The size of the images of an object, formed by a thin lens are equal when the object is placed at two different positions 8 cm and 24 cm from the lens. The focal length of the lens is $\_\_\_\_$ cm.

The magnification of a thin lens is given by $m = \frac{f}{f + u}$.
Given that the size of the images is equal for two different object positions, the magnitudes of the magnifications must be equal: $|m_1| = |m_2|$.
For a single lens to produce images of the same size at two different object distances, one image must be real and the other must be virtual.
Let $u_1 = -8$ cm and $u_2 = -24$ cm.
Then, $m_1 = -m_2$ (since one is real and one is virtual).
$\frac{f}{f + u_1} = -\frac{f}{f + u_2}$
$\frac{1}{f - 8} = -\frac{1}{f - 24}$
$f - 24 = -(f - 8)$
$f - 24 = -f + 8$
$2f = 32$
$f = 16$ cm.
The focal length of the lens is 16 cm.