# A thin rod of length $\large\frac{f}{3}$ is placed along optical axis a concave mirror of focal length f such that its image which is real and elongated just touches rod . calculate the magnification.

$(a)\;\frac{3}{4} \\ (b)\;\frac{-3}{2} \\ (c)\;\frac{3}{2} \\ (d)\;\frac{-3}{4}$

Let l be length of image
Then $m= \large\frac{l}{\Large\frac{f}{3}}$$=l=\large\frac{mf}{3} Also image of one end coincides with object, => u'=2f u'= u+ \large\frac{f}{3} =>$$ u=2f -\large\frac{f}{3}=\frac{5f}{3}$
$v= -(u+\large\frac{f}{3} +\frac{mf}{3} \bigg)$
Putting in mirror formula,
$\large\frac{1}{u+\Large\frac{f}{3}+\frac{mf}{3}}+\frac{1}{u}=\frac{1}{f}$
=> $\large\frac{3}{5f+f+mf}+\frac{3}{5f}=\frac{1}{f}$
=> $\large\frac{1}{m+6} =\frac{2}{15}$
=> $m =\large\frac{3}{2}$
Hence c is the correct answer.