Question

An infinitely long rod lies 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. Its image will have a length

• A. $\frac{uf}{u-f}$
• B. $\frac{uf}{u+f}$
• C. $\frac{f^2}{u+f}$
• D. $\frac{f^2}{u-f}$

Answer 1

The correct option is D $\frac{f^2}{u-f}$

For near end of the rod $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$
Here, u and f are negative
$\therefore$ $\left|v\right|=\frac{uf}{u-f}$
Far end of the end is at infinity. Therefore, image will be formed at focus.
$\therefore$ Length of the image = $\left|v\right|-f$
$=\frac{uf}{u-f}-f=\frac{f^2}{u-f}$