An object 5 cm high is placed at a distance of 10 cm from a convex lens of focal length 15 cm. Find the nature, position of the image.

Using Cartesian sign convention

Object distance = −10 cm

Focal length of a convex lens = 15 cm

Height of the object = h0 = 5 cm

Using lens formula 

$$\begin{gathered} \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}} \\ \frac{1}{15}=\frac{1}{\mathrm{v}}=\frac{1}{-10} \\ \frac{1}{15}=\frac{1}{\mathrm{v}}+\frac{1}{10}\Rightarrow \frac{1}{\mathrm{v}}=\frac{10-15}{15\times 10} \\ \mathrm{v}=-\frac{150}{5}\Rightarrow -30 \mathrm{cm} \end{gathered}$$

Negative sign indicates it is a virtual image

$$\begin{gathered} \mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}} \\ \mathrm{m}=\frac{-30}{-10}=3 \end{gathered}$$

Positive sign indicates the image is erect.