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

Object is placed at distance \(2 f\) from the lens. So first image \(I_1\) will be formed at distance \(2 f\) on other side. This image \(I_1\) will behave like a virtual object for mirror. The second image \(I_2\) will be formed at distance \(20 \mathrm{~cm}\) in front of the mirror, or at distance \(10 \mathrm{~cm}\) to the left hand side of the lens.


Now applying lens formula
\(
\begin{array}{rlr}
& \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\
& \frac{1}{v}-\frac{1}{+10}=\frac{1}{+15} \\
\text { or } & v=6 \mathrm{~cm}
\end{array}
\)
Therefore, the final image is at distance \(16 \mathrm{~cm}\) from the mirror. But, this image will be real.
This is because ray of light is travelling from right to left.
\(\therefore\) The correct option is (b).