The size of the images of an object, formed by a thin lens are equal when the object is placed at tw

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 .

Professor PC Thomas Classes · Answer

Since the images have the same size at two different object distances, one image must be real (inverted) and the other virtual (erect). The magnification magnitude is the same: $|m_1| = |m_2|$. Using the formula $m = \frac{f}{f+u}$, where $u$ is the object distance (negative by sign convention), we have $\frac{f}{f - 8} = -\frac{f}{f - 24}$. Cross-multiplying gives $f - 24 = -(f - 8) \implies f - 24 = -f + 8 \implies 2f = 32 \implies f = 16 	ext{ cm}$.