Worksheet 2.1 Problem 2

In order for a human being to register light with their eyes, a certain number of photons must enter. This problem examines just how powerful a light bulb must be in order for a person to see the light from a distance of 1 kilometer away in a dark cave.

Before diving into the mathematics of this problem, it is important to first consider what we know about the layout of this problem. First, we know that the number of photons that needs to reach the eye is ~10 photons. Additionally, we know that the eye has a refresh rate of approximately 10 Hz. Since the power output of the bulb will be dependent on the number of photons produced as a function of time, we will need to use both of these pieces of information to determine the output wattage.

Next we will want to consider the relevant equations to this problem.

The first equation is \( E = h\nu \). This tells us that the energy of a photon, E, is equal to the frequency, \(\nu\), times Planck's constant \(h=6.6 x 10^{-27}\frac{erg}{s} \).

The next equation is the surface area of a sphere, \(SA = 4 \pi r^{2} \). This is important because the light being produced by the bulb is isotropically radiated (going out evenly in all directions).

Now we can begin diving into the problem.

Since the light travels 1 km from the bulb and we are interested in the amount of light hitting the eye, we can set up the following proportion: \[ \frac{P_{eye}}{P_{bulb}} = \frac{SA_{eye}}{SA_{bulb}} \] Putting this in terms of our desired quantity of \(P_{bulb} \), our equation becomes: \[ P_{bulb} = \frac{P_{eye} SA_{bulb}}{SA_{eye}} \] Now, we will need to fully define each of the terms in the equation to solve for the Power of the bulb.

We can further define the power of the eye ( \(P_{eye} \) ) as the amount of energy reaching it per unit time, which we have declared to be 10 photons ( \(E_{eye} \) ) within the refresh rate of the eye ( \(T_{eye} \) ). This equation would be:\[ P_{eye} = \frac{E_{eye}}{T_{eye}} \] Additionally, we can define the energy of the eye using our initial energy equation. This gives:\[ E_{eye} = h\nu \] This, however, only provides the energy of a single photon. Since we want to know the value of 10 photons, we will multiply this by 10, and use the equation: \[ E_{eye} = 10 h\nu \]
To find the frequency of the light, we can use the following relationship:\[ \nu = \frac{c}{\lambda} \]
The surface area of the eye we can approximate to be \( 1 cm^2 \).

Using the surface area of a sphere, we can also determine the surface area of the bulb's light output with a 1 km radius. This is given by the equation: \[SA_{bulb} = 4 \pi r^2 \]
Inserting these into our equation for the power of the bulb, we now have a more complete equation of:
\[ P_{bulb} = \frac{10 h c 4 \pi r^2}{SA_{eye} \lambda T_{eye}} \]
The last step is to plug in numbers for for each term, as we have defined them.
\[h = 6.6 x 10^{-27}\frac{erg}{s} \]
\[c = 3.0 x 10^{10}\frac{cm}{s} \]
\[r = 10^{5} cm^2 \]
\[SA_{eye} = 1 cm^2 \]
\[\lambda = 500 x 10^{-7} cm \,(visible\, light) \]
\[T_{eye} = \frac{1}{10} {s} \]
This ultimately produces a result of: \[ P_{bulb} = 48 \frac{erg}{s} \]
I found this problem interesting, because it turned out to be quite straight forward determining the power output of the bulb at such a distance using basic physics concepts!



Collaborators on this problem included Willie Pirc and Johnathan Budd.