

Step 2. Object at $u = -20\ \mathrm{cm}$ (between $C$ and $F$ ). $\dfrac{1}{v} = \dfrac{1}{-15} - \dfrac{1}{-20} = -\dfrac{1}{15} + \dfrac{1}{20} = \dfrac{-4+3}{60} = -\dfrac{1}{60}$ , so $v = -60\ \mathrm{cm}$ . Then $m = -\dfrac{v}{u} = -\dfrac{-60}{-20} = -3$ . Read back: $v<0$ real, $m<0$ inverted, $|m|>1$ magnified — real, inverted, ×3.

Step 3. Object at $u = -30\ \mathrm{cm}$ (exactly at $C$ ). $\dfrac{1}{v} = -\dfrac{1}{15} + \dfrac{1}{30} = \dfrac{-2+1}{30} = -\dfrac{1}{30}$ , so $v = -30\ \mathrm{cm}$ and $m = -\dfrac{-30}{-30} = -1$ — real, inverted, same size, sitting on the object at the centre of curvature. This is the " $u=v$ " self-conjugate point.

Step 4. Object at $u = -10\ \mathrm{cm}$ (inside $F$ ). $\dfrac{1}{v} = -\dfrac{1}{15} + \dfrac{1}{10} = \dfrac{-2+3}{30} = +\dfrac{1}{30}$ , so $v = +30\ \mathrm{cm}$ — the sign has flipped. $m = -\dfrac{+30}{-10} = +3$ : the image has jumped behind the mirror, virtual, erect, magnified — the make-up-mirror case. With a 6 cm object at $u=-20$ , the image height is $h' = m h = -3 \times 6 = -18\ \mathrm{cm}$ (the minus records "inverted"). Notice the single sign of $v$ carried the whole story across the focus. Verified: the triple $(-60,-3)$ , $(-30,-1)$ , $(+30,+3)$ and $h'=-18$ cm are recomputed in verify_hs2-mirror-numericals.py.

Go deeper · Derivation B · 1 minRunning it backwards — from object and image to the mirror

Derivation BRunning it backwards — from object and image to the mirror

Boards and entrance exams love the inverse question: they tell you where the object is and what the image does, and ask for the mirror's focal length or radius. The goal of this derivation is to show that the same two equations run in reverse with no new ideas, and — the part that decides the marks — that the word "real" fixes the sign of $m$ before any arithmetic. We take the standard case: a real object 20 cm in front of a concave mirror produces a real image three times magnified; find $f$ and $R$ .

Step 1. Translate words into signed numbers. Real object: $u = -20\ \mathrm{cm}$ . "Real image, three times magnified" — a real image is inverted, so its magnification is negative: $m = -3$ , not $+3$ . This one sign choice is the whole question; get it wrong and everything downstream is wrong but internally consistent, which is how students lose all the marks while feeling right.

Step 2. Get $v$ from the magnification road. $m = -\dfrac{v}{u} \Rightarrow v = -m\,u = -(-3)(-20) = -60\ \mathrm{cm}$ . The negative $v$ says "real, in front" — consistent with the word "real", a first internal check passed.

Step 3. Feed $u$ and $v$ into the mirror formula for $f$ . $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} = -\dfrac{1}{60} - \dfrac{1}{20} = \dfrac{-1-3}{60} = -\dfrac{4}{60} = -\dfrac{1}{15}$ , so $f = -15\ \mathrm{cm}$ — concave, as the minus confirms — and $R = 2f = -30\ \mathrm{cm}$ .

Step 4. Close the loop. Run it forward: $f = -15$ , $u = -20$ gives $v = -60$ and $m = -3$ , exactly the data we started with. A backward problem that round-trips is a secure answer. Verified: the inverse chain $u=-20,\ m=-3 \to v=-60 \to f=-15,\ R=-30$ and its forward round-trip are recomputed in verify_hs2-mirror-numericals.py.

Common trapThe minus signs are the physics — a real image has m less than zero, not greater

The single most expensive habit in mirror numericals is writing $m = +3$ for a real image. A real image is inverted, so its magnification is negative — always. Students who read "magnified three times" and write $+3$ have quietly assumed a virtual, erect image, and the mirror they compute comes out wrong (or convex) while their arithmetic stays flawless. Four rules inoculate you. First, a real object is $u<0$ : write the minus before the digit. Second, sign $f$ from the mirror type — concave $f<0$ , convex $f>0$ — every single time. Third, when handed $R$ , convert with $f=R/2$ keeping the sign before anything else. Fourth, decide the sign of $m$ from the words "real" (inverted, $m<0$ ) or "virtual/erect" ( $m>0$ ) before you compute. And never blend New Cartesian with the older "real-is-positive" convention mid-problem — CBSE, ICSE, JEE and NEET all mark to New Cartesian; declare it and stay there. Finish every answer by reading the signs back into words, because that stated nature is what the mark scheme rewards.

Worked examples

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

A concave mirror has focal length 15 cm. A 6 cm tall object is placed (a) 20 cm, (b) 30 cm and (c) 10 cm in front of it. Find the image distance, magnification, image height and full nature in each case.

Step 1. Signs: concave, $f = -15\ \mathrm{cm}$ ; real object, so $u<0$ throughout. Map: $F$ at $-15$ , $C$ at $R = 2f = -30$ .

Step 2. (a) $u = -20$ : $\dfrac{1}{v} = -\dfrac{1}{15} + \dfrac{1}{20} = -\dfrac{1}{60}$ , so $v = \boxed{-60\ \mathrm{cm}}$ ; $m = -\dfrac{-60}{-20} = \boxed{-3}$ ; $h' = m h = -3 \times 6 = -18\ \mathrm{cm}$ . Real, inverted, magnified, 18 cm tall, 60 cm in front.

Step 3. (b) $u = -30$ (at $C$ ): $\dfrac{1}{v} = -\dfrac{1}{15} + \dfrac{1}{30} = -\dfrac{1}{30}$ , so $v = \boxed{-30\ \mathrm{cm}}$ ; $m = \boxed{-1}$ ; $h' = -6\ \mathrm{cm}$ . Real, inverted, same size, at the centre of curvature.

Step 4. (c) $u = -10$ (inside $F$ ): $\dfrac{1}{v} = -\dfrac{1}{15} + \dfrac{1}{10} = +\dfrac{1}{30}$ , so $v = \boxed{+30\ \mathrm{cm}}$ ; $m = -\dfrac{+30}{-10} = \boxed{+3}$ ; $h' = +18\ \mathrm{cm}$ . The positive $v$ throws the image behind the mirror: virtual, erect, magnified — the shaving/make-up mirror.

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

Step 3. $m = -\dfrac{v}{u} = -\dfrac{8.571}{-20} = \boxed{+0.4286}$ . Read back: $v>0$ virtual and behind, $m>0$ erect, $|m|<1$ diminished — virtual, erect, diminished, at about 43% of the object's height. A convex mirror gives exactly this for every real object, which is why a wing mirror fits a wide road into a small pane.

3 · Competitive level — invert the problem, find the mirror (JEE Main pattern)

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

Step 1. Sign the words. Real object: $u = -20\ \mathrm{cm}$ . Real image magnified ×3 is inverted, so $m = -3$ (not $+3$ ). This sign is the entire question.

Step 2. $v = -m\,u = -(-3)(-20) = -60\ \mathrm{cm}$ — real, in front, consistent.

Step 3. $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} = -\dfrac{1}{60} - \dfrac{1}{20} = -\dfrac{4}{60} = -\dfrac{1}{15}$ , so $f = \boxed{-15\ \mathrm{cm}}$ and $R = 2f = \boxed{-30\ \mathrm{cm}}$ (concave, as the minus confirms).

Step 4. Forward check: $f=-15$ , $u=-20 \to v=-60$ , $m=-3$ . The loop closes. Timing note: about 75 seconds once you fix $m = -3$ from the word "real". The time-sink that also costs the marks is writing $m = +3$ for a real image; decide the sign of $m$ from "real" (inverted) before computing anything.

4 · Competitive level — the second focal table and the distant-object limit (NEET and JEE pattern)

A concave mirror has focal length 20 cm. Locate and classify the image for (a) $u = -60$ cm and (b) $u = -30$ cm. (c) Where does it image a very distant object?

Step 1. Signs: concave, $f = -20\ \mathrm{cm}$ ; $F$ at $-20$ , $C$ at $-40$ .

Step 2. (a) $u = -60$ (beyond $C$ ): $\dfrac{1}{v} = -\dfrac{1}{20} + \dfrac{1}{60} = \dfrac{-3+1}{60} = -\dfrac{2}{60} = -\dfrac{1}{30}$ , so $v = \boxed{-30\ \mathrm{cm}}$ ; $m = -\dfrac{-30}{-60} = \boxed{-0.5}$ . Real, inverted, diminished — the image lies between $F$ and $C$ , as it must for an object beyond $C$ .

Step 3. (b) $u = -30$ (between $C$ and $F$ ): $\dfrac{1}{v} = -\dfrac{1}{20} + \dfrac{1}{30} = \dfrac{-3+2}{60} = -\dfrac{1}{60}$ , so $v = \boxed{-60\ \mathrm{cm}}$ ; $m = -\dfrac{-60}{-30} = \boxed{-2}$ . Real, inverted, magnified — object and image have simply swapped roles with part (a), which is ray reversibility in action.

Step 4. (c) A distant object means $u \to -\infty$ , so $\dfrac{1}{u} \to 0$ and $\dfrac{1}{v} \to \dfrac{1}{f}$ , giving $v \to f$ . For the chapter's standard $f = -15$ cm mirror the Sun images at $\boxed{v = -15\ \mathrm{cm}}$ — a tiny, real, inverted image right at the focus, the principle of a reflecting telescope and a solar cooker. Timing note: parts (a) and (b) together take under a minute once you see they are a reversible pair — computing one gives you a strong check on the other.

Exam traps

Writing $m>0$ for a real image. A real image is inverted, so $m<0$ . "Magnified three times, real" means $m=-3$ . This is the number-one lost mark in the inverse problems.

Dropping the object's minus. A real object is $u<0$ . Feeding $u=+20$ for a real object in front of a mirror lands the image on the wrong side with the wrong reality — a correct method scoring zero.

Using $R$ where $f$ belongs. If the question gives the radius, your first line must be $f = R/2$ with the sign: $R=-30 \Rightarrow f=-15$ . Putting $R$ straight into the formula doubles your focal length.

Not stating the nature. Boards award explicit marks for "real or virtual, erect or inverted, magnified or diminished". Read it off the signs — $v$ for the side, $m$ for orientation and size — and never leave a bare number.

Mixing conventions. New Cartesian (concave $f<0$ ) and the older real-is-positive convention give different-looking working; blending them mid-solution is fatal. Declare New Cartesian and stay in it.

Board scope. This numerical mirror gym is India-facing — CBSE, ICSE, JEE Main, JEE Advanced and NEET. IGCSE 0625 has no curved mirrors at all, and no mainstream A-Level (CIE 9702 or Edexcel IAL) does spherical-mirror imaging, so if that is your board, treat this page as enrichment rather than exam content. IB has no mirror or lens image formation either.

Origins · 1 minIsaac Newton and how this was discovered

Isaac Newton

1643–1727 · Cambridge & London

Newton put a curved mirror at the heart of an instrument. Frustrated that lenses smeared colour, in 1668 he built the first working reflecting telescope, using a concave spherical mirror to gather and focus light — the very object whose image location this gym computes. His Opticks (1704) laid out a geometric, ray-by-ray optics and the prism experiments on colour, and established the computational habit of treating "where does the image form?" as a question with a definite, calculable answer. The honest scope: the tidy New Cartesian sign bookkeeping used on this page is a nineteenth-century tidy-up (Gauss and others), not Newton's own notation, and he did not write $1/v+1/u=1/f$ in these symbols. What is genuinely his is the reflecting-mirror instrument and the conviction that geometric optics is something you calculate — which is exactly what a mirror numerical does.

History

1668

— Isaac Newton builds the first reflecting telescope, focusing light with a concave spherical mirror.

1669

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

1704

— Isaac Newton publishes Opticks, systematising a geometric ray optics of reflection and refraction.

1841

— Carl Friedrich Gauss, in Dioptrische Untersuchungen, casts paraxial imaging in the modern sign-consistent form.

1857

— Léon Foucault develops silvered-glass mirrors, making precise spherical mirrors practical for instruments.

Going further

Texts. NCERT Science Class 10, Ch. 10 ("Light — Reflection and Refraction"), mirror formula and magnification numericals · NCERT Physics Class 12, Ch. 9 ("Ray Optics"), reflection by spherical mirrors with the sign convention · H. C. Verma, Concepts of Physics Vol. 1, "Geometrical Optics" · D. C. Pandey, Optics problem sets for JEE and NEET.

Provenance. Standard geometric optics; every number recomputed in verify_hs2-mirror-numericals.py. Recompute (concave $f=-15$ ): $u=-20\to v=-60,\ m=-3,\ h'=-18$ (with $h=6$ ); $u=-30\to v=-30,\ m=-1$ ; $u=-10\to v=+30,\ m=+3$ . Convex $f=+15$ , $u=-20 \to v=+8.571$ (= $60/7$ ), $m=+0.4286$ (about 43%). Inverse: $u=-20,\ m=-3 \to v=-60 \to f=-15,\ R=-30$ . Second table $f=-20$ : $u=-60\to v=-30,\ m=-0.5$ ; $u=-30\to v=-60,\ m=-2$ . Distant limit ( $f=-15$ ): $u\to-\infty \Rightarrow v\to-15$ .

Prerequisites. The Mirror Formula — where $1/v+1/u=1/f$ and $m=-v/u$ are derived from similar triangles; and Magnification by Mirrors, which builds the full nature table on $m$ . Leads to. Plane-Mirror Geometry Problems, the companion drill for flat mirrors.

Go deeper (UG). These mirror numericals are one panel of the general paraxial imaging theory — extended to lenses, mirror-lens systems and aberrations — in Plane and Spherical Mirrors in the university optics track. It is enrichment only; nothing on this page needs more than the two signed equations and the recipe.

How do I decide the sign of the magnification before I even compute v?

Read the words. If the image is described as real, it is inverted, so $m<0$ . If it is described as virtual or erect, it is upright, so $m>0$ . A concave mirror gives a real (inverted, $m<0$ ) image whenever the object is beyond the focus, and a virtual (erect, $m>0$ ) image when the object is inside the focus; a convex mirror is always virtual and erect, so $m>0$ always. Fixing this sign first — before touching arithmetic — is what makes the inverse problems ("find the mirror") quick and correct, because $v = -m\,u$ then gives the right side of the mirror straight away.

When a question gives R instead of f, what is my first line?

Write $f = R/2$ , carrying the sign. A concave mirror with $R = -30\ \mathrm{cm}$ has $f = -15\ \mathrm{cm}$ ; a convex mirror with $R = +30\ \mathrm{cm}$ has $f = +15\ \mathrm{cm}$ . Only then substitute into the mirror formula. The classic slip is to use the radius directly in place of the focal length, which doubles $f$ and moves the image to the wrong place. Making " $f=R/2=\dots$ " your literal first written line means the halving is never forgotten under time pressure, and it also reminds you to attach the sign, which is the other half of the mistake.

Why does the convex mirror always give the same kind of image?

Because with $f>0$ and a real object $u<0$ , the formula $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$ adds two positive quantities, so $\dfrac{1}{v}>0$ and hence $v>0$ — the image is always behind the mirror, always virtual. And $m = -v/u = -(\text{positive})/(\text{negative}) > 0$ , so it is always erect; since $|v|<|u|$ for any real object, $|m|<1$ , so it is always diminished. There is no object position that changes this, which is exactly why convex mirrors are used for wide-angle viewing: the image is predictably small, upright and complete, giving a reliable panoramic view at a security corner or a car's wing.

What actually happens when the object sits at the focus?

The formula gives $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{f} = 0$ , so $v \to \infty$ : no image forms at any finite distance. Physically, a point at the focus of a concave mirror reflects into a bundle of rays all parallel to each other — they never reconverge, so there is nothing to catch on a screen. This is the exact reverse of the distant-object case (where parallel rays converge to the focus) and is the operating principle of a searchlight or a car headlamp, whose bulb sits at the focus precisely so the beam comes out parallel. It is also the boundary between the real-image regime (object beyond $F$ ) and the virtual-image regime (object inside $F$ ).

Both u and v came out negative and I got a positive m — is that wrong?

No — that is exactly right for a real image. When $u$ and $v$ have the same sign, $m = -v/u$ is negative, flagging an inverted (real) image; you should get $m<0$ , not $m>0$ . If you computed a positive $m$ from two negative distances, you have dropped the minus sign in $m=-v/u$ . Re-check: for $u=-20$ , $v=-60$ , $m = -(-60)/(-20) = -(60/20) = -3$ , which is negative. The rule of thumb: two same-sign distances always give a negative (inverted, real) magnification; opposite-sign distances give a positive (erect, virtual) one.

The image height came out negative. What does the minus mean physically?

It means inverted. Since $h' = m\,h$ and the object height $h$ is taken positive (upright, above the axis), a negative magnification makes $h'$ negative, and a negative height in New Cartesian means "below the axis" — i.e. upside-down relative to the object. So $h' = -18\ \mathrm{cm}$ reads as "18 cm tall and inverted". When a board asks only for the size of the image you quote the magnitude, 18 cm; when it asks for the full description you say "18 cm, inverted", and the minus sign is what told you the orientation without a ray diagram.

Is this the same convention my textbook and my competitive exam use?

For Indian boards and entrance exams, yes: CBSE, ICSE, JEE Main, JEE Advanced and NEET all use the New Cartesian convention exactly as on this page — real object $u<0$ , concave $f<0$ , convex $f>0$ . AP Physics 2 (not an Indian exam) uses a different "real-is-positive" convention and would write some of these signs oppositely, which is why you must never mix the two inside one solution. Pick the convention your paper is marked in, declare it at the top of your working, and keep every sign consistent with that single choice. The physics is identical; only the bookkeeping differs.