A candle flame 1.6 cm high is imaged in a ballbearing of diameter 0.4 cm. If the ball bearing is 20 cm away from the flame, find the location and the height of the image. 

The smooth surface of ball bearing acts like a convex mirror of focal length
$$\begin{gathered} \mathrm{f}=\frac{\mathrm{R}}{2}=\frac{0.2}{2}=0.1\mathrm{cm} \\ \frac{1}{\mathrm{V}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\Rightarrow \frac{1}{\mathrm{V}}+\frac{1}{-20}=\frac{1}{0.1}\Rightarrow \frac{1}{\mathrm{V}}=\frac{1}{20}+\frac{1}{0.1} \\ \Rightarrow \mathrm{V}=\frac{2}{20.1}\approx +0.1\mathrm{cm}=+1\mathrm{mm} \end{gathered}$$

As V is +ve, the image is formed towards rigtht of ‘P’ i.e. inside the ball bearing 
$$\begin{gathered} \frac{{\mathrm{h}}_{\mathrm{i}}}{{\mathrm{h}}_0}=-\frac{\mathrm{V}}{\mathrm{u}}\Rightarrow \frac{{\mathrm{h}}_{\mathrm{i}}}{1.6}=-\frac{(+0.1)}{-20} \\ \Rightarrow {\mathrm{h}}_{\mathrm{i}}=0.8\times {10}^{-2}\mathrm{cm}=0.08\mathrm{mm} \end{gathered}$$