A pin is placed at $10 cm$ in front of a convex lens of focal length $20 cm$ and refractive index $1.5$. The surface of the lens farther away from the pin is silvered and has a radius of curvature of $22 cm$. How far from the lens is the final image formed?

The value of the effective focal length $F$ is given by
$$\begin{gathered} \left|\frac{1}{\mathrm{F}}\right|=\frac{1}{{\mathrm{f}}_{\mathrm{L}}}+\frac{1}{{\mathrm{f}}_{\mathrm{m}}}+\frac{1}{{\mathrm{f}}_{\mathrm{L}}} \\ =\frac{2}{{\mathrm{f}}_{\mathrm{L}}}+\frac{1}{{\mathrm{f}}_{\mathrm{m}}} \\ =\frac{2}{20}+\frac{2}{22} \\ \Rightarrow \left|\mathrm{F}\right|=\frac{110}{21} \mathrm{cm} \end{gathered}$$

Since the convex lens with a silvered surface behaves as a concave mirror of effective focal length $F$, we have

$F=-\frac{110}{21}cm$ and $u=-10 cm$
Substituting these values in the mirror formula
$\frac{1}{v}+\frac{1}{u}=\frac{1}{F}$
we have
$\frac{1}{v}=-\frac{21}{110}+\frac{1}{10}$

$\Rightarrow \mathrm{v}=-11 \mathrm{cm}$.
The negative sign shows that the image is in front of the effective mirror and hence is real.