A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is

 First refraction through lens,

$$\begin{gathered} \begin{array}{r}\frac{1}{v\text{'}}-\frac{1}{-30}=\frac{1}{+15}\\ \Rightarrow \frac{1}{v\text{'}}=-\frac{1}{30}+\frac{1}{15}\end{array} \\ \Rightarrow v\text{'}=+30\mathrm{cm} \end{gathered}$$

This image acts as virtual object for plane mirror. Image by the mirror is formed at 20 cm before mirror.

Second refraction through lens,

$$\begin{gathered} \frac{1}{\mathrm{v}}-\frac{1}{+10}=\frac{1}{+15} \\ \Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{10}+\frac{1}{15}=\frac{25}{150} \\ \Rightarrow \mathrm{v}=+6 \mathrm{cm} \end{gathered}$$

Hence, final image is real and is situated at a distance of 16 cm from the mirror.