A rod of infinite length is placed along the axis of a concave mirror of focal length f. The near end of the rod is at a distance u > f from the mirror. The length of it's image is
Answer & explanation
Correct answer: option 1
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
$v = \frac{uf}{u - f}$
$\text{For far end: } u \rightarrow \infty \Rightarrow v = f$
$\text{Image extends from } v = \frac{uf}{u - f} \text{ to } v = f$
$\text{Length of image} = \frac{uf}{u - f} - f$
$= f\left(\frac{u}{u - f} - 1\right)$
$= f\left(\frac{u - (u - f)}{u - f}\right)$
$= f\left(\frac{f}{u - f}\right)$
$= \frac{f^2}{u - f}$
The length of the image is $\frac{f^2}{u - f}$.