Mirror formula

Point a torch at a spoon and you get two different reflections, one on each side — proof that a curved mirror does something a flat one never does: it can shrink, magnify, flip, or even project an image onto a screen. The mirror formula is the one equation that predicts exactly where that image lands and how big it is, for any concave or convex mirror, without a single ray diagram. You have already met the pieces separately — the laws of reflection, plane mirrors, and how concave and convex mirrors bend rays toward or away from a focus. This section assembles them into a calculation tool. Fix the Cartesian sign convention once, and the formula $1/v + 1/u = 1/f$ plus the magnification $m = -v/u$ answer every "where, how big, real or virtual" question a numerical can ask — which is exactly how ICSE Class 9 uses it: apply, don't derive.

The core insight

With the Cartesian sign convention fixed, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ and $m=-\frac{v}{u}$ together pin down an image's position, size, and nature for any spherical mirror.

The core idea

Everything starts with how distances are measured — get this wrong and every later sign is wrong too.

All distances are measured from the mirror's pole ( $P$ ), along the principal axis.
Distances in the direction the incident light travels (conventionally the mirror's left, since light is drawn arriving left to right) are negative; distances measured against that direction (to the right, behind the mirror) are positive. A real object placed in front of any mirror therefore always has $u < 0$ .
Heights measured upward from the principal axis are positive; heights measured downward are negative.

With that convention fixed, the syllabus states the mirror formula as a ready-made tool — no derivation is required or expected at this level:

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

where: $u$ is the object distance, $v$ is the image distance, and $f$ is the focal length, all measured from the pole with sign. The sign of $f$ is fixed by the mirror type, not by the problem: a concave mirror's focus sits in front of it (same side as the incoming light), so $f$ is negative; a convex mirror's focus sits behind it, so $f$ is positive. This single fact is the difference between the two worked examples below. The magnification compares image size to object size:

m = \frac{h_i}{h_o} = -\frac{v}{u}

where: $h_i$ is the image height and $h_o$ is the object height. Read the sign of $m$ like a small checklist: positive $m$ means erect and (given the convention) virtual; negative $m$ means inverted and real. Read the magnitude separately: $|m|>1$ magnified, $|m|<1$ diminished, $|m|=1$ same size. The two readings — sign and magnitude — are independent, so always state both.

Figure

A concave mirror forming a real inverted image with rays drawn from an object beyond the focus to a smaller image between the focus and the centre of curvature

Derivation AWhere the magnification formula comes from

The goal is to show that $m = -v/u$ is not an arbitrary rule but a direct consequence of similar triangles formed by a ray that leaves the object, strikes the mirror at its pole, and reflects to form the image, so that the size ratio of image to object can be read straight off the geometry.

Step 1. Draw one ray from the tip of the object, travelling parallel to nothing in particular but straight to the pole $P$ of the mirror, and let it reflect to the tip of the image. By the law of reflection, this ray makes equal angles with the principal axis on the way in and on the way out.

Step 2. Because the incident and reflected rays make equal angles with the axis at $P$ , the object-side triangle (object height $h_o$ , object distance $|u|$ ) and the image-side triangle (image height $h_i$ , image distance $|v|$ ) are similar right triangles, both with a vertex angle at $P$ . Similar triangles give a direct ratio of sides:

\frac{h_i}{h_o} = \frac{|v|}{|u|}

Step 3. Fix the sign. When the image is real and inverted, $h_i$ points downward while $h_o$ points upward, so the ratio $h_i/h_o$ is actually negative even though the triangle geometry only gave magnitudes; under the Cartesian convention $u$ is negative and $v$ carries its own sign, so writing the ratio with signed distances instead of magnitudes absorbs this automatically:

m = \frac{h_i}{h_o} = -\frac{v}{u}

Step 4. Check it against a known case. For a plane mirror, $v = -u$ (image as far behind as the object is in front), so $m = -(-u)/u = 1$ : erect, same size — exactly what a plane mirror should give, confirming the sign in Step 3 is the correct one.

Worked example

An object $2\text{ cm}$ tall is placed $20\text{ cm}$ in front of a concave mirror of focal length $15\text{ cm}$ . Find the image position, height, and nature — then compare it with the same object placed in front of a convex mirror of focal length $6\text{ cm}$ at a distance of $12\text{ cm}$ .

Step 1. Assign signs for the concave mirror. Object distance $u = -20\text{ cm}$ (real object, negative by convention). Focal length $f = -15\text{ cm}$ (concave mirror, negative). Object height $h_o = +2\text{ cm}$ .

Step 2. Apply the mirror formula.

\frac{1}{v}+\frac{1}{u}=\frac{1}{f} \implies \frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{-15}-\frac{1}{-20}=-\frac{1}{15}+\frac{1}{20}

Common denominator $60$ : $\frac{1}{v}=-\frac{4}{60}+\frac{3}{60}=-\frac{1}{60}$ , so $v=-60\text{ cm}$ .

Step 3. Interpret and find magnification. $v$ is negative, so the image forms $60\text{ cm}$ in front of the mirror — a real image. Magnification: $m=-\dfrac{v}{u}=-\dfrac{(-60)}{(-20)}=-3$ , so $h_i = m\times h_o = -3\times 2\text{ cm}=-6\text{ cm}$ .

Step 4. State the answer for the concave case, then repeat Steps 1-2 for the convex mirror ( $u=-12\text{ cm}$ , $f=+6\text{ cm}$ ): $\frac{1}{v}=\frac16-\frac{1}{-12}=\frac{2}{12}+\frac{1}{12}=\frac{3}{12}=\frac14$ , so $v=4\text{ cm}$ , and $m=-\dfrac{4}{-12}=\dfrac13$ , giving $h_i=\dfrac13\times 2\text{ cm}\approx 0.67\text{ cm}$ .

Answer: Concave mirror — image $60\text{ cm}$ in front of the mirror, real, inverted, magnified $3\times$ , height $6\text{ cm}$ . Convex mirror — image $4\text{ cm}$ behind the mirror, virtual, erect, diminished, height about $0.67\text{ cm}$ .

Competitive stretch

If instead $v$ came out as $+60\text{ cm}$ for the same concave mirror (a common sign slip), a student who forgets to check the sign would wrongly report a virtual image behind a concave mirror producing a magnified, erect image — a result that only occurs for objects placed inside the focus. Always sanity-check: for a concave mirror with the object beyond the focus, the image must be real ( $v<0$ ) and inverted ( $m<0$ ); if your arithmetic disagrees, recheck the signs substituted in Step 2, not the final formula.

Common traps

Forgetting to give the object distance $u$ a negative sign — a real object in front of any mirror is always $u<0$ under this convention.
Mixing up the sign of $f$ : concave is negative, convex is positive. Getting this backwards flips every subsequent sign.
Reading a negative $v$ as "no image exists" rather than "real image, formed in front of the mirror."
Trying to re-derive the mirror formula itself from ray geometry in an exam — the syllabus does not require it; apply it directly with correct signs. (The magnification formula above is worth knowing how to derive, but the mirror formula is a given tool.)

Ask the tutor: Want to try a few more mirror-formula numericals, or check whether your sign choice for $f$ is correct for a specific problem — ask the AI tutor to work through it with you.

Why is the object distance u always negative, even though the object is clearly real and in front of the mirror?

The negative sign is not saying the object is unreal — it is a bookkeeping rule of the Cartesian convention, where distances measured in the direction light travels toward the mirror are taken as negative. Since a real object always sits on the side the incident light comes from, $u$ is negative by definition every single time, for both concave and convex mirrors. Once you accept this as a fixed rule rather than something to re-derive per question, substituting signs becomes automatic.

Do I need to memorise separate formulas for concave and convex mirrors?

No — one formula, $1/v+1/u=1/f$ , and one magnification relation, $m=-v/u$ , cover both mirror types. The only thing that changes between concave and convex is the sign you plug in for $f$ (negative for concave, positive for convex) and, for concave mirrors, where the object sits relative to the focus. Memorise the sign convention once instead of memorising multiple formulas.

If I get a fraction like 1/v = -1/60, do I flip it directly to get v, or do something else first?

Flip it directly: $1/v = -1/60$ means $v = -60$ . There is no extra step. The most common slip here is arithmetic on the fractions before the flip — always find a common denominator carefully (as in Step 2 of the worked example) before inverting, and double-check the final sign matches what you expect physically (real image should give a negative $v$ for a mirror).

Mirror formula

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads