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:

0.25 m

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0.33 m

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0.5 m

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0.67 m

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Solution

The correct option is C $0.5 m$
Let f be the focal length of a convex mirror.
According to new cartesian sign convention,
Object distance u = -f, focal length = +f
Using mirror formula, $\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$
$\frac{1}{- f} + \frac{1}{v} = \frac{1}{f} o r \frac{1}{v} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f} o r v = \frac{f}{2}$
The image is formed at a distance $\frac{f}{2}$ behind the mirror.
Magnification m = $- \frac{v}{u} = - \frac{f / 2}{- f} = \frac{1}{2}$
As magnification is positive so It is a virtual image.
Also, m = $\frac{H e i g h t o f i m a g e (h_{I})}{H e i g h t o f o b j e c t (h_{0})}$
$∴ h_{I} = m h_{O} = \frac{1}{2} (1 m) = 0.5 m$

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

A boy of height 1m stand infront of convex mirror. His distance from the mirror is equal to its focal length. What is the height of his image?

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