# A convex mirror in an amusement park has a radius of curvature of magnitude 3m. A man stands in...

## Question:

A convex mirror in an amusement park has a radius of curvature of magnitude 3m. A man stands in front of the mirror, so that his image is half as tall as his actual height. At what distance must the man focus his eyes, in order to see his image? Explain thoroughly.

## Mirror Formula and Magnification:

Mirror equation is given by the equation;

{eq}\dfrac {1}{v}+\dfrac {1}{u} = \dfrac{1}{f} {/eq}

where, {eq}v {/eq} is image distance, {eq}u {/eq} is object distance and {eq}f {/eq} is the focal length of the mirror.

Magnification of the image is given by the relation {eq}m = \dfrac {-v}{u} {/eq}

Sign conventions:

The image formation by the spherical mirror follows some sign conventions. These are the followings;

• Distances measured in the direction of the incident ray are taken as positive.
• Distances measured in the direction opposite to the incident ray are taken as negative.
• Distances measured above the principal axis are taken as positive.
• Distances measured below the principal axis are taken as negative.

Given

• The radius of the curvature is {eq}R = + 3\ m {/eq}
• The magnification of the image formed is {eq}m = \dfrac{1}{2} {/eq}

Let

• The distance of the man from the mirror is u.
• The distance of the image formed from the mirror is v

As we know that the magnification of the image is given by;

{eq}\begin{align} m &= -\dfrac {v}{u}\\ \implies \dfrac{1}{2}&=-\dfrac {v}{u}\\ \implies u &= -2v\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{align} {/eq}

From the Mirror formula,

{eq}\begin{align} \dfrac {1}{f}&=\dfrac{1}{v}+\dfrac {1}{u}\\ \implies \dfrac{1}{1.5}&= \dfrac{1}{v}+\dfrac{1}{-2v}\\ \implies v &=0.75\ m\\ \end{align} {/eq}

Thus, the man must focus his eyes at a distance of {eq}0.75\ m {/eq} behind the mirror. 