TECHNIQUE: Mirror & Lens Sign-Convention Discipline (+ the Two-Ray Diagram)

The habit of getting a signed coordinate right is not confined to exam halls. An optical engineer designing a camera or microscope lays every lens and mirror on a signed axis and lets ray-tracing software propagate distances with exactly these conventions; a flipped sign there sends a multi-element design's image to the wrong plane. A surveyor or CAD system places every feature by signed coordinates from an origin — Descartes' axes, industrialised. The New Cartesian discipline you drill for a two-mark mirror problem is the same reflex that keeps a telescope's focal plane where the detector is, that positions the elements of a phone's tiny camera stack to within microns, and that lets an autofocus motor solve the imaging equation dozens of times a second without ever putting the image behind the sensor. Signs are not bookkeeping; they are the difference between an instrument that works and one that images into empty air.

A distance like "20 cm" is useless in optics until you say which way, and the New Cartesian sign convention — the one CBSE, ICSE, JEE and NEET all mark to — fixes that once and for all. The whole convention is one rule:

The origin is the pole $P$ (for a mirror) or the optical centre $O$ (for a lens). The incident light travels in the positive $x$ direction. Any distance measured against the incident light is negative; heights above the axis are positive, below negative. Two consequences become instinct. A real object sits on the incoming-light side, so its distance is measured against the light and $u < 0$ — real objects always have negative object distance. And a concave mirror's focus lies in front (where the light is), so $f < 0$; a convex mirror's lies behind, so $f > 0$; a converging lens has $f > 0$, a diverging lens $f < 0$. With the origin rule fixed, the two working relations are:

where: $u$ is the object distance, $v$ the image distance, $f$ the focal length and $m$ the magnification, all signed by the one origin rule. The two formulas share the origin and differ in just two places — the sign of the $1/u$ term and the sign in the magnification. Everything on this page is about placing those signs without a slip, and then reading them back: for a mirror $v < 0$ is a real image in front, for a lens $v > 0$ is a real image on the far side; and $m < 0$ is inverted, $m > 0$ erect, for both. The three moves never change from problem to problem.

Move 1 — draw the axis and sign the object. Mark the origin, the incident light's positive direction, and the real object at $u < 0$. Sign $f$ from the element type.

Move 2 — write the correct formula. Mirror or lens? Use (1) or (2). The only decision is which sign the $1/u$ and the $m$ wear.

Move 3 — substitute, solve, and read back. Put in the signed numbers, solve for $v$, compute $m$, then translate the signs into "real or virtual, erect or inverted, magnified or diminished" — and check against a quick two-ray sketch.

The fastest check on a sign is a ray sketch, and you only ever need two rays — chosen because you already know where each one goes. The craft differs slightly between a mirror and a lens, and getting the pairing right is the folded skill that saves marks. For a mirror, draw (i) a ray from the object's tip parallel to the axis, which reflects through the focus F, and (ii) a ray through the centre of curvature C, which strikes the mirror head-on and reflects straight back on itself. Where the two reflected rays cross is the image tip. For a lens, draw (i) a ray parallel to the axis, which refracts through the far focus F′, and (ii) a ray through the optical centre, which passes straight through undeviated. Where the two emergent rays cross is the image tip. The dashed-line discipline is what handles virtual images. If the two rays actually cross, the image is real and you draw solid rays to the crossing. If they diverge after the element, they never meet in front — so you extend them backward as dashed lines to the point they appear to come from, and that apparent crossing is the virtual image. A concave mirror with the object inside the focus, or a converging lens used as a magnifier, both need those dashed backward extensions; forgetting them is how students "lose" a virtual image they have actually computed. The sign of $v$ and the dashed-ray sketch must agree — that agreement is the check.

1 · Board level — the concave mirror, rote versus disciplined (CBSE and NEET pattern)

A concave mirror of focal length $15\ \mathrm{cm}$ has a real object $20\ \mathrm{cm}$ in front. Find the image distance, the magnification and the nature.

The rote attempt (wrong). A student writes $u = 20$, $f = 15$ (no minus signs) and gets $\dfrac{1}{v} = \dfrac{1}{15} - \dfrac{1}{20} = \dfrac{1}{60}$, so $v = +60\ \mathrm{cm}$ — an image behind a concave mirror, which is impossible for a real object beyond the focus. The unsigned data produced an unphysical answer.

Step 1 (the discipline). Sign by the origin rule: concave, so $f = -15\ \mathrm{cm}$; real object, so $u = -20\ \mathrm{cm}$. Use the mirror formula, $m = -v/u$.

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

Step 3. $m = -\dfrac{v}{u} = -\dfrac{-60}{-20} = \boxed{-3}$.

Step 4 (read back). $v < 0$: real, in front. $m < 0$: inverted. $|m| > 1$: magnified. A real, inverted, magnified image $60\ \mathrm{cm}$ in front — and $v$ negative for a mirror is what "real" looks like, the opposite of the rote answer. The signs turned a plausible-looking mistake into an obvious contradiction.

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

A convex mirror of focal length $15\ \mathrm{cm}$ has a real object $20\ \mathrm{cm}$ in front. Find the image and its nature.

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

Step 2. $\dfrac{1}{v} = \dfrac{1}{15} + \dfrac{1}{20} = \dfrac{4 + 3}{60} = \dfrac{7}{60}$, so $v = \dfrac{60}{7} = \boxed{+8.57\ \mathrm{cm}}$.

Step 3. $m = -\dfrac{v}{u} = -\dfrac{8.57}{-20} = \boxed{+0.43}$.

Step 4 (read back). $v > 0$: virtual, behind. $m > 0$: erect. $|m| < 1$: diminished. A virtual, erect, diminished image behind the glass — what a convex mirror gives for every real object, which is why it is the wide-angle mirror on a car. Note that here $v > 0$ means virtual (behind), the reverse of a real image's sign; the mirror's "real" is negative, its "virtual" positive.

3 · Competitive level — the one sign that flips: lens versus mirror (JEE pattern)

A converging lens of focal length $20\ \mathrm{cm}$ has a real object $30\ \mathrm{cm}$ in front. Find the image and its nature, and compare the sign bookkeeping with a mirror of the same arithmetic.

Step 1. Signs: converging lens, $f = +20\ \mathrm{cm}$; real object, $u = -30\ \mathrm{cm}$. Use the lens formula, $m = +v/u$.

Step 2. $\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{20} - \dfrac{1}{30} = \dfrac{3 - 2}{60} = \dfrac{1}{60}$, so $v = \boxed{+60\ \mathrm{cm}}$ — for a lens, positive $v$ means a real image on the far side.

Step 3. $m = +\dfrac{v}{u} = \dfrac{60}{-30} = \boxed{-2}$: real, inverted, magnified.

Step 4 (the contrast). The magnitude arithmetic here ($|v| = 60$, ratio $2$) is the same shape as a mirror problem — but the signs speak a different language. A lens's real image has $v > 0$ (a mirror's has $v < 0$), and the magnification is $m = +v/u$ (a mirror's is $-v/u$). Use the mirror's $m = -v/u$ here and you would get $m = +2$, wrongly calling this inverted image erect. That single sign is the whole difference, and it is exactly the mark examiners set the trap for. Timing note: about forty-five seconds once the origin rule is a reflex. The only decision is mirror-or-lens, which fixes the sign of $1/u$ and of $m$; everything else is arithmetic. The five-second insurance is reading $v$ back — a lens real image is positive, a mirror real image negative — before you write the nature.

4 · Competitive level — the dashed-ray virtual image (JEE and NEET pattern)

A concave mirror of focal length $15\ \mathrm{cm}$ has a real object only $10\ \mathrm{cm}$ in front, inside the focus. Locate the image by rays and by formula, and state its nature.

Step 1. By rays: draw the parallel ray (reflects through F) and the ray through C (returns on itself). After reflection they diverge, so extend both backward as dashed lines behind the mirror — they meet there, giving a virtual, erect image.

Step 2. By formula: $f = -15\ \mathrm{cm}$, $u = -10\ \mathrm{cm}$, so $\dfrac{1}{v} = -\dfrac{1}{15} + \dfrac{1}{10} = \dfrac{-2 + 3}{30} = \dfrac{1}{30}$, giving $v = \boxed{+30\ \mathrm{cm}}$.

Step 3. $m = -\dfrac{v}{u} = -\dfrac{+30}{-10} = \boxed{+3}$.

Step 4 (read back and check). $v > 0$: virtual, behind — exactly where the dashed rays met. $m > 0$, $|m| > 1$: erect and magnified. A virtual, erect, magnified image, the shaving-mirror view. The ray sketch and the sign of $v$ agree, which is the confirmation; had you forgotten the dashed backward extension you would have "lost" the image the formula clearly locates at $+30\ \mathrm{cm}$. Timing note: under a minute. The discipline is to draw the dashed rays the moment the reflected rays diverge — do not hunt for a real crossing that is not there. The positive $v$ is your signal that the image is virtual and the rays go backward.

Dropping the minus on a real object. A real object is $u < 0$, always — the single most lost mark. Write the sign before the number, for both mirrors and lenses.

Swapping the magnification formulas. Mirror $m = -v/u$, lens $m = +v/u$. Carry one into the other and every orientation flips while the distance stays right — a silent, expensive error.

Reading $v$'s sign with the wrong element in mind. For a mirror, real is $v < 0$; for a lens, real is $v > 0$. Check the sign against the element you actually have before naming the nature.

Using $R$ as $f$. $f = R/2$, with the sign. A radius put where a focal length belongs doubles $f$ and moves the image.

Losing virtual images. When the reflected or refracted rays diverge, extend them backward as dashed lines to find the virtual image; the positive $v$ (mirror) or negative $v$ (lens) is your cue it is virtual.

Mixing conventions. New Cartesian (concave $f < 0$) versus "real-is-positive" (concave $f > 0$) must never be blended mid-problem. AP Physics 2 texts often use the latter — pick one, declare it, stay in it.

Board scope. This sign-discipline technique is aimed at JEE-Main, JEE-Advanced, NEET and AP Physics 2, where the speed and accuracy of sign assignment decide marks on MCQs. The New Cartesian convention taught here is the one CBSE, ICSE, JEE and NEET mark to; AP Physics 2 examines the same physics but many American textbooks use a "real-is-positive" sign convention, so an AP candidate should apply their board's single rule without switching. IB SL/HL and A-Level 9702 carry no geometric optics of this kind, so for them the technique is enrichment, not exam material.

Texts. NCERT Physics Class 12, Ch. 9 ("Ray Optics and Optical Instruments"), the New Cartesian sign convention for mirrors and lenses · H. C. Verma, Concepts of Physics Vol. 1, "Geometrical Optics" — sign discipline and ray diagrams · D. C. Pandey / Cengage optics problem sets for the JEE and NEET sign-trap MCQs · AP Physics 2 (College Board CED, Unit 13) — note the real-is-positive convention some texts use.

Provenance. Standard geometric optics. Recompute: concave $f = -15$, $u = -20 \to v = -60$, $m = -3$; convex $f = +15$, $u = -20 \to v = +8.57$, $m = +0.43$; converging lens $f = +20$, $u = -30 \to v = +60$, $m = +v/u = -2$ (contrast mirror $m = -v/u = +2$, wrong here); concave inside focus $f = -15$, $u = -10 \to v = +30$, $m = +3$ (virtual, erect); and $f = R/2$ sign consistency.

Prerequisites. The Mirror Formula and Sign Convention and Magnification by Mirrors — the live mirror articles this technique is consistent with, where $1/v + 1/u = 1/f$ and $m = -v/u$ are proved. Leads to. Lens and Combination Numericals, which applies this sign discipline to lenses at speed; and Lens–Mirror Systems and the Silvered Lens, which needs both formulas at once.

Go deeper (UG). The sign convention generalises to the ray-transfer (matrix) method, where a signed distance and a signed curvature propagate through any stack of surfaces automatically — developed in Plane and Spherical Mirrors in the university optics track. It is enrichment only; every result here follows from the one origin rule and the two formulas, and no exam on this page asks for the matrix method.