# The radius of curvature of a convex mirror is 1.60 x 10^2 cm. An object that is 11.0 cm high is...

## Question:

The radius of curvature of a convex mirror is 1.60 x 10{eq}^{2} {/eq} cm. An object that is 11.0 cm high is placed 40.0 cm in front of this mirror.

Determine (a) the location and (b) the height of the image.

## Mirror Formula and Magnification of image:

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,

{eq}\begin{align} m ****= \dfrac {v}{u}\\ \end{align} {/eq}

Sign conventions used for the spherical mirror 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 mirror is a convex mirror.
• Radius of curvature of the mirror is {eq}R= 1.60\times 10^2\ cm {/eq}
• Height of the object is {eq}h_o = 11\ cm {/eq}
• Distance of object from the mirror is {eq}u= -40\ cm {/eq}

Let

• Distance of image from the mirror is {eq}v {/eq}
• Height of the image is {eq}h_i {/eq}
• The magnification of the image is m

Part (a)

Focal length of the mirror is given by;

{eq}\begin{align} f&=\dfrac{R}{2}\\ &=\dfrac{1.6\times 10^2\ cm }{2 }\\ &= 80\ cm\\ \end{align} {/eq}

From the Mirror formula,

{eq}\begin{align} \dfrac {1}{v}+\dfrac {1}{u} &= \dfrac {1}{f}\\ \implies {1}{v}&= \dfrac {1}{f}-\dfrac {1}{u} \\ \implies {1}{v}&= \dfrac {1}{80\ cm}-\dfrac {1}{-40\ cm} \\ \implies v &= 26.67\ cm\\ \end{align} {/eq}

Thus, the image is located at a distance of {eq}26.6667\ cm {/eq} from the mirror along the principal axis.

Part (b)

Magnification of the image is given by,

{eq}\begin{align} m &= \dfrac {h_i}{h_o}\\ \implies \dfrac {-v}{u} & = \dfrac {h_i}{h_o}\\ implies h_i &= \dfrac {-v}{u}\times h_o\\ &=\dfrac {-26.6667\ cm}{-40\ cm}\times (11\ cm)\\ &=7.33 \ cm\\ \end{align} {/eq}

Thus, the height of the image is {eq}h_i = 7.33 \ cm\\ {/eq}