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Homework Help: Concave Spherical Mirrors

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

    A concave spherical mirror has a radius of curvature of magnitude 27.1 cm. Determine the object position for which the resulting image is inverted and larger than the object by a factor of 4.00.

    2. Relevant equations

    Mirror equation in terms of focal length: 1/p + 1/q = 1/f

    3. The attempt at a solution

    I'm not getting the correct answer for this question. Here's what I've done so far:

    First we find the focal length

    [tex]f=\frac{R}{2} \implies \frac{27.1}{2}=13.55 \ cm[/tex]

    [tex]M = \frac{-q}{p}=4 \implies q=-4p[/tex]

    Substituting in

    [tex]\frac{1}{p} - \frac{1}{4p} = \frac{1}{13.55}[/tex]

    [tex]\frac{3}{4p} = \frac{1}{13.55} \implies p =10.162[/tex]

    I don't know why the answer is wrong. I know that image is inverted when Magnification is negative. Here it is not -ve because the questions says "larger" so I took it as positive...
     
    Last edited: Aug 11, 2011
  2. jcsd
  3. Aug 11, 2011 #2

    PeterO

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    Homework Helper

    What are you trying to find??? You haven't said.
     
  4. Aug 11, 2011 #3
    Oops. The question I was trying to ask was:

    Determine the object position for which the resulting image is inverted and larger than the object by a factor of 4.00.
     
  5. Aug 11, 2011 #4

    PeterO

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    Homework Helper

    You are having trouble with the minus signs.

    There are two conventions out there one says magnification is -p/q the other says it is p/q

    one of them has 1/p + 1/q = 1/r , the other has 1/p - 1/q = 1/r

    make everything positive, and work on a positive magnification meaning inverted.
     
  6. Aug 12, 2011 #5
    So, are you saying I should have used M=q/p so that q=4p? Because I am using the +ve equation 1/p+1/q=1/f.
     
  7. Aug 12, 2011 #6

    PeterO

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    Try it and see.
     
  8. Aug 12, 2011 #7
    YES! It worked. Thanks for clarifying this, I REALLY appreciate all your help.
     
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