

Light Intensity to Space
Name: Brian
Status: student
Grade: 912
Location: MO
Country: N/A
Date: 6/26/2005
Question:
How powerful must a light on earth be to be seen in
space? For example, how far away could you see a 1 kilowatt light versus
a 350,000 kilowatt light?
Replies:
Dear Brian,
This is a straightforward calculation except for the precise definition of
what you mean by "seen". I will assume the detector is a normal human eye
fully dilated to a diameter of 8 mm and sensitive to a light flux of 10
photons/second. The human eye at its most sensitive can sometimes detect a
single photon, so a sensitivity to 10/s seems reasonable. I assume you know
that light is made up of photons as postulated by Max Planck around the turn
of the century and proven by Einstein's theory of the photoelectric effect
published during his magic year of 1905. Each photon carries an energy E =
hf where h is Planck's constant (6.6 E34 Js) and f is the frequency of the
light wave. Visible light is a spectrum centered near a wavelength of 500
nm (nm = nanometers; 1 nm = 1E9 m) and a frequency of 6 E14/s. The energy
carried by a single photon of that frequency is 4.0 E19 J.
Note that I am writing large (and small) numbers using the old computer
notation 100 = 1E2 and 0.001 = 1E3. I will show powers by rr = r^2.
Assuming the light source sends the light out equally in all directions, at
a distance R it will uniformly illuminate a sphere of radius R, which has an
area of 4 pi R^2. An eye of radius r will have an area of pi r^2 so the
fraction of photons hitting the eye is (r/R)^2/4. The number of photons of
energy 4.0 E19 J produced per second by a P kW = 1000P J/s light is 2.5 P
E21 Ph/s. Finally, where N is the number of photons detected by the eye per
second:
N Ph/s = 2.5 P E21 Ph/s (r/R)^2 /4 or
R^2 = 6.2 P
E20 ph/s r^2 m^2 / (N Ph/s)
Plugging in P = 1 kW, r = 0.004 m = 4 mm, we find R = 3.2 E7 m = 3.2E4 km =
20,000 mi.
Notice that the distance increases linearly with the radius of the detector
and with the square root of the power and decreases like the inverse of the
square root of the sensitivity of the detector. If you plug in a bunch of
numbers, I think you will develop a good feeling for how this works. For
example, using a 350,000 kW light (P = 350,000) increases R to 1.9 E10 m =
11,400,000 miles.
Best, Dick Plano, Professor of Physics emeritus, Rutgers University
This is mostly dependent upon the sensitivity of the detector and the
length of time that the detector can be kept operational and pointed
accurately pointed at the source. The smallest unit of "light" I use the
quotes because light is just one small part of the electromagnetic
spectrum  is one quantum. Assuming that the detector is 100%
efficient  that is, it can "count" every photon that strikes it  then
it is (only!!) a matter of how many photons it takes to generate a signal
in the detector. In the reverse configuration the Hubble telescope can
"watch" a distant source for weeks to compile a picture of the source.
Quite remarkable.
This remarkable telescope has produced more data about the Universe than
all telescopes in all prior history. Let us hope it meets a better fate than
to let it "die", which is the present intent of NASA and the federal
government.
Vince Calder
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Update: June 2012

