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Spherical mirror radius of curvature

  1. Aug 11, 2011 #1
    1. The problem statement, all variables and given/known data

    A dentist uses a spherical mirror to examine a tooth. The tooth is 1.13 cm in front of the mirror, and the image is formed 10.8 cm behind the mirror. Determine the mirror's radius of curvature.

    2. Relevant equations

    1/p+1/q=1/f

    f=R/2

    3. The attempt at a solution

    Since the object is in front of the mirror, p > 0. With the image behind the mirror q < 0. So the radius of curvature is

    [tex]\frac{2}{R}=\frac{1}{p} + \frac{1}{q} = \frac{1}{1.13} - \frac{1}{10.8} = \frac{10.8-1.13}{10.8}[/tex]

    So [tex]R = 2 \frac{10.8}{9.67} = 2.233[/tex]

    Why is my answer is wrong? I think I used the correct equations. I tried the same question with different numbers and the computer still marks my answer wrong. Any explanation would be appreciated.
     
  2. jcsd
  3. Aug 11, 2011 #2

    kuruman

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    Gold Member

    [itex]\frac{1}{1.13}-\frac{1}{10.8}\neq\frac{10.8-1.13}{10.8}[/itex]
     
  4. Aug 11, 2011 #3

    PeterO

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    You mucked up the subtraction of two fractions - in particular, the denominator.
     
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