$u=-60\ \text{cm}\;\text{ and }\; f=20\ \text{cm}$ From the lens formula, we have: $\;\frac{1}{v}-\;\frac{1}{u}\;=\;\frac{1}{f}$ $\;\frac{1}{v}\;=\;\frac{1}{f}+\;\frac{1}{u}$ $\;=\;\frac{1}{20}+\;\frac{1}{-60}$ $\;=\;\frac{3-1}{60}=\;\frac{2}{60}=\;\frac{1}{30}$ $v\;=+30\ \text{cm}$ The positive sign indicates that the image is formed at the right of the lens. The image $I_1$ is formed behind the mirror and acts as a virtual oject for the mirror. The convex mirror forms the image $I_2$, whose distance from the mirror can be determined as: $\;\frac{1}{v}+\;\frac{1}{u}\;=\;\frac{1}{f}$ Here, $u\;=15\ \text{cm}$ and, $f\;=\;\frac{R}{2}=10\ \text{cm}$ $\;\frac{1}{v}\;=\;\frac{1}{f}-\;\frac{1}{u}$ $v\;=30\ \text{cm}$ Hence, the final virtual image is formed at a distance $30\ \text{cm}$ from the convex mirror.