A thin rod of length f/3 lies along the axis of a concave mirror of focal length f. One end of its magnified image touches an end of the rod. The length of the image is

$\frac{1}{2}f$

One end of its image touches an end of the rod. So, one end of the rod is present at the centre of curvature of mirror. Image is magnified. So, the other end A is between C and focus.

Lets assume that end A of rod has its image at A'.

For it, $u=-\left(2 f-\frac{f}{3}\right)=\frac{-5 f}{3}$

So, using $\frac{1}{f}=\frac{1}{v}+\frac{1}{u} \Rightarrow \frac{1}{-f}=\frac{1}{v}+\frac{1}{\frac{-5 f}{3}} \Rightarrow v=-\frac{5}{2} f$

∴ Length of image $=\frac{5}{2} f-2 f=\frac{f}{2}$