Blog #5, Worksheet #2.1, Black Bodies and Flux

*Worked with Tim Mcnamara 

2. You are floating space with Sandra Bullock. It’s dark and there is a barely discernible 1 Watt light bulb ahead of you. You know the eye must receive ~ 10 photons in order to send a signal to the brain that says, “Yep, I see that.” How far away is it? 

Note: Assume the bulb is emitting light isotropically (the same in all directions). 
Note: 1 Watt is \(10^7 \frac{erg}{s}\) 
Note: The energy of a photon is E = hν, for a frequency ν (Greek letter ‘nu’), and Planck’s constant
h = \(6.67x10^{-27} erg \cdot s\) 

We start by solving for the energy in one photon. We will approximate the wavelength of a photon as \(\lambda = 500 nm = 5x10^{-5} cm\). We then use quantum equations for light to solve for energy, E. \[E = h \cdot v\] and \[v = \frac{c}{\lambda}\] thus \[E = \frac{h \cdot c}{\lambda}\]When we solve this with our constants we find that the energy in 1 photon, \(E = 4x10^{-12}\) but we need 10 photons to see the light and end up with \(E = 4x10^{-11}\). Next we must approximate how long it will take that many photons to enter the eye. We will approximate that the speed at which we can process light or the eye's "refresh rate" as 24 hz which converts to 40ms or .04s between. Now we have the conversion for the amount of energy our eye needs to take in per second\[\frac{4x10^{-11}}{.04} = 1x10^{-9} \frac{erg}{s}\]The last step is to find how far away the lightbulb. We can do this by comparing the amount of power our eye is receiving, the amount of power given off by the light bulb, and the sizes of our eye and the light. The amount of light entering our eye will differ proportionally based on how far away from the bulb we are. We approximate the radius of our eye as .5cm which gives us the area as a .25cm^2 circle receiving the light from a spherical bulb. So the equation becomes \[\frac{A_{eye}}{A_{light}} = \frac{E_{eye}}{E_{light}}\] where \[a_{eye} \thickapprox .25 cm^2 \qquad a_{light} = 4 d^2 \qquad E_{eye} = 1x10^{-9} \qquad E_{light} = 1x10^7\] Plugging in a rearranging for d we get \[4d^2 = \frac{1x10^7 \frac {erg}{s} \cdot .25cm^2}{1x10^{-9} \frac {erg}{s}}\] which gives us a final answer of \[d = 250 km\]