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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} $

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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{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.
answered Feb 11, 2014 by meena.p

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