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Concave mirror - finding object and image distance

  1. Mar 24, 2009 #1
    1. The problem statement, all variables and given/known data
    A concave mirror has a focal length of 27.2 cm. The distance between an object and its image is 50.2 cm. Find (a) the object and (b) the image distance assuming that the object lies beyond the center of curvature.


    2. Relevant equations
    f=1/2 R
    1/object distance + 1/image distance = 1/f


    3. The attempt at a solution
    image distance - object distance = 50.2 cm
    (image distance + object distance)/(image distance*object distance) = 1/f
    50.2/(image distance*object distance)=1/27.2
    image distance*object distance=1365.44

    image distance=50.2 + object distance

    substitute in and get: (50.2 + object distance)object distance = 1365.44
    quadratic equation solves for object distance = 69.77 or 19.57 cm


    Am I even doing this right?? If so, I'm stuck from here.. Thanks for any help you can give!!
     
  2. jcsd
  3. Mar 24, 2009 #2

    Redbelly98

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    Whoops.
    image distance - object distance = 50.2 cm
    But you substituted 50.2 cm for image distance + object distance

    Try substituting for "image distance" in
    1/object distance + 1/image distance = 1/f​
     
  4. Mar 24, 2009 #3
    Can anyone help me figure out if I'm even on the right track?
     
  5. Mar 24, 2009 #4
    so I tried saying image distance = 50.2 + object distance

    substituting that into 1/object distance + 1/image distance = 1/f

    I got object distance = -620.65 cm
    then image distance would = -620.65 + 50.2 = -570.45 cm

    This is incorrect... Have I done this right, and the 620.65 simply needs to be positive? Or am I completely off track?
     
  6. Mar 24, 2009 #5

    Redbelly98

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    Something is wrong there. Those numbers definitely don't satisfy the equation,

    1/object distance + 1/image distance = 1/f

    1/(-620.65) + 1/(-570.45) = -0.00336
    While 1/f = 1/27.2 = +0.0367

    Since f (and also 1/f) is positive, the object and image distances cannot both be negative.
     
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