Sign convention for mirrors and lenses: Question bank

Definition Two reference points, then five signed quantities

Everything is measured along the principal axis (the straight line through the centre of the mirror/lens, perpendicular to its surface). First fix where you measure from:

The pole $P$ — for a mirror, the point where the principal axis meets the reflecting surface (the dead centre of the mirror).
The optical centre $O$ — for a thin lens, the middle point of the lens where a ray passes straight through undeviated.

$P$ (mirrors) or $O$ (lenses) is the origin, not a distance. From that origin we read off five signed quantities:

$u$ = object distance — coordinate of the object from $P / O$ .
$v$ = image distance — coordinate of the image from $P / O$ .
$f$ = focal length — coordinate of the focus (where parallel rays meet) from $P / O$ .
$R$ = radius of curvature — coordinate of the centre of curvature $C$ (centre of the sphere the mirror is cut from) from $P$ ; and $f = R /2$ for a mirror.
$m$ = magnification = $\frac{h ^{'}}{h}$ = (signed image height) ÷ (signed object height) — a pure ratio whose sign gives orientation.

Figure — Sign convention for mirrors and lenses

Figure s01 — the master picture: origin at $P / O$ , orange arrow = incident-light direction (our $+ x$ ), the magenta region upstream is $-$ , the violet region downstream is $+$ . Every sign on this page is just "which coloured region is this point in?"

Figure s02 — MIRROR. Reflection sends light back, so a real object (left) and its real image (also left) sit on the same side of $P$ . Paraxial similar triangles give the magnitude law $\frac{1}{a} + \frac{1}{b} = \frac{1}{F}$ . Now sign it: object left $\Rightarrow u = - a$ ; real image left $\Rightarrow v = - b$ ; concave focus left $\Rightarrow f = - F$ . Substitute $a = - u, b = - v, F = - f$ into $\frac{1}{a} + \frac{1}{b} = \frac{1}{F}$ and every reciprocal picks up the same $- 1$ , which cancels, leaving $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ — a plus, because both points share a side.

Figure s03 — LENS. Light passes through, so a real object (left) and its real image (right) sit on opposite sides of $O$ . The magnitude law is still $\frac{1}{a} + \frac{1}{b} = \frac{1}{F}$ . Sign it: object left $\Rightarrow u = - a$ so $a = - u$ ; real image right $\Rightarrow v = + b$ so $b = + v$ ; converging focus right $\Rightarrow f = + F$ so $F = + f$ . Substitute: $\frac{1}{- u} + \frac{1}{v} = \frac{1}{f}$ , i.e. $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ — a minus, forced purely by the image and object being on opposite sides.

True or false — justify

A concave mirror always has a negative focal length.

True. Its focus sits on the incident-light side (left, in front of the mirror), which is against the light direction, so

f < 0

regardless of where the object is placed.

A convex mirror and a convex lens are both converging, so they share the same focal-length sign.

False on the premise and the conclusion. A convex mirror diverges; a convex lens converges. They do happen to share

f > 0

, but for unrelated geometric reasons (mirror focus behind = downstream; lens focus to the right = downstream), not because of "converging."

For a real object the object distance

u

is always negative.

True. A real object sits upstream of the pole/optical centre (to the left), opposite the incident-light direction, so its coordinate

u < 0

every time.

A positive image distance always means the image is real.

False. For a lens

v > 0

means real (image is downstream, right). For a mirror

v > 0

means virtual (image is behind the mirror). The same sign reads oppositely because a mirror sends light backward.

Magnification

m > 0

guarantees the image is virtual.

False.

m > 0

only means the image is erect (image height and object height share a sign). Erect images are usually virtual, but the sign of

m

reports orientation, not reality — check the sign of

v

for reality.

The radius of curvature

R

of a concave mirror is negative.

True. The centre of curvature lies on the same side as the object (front, left), upstream of the light, so

R < 0

, consistent with

f = R /2 < 0

Swapping which side the object sits on would flip all the standard signs.

True. The whole convention is anchored to "incident light travels left→right." If you (unusually) sent light right→left, every

+

and

-

above would invert. The signs are relative to the chosen light direction, not absolute.

A plane mirror has

f = 0

False. A plane mirror has an infinite radius of curvature, so

f = R /2 = \infty

, not zero. Plug

f \to \infty

into

\frac{1}{v} + \frac{1}{u} = \frac{1}{f} = 0

to get

v = - u

: the image sits as far behind as the object is in front.

Spot the error

"Object 30 cm from a concave mirror, so I'll use

u = + 30

cm."

Wrong sign. The object is upstream of the pole →

u = - 30

cm. Treating

u

as a plain positive length is the single most common convention slip;

u

is a coordinate, not a distance.

"Convex lens, so I'll write

\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

Wrong equation. Lenses use a minus:

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

. The reason is derived in figure s03: object left (

u = - a

) and real image right (

v = + b

) lie on opposite sides of

O

, so substituting into the magnitude law

\frac{1}{a} + \frac{1}{b} = \frac{1}{F}

gives

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

. The mirror keeps a plus (figure s02) because reflection puts object and image on the same side.

"I got

v = - 15

cm for a lens, so the image is virtual and on the object's side — but I'll report it as real since lenses make real images."

The report contradicts the sign. For a lens

v < 0

genuinely means a virtual image on the same side as the object (e.g. object inside the focus of a converging lens). Never override the sign with a memorised habit.

"Concave mirror is the strong converging one, so its focal length must be positive."

The "strength/converging = positive" instinct is the trap. Sign is set by geometry of where the focus lies, not by whether the element converges. Concave mirror focus is upstream ⇒

f < 0

"Convex mirror, real object, so I'll expect

v < 0

like usual."

A convex mirror always makes a virtual image behind itself ⇒

v > 0

. Expecting the "usual" front-image sign forgets that this mirror can never form a real image of a real object.

"For the magnification I used

m = + v / u

for a mirror."

Wrong sign for a mirror. Mirrors use

m = - v / u

; lenses use

m = + v / u

. The extra minus for mirrors follows from the reflection geometry of the similar triangles at the pole.

Why questions

Why does a single signed equation replace a whole pile of case-by-case formulas?

Because the signs of

u

v

f

encode which side each object/image/focus is on. Once side-information lives in the sign, one algebraic relation covers concave/convex, real/virtual, inside/outside focus — no redrawing needed.

Why is the mirror formula a "

+

" but the lens formula a "

-

A mirror reflects, so object and image lie on the same side (figure s02) and signing the magnitude law leaves a plus. A lens transmits, so image and object are on opposite sides (figure s03), which forces a minus. The connecting sign is a direct fingerprint of same-side vs opposite-side geometry.

Why does the sign of the answer

v

tell you the image's nature for free?

Because

v

is a coordinate: its sign says which side of the surface the image lands on, and each side is a nature (front/real vs behind/virtual for mirrors; opposite/real vs same-side/virtual for lenses). The algebra places the image; the sign labels it.

Why must the mirror formula be symmetric in

u

and

v

Because light paths are reversible — trade the object and image and the ray diagram is unchanged. The formula

\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

is symmetric precisely to honour that physical reversibility.

Why is a convex mirror's focus taken as positive when its object still gives

u < 0

The object sits upstream (left) →

u < 0

, but the convex mirror's focus lies behind the mirror, which is downstream of the reflected light →

f > 0

. Two different points on two different sides, so opposite signs.

Why can't you decide

f

's sign just from the words "converging" or "diverging"?

Because "converging" describes the behaviour of rays, not the location of the focus relative to the light. Only the focus's side (upstream/downstream) fixes the sign, and that side differs between mirrors and lenses.

Edge cases

An object placed exactly at the focus of a concave mirror — what does the formula say?

Set

u = f

: then

\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = 0

, so

v \to \infty

. The reflected rays come out parallel and meet only at infinity — no finite image forms.

A concave mirror with the object inside the focal length (

∣ u ∣ < ∣ f ∣

) — real or virtual image?

Virtual. Take

u = - 6

cm,

f = - 10

cm:

\frac{1}{v} = \frac{1}{- 10} - \frac{1}{- 6} = + \frac{1}{15}

, so

v = + 15

> 0

— a positive

v

for a mirror means the image is behind the mirror, i.e. virtual, erect, magnified. This is the shaving/make-up-mirror regime, and it shows a concave mirror does not always give a real image.

A diverging (concave) lens with a real object — what image does it always make?

Always virtual, erect, reduced, on the object's side. Take

u = - 20

cm,

f = - 10

cm:

\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{- 10} + \frac{1}{- 20} = - \frac{3}{20}

, so

v = - \frac{20}{3} \approx - 6.67

< 0

(same side as object ⇒ virtual), and

m = v / u = (- 20/3) / (- 20) = + \frac{1}{3} > 0

(erect) and

< 1

(reduced). A concave lens can never give a real image of a real object.

An object at the centre of curvature of a concave mirror (

u = R = 2 f

) — what happens to the sign of

m

v = u = 2 f

, so

m = - v / u = - 1

: the image is real, same size, and inverted (negative

m

), sitting back at

C

on the object's side.

An object at infinity (parallel incident rays) — where does the image form for a mirror and for a lens?

u \to - \infty

\frac{1}{u} \to 0

, so the formula collapses to

\frac{1}{v} = \frac{1}{f}

, i.e.

v = f

. The image forms exactly at the focus: for a concave mirror at

f < 0

(real, in front), for a convex lens at

f > 0

(real, on the far side). This is the very definition of the focal point.

A virtual object (converging light hitting a lens before it focuses) — how does its sign differ?

A virtual object lies downstream of the surface (to the right), with the light direction, so

u > 0

— the opposite sign to an ordinary real object. This arises in Combination of thin lenses where one lens hands converging rays to the next.

As a concave mirror is flattened toward a plane (

R \to \infty

), what happens to

f

and

v

f = R /2 \to - \infty

, so

\frac{1}{f} \to 0

and

\frac{1}{v} + \frac{1}{u} = 0

gives

v = - u

: the image approaches the plane-mirror result — same distance behind, virtual, erect, same size.

A convex lens with the object at

2 f

— why is the answer sign-neutral about which "

2 f

With

u = - 2 f

and

f > 0

, you get

v = + 2 f

: real image on the far side, inverted, unit magnification. The symmetry of the two

2 f

points is a direct consequence of the formula's structure, not a coincidence.

The object placed right against the mirror (

u \to 0^{-}

) — where is the image, as a proper limit?

Rearrange to

v = \frac{f u}{u - f}

. Taking the limit

u \to 0

: numerator

f u \to 0

while the denominator

u - f \to - f \neq = 0

, so

v \to \frac{0}{- f} = 0

. The image sits at the surface with

m = - v / u \to 1

(unit, erect) — the object and image coincide at the pole. This is an idealised limit; the thin-element/paraxial model stops describing reality this close.

Recall

Recall Beyond memorisation — reason each one out

Given only $u < 0$ and $f < 0$ , without a diagram, decide if a concave-mirror image can ever be virtual. ::: Yes — when $∣ u ∣ < ∣ f ∣$ the algebra yields $v > 0$ , which for a mirror is a virtual image behind it; the sign convention predicts the shaving-mirror case with no drawing. Why does $v = f$ pop out whenever the object is at infinity, for both mirror and lens? ::: Because $\frac{1}{u} \to 0$ kills the object term, leaving $\frac{1}{v} = \frac{1}{f}$ ; "parallel rays meet at the focus" is literally what defines $f$ . A student reports $m = + 2$ and $v > 0$ for a lens — consistent? ::: No. $m > 0$ demands erect (virtual for a single lens), but $v > 0$ for a lens means real (inverted, $m < 0$ ). The two claims contradict; one sign is wrong. How could you tell mirror-vs-lens apart from a bare formula with no labels? ::: The connecting sign: a " $+$ " between the $1/ v$ and $1/ u$ terms is a mirror (same-side reflection); a " $-$ " is a lens (opposite-side transmission).

Cross-links: Mirror formula derivation, Lens formula and lensmaker's equation, Magnification in optics, Real and virtual images, Refraction at spherical surfaces.