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

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A

$0.25\,m$

6%

B

$0.33\,m$

9%

C

$0.5\,m$

81%

D

$0.67\,m$

4%

Solution:

Let $f$ be focal length of the convex mirror.
According to new cartesian sign convention
Object distance, $ u = -f$, focal length $= +f$
Using mirror formula, $\frac{1}{u}+\frac{1}{v}= \frac{1}{f}$
$\frac{1}{-f}+\frac{1}{v} = \frac{1}{f} $ or $\frac{1}{v} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f} $ or $v = \frac{f}{2}$
The image is formed at a distance $\frac{f}{2}$ behind the mirror. It is a virtual image.
Magnification, $m = - \frac{v}{u} = -\frac{(f/2)}{-f} = \frac{1}{2}$
Also, $m = \frac{\text{Height of image (h_I)}}{\text{Height of object (h_0 )}}$
$\therefore h_1 = mh_0 $
$= \frac{1}{2}(1\,m) = 0.5$

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