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10. A 20.0 cm tall object is placed 50.0 cm in front of a convex mirror with a radius of curvature of 34.0 cm. Where will the image be located, and how tall will it be? Please show all work. (ref: p.471-481)

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lucic

Answer:

  1. The image will be located at -13 cm
  2. The height of image is 5.2 cm

Explanation:

First you should remember that the images produced by convex mirror is virtual.

In this question, you should find the focal length using the radius of curvature, then apply the focal length relationship to find the image distance which is a virtual distance and finally use the magnification relationship to find the image height.

Let;

  • height of object=h
  • distance of object from mirror=o
  • radius of curvature=r
  • image distance=i
  • height of image=h'
  • focal length=f

Find the focal length

f=focal length is half the radius of curvature

f=r/2=34/2 =17cm =0.17m (This distance is virtual thus include a -ve sign)

f= -0.17m

Apply the focal length relationship

[tex]\frac{1}{o} +\frac{1}{i} =\frac{1}{f} \\\\\\\frac{1}{0.5} +\frac{1}{i} =-\frac{1}{0.17} \\\\\\2+\frac{1}{i} =-5.88\\\\\\\frac{1}{i} =-5.88-2\\\\[/tex]

Solve for the reciprocal

[tex]\frac{1}{i} =-7.88\\\\i=-0.13m[/tex]

This is a virtual distance for the virtual image formed

Apply the magnification relationship

Magnification = height of image÷height of object

or

Magnification= - image distance÷object distance

[tex]\frac{h'}{0.20} =-\frac{-0.13}{0.5} \\\\\\0.5h'=0.13*0.20\\\\\\h'=\frac{0.13*0.20}{0.5} =0.052m=5.2cm[/tex]

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