Question

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.

Hint:

If magnification magnitudes are equal for object distances $d_1$ and $d_2$, then $f = \frac{d_1+d_2}{2}$.

Answer 1

Correct Answer: 16

Let the focal length be $f$. The magnification $m$ is given by $m = \frac{f}{f+u}$.
For the image sizes to be equal at different object positions, one image must be real and the other virtual (magnifications are $m$ and $-m$).
Let $u_1 = -8$ cm and $u_2 = -24$ cm.
$|m_1| = |m_2| \implies \frac{f}{f-8} = - \frac{f}{f-24}$.
$f - 24 = -(f - 8) \implies f - 24 = -f + 8$.
$2f = 32 \implies f = 16 \text{ cm}$.