A thin rod of length $\frac{f}{3}$ is lying along the principal axis of a concave mirror of focal length $f$. Image is real, magnified and inverted and one of the end of the rod coincides with its image itself. Find length of the image.

Image is real, magnified and inverted. So, the given rod lies between $F$ and $C$. Further, one end of the rod is coinciding with its image itself. Therefore, it is lying at $C$. So, the thin rod $CR$ is kept as shown below.

 $$\begin{gathered} CR=\frac{f}{3} \\ PR=2f-\frac{f}{3} \\ PR =\frac{5f}{3} \end{gathered}$$

we have to apply mirror formula only for point $R$.

$u=-\frac{5f}{3}, Focal length=-f$

Using the mirror formula, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f},$

we have, $\frac{1}{v}+\frac{1}{{\displaystyle \frac{-5f}{3}}}=\frac{1}{-f}$

Solving this equation, we get $v=\frac{-5f}{2} or -2.5f$

So, the image of rod $CR$ is $C\text{'}R\text{'}$ as shown below.

So, image length = $C\text{'}R\text{'} = 0.5f or \frac{f}{2}$