Given, $A=-40\ \mathrm{cm}$ and, $f=20\ \mathrm{cm}$ From the lens formula, we have: $\;\frac{1}{v}\;-\;\frac{1}{u}\;=\;\frac{1}{f}\;$ $\Rightarrow \;\frac{1}{v}\;=\;\frac{1}{f}+\;\frac{1}{u}\;$ $\Rightarrow \;\frac{1}{v}\;=\;\frac{1}{20}+\;\frac{1}{(-40)}\;$ $\Rightarrow \;\frac{1}{v}\;=\;\frac{2-1}{40}=\;\frac{1}{40}\;$ $\;v\;=40\ \mathrm{cm}$ The positive sign describes that the image is formed to the right of the lens. The image $I_1$ is formed behind the mirror and thus acts as a virtual source for the mirror. The convex mirror forms the image $I_2$, whose distance from the mirror is given by: $\;\frac{1}{v}+\;\frac{1}{v}=\;\frac{1}{f}$ Here: $\;u=25\ \mathrm{cm}$ $\;f=\;\frac{R}{2}=10\ \mathrm{cm}$ $\;\frac{1}{v}+\;\frac{1}{v}\;=\;\frac{1}{f}$ $\Rightarrow \;\; \;\frac{1}{v}\;=\;\frac{1}{f}-\;\frac{1}{u}$ $\Rightarrow \;\frac{1}{v}=\;\frac{1}{10}-\;\frac{1}{25}$$\Rightarrow \;\frac{1}{v}=\;\frac{5-2}{50}=\;\frac{3}{50}$$\Rightarrow v=+16.67\ \mathrm{cm}$Hence, the final image is formed at a distance of $16.67\ \mathrm{cm}$ behind the convex mirror.