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?
P=?
λ=532nm
f=50hz
h=6.6×10-27erg×s
c=3
Aeye= 1cm2
To begin solving this problem we must first understand what is being asked. A lightbulb is a distance d away from an eye. The eye must receive at least 10 photons before it refreshes to see the object. Thus, we are solving for the power output of the lightbulb such that the area occupied by the eye will receive the energy equivalent of 10 photons per refresh cycle.
First we should find how much energy 10 photons is.
\[E=hf=\frac{hc}{\lambda}\]
We know that an average visible photon of green light has a wavelength of 532nm,so:
\[E=\frac{hc}{\lambda}=\frac{6.6\times 10^{-27} erg\cdot s\times 3 \times 10^{10} cm/s}{532nm}=\]
\[=\frac{6.6\times 10^{-27} erg\cdot s\times 3 \times 10^{10} cm/s}{5.32\times 10^{-5}}=\]
\[3.7\times 10^{-12}erg\]
So, for ten photons this is:
\[\boxed{3.7\times 10^{-11}erg}\]
The eye needs to receive this many photons in one refresh cycle. So:
\[P_{eye}=Energy/cycle \times f=3.7\times 10^{-11}erg \times 50hz=\]
\[\boxed{P_{eye}=1.9\times 10^{-9}erg/sec}\]
Next, let's calculate the power of the bulb. Since the lightbulb emits photons in all directions, if we approximate the bulb as a point source, once can imagine a large sphere of radius D whose surface has the energy of the lightbulb spread evenly across it. Due to this even distribution of energy, the ratio of the surface areas of he eye and sphere will equal the ratio of the powers of the eye and sphere. Using this, relation, we can solve for \(P_{sphere}\).
\[\frac{A_{sphere}}{A_{eye}}=\frac{P_{sphere}}{P_{eye}}\]
\[P_{sphere}=\frac{A_{sphere}}{A_{eye}} \times P_{eye}\]
\[P_{sphere}=\frac{4 \pi D^2}{1cm^2} \times 1.9 \times 10^{-9} erg/sec\]
\[P_{sphere}=\frac{4 \pi (1 \times 10^5 cm)^2}{1cm^2} \times 1.9 \times 10^{-9} erg/sec\]
\[\boxed{P_{sphere}=2.4 \times 10^{2} erg/sec}\]
Solved with collaboration from B. Beak, G. Grell, and F. Young.
D=1km
P=?
λ=532nm
f=50hz
h=6.6×10-27erg×s
c=3
Aeye= 1cm2
To begin solving this problem we must first understand what is being asked. A lightbulb is a distance d away from an eye. The eye must receive at least 10 photons before it refreshes to see the object. Thus, we are solving for the power output of the lightbulb such that the area occupied by the eye will receive the energy equivalent of 10 photons per refresh cycle.
First we should find how much energy 10 photons is.
\[E=hf=\frac{hc}{\lambda}\]
We know that an average visible photon of green light has a wavelength of 532nm,so:
\[E=\frac{hc}{\lambda}=\frac{6.6\times 10^{-27} erg\cdot s\times 3 \times 10^{10} cm/s}{532nm}=\]
\[=\frac{6.6\times 10^{-27} erg\cdot s\times 3 \times 10^{10} cm/s}{5.32\times 10^{-5}}=\]
\[3.7\times 10^{-12}erg\]
So, for ten photons this is:
\[\boxed{3.7\times 10^{-11}erg}\]
The eye needs to receive this many photons in one refresh cycle. So:
\[P_{eye}=Energy/cycle \times f=3.7\times 10^{-11}erg \times 50hz=\]
\[\boxed{P_{eye}=1.9\times 10^{-9}erg/sec}\]
Next, let's calculate the power of the bulb. Since the lightbulb emits photons in all directions, if we approximate the bulb as a point source, once can imagine a large sphere of radius D whose surface has the energy of the lightbulb spread evenly across it. Due to this even distribution of energy, the ratio of the surface areas of he eye and sphere will equal the ratio of the powers of the eye and sphere. Using this, relation, we can solve for \(P_{sphere}\).
\[\frac{A_{sphere}}{A_{eye}}=\frac{P_{sphere}}{P_{eye}}\]
\[P_{sphere}=\frac{A_{sphere}}{A_{eye}} \times P_{eye}\]
\[P_{sphere}=\frac{4 \pi D^2}{1cm^2} \times 1.9 \times 10^{-9} erg/sec\]
\[P_{sphere}=\frac{4 \pi (1 \times 10^5 cm)^2}{1cm^2} \times 1.9 \times 10^{-9} erg/sec\]
\[\boxed{P_{sphere}=2.4 \times 10^{2} erg/sec}\]
Solved with collaboration from B. Beak, G. Grell, and F. Young.
I like your diagram!
ReplyDelete