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. $1$
• B. $1.5$
• C. $2$
• D. $2.5$

Answer 1

The correct option is B $1.5$

Since the image formed is real and elongated, the ray diagram is shown below.

Since the image of $B$ is formed at $B$ itself $\therefore B$ is situated at the centre of curvature that is at a distance $2f$ from the pole $P$.

rom the figure
$PA=2f-\frac{f}{3}=\frac{5f}{3}$

The image $A^{\prime }$ for the point $A$ is
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

$\frac{1}{-f}=\frac{1}{v}-\frac{1}{\left(\frac{5f}{3}\right)}$

$\frac{1}{v}=\frac{3}{5f}-\frac{1}{f}=\frac{-2}{5f}$

$v=-2.5f$

Image length $=2.5f-2f=0.5f$

Magnification, $m=\frac{\text{Image length}}{\text{object length}}=\frac{0.5f}{\left(\frac{f}{3}\right)}$
$m=0.5\times 3$
$m=1.5$