Worksheet 2, Problem 2: The Power of a Distant Star

The problem: 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 \(h = 6.6 x 10^{-27}\) erg/s.

Well, the problem is really:
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 only change here is that we have a more difficult conversion to get the distance between the star and our eye.

What we know:

• We know the speed of light is \(3 \cdot 10^{10} \frac{cm}{sec}\)
• We know there are \( 3.16 \cdot 10^7 \frac{sec}{year}\) from worksheet 2.2. \[ L = 100 \: years \cdot \frac{3.16 \cdot 10^7 \: sec}{year} \cdot  \frac{3 \cdot 10^{10} \:cm}{sec} \approx 9.5 \cdot 10^{19} \: cm \]
• E = hν
• \(P = \frac{\delta E}{\delta t}\)
• Visible light falls into a range between 390 - 700 nm, which an average could be 6.0 x \(10^{-7}\) cm.
• The radius of an eye is about 1 cm.
• The "refresh rate" of a movie is 50 Hz.

Solve:

Based on the fact that the light is emitted isotropically (shown partially in the diagram above), we can set up a ratio between the power emitted by the star over the area the light reaches to the power received by your eye to the area of your eye.\[\frac{P_{star}}{a_{light}} = \frac{P_{eye}}{a_{eye}}\]plugging in area formulas gives:\[\frac{P_{star}}{4 \pi L^2} = \frac{P_{eye}}{\pi r_{eye}^2}\] the main formula is now:\[P_{star} = \frac{P_{eye}\cdot 4 \pi L^2}{\pi r_{eye}^2 }\]

Now we all we need to find is the power reaching our eye, starting with finding the energy.  \[E = hv \: or \: E = h\frac{c}{\lambda} = 6.6 x 10^{-27} \cdot \frac{3 x 10^{10}}{6.0 x 10^{-7}} \approx 3.3x10^{-10} ergs\]

Now we need to find time. Something new I learned this week is refresh rate, or the rate at which movies emit light so that the brain does not notice any flickering in the movie. If the refresh rate of a movie is 50 Hz (\( \frac{1}{s}\)) and 10 photons must hit the eye every \( \frac{1}{50}\)s, so we need 500 photons a second. So the power would be  \[ P = \frac{500 photons}{sec} \cdot 3.3x10^{-10} ergs \approx 1.65x10^{-7} \frac{erg}{s}\] Plugging this into the original equation we get:\[P_{star} = \frac{1.65x10^{-7}\cdot 4 \pi (9.5 \cdot 10^{19})^2 }{\pi 1^2} \approx 5.9 \cdot 10^{33} \frac{erg}{s}\]

Update: The power output of a star is really just its luminosity \[L_{star} \approx 5.9 \cdot 10^{33} \frac{erg}{s}\] To check if our answer is reasonable we can compare that to the known luminsity of the sun \[L_{\odot} = 3.8 \times 10^{33} \frac{erg}{s} \] Because the luminosities are in the same order of magnitude, our answer seems reasonable.

Acknowledgements: Thanks to Sean and Eden for working on this problem with me and thanks to Fiona for explaining refresh rate to me.