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}

Answer and Explanation:

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}


Learn more about this topic:

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What is a Convex Mirror? - Definition, Uses & Equation

from ICSE Physics: Study Guide & Syllabus

Chapter 7 / Lesson 4
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