Worksheet 2.1 Problems 1 and 2: Calculating Light Source Power Output

Today in class we began working on Worksheet 2.1: Intro to Collaborative Learning and Street-Fighting Mathematics. The worksheet featured a variety of problems that required intuition and estimation in order to come up with an answer using limited information. My partner, Steven Johnson, and I began with the first problem. Although it was the first problem on the worksheet, we found it to be one of the more difficult ones because it required specific knowledge about relationships we hadn't encountered in our studies.

Problem 1: The eye must receive ~ 10 photons in order to send a signal to the brain that says, “Yep, I see that.” If you are standing in an enormous, completely dark cave and just barely discern a light bulb at a distance of 1 kilometer, what is the power output of the bulb?

Note: Assume the bulb is emitting light isotropically.
Note: The energy of a photon is E = hν, for a frequency ν (Greek letter ‘nu’), and Planck’s constant \(h\:=\:6.6\times\:10^{-27}\)

We first drew the following picture to illustrate our interpretation of the situation laid out in the problem:

Next, we compiled a list of all the known values that were given by the problem:

• The observer is 1 km away from the light bulb and can barely see the light (minimum photons)
• Eye needs a minimum of 10 photons to register light
• The bulb is emitting light isotropically (i.e. identical in all directions)
• The energy of a photon is given by E = hν, for a frequency ν (Greek letter ‘nu’), Planck’s constant \(h\:=\:6.6\times\:10^{-27}\:erg\:*\:s\) . 

In order to determine the bulb's power output, we used these known values to break the problem up into smaller steps. Clearly, since the equation was explicitly give, the energy of a photon would be necessary to solve this problem. Also, since 10 photons are perceived at once from 1 km away in this situation, we decided that it would be useful to calculate the total number of photons reaching the observer per second. Finding those two quantities would allow us to solve for the power output reaching the observer's pupil. Then, since the light is emitted isotropically, we know that the energy reaching the observer's pupil is proportional to the total energy reaching a circle with radius 1 km (the observer's distance from the light bulb). 

Calculating photons/second reaching the observer

Since the observer is at the maximal distance from the light bulb that still allows him to perceive it, we can assume that roughly 10 photons are perceived by him in any single instant.  However, photons processed by the eye are not cumulative. As Professor Johnson told us, if the photons were to accumulate in the eye it wouldn't be long before we would be only able to see an overwhelming pure white. Instead, the eye refreshes in order to perceive new photons coming in. We estimated that television uses 30 frames per second to display moving pictures. Since human observers perceive television as a constantly moving picture, and not as a series of frames, the human eye must have a refresh rate that is lower than 30 fps. We estimated that human eye has a refresh rate of ~10 fps meaning that every one-tenth of a second they eye processes a fresh batch of incoming photons. Since our observer is receiving 10 photons each interval and since intervals occur 10 times in one second, we calculated that the 100 photons are reaching the eye each second. 
\[\frac{10\: photons}{\frac{1}{10}\: seconds}=\frac{100\: photons}{1\:second}\]

Calculating the energy of a single photon

Using the given equation for the energy of a photon, \(E\:=\:h\nu\), we would easily be able to figure out our problem because we were given the value of h (Planck's constant). However, we did not know how to find out the frequency of the photons. We mistakenly assumed that the photons' frequency was related to the number of photons leaving the light bulb per second. However, we were corrected by a TF who told us that the frequency of a photon is dependent on upon its wavelength because of the following equation: \(\nu\:\times\:\lambda\:=\:c\:\rightarrow\:\nu\:=\:\frac{c}{\lambda}\). While we were unable to independently come up with this equation, we were quickly able to employ it in our calculations. Before plugging in numbers, we made sure we understood what all of the variables meant and that the variable we were solving for. The speed of light (c) is equal to 3*\(10^{10}\) cm/s and wavelength of visible light can be approximated as 500 nm 5*\(10^{-7}\) cm, which is about the wavelength of green light. We then plugged in and solved.\[E\:=\:h\:\nu\quad\quad\nu\:=\:\frac{c}{\lambda}\]

\[E\:=\:\frac{c}{\lambda}\:\times\:h\:\rightarrow\:E\:=\:\frac{3\:\times\:10^{10}\:\frac{cm}{s}}{500 nm}\:\times\:6.6\:\times\:10^{-27}\:\approx\:4\:\times\:10^{-10}\:ergs\: per\: photon\:\]

Calculating the energy of 100 photons

After determining the energy in a single photon with a wavelength of ~500 nm, we needed to figure out how much energy would be held in 100 identical photons because 100 photons were entering the observer's eye each second. We simply multiplied by 100 and came up with the following value: \(4\:\times\:10^{-8}\:ergs\). This value is the same as the power output(in ergs/second) reaching the observer's eye.

Calculating the power of the bulb

Now that we had the power output reaching the observer's eye, we were able to use a relatively simple proportion to find the total power output because of the isotropic nature of the light bulb. We used the following proportion, reasoning that if a certain amount of energy reached a small section of a uniform area it would be proportional to the total power output of the bulb. \[\frac{Power\:Reaching\:Eye}{Area\:of\:Iris}\:=\:\frac{Total\:Power}{Total\:Area}\]

We figured that, since the light was entering the eye through the iris, it would be appropriate to use the area of an iris, which we estimated to have a radius of 5 mm (.5 cm). The total area was given by the circle with radius of 1 km centered at the light source.

Radius of visible light emitted from bulb

Observer's Eye

\[\frac{4\:\times\:10^{-8}\:\frac{ergs}{s}}{\pi(1\:cm)^2}\:=\:\frac{Bulb\:Power\:Output}{\pi(10^{5})^2}\]

Solving this equations for Bulb Power Output shows that the power output of the light bulb is \(\approx\:1.6\:\times\:10^{-1}\:\frac{ergs}{s}\).

Quick Analysis

Now, at first glance, this seems like an absolutely tiny value; so tiny that it can't possibly be anywhere close to right (which might be the case). However, looking up a few numbers shows that this might make sense because the light bulb in this situation in likely not very powerful. However, this article might justify such a small answer. In the article, a reader asks what the maximum distance one can stand from a candle and still be able to see it. The writer estimates that a candle has an effective brightness of about 1 watt (\(10^7\:ergs\)). He estimates that with such a candle, it would be possible to view its light from 70 km away. Now, this light bulb is much less powerful because it only reaches 1 km away. I'm not sure if it's so weak that it justifies such a small power output as calculated through our estimates, but it will certainly be much less than 1 watt. 

Problem 2

Problem 2 is essentially the exact same question as problem 1, but on a much larger scale. Instead of asking the power output of a small light bulb in a cave, it asks: What is the power output of a star that is 100 light years away that you can barely see from a dark site at night? Assume the star emits most of its energy at the peak of the eye's sensitivity. The significance of this question is that it requires you to see that light years are a unit of distance, not time, and to convert 100 light years into centimeters. Other than that the problem follows the same steps as solving for the power output in problem 1.

Converting 100 light years to centimeters

Known: 

• The speed of light, c, is equal to \(3\:\times\:10^{3}\frac{cm}{s}\)
• 1 light year is the distance covered by something traveling at speed c for one year 

\[Distance\:=\:\frac{3\:\times\:10^{3}\:cm}{s}\:\times\:\frac{60\:s}{min}\:\times\:\frac{60\:min}{hr}\:\times\:\frac{24\:hr}{day}\:\times\:\frac{365\:days}{year}\:\times\:100\:years\]

\[Distance\:\approx\:8.1\:\times\:10^{