$A$ beam of light converges at a point $P$. Now a lens is placed in the path of the convergent beam $12 \, cm$ from $P$. At what point does the beam converge if the lens is
$(a)$ a convex lens of focal length $20 \, cm$,and
$(b)$ a concave lens of focal length $16 \, cm$?

(N/A) In the given situation,the object is virtual and the image formed is real.
Object distance,$u = +12 \, cm$.
$(a)$ Focal length of the convex lens,$f = +20 \, cm$.
Using the lens formula,$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
$\frac{1}{v} - \frac{1}{12} = \frac{1}{20} \implies \frac{1}{v} = \frac{1}{20} + \frac{1}{12} = \frac{3 + 5}{60} = \frac{8}{60}$.
$\therefore v = \frac{60}{8} = 7.5 \, cm$.
Hence,the beam converges $7.5 \, cm$ to the right of the lens.
$(b)$ Focal length of the concave lens,$f = -16 \, cm$.
Using the lens formula,$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
$\frac{1}{v} - \frac{1}{12} = -\frac{1}{16} \implies \frac{1}{v} = \frac{1}{12} - \frac{1}{16} = \frac{4 - 3}{48} = \frac{1}{48}$.
$\therefore v = 48 \, cm$.
Hence,the beam converges $48 \, cm$ to the right of the lens.