A boy of height $1m$ stands in front of a convex mirror. His distance from the mirror is equal to its focal length. The height of his image is :

The correct option is C $0.5m$

Let $f$ be the focal length of the convex mirror. According to the new cartesian sign convention

Object distance, $u=-f$, f

ocal length $=+f$

Using mirror formula,

$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$

$\Rightarrow \frac{1}{-f}+\frac{1}{v}=\frac{1}{f}$

$\Rightarrow \frac{1}{v}=\frac{1}{f}+\frac{1}{f}=\frac{2}{f}$

$v=\frac{f}{2}$

The image is formed at a distance$\frac{f}{2}$ behind the mirror. It is virtual image.

Magnification,

$m=-\frac{v}{u}=-\frac{\left(f/2\right)}{(-f)}=\frac{1}{2}$

Also, $m=\frac{\text{Height of image}\left(h_I\right)}{\text{Height of object}\left(h_0\right)}$

$\therefore h_I=m{h}_0=\frac{1}{2}\left(1m\right)=0.5m$