A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that the end close to the pole is 20 cm away from it. Find the length of the image of the rod.

Focal length f = −10 cm

∴ Radius of curvature R = 2f = −20 cm

Length of the rod AB = 10 cm

Distance of the end A = 20 cm = R

Since the near end ‘A’ of rod is at the centre of curvature, its image A¹ will also be at ‘C’.

Distance of the end B of the road= 20 + 10 = 30 cm

∴ u = −30 cm (from Cartesian sign convention) f = −10 cm v =?

From mirror formula $\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}$

$$\begin{gathered} \frac{1}{-10}=\frac{1}{-30}+\frac{1}{\mathrm{v}} \\ \frac{1}{\mathrm{v}}=\frac{-1}{10}+\frac{1}{30} \\ \Rightarrow \frac{1}{\mathrm{v}}=\frac{-30+10}{30\times 10} \\ \frac{1}{\mathrm{v}}=\frac{-20}{300} \\ \mathrm{v}=\frac{-30}{2}=-15 \mathrm{cm} \end{gathered}$$

Length of the image = A¹B¹ = 20 – 15 = 5 cm