An object is placed at a distance of 20 cm from a thin plano convex lens of focal length 15 cm. The plane surface of lens is now silvered. What is the position of image? 

The given lens is equivalent to the combination of a plano convex lens and a plane mirror. If P_L and P_M represent their powers respectively, the power of system 
$\mathrm{P}=2{\mathrm{P}}_{\mathrm{L}}+{\mathrm{P}}_{\mathrm{M}}=2\left(\frac{1}{{\mathrm{f}}_{\mathrm{L}}}\right)-\frac{1}{{\mathrm{f}}_{\mathrm{M}}}$
If the system behaves as a mirror of focal length f₀ then 
$$\begin{gathered} P=-\frac{1}{f_0}\therefore -\frac{1}{f_0}=\frac{2}{f_L}-\frac{1}{f_M} \\ \text{ here }{\mathrm{f}}_{\mathrm{L}}=15\mathrm{cm},{\mathrm{f}}_{\mathrm{M}}=\frac{\mathrm{R}}{2}=\frac{\mathrm{\infty }}{2}=\mathrm{\infty } \\ -\frac{1}{{\mathrm{f}}_0}=\frac{2}{15}-0\Rightarrow {\mathrm{f}}_0=-\frac{15}{2}\mathrm{cm}. \end{gathered}$$
The given silvered lens behaves as a concave  mirror of focal length 15/2 cm. Given that u = - 20cm for this equivalent mirror 
$\therefore \frac{1}{\mathrm{V}}+\frac{1}{\mathrm{u}}=\frac{1}{{\mathrm{f}}_0}\Rightarrow \frac{1}{\mathrm{V}}+\frac{1}{-20}=\frac{1}{\left(-\frac{15}{2}\right)}$
$\Rightarrow \mathrm{V}=-12\mathrm{cm}$
As V is -ve, the image is formed towards left of the lens and it is real