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Concave mirror equation calculation

  1. Apr 11, 2009 #1
    1. The problem statement, all variables and given/known data

    A physics student wants toobtain a real and virtual image that each is 3 times as large as the original object she has a concave mirror with a 0.50 m radius of curvature. what object distance will she require to obtain the real image?

    2. Relevant equations
    1/do + 1/di = 1/f

    3. The attempt at a solution

    m = 3 = -di / do 3do = -di

    I made do=d di=3d so...

    1/d + 1/3d =1/0.25

    [1/d + 1/3d =1/0.25]3d

    2d + 1 = 3d/0.25

    [2d + 1 = 3d/0.25]0.25

    0.5d + 0.25 = 3d

    2.5d= 0.25

    d= .10m m =do huh? I know I did something wrong

    based upon my ray diagram do= .32m ish
    and if that was the do then there would not be a real image formed becaues the object is inside of the focal point so only a virtual image would be formed right?

    since m = 3 = -di / do 3do = -di

    I tried to make di negative and that didnt turn out either I am not sure what I have done wrong
     
  2. jcsd
  3. Apr 11, 2009 #2

    rl.bhat

    User Avatar
    Homework Helper

    1/d + 1/3d =1/0.25
    Write this step as
    1/d + 1/3d = 4 and solve for d
     
  4. Apr 12, 2009 #3
    after looking at the problem today I see that I subtracted d from 3d instead of divided. what a silly mistake lol o well thx anyways.
     
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