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Energy from Light Bulb

  1. Mar 2, 2007 #1
    1. The problem statement, all variables and given/known data

    You stand 2.7 m from a 150 W light bulb that converts 5% of the electrical energy dissipated within it to light.

    (a) If the pupil of your eye is 5 mm in diameter, how much energy enters you eye per second? ___J

    2. Relevant equations
    Iave= Pave/Area

    3. The attempt at a solution

    I thought, since the total power output is 150W * 5% = 7.5, I could fine the Pave by Pave=Ptotal ^2 x 0.5 = 28.125. That did not work. I don't understand why this approach would not equal the answer, knowing that Power=Energy/sec, and we're assuming 1 sec. here. Thank you.
  2. jcsd
  3. Mar 2, 2007 #2
    How much area is the light spread out over at a distance 2.7 meters away? (What does this mean geometrically, and what is the equation for the total area that is that distance away from a point?)

  4. Mar 2, 2007 #3


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    You need to find the fraction of the 7.5W emitted by the bulb that enters the eye. First calculate the area of the pupil. Then, what do you divide by?
  5. Mar 2, 2007 #4
    I don't understand this Pave=Ptotal ^2 x 0.5.
    The power output (light) is correct. The intensity equation involves a power and an area. What area is of interest here?
  6. Mar 2, 2007 #5
    Seeing that my class had just finished a chapter on AC circuits, I assumed that I could find the average power from total power by the equation:

    (xave)^2= .5*(xmax)^2, which we would use to find rms values and what not. I may be completely flawed in my rational for this.

    I more of a bio/organic person...not physical sciences person, obviously :tongue2:

    But as far at the area of the pupil, I managed to find that using the area equation of a circle and obtained 5.3x10^-5 m^2. Now that I have the area, I doubted by dividing 7.5W/area would yield the answer, but I know that it yields average intensity.

    So, as far as finding the area from the point 2.7m away, I'm still unclear about how to find that area. I thought maybe multiply 2.7m by the diameter?
  7. Mar 2, 2007 #6
    (xave)^2= .5*(xmax)^2
    Okay at least the units work now. But that is the rms value compared to the peak value for a sinusoidal form.

    Your lightbulb is transforming 7.5 J electrical energy every second into light ouput. This energy gets spread out in each direction equally, with some assumptions. Does the intensity decrease as you move away from the source?
  8. Mar 2, 2007 #7
    As you move away from the source, the larger the area that the light emitted would cover, thus, the intensity would in fact decrease, according to the equation Average Intensity = Average Power / Area.

    Ok, so I understand that because the source is 2.7m away from the eye, the intensity would be much less that if the source was directly in front of you. With that being said, how does the area that the light covers compare to the area of the pupil? Or doesn't it?
  9. Mar 3, 2007 #8
    You know the intensity at the eye. Remember that energy per time is power.
  10. Mar 3, 2007 #9


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    Forget intensity, you are interested in power. (Your question doesn't even contain intensity!) The power is radiated by the bulb in all directionss, so you want to find the area of a sphere of radius 2.7m. The ratio of the pupil area to the sphere area gives you the fraction of total power (which you've found) that's intercepted by the eye.
  11. Mar 4, 2007 #10

    I tried doing what you suggested...finding the ratio between the two areas...however it still didn't work! Here were my results:

    Area pupil/Area sphere= 2.14x10^-7

    2.14x10^-7 * 7.5W= 1.60725E-6 J which is not the correct answer.
  12. Mar 4, 2007 #11


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    Hmm, that's the answer I get, too. Could the book's answer be wrong? Let us know what your teacher says...
  13. Mar 8, 2007 #12
    Here's how to answer the problem.

    1. Iave=Pave/Area
    Iave= (150W*.05)/(4*pi*(2.7m)^2) = 0.08187 W/m^2

    2. 0.08187 W/m^2 x (pi*(2.5mm)^2) = Power = 1.6075E-6
  14. Mar 8, 2007 #13


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    I feel better that the answer we got together above is correct! :biggrin:
    Thanks for getting back to us.
  15. Mar 8, 2007 #14


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    There are far too many significant digits in the answers given here -- there should be no more than two sig figs in the answer. Answers that are "too precise" but otherwise correct are often rejected by automated grading systems.

    - Warren
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