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Calculating Photon Energy given Luminosity

  1. Feb 15, 2012 #1
    1. The problem statement, all variables and given/known data
    NOTE: This is part two of a problem. The first part used the equation for the energy density of photons in a blackbody to find the photon energy, but that doesn't seem to apply here.

    Imagine that you are looking at a 100 W lightbulb from 1 meter away. At any instant in
    time there will be some number of photons which came from the lightbulb inside your eye. If the
    area of your pupil is taken to be A = 0.5 [itex]cm^{2}[/itex] what is the energy associated with these photons?

    2. Relevant equations
    Area of a Sphere = 4[itex]\pi[/itex][itex]r^{2}[/itex]

    3. The attempt at a solution
    So far I have found:

    First converting A to units of [itex]m^{2}[/itex]:
    A = [itex]0.5 cm^{2} * \frac{1 m^{2}}{10000 cm^{2}} = 5 x 10^{-5} m^{2}[/itex]

    Power/Area @ 1 meter = [itex]\frac{100W}{4\pi r^{2}}[/itex] = [itex]\frac{100W}{4\pi (1)^{2}}[/itex] = [itex]\frac{25}{\pi} \frac{W}{m^{2}}[/itex]

    Then the power through the pupil:

    P = [itex]A * \frac{25}{\pi} \frac{W}{m^{2}} = \frac{1.25}{\pi} mW[/itex]

    And here's where I am stuck. I know that a Watt is Joules/Sec and I am looking for units of Joules, but it says at any instant, so I do not have a time to multiply this answer by. I'm guessing I either need another equation to get to Joules, or I need to somehow take the limit of an integration as Δt → 0, however that route doesn't seem to be the intent of the question. Just need some direction at this point. Thanks.
  2. jcsd
  3. Feb 15, 2012 #2


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    Homework Helper

    The first thing that comes to my mind is a cylinder of light extending from the pupil to the retina at the back of the eye. You could estimate that distance and get your volume.
    The lens at the front of the eye will focus the light - not to a point on the retina, but to a reduced image of the bulb. This makes the cylinder volume estimate a little too large. But it is a minor consideration compared to the estimating you will have to do in regard to what range of wavelengths will effectively penetrate the lens into the eye and what portion of the bulb's energy is emitted in that range.

    Incidentally, I read somewhere in the news this week that people who get lens replacements (cataract surgery) can see near ultraviolet that is invisible to people with their natural lenses.
  4. Feb 15, 2012 #3
    Thank You! That's exactly what I needed. I was also given the radius of the eye, so by using the diameter as the length of the cylinder and dividing by the speed of light I arrive at Joules. I believe that's the correct result. Much appreciated.

    Also, that is quite interesting. Not sure why that is, as the retina is what is sensitive to certain wavelengths. Perhaps the lens replacements slow the ultraviolet light, increasing its wavelength to within the threshold of visibility?
  5. Feb 15, 2012 #4


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    The natural lens only lets "visible" light through. Not UV. But new plastic ones let UV down to 350 nm through. Yes, the retina must also be sensitive in the UV.

    Use the radius of the pupil for the cylinder of light.
    Last edited: Feb 15, 2012
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