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Finding maximum distance of a lit power source

  1. Dec 14, 2009 #1
    The minimum energy of the visible lights of wavelength 500to 600 nm that can just be detected by the eye is 10e-17Watts. Only 1% of power input to a 60W light bulb is emitted between these wavelengths. Estimate the maximum distance that it might be possible to see the lit bulb. The pupil of the eye will be at its maximum diameter say 8mm. At its maximum sensitivity the eye only detect 8% of the light incident on it.

    My answer

    Minimum power the eye can detect= 10e-17w
    1% of 60w=0.6w

    Therefore, the maximum distance to see the lit bulb,

    Power output ÷ surface area of a sphere = power per meter square
    0.6÷(4πr^2)=10e-17
    0.6÷10e-17=(4πr^2)
    (6X10^16)/4π=r^2
    so r=sqrt(6X10^16)/4π)
    r= 217080km however in answer is 140km.

    Please help me with this. Thanks.

    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 15, 2009 #2
    You're problem is units! There's a reason that you've been given more information that you're actually using. On the left you have power per square meter. But the piece of information you used on the right has units of power.

    Break the problem into a couple different stages: figure out how much of the pupil area, which I think should be safe to assume as a circle, relates to that minimum power, and then figure out how much of that quantity relates to the power radiated/m^2.
     
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