b
(b) If end \(A\) of rod acts an object for mirror then it's image will be \(A'\) and if

\(u = 2f - \frac{f}{3} = \frac{{5f}}{3}\) so by using \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\)

\( \Rightarrow \,\,\,\frac{1}{{ - f}} = \frac{1}{v} + \frac{1}{{\frac{{ - 5f}}{3}}}\) \( \Rightarrow \,\,\,v = - \frac{5}{2}f\)

\(\therefore \) Length of image \( = \,\frac{5}{2}f - 2f = \frac{f}{2}\)