Magnification by Mirrors

Hold a shiny spoon at arm's length. On the bowl side, the concave side, you are tiny and upside-down. Tip it closer to your face and — at some point you cross without noticing — you suddenly balloon up the right way round, a giant eye filling the metal. Flip to the back, the convex side, and the whole kitchen shrinks into a neat little upright picture that never changes character however you move it. One curved piece of metal has just shown you every trick in this article: it has shrunk, flipped, magnified and re-erected an image, and there is a single number that predicts which of those it is doing at any moment. That number is the magnification, $m = -\dfrac{v}{u} = \dfrac{h'}{h}$ , and it is the finale of the mirror chapter. The last article located the image with the mirror formula; this one measures it — how big, and which way up. The beauty of $m$ is that it packs two pieces of information into one signed number. Its magnitude tells you the scaling: $|m| > 1$ magnified, $|m| < 1$ diminished, $|m| = 1$ same size. Its sign tells you the whole character of the image at once: $m < 0$ means a real, inverted image; $m > 0$ means a virtual, erect one. Read $m$ correctly and you can state an image's complete nature without a single ray drawn. The traps are all about that sign. The minus in $m = -v/u$ is not decoration — drop it and every real image you meet comes out labelled "upright", which is wrong. And a positive $m$ does not mean "magnified"; it means erect and virtual, and might well be a shrunken image. This page derives $m = -v/u$ from similar triangles, gives you the handy shortcut $m = f/(f-u)$ , and drills the two full nature tables — the shape-shifting concave mirror and the ever-reliable convex one — so that a magnification question becomes three seconds of reading a sign.

Key takeaways

Magnification $m = -\dfrac{v}{u} = \dfrac{h'}{h}$ measures both the size and the orientation of a mirror image in one signed number.
Sign of $m$ gives the nature: $m < 0$ is real and inverted; $m > 0$ is virtual and erect. Magnitude gives the scale: $|m| > 1$ magnified, $|m| < 1$ diminished, $|m| = 1$ same size.
A handy second form, straight from the mirror formula, is $m = \dfrac{f}{f - u}$ — magnification from $f$ and $u$ without first finding $v$ .
Concave mirror ( $f = -20$ cm): beyond C $\to m = -0.5$ (real, inverted, diminished); at C $\to m = -1$ ; between C and F $\to m = -2$ (magnified); inside F $\to m = +2$ (virtual, erect, magnified).
Convex mirror always gives $0 < m < 1$ — virtual, erect and diminished for every real object (e.g. $m = 0.714, 0.455, 0.200$ ).
Key idea: the single signed number $m = -v/u$ tells you at a glance whether a mirror image is real or virtual, erect or inverted, and bigger or smaller — and the minus sign is the part that knows which way up.

The core insight

Magnification $m = -v/u = h'/h$ compresses a mirror image's size and orientation into one signed number: its sign distinguishes real-inverted ( $m < 0$ ) from virtual-erect ( $m > 0$ ), and its magnitude distinguishes magnified from diminished — so reading $m$ replaces drawing the whole ray diagram.

Where this lives today

Magnification by mirrors is the reason curved reflectors are chosen for particular jobs. A reflecting telescope — Gregorian, Newtonian or Cassegrain — uses a concave primary to form a small real image of the sky, then a secondary mirror to magnify it for the eye or camera; the whole design is a magnification budget in glass and metal. A make-up or shaving mirror is a concave mirror used with the face held inside its focus, exactly the $m > 1$ , erect, virtual regime, so you see an enlarged upright view to shave or apply make-up by. A dentist's mirror does the same for a tooth. On the diverging side, the passenger-side car mirror, the convex mirrors at blind road junctions, in shops and on buses, and the domed security mirrors in banks all exploit the guaranteed $0 < m < 1$ of a convex mirror: a small, upright, wide-angle image that fits a large scene into a small pane — with the honest warning that, because the image is shrunk, "objects in mirror are closer than they appear". Even the parabolic dish of a solar concentrator or a radio telescope is sized by how much it must magnify or concentrate its faint incoming signal. Wherever a mirror is curved, someone has done a magnification calculation.

The physical picture

When a mirror forms an image, two things about it can change relative to the object: its size and its orientation. A distant streetlight in a concave shaving mirror is tiny and upside-down; your own face held close in the same mirror is large and upright. Magnification is the one quantity that captures both changes together, defined as the ratio of the image height to the object height:

m = \frac{h'}{h} \qquad (1)

where: $h$ is the object's height (positive, measured upward from the axis) and $h'$ the image's height (positive if the image is upright, negative if it is inverted below the axis). Because heights carry signs in the New Cartesian convention, this ratio is automatically signed, and that sign is exactly what makes $m$ so informative. An upright image has $h' > 0$ , so $m > 0$ ; an inverted image has $h' < 0$ , so $m < 0$ . The magnitude $|m|$ is just how many times taller (or shorter) the image is than the object. The remarkable fact — the one that makes $m$ calculable rather than merely measurable — is that this height ratio equals a simple ratio of distances. The very same paraxial ray that built the mirror formula shows that the image and object heights stand in the same proportion as their distances from the mirror, and the sign works out so that:

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

where: $u$ and $v$ are the object and image distances with their New Cartesian signs. This is the workhorse. It says you never need to measure a height to find the magnification — the two distances you already computed from the mirror formula hand it to you, sign and all. The derivation below shows exactly where the minus sign is born.

Figure

The similar triangles that give the magnification: a ray from the top of the object to the pole reflects about the axis to the top of the image, making the object triangle of height h at distance u similar to the image triangle of height h prime at distance v, so h prime over h equals minus v over u

Reading the sign, reading the size

The whole skill of magnification is learning to read one signed number. Split $m$ into its two messages and it becomes automatic: The sign answers "real or virtual, erect or inverted?" A negative $m$ means the image is inverted, and for a mirror an inverted image is always a real one — so $m < 0$ reads as real and inverted. A positive $m$ means erect, and an erect mirror image is always virtual — so $m > 0$ reads as virtual and erect. There is no case of a real erect image or a virtual inverted image for a single mirror, which is why the one sign settles two questions at once. The magnitude answers "bigger or smaller?" If $|m| > 1$ the image is magnified; if $|m| < 1$ it is diminished; if $|m| = 1$ it is the same size. Crucially, size and orientation are independent: a magnification of $-0.5$ is real, inverted and diminished; a magnification of $+2$ is virtual, erect and magnified. So "the magnification is positive" tells you the image is erect and virtual — it says nothing about whether it is large or small until you look at the magnitude.

The concave mirror: a shape-shifter, position by position

Slide an object along the axis of a concave mirror and $m$ runs through its whole range. Anchor everything to a concave mirror of $f = -20\ \mathrm{cm}$ , so the focus $F$ is $20\ \mathrm{cm}$ in front and the centre of curvature $C$ is at $R = 2f = -40\ \mathrm{cm}$ , i.e. $40\ \mathrm{cm}$ in front. Every value below comes from the mirror formula and $m = -v/u$ of the last article:

Object position	$u$ (cm)	$v$ (cm)	$m$	Nature
beyond C	$-60$	$-30$	$-0.5$	real, inverted, diminished
at C	$-40$	$-40$	$-1$	real, inverted, same size
between C and F	$-30$	$-60$	$-2$	real, inverted, magnified
inside F	$-10$	$+20$	$+2$	virtual, erect, magnified

Trace the story down the column. Far out beyond C, the concave mirror shrinks the object into a small inverted picture — this is the concave mirror used as a "long shot", the way a telescope's primary throws a tiny inverted image of a galaxy. Bring the object in to C and the image is exactly life-size, still inverted, sitting right on top of the object. Cross inside C and the image swells past life-size, still real and inverted, and races outward. Reach the focus and it blows up to infinity (no image). Then, the instant the object crosses inside the focus, the magnification flips sign to $+2$ : the image jumps behind the mirror and becomes virtual, erect and magnified — the make-up mirror, the dentist's mirror, your giant eye in the spoon.

Figure

Four snapshots of a concave mirror of focal length twenty centimetres: an object beyond C gives a small inverted image, at C a same-size inverted image, between C and F a large inverted image, and inside F a large upright virtual image, with magnifications minus a half, minus one, minus two and plus two

The convex mirror: always the same, always trustworthy

A convex mirror has none of this drama, and that is precisely why it is so useful. For any real object, at any distance, a convex mirror ( $f > 0$ ) produces an image that is virtual, erect and diminished, with a magnification strictly between $0$ and $1$ . Take a convex mirror of focal length $+25\ \mathrm{cm}$ :

Object position	$u$ (cm)	$m$	Nature
close	$-10$	$0.714$	virtual, erect, diminished
moderate	$-30$	$0.455$	virtual, erect, diminished
far	$-100$	$0.200$	virtual, erect, diminished

As the object recedes, the magnification falls steadily towards zero — the further away something is, the smaller its convex-mirror image — but it never turns negative and never exceeds one. The image is always upright, always shrunken, always virtual. This unwavering behaviour is the whole reason a convex mirror is the wide-angle mirror of choice: it guarantees a small, upright, right-way-round view of a large area, whether that area is the road behind a car, a shop aisle, or a hospital corridor junction. A convex mirror never surprises you.

Derivation AMagnification is minus v over u

The magnification formula and its all-important minus sign come straight from the pole ray of the standard mirror diagram, and doing the derivation cements why an inverted image must carry a negative magnification. The goal is to prove $m = h'/h = -v/u$ using one similar-triangle pair, and to track the signs so the minus is inevitable rather than memorised. This is the result the whole nature-table reading rests on.

Step 1. Draw the object $AB$ , height $h$ , standing on the axis a distance $u$ from the pole $P$ , and its image $A'B'$ , height $h'$ , at distance $v$ . Send the one ray from the top of the object $B$ straight to the pole $P$ ; it reflects there, obeying the law of reflection with the axis as the local normal, and passes through the top of the image $B'$ .

Step 2. Identify the two triangles this pole ray makes: $ABP$ (object, from foot $A$ up to top $B$ , across to $P$ ) and $A'B'P$ (image, from foot $A'$ up to top $B'$ , across to $P$ ). Both are right-angled at the axis, and they share equal angles at $P$ because the incident and reflected rays make equal angles with the axis. So the triangles are similar.

Step 3. Write the proportionality of corresponding sides in plain magnitudes: $\dfrac{|h'|}{|h|} = \dfrac{|v|}{|u|}$ — the image is scaled from the object exactly as the image distance is scaled from the object distance.

Step 4. Restore the New Cartesian signs. The object is erect, $h > 0$ . In the diagram above the image is inverted, hanging below the axis, so $h' < 0$ ; and the real image and object are both in front, so $u < 0$ and $v < 0$ , making $v/u > 0$ . For $h'/h$ to come out negative (matching the inverted image) while $v/u$ is positive, the relation must carry a minus: $\dfrac{h'}{h} = -\dfrac{v}{u}$ . Test it on a virtual case — object inside F, where $v > 0$ , $u < 0$ , so $-v/u > 0$ — and it correctly predicts an erect image. The one formula $m = -v/u$ therefore reports orientation in every case, and the minus sign is what carries that information. Verified: $m = -v/u$ reproduces the full concave table ( $-0.5, -1, -2, +2$ ) and convex table ( $0.714, 0.455, 0.200$ ), recomputed in verify_hs2-mirror-magnification.py.

Go deeper · Derivation B · 1 minMagnification straight from f and u

Derivation BMagnification straight from f and u

Often you know the focal length and object distance and want the magnification without the detour of computing $v$ first — a real time-saver in a timed paper. The goal is to derive the compact form $m = f/(f - u)$ from the mirror formula and the definition $m = -v/u$ , and to check it against the anchor table. It is algebra, no new geometry.

Step 1. Start from the mirror formula solved for $v$ . From $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{u - f}{uf}$ , invert to get $v = \dfrac{uf}{u - f}$ .

Step 2. Substitute into $m = -\dfrac{v}{u}$ : $m = -\dfrac{1}{u}\cdot\dfrac{uf}{u - f} = -\dfrac{f}{u - f}$ .

Step 3. Tidy the sign by flipping the denominator: $m = \dfrac{f}{f - u}$ . This is the shortcut — magnification directly from the two quantities a problem usually gives you, $f$ and $u$ , with every sign built in.

Step 4. Check it against the concave anchor ( $f = -20$ ). Object at C, $u = -40$ : $m = \dfrac{-20}{-20 - (-40)} = \dfrac{-20}{20} = -1$ , correct. Inside F, $u = -10$ : $m = \dfrac{-20}{-20 - (-10)} = \dfrac{-20}{-10} = +2$ , correct. The compact form agrees with the full table row for row, and it saves you a line of working every time. Verified: $m = f/(f - u)$ agrees with $m = -v/u$ on every anchor row, recomputed in verify_hs2-mirror-magnification.py.

Limiting cases

Case	Result	Meaning
$u \to -\infty$	$m \to 0$	a very distant object images as a point — a telescope's shrunken star
concave, $u \to f^-$	$\|m\| \to \infty$	object at the focus blows the image up without limit
concave, object at C	$m = -1$	image is life-size and inverted, on top of the object
concave, object just inside F	$m$ large and positive	virtual, erect, hugely magnified — the make-up mirror
convex, any real object	$0 < m < 1$	always virtual, erect, diminished
plane mirror ( $f \to \infty$ )	$m = +1$	same size and erect — the earlier chapter's result

Common trapA positive magnification does not mean magnified — and the minus sign is not optional

Two sign confusions dominate the lost marks here. The first is reading "positive $m$ " as "magnified". It does not mean that. A positive $m$ means the image is erect and virtual; whether it is enlarged or shrunk depends on the magnitude. A convex mirror's $m = +0.714$ is positive yet the image is diminished; a concave make-up mirror's $m = +2$ is positive and magnified. Size lives in $|m|$ , orientation in the sign of $m$ , and they are independent — never let one word do both jobs. The second is dropping the minus in $m = -v/u$ . That minus is part of the definition in the New Cartesian convention, and it is the only thing that distinguishes an inverted image from an erect one. Omit it and every real image you compute comes out with a positive magnification, so you will confidently label an upside-down image "upright" and lose the nature mark. Write the minus first, then the ratio; and finish by reading the sign back as orientation, not size.

Worked examples

Board level — compute the magnification and state the nature (CBSE and NEET pattern)

A 5 cm tall object is placed in front of a concave mirror of focal length 20 cm. Find the magnification, the image height and the full nature of the image when the object is at (a) 30 cm and (b) 10 cm.

Step 1. Signs: concave, $f = -20\ \mathrm{cm}$ ; real object, so $u < 0$ . Use the shortcut $m = \dfrac{f}{f - u}$ and $h' = m h$ .

Step 2. Part (a), $u = -30\ \mathrm{cm}$ (between C at $-40$ and F at $-20$ ): $m = \dfrac{-20}{-20 - (-30)} = \dfrac{-20}{10} = \boxed{-2}$ . Image height $h' = m h = -2 \times 5 = -10\ \mathrm{cm}$ . Nature from the sign: $m < 0$ so real and inverted, $|m| = 2 > 1$ so magnified — a real, inverted image $10\ \mathrm{cm}$ tall.

Step 3. Part (b), $u = -10\ \mathrm{cm}$ (inside F): $m = \dfrac{-20}{-20 - (-10)} = \dfrac{-20}{-10} = \boxed{+2}$ . Image height $h' = +2 \times 5 = +10\ \mathrm{cm}$ . Nature: $m > 0$ so virtual and erect, $|m| = 2 > 1$ so magnified — a virtual, erect, magnified image behind the mirror. The object crossed the focus and the magnification flipped from $-2$ to $+2$ : same size, opposite character.

Board level — the convex wing mirror (CBSE and IGCSE pattern)

An object 10 cm in front of a convex mirror of focal length 25 cm. Find the magnification and describe the image.

Step 1. Signs: convex, $f = +25\ \mathrm{cm}$ ; $u = -10\ \mathrm{cm}$ . Use $m = \dfrac{f}{f - u} = \dfrac{25}{25 - (-10)} = \dfrac{25}{35}$ .

Step 2. $m = \dfrac{25}{35} = \boxed{0.714}$ . It is positive and less than one, so the image is virtual, erect and diminished — about $71\%$ of the object's height, upright, behind the mirror.

Step 3. Generalise: move the object out to $30\ \mathrm{cm}$ and $m$ falls to $0.455$ ; out to $100\ \mathrm{cm}$ and it drops to $0.200$ . A convex mirror always returns $0 < m < 1$ , which is why it shows a small, upright, wide view of everything behind you.

Competitive level — run it backwards from the image (JEE and NEET pattern)

A concave mirror of focal length 15 cm forms a real, inverted image 12 cm tall of an object 4 cm tall. Find the object distance and the image distance.

Step 1. Get the magnification from the heights, with the sign from the word "inverted": $m = \dfrac{h'}{h} = \dfrac{-12}{4} = -3$ (real and inverted, so $h'$ is negative). Deciding that sign is the crux of the question.

Step 2. Use the shortcut $m = \dfrac{f}{f - u}$ to find $u$ : $-3 = \dfrac{-15}{-15 - u}$ . Cross-multiply: $-3(-15 - u) = -15$ , so $45 + 3u = -15$ , giving $3u = -60$ and $u = \boxed{-20\ \mathrm{cm}}$ — the object is $20\ \mathrm{cm}$ in front.

Step 3. Find $v$ from $m = -v/u$ : $v = -m u = -(-3)(-20) = \boxed{-60\ \mathrm{cm}}$ — real and in front, consistent with the negative magnification.

Step 4. Check forwards: $f = -15$ , $u = -20$ gives $m = \dfrac{-15}{-15 + 20} = \dfrac{-15}{5} = -3$ , matching the data. The loop closes. Timing note: under 90 seconds. The one place it goes wrong is the sign of $m$ — a real image is inverted, so $m = -3$ , not $+3$ . Fix the sign from the words "real, inverted" before you write a single equation, and the algebra is routine.

Exam traps

Dropping the minus in $m = -v/u$ . The minus is part of the definition and carries the orientation. Omit it and every real image comes out "erect", costing you the nature mark. Write the minus first.

Reading positive $m$ as "magnified". Positive $m$ means erect and virtual, not enlarged. Whether the image is bigger or smaller is in $|m|$ : a convex mirror's $m = +0.5$ is positive but diminished. Keep sign (orientation) and magnitude (size) as separate readings.

Confusing $|m|$ with signed $m$ . When a problem says "magnified three times" it constrains $|m| = 3$ ; the sign comes from whether the image is real (inverted, $m = -3$ ) or virtual (erect, $m = +3$ ). Choosing the wrong sign puts the object on the wrong side of the focus.

Forgetting to sign the image height. $h' = m h$ inherits the sign of $m$ : a magnification of $-2$ on a $5\ \mathrm{cm}$ object gives $h' = -10\ \mathrm{cm}$ , and the minus means inverted. Reporting $+10\ \mathrm{cm}$ loses the orientation.

Board scope. Magnification by mirrors is examined by CBSE, ICSE, JEE Main, JEE Advanced, NEET, IGCSE and A-Level. IB has no geometric optics at SL or HL, so this is enrichment for an IB candidate, not exam material. In AP Physics 2 the magnification is folded into the mirror-equation objective rather than tested as a separate topic. Everyone else should be able to state an image's full nature from the sign and size of $m$ in seconds.

Origins · 1 minJames Gregory and how this was discovered

James Gregory

1638–1675 · Aberdeen and St Andrews

Gregory was a Scottish mathematician who, at just twenty-four, published Optica Promota (1663) — "The Advance of Optics" — containing the design of the first practical reflecting telescope. His Gregorian telescope used a large concave primary mirror and a small concave secondary to fold a highly magnified, upright image out through a hole in the primary to the eye. It was mirror magnification turned into an instrument: the concave primary gathers and shrinks a real image of the sky, and the secondary re-magnifies it. Gregory could not find an optician able to grind his mirrors accurately, so the design waited years to be built, but the Gregorian layout is still used today, notably in solar telescopes. A gifted analyst who also worked on infinite series close to the ideas of calculus, he became the first Professor of Mathematics at St Andrews and later at Edinburgh, dying at only thirty-six.

History

1611

— Johannes Kepler, in Dioptrice, works out how a telescope magnifies and how images invert.

1663

— James Gregory, in Optica Promota, designs the Gregorian reflecting telescope, magnifying with two concave mirrors.

1668

— Isaac Newton builds the first working reflecting telescope, with a concave primary mirror.

1672

— Laurent Cassegrain proposes his reflector, using a convex secondary mirror to magnify the image.

1845

— Lord Rosse completes the six-foot Leviathan of Parsonstown, whose giant concave mirror first resolved the spiral structure of galaxies.

Going further

Texts. NCERT Science Class 10, Ch. 10 ("Light — Reflection and Refraction"), magnification of spherical mirrors · NCERT Physics Class 12, Ch. 9 ("Ray Optics"), magnification and image nature · H. C. Verma, Concepts of Physics Vol. 1, "Geometrical Optics" · Cambridge IGCSE 0625 and A-Level 9702 on image size and curved mirrors.

Provenance. Standard geometric optics. Recompute (concave $f = -20$ cm): $u = -60 \to m = -0.5$ ; $u = -40 \to m = -1$ ; $u = -30 \to m = -2$ ; $u = -10 \to m = +2$ ; each checked both as $-v/u$ and as $f/(f - u)$ . Convex $f = +25$ cm: $u = -10 \to m = 0.714$ ; $u = -30 \to m = 0.455$ ; $u = -100 \to m = 0.200$ . Image heights: $h' = m h = -3 \times 4 = -12$ cm, and $-2 \times 5 = -10$ cm, $+2 \times 5 = +10$ cm. Inverse problem: real, inverted, $h' = -12$ , $h = 4 \to m = -3$ , and with $f = -15$ this gives $u = -20$ , $v = -60$ cm. Limits: $u \to -\infty \Rightarrow m \to 0$ ; plane mirror $f \to \infty \Rightarrow m = +1$ .

Prerequisites. The Mirror Formula and Sign Convention — the source of $m = -v/u$ and the signs — and Spherical Mirrors: Concave and Convex, for the anatomy. Leads to. the mirror-numericals archetype, which drills every combination of $u$ , $v$ , $f$ and $m$ into exam speed.

Go deeper (UG). Transverse, longitudinal and angular magnification, and how they combine in multi-element instruments like telescopes and microscopes, are developed in the university optics track. It is enrichment only; every result on this page follows from $m = -v/u$ and the sign convention, and no exam here will ask for the longitudinal or angular forms.

What is the difference between the sign and the magnitude of the magnification?

They answer different questions, and keeping them apart is the whole skill. The sign of $m$ tells you the image's character: a negative $m$ means real and inverted, a positive $m$ means virtual and erect. The magnitude $|m|$ tells you the image's size: greater than one is magnified, less than one is diminished, equal to one is the same size. The two are independent, so all four combinations occur — for example $m = -0.5$ is real, inverted and diminished, while $m = +2$ is virtual, erect and magnified. When you read a magnification, read it twice: once for its sign to get the nature, once for its magnitude to get the size.

Why does a real image always have a negative magnification for a mirror?

Because a mirror's real image is always inverted, and an inverted image has a negative height, which makes $m = h'/h$ negative. Physically, when a concave mirror actually converges rays to form a real image, the ray from the top of the object crosses the axis and lands below it, flipping the image upside-down. In the formula, a real image has an image distance $v$ of the same sign as $u$ (both negative, both in front), so $-v/u$ comes out negative. There is no single-mirror arrangement that makes a real and erect image, which is exactly why "negative magnification" and "real, inverted" are two names for the same situation.

If the magnification is positive, is the image always bigger?

No — this is the most common misreading. A positive magnification means only that the image is erect and virtual; it says nothing about size. A convex mirror gives a positive magnification for every object, yet its image is always smaller than the object ( $0 < m < 1$ ). A concave mirror used inside its focus also gives a positive magnification, but there the image is larger ( $m > 1$ ). So positive tells you "upright and behind the mirror", and you must look at whether $|m|$ is above or below one to know if it is enlarged or shrunk. Sign is orientation; magnitude is size.

How do I get the image height from the magnification?

Multiply: $h' = m \times h$ , and let the sign of $m$ carry through. If a $5\ \mathrm{cm}$ object has magnification $m = -2$ , the image height is $h' = -2 \times 5 = -10\ \mathrm{cm}$ , where the minus sign means the image is inverted and the "10" means it is $10\ \mathrm{cm}$ tall. If instead $m = +2$ , then $h' = +10\ \mathrm{cm}$ : same size, but erect. Always keep the sign on $h'$ — reporting a bare " $10\ \mathrm{cm}$ " throws away the orientation information the question is usually testing. Read $h'$ back as "so many centimetres tall, inverted or erect" depending on the sign.

Is there more than one kind of magnification?

The magnification in this chapter is the transverse (or lateral) magnification — the ratio of image height to object height, measured perpendicular to the axis. It is the one every school problem means. For completeness, there is also a longitudinal magnification describing how depth along the axis is scaled (it goes as $m^2$ for small objects), and an areal magnification of $m^2$ for how a small patch's area changes; both are university-level refinements. For your exams, "magnification" means $m = -v/u = h'/h$ , the transverse one, unless a question explicitly says otherwise — which at this level it will not.

Why does the make-up mirror magnify but the car mirror shrink?

Because they are opposite kinds of mirror used in opposite regimes. A make-up or shaving mirror is concave, and you hold your face inside its focus, which is the one situation where a concave mirror gives a virtual, erect, magnified image ( $m > 1$ ) — so you see yourself enlarged and the right way up. A car wing mirror is convex, which for every object gives a virtual, erect, diminished image ( $0 < m < 1$ ) — so it shrinks a wide scene into a small upright picture, trading magnification for field of view. Same family of physics, opposite curvature and opposite object placement, opposite result.