Derive mirror formula 1/v + 1/u = 1/f

Question

Derive the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ for a concave mirror, using the standard sign convention.

Solution — Step by Step

Set up the geometry

Consider a concave mirror with pole P, centre of curvature C, and principal focus F. An object AB is placed beyond C on the principal axis. A ray parallel to the principal axis reflects through F. A ray through C reflects back on itself. These two rays meet at A'B', the real inverted image.

We use the New Cartesian Sign Convention: all distances measured from the pole P. Distances in the direction of incident light (left to right) are positive; against incident light are negative. For a concave mirror, $f$ and $R$ are negative.

Identify the two similar triangles

Triangle ABP and triangle A'B'P are similar (both right-angled, sharing the angle at P). So:

$\frac{A'B'}{AB} = \frac{PA'}{PA}$

Triangle ABF and triangle MPF are also similar (where M is the point where the parallel ray hits the mirror). Since MP = AB (for paraxial rays, MP ≈ AB), we get:

$\frac{A'B'}{MP} = \frac{FA'}{FP} \implies \frac{A'B'}{AB} = \frac{FA'}{FP}$

Equate the two expressions for magnification

From the two similarity results:

$\frac{PA'}{PA} = \frac{FA'}{FP}$

Now $FA' = PA' - PF$ . Substituting:

$\frac{PA'}{PA} = \frac{PA' - PF}{PF}$

Apply sign convention and simplify

Using sign convention: $PA = -u$ , $PA' = -v$ , $PF = -f$ .

$\frac{-v}{-u} = \frac{-v - (-f)}{-f} = \frac{-v + f}{-f}$

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

Cross-multiplying: $vf = u(f - v)$

$vf = uf - uv$

Divide throughout by $uvf$ :

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

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

Verify with the relation between f and R

Since $f = R/2$ , we can also write:

$\frac{1}{v} + \frac{1}{u} = \frac{2}{R}$

This is consistent with the geometry: a ray through C retraces its path, so $R = 2f$ .

Why This Works

The mirror formula comes purely from the geometry of similar triangles formed by paraxial rays (rays close to the principal axis). The paraxial approximation is key — for rays far from the axis, we get spherical aberration and the formula breaks down.

The sign convention is just a bookkeeping system. The physics is in the similar triangles. Once you understand that the two triangles share the same linear magnification ratio, the algebra is straightforward.

Alternative Method

You can also derive the formula using the focal length approach directly from ray diagrams without explicitly naming the triangles — just state $\frac{\text{image height}}{\text{object height}}$ is the same ratio whether measured from pole or from focus. The algebra works out identically.

For numerical problems, it is faster to just plug values into $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ with careful attention to signs.

Common Mistake

⚠️ Common Mistake

The most common error is forgetting sign convention when substituting. For a concave mirror with object at 30 cm in front: $u = -30$ cm, $f = -15$ cm (concave mirror). Students often write $f = +15$ and get the wrong image position. Always assign signs before plugging into the formula, not after.

💡 Expert Tip

In CBSE derivation questions (typically 3 marks), examiners award marks for: (1) drawing the correct ray diagram with labelled points, (2) identifying the similar triangles with reason, (3) final algebraic steps. Even if your algebra goes wrong, you can score 2/3 for the diagram and triangle identification.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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