The object is midway between the lens and the mirror. The mirror’s radius of curvature is 20.0cm and the lens has a focal length of -16.7.Considering only the rays that alevees the object and travels first toward the mirror, the magnification of the system is

Given for mirror R = 20 cm
$$\begin{gathered} \text{f=-10cm} \\ \text{u=}\frac{25}{2}=-12.5\mathrm{cm} \end{gathered}$$
Let the image distance from mirror is v₁
$\begin{array}{l}\frac{1}{{\mathrm{v}}_1}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\\ \frac{1}{{\mathrm{v}}_1}=\frac{-1}{10}+\frac{1}{12.5}\\ \Rightarrow \frac{1}{{\mathrm{v}}_1}=\frac{-2.5}{12.5}\\ {\mathrm{v}}_1=-50\mathrm{cm}\end{array}$
 

Magnification by mirror$\mathrm{m}=\frac{-\mathrm{v}}{\mathrm{u}}=\frac{-\left(-50\right)}{\left(-12.5\right)}$
Image formed by lens
Image formed by mirror acts as a object for lens
Lens $\mathrm{Cl}=+25\mathrm{cm}$
Image formed by mirror
$$\begin{gathered} \begin{array}{l}\mathrm{f}=-16.7\mathrm{cm}\\ {\mathrm{v}}_2=?\\ \frac{1}{{\mathrm{v}}_2}-\frac{1}{25}=\frac{1}{-16.7}\end{array} \\ {\mathrm{v}}_2=-50.3\mathrm{cm} \end{gathered}$$
magnification ${\mathrm{m}}_2=\frac{{\mathrm{v}}_2}{{\mathrm{u}}_1}=\frac{-50.3}{25}=-2.012$
Total magnification is $\mathrm{m}=\mathrm{m}\times {\mathrm{m}}_2$
$\begin{array}{r}=-4\times -2.012\\ \mathrm{m}=8.048\end{array}$