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.
Answer & explanation
Correct answer: option 2
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,
$\frac{1}{u+\frac{f}{3}+\frac{mf}{3}}+\frac{1}{u}=\frac{1}{f}$
$⇒\frac{3}{5f+f+mf}+\frac{3}{5f}=\frac{1}{f}⇒\frac{1}{m+6}=\frac{2}{15}$
$⇒m=\frac{2}{3}=1.5$