Explanation

(b) The point on the right side of the lens at will rays converge will behave as virtual object of the lens.
$\therefore\ u = \ + 12\ cm,\ f = 20\ cm$
From the relation
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\therefore\ \frac{1}{20} = \frac{1}{v} - \frac{1}{12}$
$\Longrightarrow \frac{1}{v} = \frac{1}{20} + \frac{1}{12}$
$= \frac{3 + 5}{60} = \frac{8}{60}$
$\therefore\ u = \frac{60}{8} = 7.5\ cm$
So, image will be formed on same side of the virtual object at a distance of 7.5 cm from the lens.