A thin biconvex lens is prepared from glass of refractive index ${\mu }_2=\frac{3}{2}$. The two convex surfaces have equal radii of 20 cm each. One of the surfaces is silvered from  outside to make it reflecting. The lens now is placed in a large medium of refractive index ${\mu }_1=5/3$ . It acts as a 

For refraction at $P_1$  :
  $\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}\Rightarrow \frac{3/2}{v}-\frac{5/3}{\infty }=\frac{3/2-5/3}{20}$
 $\Rightarrow \frac{3}{2v}-0=-\frac{1}{120}\Rightarrow v=-180cm$
For reflection at  $P_2$:-  
 $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{v}+\frac{1}{-180}=\frac{1}{(-20/2)}$
 $\frac{1}{v}=\frac{1}{180}-\frac{1}{10}\Rightarrow v=-\frac{180}{17}cm$
Again refraction at $P_1$  :-
 $\frac{{\mu }_2}{v}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}\Rightarrow \frac{5/3}{v}-\frac{3/2}{(180/17)}=\frac{5/3-3/2}{-20}$
 
$\frac{5}{3v}-\frac{51}{360}=-\frac{1}{120}\Rightarrow \frac{5}{v}=\frac{51}{120}-\frac{1}{40}\Rightarrow v=12.5cm$
Final real image is formed at 12.5 cm from given silvered lens. This distance is focal length of the effective mirror. As the focus is real, it behaves like a converging or concave mirror of focal length 12.5 cm.