A thin rod of length f/3 is placed along the optical axis of a concave mirror of focal length f such that its image which is real and elongated just touches the rod. Calculate the magnification.

1.5

Let $l$ be the length of the image.

Then, $m=\frac{l}{\frac{f}{3}}⇒l=\frac{mf}{3}$

Also image of one end coincides with the object,

$⇒ u' = 2f$

$u'=u+\frac{f}{3}⇒u=2f-\frac{f}{3}=\frac{5f}{3}$

$v=-(u+\frac{f}{3}+\frac{mf}{3})$. Putting in mirror formula,

$\frac{1}{u+\frac{f}{3}+\frac{mf}{3}}+\frac{1}{u}=\frac{1}{f}$

$⇒\frac{3}{5f+f+mf}+\frac{3}{5f}=\frac{1}{f}⇒\frac{1}{m+6}=\frac{2}{15}$

$⇒m=\frac{2}{3}=1.5$