$(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.

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