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  

 For the image formed by the lens, for the object kept at 30 cm from the lens, 

f = 15 cm (given)

From the lens formula, 

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

$\Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{15}-\frac{1}{30}$

$\Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{30}$

$\Rightarrow \mathrm{v} = 30 \mathrm{cm}$

The image is formed at a distance of 30 cm from the lens, i.e 20 cm behind the plane mirror, and this image acts as a virtual object for the plane mirror. The mirror forms an image of this object at a distance of 20 cm in front of the mirror, i.e 10 cm behind the lens. This image acts as an object for the lens.

According to the sign convention, 

u = 10 cm, f = 15 cm

Using the lens formula, 

 $\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$

$\Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{15}+\frac{1}{10}$$\Rightarrow \mathrm{v} = 6 \mathrm{cm}$

Hence, the image is formed 6 cm from the lens, or 16 cm (10 + 6 cm) in front of the mirror and is real image formed.