Question

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

• A. 1.5
• B. 2.5
• C. 3.5
• D. 4.5

Answer 1

Answer: (A)

$u=P A=P C-A C=2 f-\frac{f}{3}=\frac{5 f}{3}$

From mirror formula
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
$\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{f}-\frac{3}{5 f}=\frac{2}{5 f}$
Length of image
$CA '= PA '- PC =\frac{5 f}{2}-2 f=\frac{f}{2}$
$\therefore $ Magnification $(m)=\frac{C A^{\prime}}{C A}=\frac{f / 2}{f / 3}=1.5$