# A concave spherical mirror has a radius of curvature of 29.4 cm.The object distance is 5.7 cm .a)...

## Question:

A concave spherical mirror has a radius of curvature of {eq}29.4 \ cm {/eq}.The object distance is {eq}5.7 \ cm {/eq} .

a) Find the magnitude of the image distance. Answer in units of cm.

b) Find the magnification.

## Mirror Formula and Magnification:

Mirror formula gives a relationship between the focal length of the mirror and the object and image distances from the mirror. It is expressed by the formula;

{eq}\displaystyle\frac {1}{f}= \frac {1}{v}+\frac {1}{u}\\ {/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 is given by the formula;

{eq}m=\dfrac {h_i}{h_o} =- \dfrac {v}{u} {/eq}

where, {eq}h_i {/eq} represents height of the image and {eq}h_o {/eq} represents height of the object.

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

• 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:

• Distance of object from the mirror is {eq}u= -5.7\ cm {/eq}
• Radius of curvature of the concave mirror {eq}R= -29.4 \ cm {/eq}
• Focal length of the concave mirror is {eq}\displaystyle f= \frac{R}{2}=\frac{-29.4 }{2}=-14.7\ cm {/eq}

Part (a):

From the Mirror formula,

{eq}\begin{align} \frac {1}{v}+\frac {1}{u} &= \frac {1}{f}\\ \implies \frac {1}{v}+\frac {1}{-5.7} &= \frac {1}{-14.7}\\ \implies v &= 9.31\ cm\\ \end{align} {/eq}

Therefore, the magnitude of the image distance is {eq}v=9.31\ cm {/eq}

Part (b)

Magnification of the image is given by,

{eq}\begin{align} m &= - \dfrac {v}{u} \\ &= - \dfrac {9.31\ cm }{-5.7\ cm} \\ &= 1.633\ cm \\ \end{align} {/eq}