Concave mirror — 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. An object is placed 20 cm from a concave mirror of focal length 15 cm. Find the image position and magnification.

(NCERT Class 10 / CBSE 12th board standard)

Solution — Step by Step

Derivation

Set up the geometry

Consider a concave mirror with pole $P$ , centre of curvature $C$ , focus $F$ , and principal axis. An object $AB$ is placed beyond $F$ . The image $A'B'$ is formed by the intersection of reflected rays.

Using the New Cartesian Sign Convention: distances measured from pole, against the incident light are negative.

Use similar triangles

Consider two pairs of similar triangles:

Pair 1: $\triangle ABP \sim \triangle A'B'P$ (formed by the ray through $P$ ): $\frac{A'B'}{AB} = \frac{-v}{-u} = \frac{v}{u}$

Pair 2: $\triangle A'B'F \sim \triangle MPF$ (formed by the ray parallel to axis, reflected through $F$ ; $M$ is at pole height): $\frac{A'B'}{MP} = \frac{v - f}{f}$

Since $MP = AB$ (for small apertures): $\frac{A'B'}{AB} = \frac{v - f}{f}$

Equate and simplify

From both ratios: $\frac{v}{u} = \frac{v - f}{f}$

$vf = u(v - f) = uv - uf$

$vf + uf = uv$

Divide throughout by $uvf$ :

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

Numerical Problem

Apply the mirror formula

Object at $u = -20$ cm (negative by convention), $f = -15$ cm (concave mirror).

$\frac{1}{v} + \frac{1}{-20} = \frac{1}{-15}$

$\frac{1}{v} = -\frac{1}{15} + \frac{1}{20} = \frac{-4 + 3}{60} = -\frac{1}{60}$

$v = -60 \text{ cm}$

Image is 60 cm in front of the mirror — real and inverted.

Magnification: $m = -\frac{v}{u} = -\frac{-60}{-20} = -3$

The image is 3 times magnified and inverted (negative $m$ ).

Why This Works

The mirror formula connects three quantities: object distance, image distance, and focal length. It works because light follows the law of reflection, and for mirrors with small aperture (paraxial rays), the geometry produces consistent similar triangles.

The formula $1/v + 1/u = 1/f$ is valid for all mirrors (concave and convex) and all object positions, as long as you apply the sign convention correctly. For convex mirrors, $f$ is positive; for concave, $f$ is negative.

Alternative Method

You can also derive using the relation $R = 2f$ and the formula from the geometry of a concave mirror with centre of curvature. The geometry gives $\frac{1}{v} + \frac{1}{u} = \frac{2}{R} = \frac{1}{f}$ directly.

💡 Expert Tip

For quick image identification with concave mirrors: if the object is beyond $C$ ( $|u| > 2|f|$ ), image is between $F$ and $C$ (diminished). If between $F$ and $C$ , image is beyond $C$ (magnified). If between $F$ and $P$ , image is virtual (behind mirror). Memorise these three cases — they cover all CBSE and NEET MCQs.

Common Mistake

⚠️ Common Mistake

The most common derivation error: not applying sign convention when using similar triangle ratios. The distances $u$ , $v$ , and $f$ are all negative for a concave mirror (measuring against incident light). If you use magnitudes, you get $1/v + 1/u = 1/f$ , but if you use signed values, the signs must be consistent. CBSE examiners check this carefully.