An object is kept at a distance of 16cm from a thin lens and the image formed is real. If the object is kept at a distance of 6cm from the same lens the image formed is virtual. If the size of the images formed are equal, then find the focal length of the lens?

When object is kept 16cm away
$\frac{1}{v_1}-\frac{1}{(-16)}=\frac{1}{f}\mathrm{..............}\left(1\right)$ 
Magnification $=\frac{v_1}{(-16)}$  
When object is kept 6 cm away
$\frac{1}{v_2}-\frac{1}{(-6)}=\frac{1}{f}\mathrm{................}\left(2\right)$ 
Magnification $=\frac{v_2}{-6}$  
Since the sizes of the images is same,  $M_1=-M_2$ 
Hence, $v_2=\frac{3v_1}{8}$  
Using this relation in Equation 1 equation 2, we get f = 11cm