(adsbygoogle = window.adsbygoogle || []).push({}); 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.

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# Calculating Photon Energy given Luminosity

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