What will be the height of the image when an object of $2\text{mm}$ is placed at a distance $20\text{cm}$ in front of the axis of a convex mirror of radius of curvature $40\text{cm}$?

The correct option is D $1\text{mm}$
Height of object, $h=2\text{mm}$
Radius of curvature, $R=40\text{cm}$
$\therefore$ Focal length: $f=\frac{R}{2}=20\text{cm}$
Object distance, $u=-20\text{cm}$

Mirror formula:
$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$

$\Rightarrow \frac{1}{v}=\frac{1}{20}-\left(\frac{1}{-20}\right)$

$\Rightarrow \frac{1}{v}=\frac{2}{20}=\frac{1}{10}$
Image distance, $v=10\text{cm}$

Magnification, $m=\frac{-v}{u}$
$m=\frac{-10}{-20}=\frac{1}{2}=0.5$

Also, $\frac{h^{\prime }}{h}=m=0.5$

$\Rightarrow h^{\prime }=mh=0.5\times 2=1\text{mm}$