Mirror equation

Point a torch at a bent spoon and the reflection warps: your face appears tiny and upright on the back of the bowl, huge and upside down on the front. No two spherical mirrors seem to behave alike — a shaving mirror magnifies, a car's side mirror shrinks and pushes objects "farther than they appear," a torch reflector throws a sharp beam from a single bulb. It looks like each mirror needs its own separate rule. It doesn't. Every one of these behaviours, without exception, falls out of one short algebraic sentence relating three distances measured from the mirror's pole: the mirror equation, $\frac1v+\frac1u=\frac1f$ . This is the payoff of everything you learned about spherical mirrors, focal length, and the Cartesian sign convention. Once you commit to measuring $u$ (object distance), $v$ (image distance), and $f$ (focal length) with consistent signs, the equation stops caring whether the mirror is concave or convex, whether the image is real or virtual, upright or inverted, magnified or shrunk. Feed it two known quantities and it hands back the third — no ray diagram, no protractor, no guesswork. Pair it with the magnification formula $m=-v/u$ and you can fully describe an image — where it forms, how big it is, which way up it is — using nothing but arithmetic. This article rebuilds the equation from similar triangles so you see why it works, not just that it works, then drills the two worked-example patterns CBSE loves: a concave mirror forming a real image, and a convex mirror forming a virtual one. By the end you should be able to solve any mirror numerical CBSE can throw at you, and — just as importantly — catch the sign-convention slip that costs the most marks in this chapter.

Key takeaways

The mirror equation $\dfrac1v+\dfrac1u=\dfrac1f$ links object distance $u$ , image distance $v$ , and focal length $f$ for any spherical mirror, concave or convex.
All distances are measured from the mirror's pole, along the principal axis, using the Cartesian sign convention: distances measured against the incident light (behind the mirror surface, toward the object side... more precisely, opposite to the direction of incident light) are negative, distances measured in the direction of incident light are positive.
For a real object placed in front of any mirror, $u$ is always negative; $f$ is negative for concave mirrors and positive for convex mirrors.
The sign of the computed $v$ tells you everything about the image: negative $v$ means a real image in front of the mirror; positive $v$ means a virtual image behind it.
Linear magnification $m=\dfrac{h'}{h}=-\dfrac{v}{u}$ gives size and orientation together: negative $m$ is inverted, positive $m$ is erect; $|m|>1$ magnified, $|m|<1$ diminished.
The equation is derived once, from similar triangles in a ray diagram, and then applies to every object position — near, far, between focus and pole, beyond the centre of curvature — without redrawing anything.
Key idea: the single relation 1/v + 1/u = 1/f, used with a consistent sign convention, converts any spherical-mirror ray diagram into pure arithmetic that gives image position, nature, and (via m = −v/u) size and orientation.

The core insight

The mirror equation $\dfrac1v+\dfrac1u=\dfrac1f$ , applied with signed distances from the pole, turns every spherical-mirror problem into arithmetic: the sign of $v$ reveals whether the image is real or virtual, and $m=-v/u$ reveals its size and orientation.

Where it fits in the syllabus

In CBSE Class 12 Unit IX, "Ray Optics and Optical Instruments," the mirror equation follows directly after spherical mirrors, the pole–focus–centre-of-curvature geometry, and the sign convention. It is one of the two headline results the board expects verbatim — the other being magnification $m=-v/u$ — and both are prescribed exactly as written here, with no aberration corrections or lens-combination complications attached to mirrors. The equation is examined almost every year, both as a short numerical (find $v$ given $u$ and $f$ ) and folded into optical-instrument questions later in the same chapter.

The physical picture

Consider a concave mirror of small aperture (so that rays stay close to the principal axis — the "paraxial" approximation every board-level mirror formula silently assumes) and pole $P$ . An object of height $h$ stands at distance $u$ in front of it. A ray from the tip of the object travels parallel to the axis, reflects through the focus $F$ ; a second ray passes through $F$ on the way in and reflects parallel to the axis. Where the two reflected rays cross is the image tip, at distance $v$ , height $h'$ .

\frac1v+\frac1u=\frac1f

where: $u$ is the object distance from the pole, $v$ is the image distance from the pole, and $f$ is the focal length, all measured along the principal axis with the Cartesian sign convention — light is taken to travel from the object toward the mirror in the negative-to-positive sense, so a real object in front of the mirror always has $u<0$ ; a concave mirror has its focus in front of it, so $f<0$ , while a convex mirror has its focus behind it, so $f>0$ . The physical reading is compact: once $u$ and $f$ are known with their correct signs, the equation is one linear relation in the single unknown $v$ , and the sign that pops out of that arithmetic is itself the answer to "is the image real or virtual" — a negative $v$ places the image in front of the mirror (real, where reflected rays actually meet), a positive $v$ places it behind the mirror (virtual, where rays only appear to diverge from). Linear magnification compares image height to object height:

m=\frac{h'}{h}=-\frac{v}{u}

where: $h$ is object height, $h'$ is image height (signed — positive if the image is erect, negative if inverted, following the convention that heights above the axis are positive), and $u,v$ carry the same signs as in the mirror equation. The physical reading: because $u$ is always negative for a real object, the minus sign in $m=-v/u$ means that whenever $v$ and $u$ share the same sign pattern the image comes out inverted, and whenever $v$ is positive while $u$ is negative the image comes out erect — which is exactly the algebra behind the everyday facts that concave mirrors invert real images and convex mirrors never invert anything for a real object.

Figure

Ray diagram for a concave mirror with an object beyond the centre of curvature, showing the pole, focus, and centre of curvature marked on the principal axis, a parallel ray reflecting through the focus, a focal ray reflecting parallel to the axis, and the two rays crossing to form a real inverted image between the focus and the centre of curvature

Deriving the mirror equation

Derivation AGetting 1/v + 1/u = 1/f from similar triangles

The goal of this derivation is to show that the mirror equation is not a memorised fact handed down from nowhere but a direct algebraic consequence of two pairs of similar triangles formed by a single ray diagram — the same diagram used to locate images qualitatively, now read quantitatively. We use a concave mirror with a real, inverted image, small-aperture (paraxial) rays so the mirror surface can be treated as effectively flat near the axis for the triangle argument, and the standard Cartesian sign convention throughout; the final relation, once derived for this one configuration, is proved in every CBSE textbook treatment to hold for all mirror types and object positions provided signs are respected consistently.

Step 1. Set up the ray diagram. Let the object $AB$ stand upright on the principal axis at distance $u$ from the pole $P$ , with $B$ on the axis and $A$ the tip. A ray from $A$ travels parallel to the axis, strikes the mirror, and reflects through the focus $F$ to meet a second ray — one that leaves $A$ , passes through $F$ on its way to the mirror, and reflects out parallel to the axis — at the image tip $A'$ , directly above $B'$ on the axis at distance $v$ from $P$ .

Step 2. Extract the first pair of similar triangles. Triangles $A'B'F$ and $MPF$ (where $M$ is the point on the mirror struck by the ray through $F$ , and $MP$ equals the object height $AB$ because that ray left the mirror parallel to the axis at the same height it entered) are similar, because their angles match. This gives $\dfrac{A'B'}{MP}=\dfrac{FB'}{FP}$ , and since $MP=AB$ , this is $\dfrac{A'B'}{AB}=\dfrac{FB'}{FP}$ .

Step 3. Extract the second pair of similar triangles. Triangles $A'B'P$ and $ABP$ are also similar (both right-angled at the axis, sharing the vertex angle at $P$ ), giving $\dfrac{A'B'}{AB}=\dfrac{PB'}{PB}$ .

Step 4. Equate the two magnification expressions. From Steps 2 and 3, $\dfrac{FB'}{FP}=\dfrac{PB'}{PB}$ . Writing $FB'=PB'-PF$ (using paraxial distances along the axis) and substituting the sign-convention distances $PB=-u$ , $PB'=-v$ , $PF=-f$ , this ratio becomes, after algebraic simplification, $\dfrac{v-f}{f}=\dfrac{v}{u}$ in magnitude form; carrying the signs through carefully and cross-multiplying converts this into $uf+vf=uv$ .

Step 5. Divide through by $uvf$ . $uf+vf=uv \implies \dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$ . This is the mirror equation. Because the sign convention was used consistently throughout the triangle algebra rather than bolted on afterward, the same relation — proved here for one concave-mirror, real-image configuration — continues to hold for convex mirrors and for virtual images; the textbook argument (extending it to other cases) simply repeats these same similar-triangle steps with the appropriate ray diagram, always landing on the identical relation. Verified: setting $u=-30\ \text{cm}$ , $f=-15\ \text{cm}$ into $1/v=1/f-1/u$ gives $v=-30\ \text{cm}$ , matching the known result that an object at the centre of curvature ( $R=2f=30\ \text{cm}$ ) images onto itself.

Worked examples

Board level - concave mirror, object beyond the focus

An object of height 2 cm is placed 30 cm in front of a concave mirror of focal length 15 cm. Find the position, nature, and magnification of the image, and its height.

Step 1. Write down the sign-convention values. Real object in front of a concave mirror: $u=-30\ \text{cm}$ ; concave mirror, focus in front: $f=-15\ \text{cm}$ ; object height $h=+2\ \text{cm}$ .

Step 2. Apply the mirror equation to solve for $v$ . $\dfrac1v=\dfrac1f-\dfrac1u=\dfrac{1}{-15}-\dfrac{1}{-30}=-\dfrac{1}{15}+\dfrac{1}{30}=\dfrac{-2+1}{30}=-\dfrac{1}{30}$ , so $v=-30\ \text{cm}$ .

Step 3. Interpret the sign of $v$ . Negative $v$ places the image 30 cm in front of the mirror, on the same side as the object — this is a real image, formed where reflected rays genuinely converge, and it could be caught on a screen.

Step 4. Compute magnification and image height. $m=-\dfrac{v}{u}=-\dfrac{(-30)}{(-30)}=-1$ , so $h'=m\times h=-1\times 2=-2\ \text{cm}$ .

Step 5. State what $m=-1$ means physically. The image is inverted (negative sign) and exactly the same size as the object (magnitude 1) — 2 cm tall, upside down. This matches the geometric fact that an object placed exactly at the centre of curvature ( $R=2f=30\ \text{cm}$ ) always forms a real, inverted, same-size image at that same point. Answer: $v=-30\ \text{cm}$ (real image), $m=-1$ , image height $2\ \text{cm}$ , inverted.

Competitive level - convex mirror with a follow-up displacement check

An object of height 4 cm is placed 20 cm in front of a convex mirror of focal length 10 cm. Find the image position, magnification, and height. Then find how the image distance changes if the object is moved to 40 cm from the mirror, and comment on the trend.

Step 1. Sign-convention values for the first position. Real object: $u=-20\ \text{cm}$ ; convex mirror, focus behind the mirror: $f=+10\ \text{cm}$ .

Step 2. Solve the mirror equation. $\dfrac1v=\dfrac1f-\dfrac1u=\dfrac{1}{10}-\dfrac{1}{-20}=\dfrac{1}{10}+\dfrac{1}{20}=\dfrac{2+1}{20}=\dfrac{3}{20}$ , giving $v=\dfrac{20}{3}\approx 6.67\ \text{cm}$ .

Step 3. Interpret and find magnification and height. Positive $v$ places the image about 6.67 cm behind the mirror — a virtual image. Magnification $m=-\dfrac{v}{u}=-\dfrac{20/3}{-20}=\dfrac{1}{3}\approx 0.33$ , so $h'=0.33\times 4\approx 1.33\ \text{cm}$ : erect and diminished.

Step 4. Repeat with $u=-40\ \text{cm}$ . $\dfrac1v=\dfrac1{10}-\dfrac1{-40}=\dfrac{4+1}{40}=\dfrac{5}{40}$ , so $v=8\ \text{cm}$ , and $m=-\dfrac{8}{-40}=0.2$ .

Step 5. Compare and state the trend. As the object moves from 20 cm to 40 cm (farther away), $v$ increases slightly from 6.67 cm to 8 cm, while $|m|$ decreases from 0.33 to 0.2 — the image shrinks and creeps toward the focus (which sits at $v=+10\ \text{cm}$ ) as the object recedes, never quite reaching it. This is the general convex-mirror behaviour: for any real object, the image stays virtual, erect, diminished, and confined between the pole and the focus. Answer: at 20 cm, $v\approx 6.67\ \text{cm}$ , $m\approx 0.33$ , height $\approx 1.33\ \text{cm}$ ; at 40 cm, $v=8\ \text{cm}$ , $m=0.2$ — the image shrinks and drifts toward $F$ as the object recedes. Timing note: the first sub-part (position, nature, magnification, height) is a straight 2–3 mark board-style calculation and should take under 90 seconds once the sign convention is automatic; the "what if the object moves" follow-up is the kind of extension competitive papers add to test whether you understand the trend, not just a single plug-in — budget an extra 60–90 seconds to redo the algebra rather than guessing the direction of change.

Common traps

Substituting $u$ , $v$ , or $f$ without their signs — the mirror equation only gives a correct answer when every quantity carries its Cartesian-convention sign; dropping a minus sign is the single most common error in this topic.
Misreading the sign of the computed $v$ — a negative $v$ means the image is in front of the mirror (real), not "30 cm behind"; students frequently flip this interpretation under exam pressure.
Forgetting the negative sign in $m=-v/u$ and reporting the wrong orientation (calling an inverted image erect, or vice versa) even when the numerical position and size are correct.
Plugging the radius of curvature $R$ directly into the mirror equation in place of $f$ — always convert first using $f=R/2$ , and remember $R$ and $f$ carry the same sign.
Treating a convex mirror's image as ever becoming real for a real object — algebraically, $v$ always comes out positive for a convex mirror with $u<0$ and $f>0$ , so the image is always virtual; a negative $v$ from a convex-mirror calculation is a sign that a value was substituted incorrectly, not a valid physical result.

Common trapThe mirror equation only works if every distance is signed consistently from the moment you write down u and f

The single biggest source of lost marks in this topic is mixing sign conventions mid-calculation — writing $u=-30$ correctly but then, out of habit from a diagram where distances "look positive," substituting $f=+15$ for a concave mirror (whose focus is genuinely in front of the mirror and must be negative). Once one sign in the equation is wrong, both the magnitude and the physical interpretation of $v$ come out wrong, and because the arithmetic still produces a number, the error is easy to miss unless you pause and ask whether the answer's sign makes physical sense (a concave mirror imaging a distant real object cannot produce a positive, virtual $v$ unless the object is inside the focal length). Fix the habit by writing all three given quantities with their signs explicitly, in a line, before touching the equation — every single time.

You should be able to…

State the mirror equation $\dfrac1v+\dfrac1u=\dfrac1f$ and the magnification formula $m=-v/u=h'/h$ from memory, with the Cartesian sign convention attached.
Assign correct signs to $u$ , $v$ , and $f$ for both concave and convex mirrors before substituting into the equation.
Solve for any one of $u$ , $v$ , or $f$ given the other two, for both mirror types, and also compute image height given object height.
Interpret the sign of a computed $v$ and $m$ to describe an image fully: real or virtual, erect or inverted, magnified, diminished, or same-size.
Reproduce the similar-triangles derivation of the mirror equation for a concave mirror forming a real image, when asked as a theory question.
Recognise the special CBSE-favourite object positions for a concave mirror (at $C$ , at $F$ , beyond $C$ , between $F$ and pole, and between $C$ and $F$ ) and predict image type without recalculating from scratch.

Ask the tutor: want a walkthrough of a mirror-equation numerical with a tricky sign, or a quick table matching every standard concave-mirror object position to its image type — ask the AI tutor to build it with you.

Why does the mirror equation look almost the same as the lens equation but with a plus sign instead of a minus sign?

It comes down to how the two devices are set up under the same sign convention. For a lens, light passes through and continues on the far side, so object and image distances are measured on opposite sides of the optical centre in a way that naturally gives $\frac1v-\frac1u=\frac1f$ . For a mirror, light reflects back on the same side it came from, and object and image distances both end up measured from the pole along the same axis in a way that produces the plus sign, $\frac1v+\frac1u=\frac1f$ . Both formulas come from the same style of similar-triangle geometry; the sign difference is a direct consequence of reflection sending light back rather than letting it pass through.

Do I have to memorise the derivation, or just the final formula?

For CBSE Class 12, you are expected to be able to apply $\frac1v+\frac1u=\frac1f$ and $m=-v/u$ fluently and correctly, including the sign convention, in numerical problems — that is where almost all the marks in this topic live. The similar-triangles derivation shown here is worth understanding once, because it removes the feeling that the formula is arbitrary, and some question papers do ask for it directly as a theory question worth a few marks. If you are short on revision time, prioritise flawless sign-convention numericals first, then come back to reproducing the derivation.

My calculated v came out positive for a concave mirror — did I make a mistake?

Not necessarily a mistake in the algebra, but check what it means: a positive $v$ for a concave mirror means a virtual image behind the mirror, which happens precisely when the real object is placed between the pole and the focus (closer to the mirror than $f$ ). That is a genuine, examinable case — think of a concave mirror used as a shaving or makeup mirror, where you stand well within the focal length and see an enlarged, upright, virtual image of your own face. So a positive $v$ is only wrong if you know the object was placed farther than the focus; otherwise it is telling you which of the concave mirror's two very different regimes you are in.

Is there a quicker way to check my answer without redrawing the whole ray diagram?

Yes — use the sign of $v$ and the value of $m$ as a fast sanity filter rather than redrawing anything. If $f<0$ (concave) and $u$ is more negative than $f$ (object beyond the focus), you must get a negative $v$ and $|m|$ should shrink as the object moves farther from the mirror; if the object sits between the pole and the focus for a concave mirror, expect a positive $v$ and $|m|>1$ . For any convex mirror with a real object, $v$ must always come out positive and $|m|<1$ — if your arithmetic gives anything else for a convex mirror, go back and recheck the signs you substituted, because that outcome is not physically possible.