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Concave Mirror

  1. May 3, 2012 #1
    1. The problem statement, all variables and given/known data

    Suppose the radius of curvature of a concave mirror is 5.0 cm
    a) Find the object distance that gives an upright image with a magnification of +1.51.


    2. Relevant equations

    1/Do + 1/Di= 1/f

    m=-di/do=hi/ho


    3. The attempt at a solution

    First i convered all cm into m.

    f=R/2, so .05/2 = .025.
    1/.025=40

    then I used m=-Di/Do. Since don't know Do or Di, I substituted for the unknowns using 1/Do + 1/Di= 1/f

    I came up with...
    m=-[40-(1/Do)]/[40-1/(40-1/Do)] and solved for Do.

    I got Do=.02506m.

    Online HW says it's wrong.

    Any help is GREATLY appreciated. What I did makes sense to me but obviously there's an error or just some major concept I don't understand. Thank you so much for taking the time!
     
  2. jcsd
  3. May 3, 2012 #2

    collinsmark

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    Okay, everything seems right so far. :approve:
    I'm not sure why you would want to invert that at this point in the process, but okay.
    Substitution is the right idea. :approve:
    Now you've lost me. :uhh:

    So far, you've already figured out
    [tex] \frac{1}{D_o} + \frac{1}{D_i} = \frac{1}{f} [/tex]
    and
    [tex] m = -\frac{D_i}{D_o}. [/tex]
    Rather than going the other way around, try substituting [itex] D_i = -mD_0 [/itex] into the first equation. Also substitute [itex] f = R/2 [/itex] into the first equation. It might make it a little easier to solve for [itex] D_o [/itex] that way. (Hint: then find a common denominator for the left side of the equation :wink:)
     
  4. May 3, 2012 #3
    Oh god that's SO much simpler than what I was trying to do. Thanks a million!
     
  5. May 4, 2012 #4

    Redbelly98

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    Glad it worked out. By the way, I find it easier to work in cm rather than meters for this problem. For example, "1/2.5" is easier (for me) to work with than "1/0.025". In the end you get the same answer, of course.
     
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