Focal length of a mirror

  1. [SOLVED] Focal length of a mirror

    1. The problem statement, all variables and given/known data
    A 2.0-cm-tall object is placed in front of a mirror. A 1.0-cm-tall upright image is formed behind the mirror, 150 cm from the object.

    What is the focal length of the mirror?


    2. Relevant equations
    1/f=1/s+1/s'
    m=h'/h
    m=-s'/s


    3. The attempt at a solution
    .5=-s'/s
    s+s' = 1.5m

    Combining those 2 equations, I get:
    s'=3
    s=-1.5

    1/f=(1/-1.5)+(1/3)
    f=3m

    When I calculate it the focal length, f, it comes out to 3m, which is obviously way too big.
     
    Last edited: Oct 15, 2009
  2. jcsd
  3. rl.bhat

    rl.bhat 4,435
    Homework Helper

    If image is virtual, magnification is also negative.
    So -m = -s'/s
    Now try to solve the problem.
     
  4. Setting m negative gave me the same answer
    -.5=s'/s
    s=s'/-.5
    s+s'=1.5
    s'/.5+s'=1.5
    -(1/2)s'+s'=1.5
    s'/2=1.5
    s'=3

    According to my book, s' should be negative, though, because we have a virtual image on the opposite side of the object.

    Continuing anyways...
    s=1.5-s'
    =-1.5

    1/f=(1/-1.5)+(1/3)=-.3
    f=-3m

    Still not correct.
     
  5. rl.bhat

    rl.bhat 4,435
    Homework Helper

    -0.5 = - s'/s
    Or 0.5 = s'/s
     
  6. -.5=-s'/s
    s=-s'/-.5

    s+s'=1.5

    -s'/-.5 + s' = 1.5
    (1/2)s' + s' = 1.5
    (3/2)s'=1.5
    s'=1

    s+s'=1.5
    s=1.5-s'
    s=.5

    1/f=(1/.5)+(1/1)=3
    f=1/3=.33m

    Which is still wrong.
     
  7. rl.bhat

    rl.bhat 4,435
    Homework Helper

    Image is virtual.
    so 1/f = 1/s - 1/s'
     
  8. Solved:

    m=-s'/s=1/2
    -s'=s/2
    |s'|=s/2

    |s|+|s'|=1.5
    |s|+|s/2|=1.5
    3/2s=1.5
    s=1

    |s|+|s'|=1.5
    |s'|=1.5-|s|
    s'=1.5-1
    s'=.5
    but since s' is negative for virtual, upright images
    s'=-.5

    1/f=(1/1)+(1/-.5)=-1
    f=-1m
     
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