Calculating Photon Energy given Luminosity

In summary: The energy associated with a photon is proportional to its energy and the area of the photon's impact. Converting from m^2 to J^2:E = 5 x 10^{-5} J^2
  • #1
WRoach
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Homework Statement


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?


Homework Equations


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

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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  • #2
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.
 
  • #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?
 
  • #4
The natural lens only let's "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.
 
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  • #5


Dear student,

It seems like you are on the right track with your calculations so far. To calculate the energy associated with the photons, you can use the relationship between power (Watts) and energy (Joules) which is:

Energy = Power x Time

In this case, the time would be the time it takes for the photons to reach your eye, which is essentially the time it takes for the light to travel from the lightbulb to your eye. This can be calculated using the speed of light (c) which is approximately 3 x 10^8 m/s. Therefore, the time it takes for the photons to reach your eye from a distance of 1 meter would be:

Time = Distance/Speed = 1m/3 x 10^8 m/s = 3.33 x 10^-9 seconds

Now, you can plug this value into the equation for energy to calculate the energy associated with the photons:

Energy = Power x Time = (1.25/π) mW x 3.33 x 10^-9 seconds = 4.16 x 10^-12 Joules

Therefore, the energy associated with the photons from the 100 W lightbulb at a distance of 1 meter would be approximately 4.16 x 10^-12 Joules.

Hope this helps!
 

1. How do you calculate the photon energy given luminosity?

To calculate the photon energy given luminosity, you can use the formula E = hν, where E is the energy of the photon, h is Planck's constant, and ν is the frequency of the photon. Luminosity can be calculated by multiplying the power of the source by the inverse square of the distance from the source.

2. What is the unit of measurement for photon energy?

The unit of measurement for photon energy is the electron-volt (eV). This unit is commonly used in atomic and nuclear physics, and is equal to the amount of energy gained or lost by an electron when it moves through a potential difference of one volt.

3. Can you calculate photon energy without knowing the frequency?

No, the frequency of the photon is a crucial component in calculating its energy. Without knowing the frequency, you cannot accurately determine the energy of the photon using the formula E = hν.

4. How does luminosity affect photon energy?

Luminosity is directly proportional to photon energy. This means that as the luminosity of a source increases, the energy of the photons it emits also increases. This relationship is due to the fact that luminosity is directly related to the power of the source, which is a factor in the photon energy equation.

5. What other factors can affect the calculation of photon energy?

In addition to luminosity and frequency, other factors that can affect the calculation of photon energy include the type of source emitting the photons, the distance from the source, and any interactions or absorption that may occur between the photons and other particles in the medium.

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