Sign convention for mirrors

Look into a spoon. Turn it one way and your face swells into a huge, upside-down blob; turn it the other way and a small, upright, cheerful version of you sits inside the bowl. Both images are real physics, both come from the very same curved surface, and both must be predicted by the very same pair of formulas — the mirror equation $\dfrac1v+\dfrac1u=\dfrac1f$ and the magnification $m=-v/u$ . The only way one compact formula can describe an inverted real image and an upright virtual image, a concave mirror and a convex one, is if every distance and every height carries a sign, not just a magnitude. That bookkeeping rule is the sign convention for mirrors. Physicists did not invent this to make Class 12 harder. Before a standard convention existed, every textbook wrote a slightly different formula for each of the four or five standard cases (real object outside centre of curvature, real object between focus and pole, convex mirror, and so on), and students had to memorise which formula went with which picture. The Cartesian sign convention collapses all of that into one formula plus one rule for assigning plus and minus signs, borrowed straight from the $x$ – $y$ coordinate geometry you already know — the mirror's pole sits at the origin, and the direction the light actually travels is positive. This article sets up that rule with total precision, ties it to a labelled ray diagram, derives how it forces the familiar $f = R/2$ relation to carry a sign, and then works two full numericals — one board-style, one competitive-style — using nothing beyond the convention itself and the mirror formula. Get the signs right here and the rest of ray optics, including lenses later in the chapter, becomes mechanical arithmetic instead of guesswork.

Key takeaways

The Cartesian sign convention places the origin at the mirror's pole, takes the direction of the incident light as positive $x$ , and measures heights upward as positive.
A real object in front of any mirror always has negative object distance, $u<0$ , because it lies against the direction of incident light.
For a concave mirror, the centre of curvature $C$ and focus $F$ lie in front of the mirror, so $R<0$ and $f<0$ ; for a convex mirror, $C$ and $F$ lie behind it, so $R>0$ and $f>0$ .
The mirror formula $\dfrac1v+\dfrac1u=\dfrac1f$ and $f=\dfrac{R}{2}$ (both signed) are the only two equations needed; every case — real, virtual, magnified, diminished — falls out of the arithmetic automatically once signs are entered correctly.
Magnification $m=-\dfrac{v}{u}=\dfrac{h'}{h}$ : a negative $m$ means an inverted, real image; a positive $m$ means an erect, virtual image; $|m|>1$ is magnified and $|m|<1$ is diminished.
The single most common exam error is entering $u$ as a positive number, or reversing the sign of $f$ for concave versus convex — both quietly wreck an otherwise correct calculation.
Key idea: because the pole is the origin and the incident-light direction is positive, a real object always gives negative u, a concave mirror always gives negative f, a convex mirror always gives positive f, and the signed mirror formula then automatically tells you real or virtual, erect or inverted, magnified or diminished.

The core insight

Every distance in mirror formulas is measured from the pole along the incident-light direction (positive) or against it (negative), so a real object always has $u<0$ , a concave mirror always has $f<0$ , a convex mirror always has $f>0$ , and the sign of the calculated $v$ and $m$ then reads off real/virtual and erect/inverted automatically.

Where it fits in the syllabus

Within CBSE Unit IX, "Ray Optics and Optical Instruments," the sign convention is the mandatory setup step between the qualitative ray-diagram treatment of spherical mirrors and the quantitative mirror formula $\dfrac1v+\dfrac1u=\dfrac1f$ with magnification $m=-v/u$ , both explicitly listed in the prescribed scope. It is examined every year, almost always folded silently into numerical problems rather than asked as a standalone definition question, which is exactly why getting it automatic matters. The same Cartesian logic, with an adapted rule for refraction, carries forward unchanged to the thin-lens formula later in this chapter, so mastering it here pays twice.

The physical picture

Picture the mirror as a vertical line (the pole is the point where the principal axis pierces the mirror surface) with the principal axis running horizontally through it, and imagine light always travelling from left to right, striking the mirror and reflecting back to the left. The pole $P$ is the origin of the coordinate system. Distances measured in the direction the incident light travels — to the right of $P$ in this standard layout — are taken as positive; distances measured against that direction — to the left of $P$ , where every real object and every concave mirror's focus sit — are taken as negative. Heights measured upward from the principal axis are positive; heights measured downward are negative. This is nothing more than ordinary $x$ – $y$ coordinate geometry with the mirror at the origin.

Figure

A concave mirror with its pole at the origin, the principal axis running left to right, a real object to the left of the pole marked with a negative distance u, and the centre of curvature and focus also to the left marked negative, illustrating that light travels left to right so leftward distances are negative

For a real object, which sits in front of any mirror so that light genuinely travels from the object to the mirror, the object distance $u$ is measured from the pole back toward the object, against the direction of incident light — so $u$ is always negative, for both concave and convex mirrors, with essentially no exception at this level. For a concave mirror, the reflecting surface curves toward the incoming light, and its centre of curvature $C$ and focus $F$ lie in front of the mirror, on the same side as the object — so $R<0$ and $f<0$ . For a convex mirror, the surface bulges toward the incoming light, and $C$ and $F$ lie behind the mirror, in the direction the light is travelling — so $R>0$ and $f>0$ . The image distance $v$ is not fixed by the type of mirror the way $u$ , $R$ , $f$ are; it is whatever the mirror formula computes, and its sign then tells you the nature of the image: $v<0$ places the image in front of the mirror (a real image, since reflected rays genuinely converge there and could be caught on a screen), while $v>0$ places the image behind the mirror (a virtual image, where rays only appear to diverge from). Magnification ties heights to distances:

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

where: $h$ is the (signed, usually positive) object height measured upward from the axis, $h'$ is the signed image height, $v$ is the signed image distance, and $u$ is the signed object distance. Read physically: $m>0$ means $h'$ and $h$ have the same sign, so the image is erect (same orientation as the object) — this always pairs with $v>0$ , a virtual image; $m<0$ means the image is inverted, always pairing with $v<0$ , a real image. The magnitude $|m|$ separately tells you magnified ( $|m|>1$ ) versus diminished ( $|m|<1$ ), completely independent of the erect/inverted question.

Derivation AWhy f must carry a sign, and why f = R/2 forces concave mirrors negative

The goal of this derivation is to show, from the geometry of a spherical mirror alone, why the relation between focal length and radius of curvature, $f=R/2$ in magnitude, must be written as a signed equation once the Cartesian convention is adopted, and why that signed equation automatically makes every concave mirror's focal length negative and every convex mirror's focal length positive — without needing to memorise the two cases as separate rules.

Step 1. Recall the geometric origin of $f=R/2$ : for a ray travelling parallel to the principal axis and striking a spherical mirror close to the axis (the paraxial approximation used throughout this chapter), the reflected ray crosses the axis at a point that bisects the segment from the pole to the centre of curvature. That is a purely geometric fact about a sphere of radius $|R|$ — it fixes only the magnitude of the focus's distance from the pole, $|f|=|R|/2$ , and says nothing yet about sign.

Step 2. Now place the coordinate origin at the pole $P$ and orient the positive axis along the incident-light direction, as the convention demands. For a concave mirror, the reflecting surface curves toward the source of light, so both the centre of curvature $C$ and the point where reflected parallel rays converge (the focus $F$ ) lie on the same side as the incoming light source — that is, against the incident direction, hence at negative coordinates. So $R<0$ and, since $F$ sits between $P$ and $C$ on that same negative side, $f<0$ too.

Step 3. For a convex mirror, the surface bulges away from the light source, so the reflected rays diverge as though coming from a point behind the mirror. Both $C$ and the virtual focus $F$ lie behind the pole, in the direction the incident light is travelling — that is, at positive coordinates. So $R>0$ and $f>0$ .

Step 4. Because Steps 2 and 3 place $C$ and $F$ on the same side of the pole in both cases (never split between them), the ratio $f/R$ is always $+1/2$ , never $-1/2$ . This is the payoff: the single signed formula $f=\dfrac{R}{2}$ holds for both mirror types without modification. You never need to memorise " $f$ is negative for concave, positive for convex" as a separate fact — it falls straight out of applying the pole-origin, incident-direction-positive rule consistently to $R$ , and then halving.

Worked examples

Board level - Signed R and f for a concave mirror, then image position

Question: A concave mirror has a radius of curvature of 30 cm. An object is placed 20 cm in front of it. Find the signed focal length, the image distance, and describe the image.

Step 1. Assign signs before touching any formula. The mirror is concave, and both $C$ and $F$ lie in front of it (against the incident-light direction), so $R=-30\ \text{cm}$ . The object is real and in front of the mirror, so $u=-20\ \text{cm}$ .

Step 2. Get $f$ from the signed relation: $f=\dfrac{R}{2}=\dfrac{-30}{2}=-15\ \text{cm}$ .

Step 3. Substitute into the mirror formula and solve for $v$ : $\dfrac1v=\dfrac1f-\dfrac1u=\dfrac{1}{-15}-\dfrac{1}{-20}=-\dfrac{1}{15}+\dfrac{1}{20}$ . With a common denominator of 60: $-\dfrac{4}{60}+\dfrac{3}{60}=-\dfrac{1}{60}$ , so $v=-60\ \text{cm}$ .

Step 4. Read the sign. $v=-60\ \text{cm}$ is negative, so the image forms 60 cm in front of the mirror — a real image. Magnification $m=-v/u=-(-60)/(-20)=-3$ ; the negative sign confirms it is inverted, and $|m|=3>1$ means it is magnified, three times the object's height. The image is real, inverted, magnified 3×, formed 60 cm in front of the mirror.

Competitive level - Working backward from a described image to find R

Question: An object 4 cm tall is placed 15 cm from a spherical mirror. The image formed is virtual, erect, and 8 cm tall. Find the signed magnification, the image distance, the focal length, and the mirror's radius of curvature; identify the mirror type.

Step 1. Fix the known signed quantity first: the object is real, so $u=-15\ \text{cm}$ , and the object height is $h=+4\ \text{cm}$ (measured upward). The image is described as erect and 8 cm tall, so $h'=+8\ \text{cm}$ (upward, same side as the object, since erect means same orientation).

Step 2. Compute magnification directly from heights: $m=\dfrac{h'}{h}=\dfrac{8}{4}=+2$ . A positive magnification greater than 1 signals an erect, virtual, enlarged image.

Step 3. Use $m=-v/u$ to get $v$ : $2=-\dfrac{v}{-15}=\dfrac{v}{15}$ , so $v=+30\ \text{cm}$ . The positive sign confirms virtual (behind the mirror), consistent with the "virtual" description given.

Step 4. Substitute $u=-15\ \text{cm}$ and $v=+30\ \text{cm}$ into the mirror formula to get $f$ : $\dfrac1f=\dfrac1v+\dfrac1u=\dfrac{1}{30}+\dfrac{1}{-15}=\dfrac{1}{30}-\dfrac{2}{30}=-\dfrac{1}{30}$ , so $f=-30\ \text{cm}$ .

Step 5. Since $f<0$ , this must be a concave mirror (a convex mirror can never produce a negative $f$ , and can never magnify a real object beyond life-size — it always gives $|m|<1$ ). Then $R=2f=2(-30)=-60\ \text{cm}$ . The mirror is concave with $f=-30\ \text{cm}$ and $R=-60\ \text{cm}$ ; the object sits between the pole and the focus, giving the classic shaving-mirror image: virtual, erect, magnified 2×, 30 cm behind the mirror. Timing note: In a competitive setting, do not solve for $v$ from the mirror formula first — that requires knowing $f$ , which you don't have yet. Instead extract $m$ straight from the given heights (10 seconds), get $v$ from $m=-v/u$ (10 seconds), and only then plug $u,v$ into the mirror formula for $f$ . Reversing this order wastes a full minute hunting for $f$ from insufficient data.

Common traps

Writing $u$ as a positive number "because distance can't be negative" — in this convention distance from the pole is a signed coordinate, not a physical length, and a real object's $u$ is essentially always negative.
Assuming concave mirrors always have positive $f$ "because they converge light" — under the Cartesian convention it is exactly the opposite: concave gives $f<0$ , convex gives $f>0$ , and this reversal from casual intuition is the single most repeated exam slip.
Reading the sign of $v$ as telling you erect/inverted, or the sign of $m$ as telling you real/virtual — they answer different questions: $v$ 's sign gives real/virtual (and, tied to it, the position of the image), while $m$ 's sign gives erect/inverted; always check both separately rather than assuming one implies the other.
Forgetting that the derived formula $f=R/2$ is only valid for rays close to the principal axis (the paraxial approximation) — it is not an exact statement for rays striking far from the pole, though this refinement is not required at CBSE numerical level.
Plugging in a radius of curvature with the wrong sign when a problem states "radius of curvature 20 cm" without saying which mirror — always determine concave-versus-convex from the problem's description first, then assign the sign to $R$ , never the other way round.

You should be able to…

State the Cartesian sign convention: origin at the pole, incident-light direction positive, upward heights positive.
Correctly assign signs to $u$ , $R$ , and $f$ before substituting into the mirror formula, for both concave and convex mirrors, without needing to recall the "opposite of intuition" trap each time.
Solve the mirror formula for an unknown ( $v$ or $f$ ) and correctly read its sign to decide real versus virtual and the image's location relative to the pole.
Compute magnification from either $m=-v/u$ or $m=h'/h$ and correctly read its sign for erect versus inverted and its magnitude for magnified versus diminished.
Work a numerical "backward," starting from a described image (as in the competitive example) and recovering the mirror's focal length and type.

Ask the tutor: Still second-guessing whether R and f should be negative or positive for a given mirror — ask the AI tutor to walk through two or three fresh mirror scenarios with you until assigning the signs becomes automatic.

Why does the sign convention even matter — cant I just remember which case gives which kind of image from the ray diagram?

You can, for the four or five standard cases every textbook shows, but numericals rarely hand you a clean standard case — they give you a random object distance and ask you to compute where the image lands and what it looks like, without telling you in advance whether it will be real or virtual. The sign convention lets one formula produce the right answer automatically for any input, instead of you first classifying which of five diagrams applies and then picking the matching formula. It is the same reason coordinate geometry replaced case-by-case geometric proofs — one algebraic rule, applied consistently, replaces memorised special cases.

Is the sign of f really opposite to what feels natural for a concave mirror?

Yes, and this is the single most-tested trap in this topic. Intuitively you might think "concave mirrors converge light to a real focus, so that should be a positive thing." But the convention has nothing to do with whether light converges or diverges — it only cares about which side of the pole the focus sits on, relative to the direction the incident light is travelling. A concave mirror's focus sits in front of the mirror, on the same side as the object, which is against the incident-light direction, so it is negative. Say it out loud a few times during practice until "concave is negative" feels automatic rather than backwards.

What happens if I mix up the direction of positive x, say I draw light travelling right to left instead of the textbook's usual left to right?

Nothing breaks, as long as you are internally consistent — the sign convention is a choice of coordinate system, and physics does not care which way you draw your axis. If you draw light travelling right to left, then "positive" becomes rightward-relative-to-that-direction, and every sign in your working flips consistently, giving the same final physical answer (real/virtual, erect/inverted, magnified/diminished are all coordinate-independent facts). The trouble only starts if you mix conventions mid-problem — say, calling an object distance negative under one orientation and then plugging in a focal length you signed under the opposite orientation. Pick one orientation, usually left-to-right as in your textbook, and stay with it for the whole problem.

Does this same idea apply to lenses, or is it a totally different rule there?

The same coordinate philosophy carries over — pole-equivalent origin, incident-light direction positive, upward heights positive — but the details differ because a lens has light passing through it rather than bouncing off it, so both a real object side and a real image side exist simultaneously on opposite faces of the lens. You will meet the lens version, with its own formula $\dfrac1v-\dfrac1u=\dfrac1f$ , later in this same chapter; getting completely comfortable with the mirror convention here is exactly the preparation that makes the lens convention feel like a small variation rather than a brand-new rule.