A thin convex lens of focal length ‘f’ is put on a plane mirror as shown in the figure. When an object is kept at a distance ‘a’ from the  lens mirror combination, its image is formed at a distance $\frac{\mathrm{a}}{3}$ in front of the combination. The value of ‘a’ is 

$$\begin{gathered} \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\ \frac{1}{v}-\frac{1}{-a}=\frac{1}{f}\Rightarrow v=\frac{af}{a-f} \\ The image formed will be on left side of lens provided a>f The mirror will form the image at right side of the lens with dis\tan ce equal to \\ \frac{af}{a-f} \\ Now the third image will be formed by lens again for which the coordinate system will be reversed of earlier as the rays are travelling from left to right: \\ u=\frac{\text{ af }}{a-f},f=+f\text{ and }v=\frac{a}{3} \\ \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\dots \text{ (i) } \\ \begin{array}{r}\frac{3}{a}-\frac{a-f}{af}=\frac{1}{f}\dots \\ \frac{3f}{af}-\frac{a-f}{af}=\frac{a}{af}\end{array} \\ 3f-a+f=a\Rightarrow a=2f \end{gathered}$$