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 its image is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{f^2}{u-f} \)
The image of the rod stretches between the images of its two ends, so find each with the mirror equation (distances taken as magnitudes, all on the reflecting side of the concave mirror, and \( u \gt f \) so the image is real).
\( \displaystyle \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow v = \frac{uf}{u-f} \)
The far end is at infinity, so its image is at the focus: \( v_\infty = f \).
\( \displaystyle \text{Length} = \frac{uf}{u-f} - f = \frac{uf - f(u-f)}{u-f} \)
\( \displaystyle \text{Length} = \frac{f^{2}}{u-f} \)