Question

A thin rod of length $\frac{f}{3}$, where $f$ is focal length, lies along the axis of a concave mirror. One end of its magnified image touches an end of the rod. The length of the image will be

• A. $0.5f$
• B. $1.5f$
• C. $2.5f$
• D. $3.5f$

Answer 1

The correct option is A $0.5f$

As the image is touching the object, the image is real. It is magnified, thus it is kept between centre of curvature and focus. Also, one end of the image touches the one end of the object, therefore, one end of the object is at centre of curvature.
For end $A$,
$u=-2f$
So, by ray diagram, $v=-2f$ ...........$\left(1\right)$
For end $B$,
$u=-(2f-\frac{f}{3})=-\frac{5f}{3}$
So, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{v}-\frac{3}{5f}=\frac{-1}{f}$
$\Rightarrow \frac{1}{v}=\frac{-2}{5f}$
$\Rightarrow v=-2.5f$ ...........$\left(2\right)$
So, length of image $=|-2.5f|-|-2f|=0.5f$
$[$from $\left(1\right)$ and $\left(2\right)]$

$\begin{array}{c}\text{Why this question?}\\ \text{Such problems will create confusion, if you don't}\\ \text{set or draw the boundary segment of object.}\\ \text{Tips: The object is confined within its boundary segment}\\ \text{thus applying mirror formula will give image boundaries.}\\ \text{Length of image is distance between boundary segment of image.}\end{array}$