A lens of focal length 12 cm forms an erect image three times the size of the object. The distance between the object and image is:(a) 8 cm (b) 16 cm (c) 24 cm (d) 36 cm

 $v=3u$                        --------------(i)

From the lens formula we know that-

$\frac {1}{f}=\frac {1}{v}-\frac {1}{u}$

Putting the values of $f$ and $v$ in the formula, we get-

$\frac {1}{12}=\frac {1}{3u}-\frac {1}{u}$

$\frac {1}{12}=\frac {1-3}{3u}$

$\frac {1}{12}=\frac {-2}{3u}$

$3u=12\times {(-2)}$

$u=\frac {-24}{3}$

$u=-8cm$

Now, putting the value of $u$ in equation (i), we get-

 $v=3\times {(-8)}$   

 $v=-24cm$   

Thus, the image distance $v$ is $-$24 cm from the lens, and the negative sign implies that the image is formed on the left side of the lens.

Therefore,

The distance between image and object = $24-8=16cm$

Thus, (b) 16cm is the correct option.