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Topic: Light Physics (Read 6305 times) |
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Kozo Morimoto
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Am I missing something here?
Isn't E = h v
where
E energy
h Planck const
v frequency
so the question you need to ask to answer is "What colour is the light?" or "what's the frequency of the ray of light?" or something to that effect.
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Mentor
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Also, you need to consider the follow things:
How many photons(light quanta) it takes for the human eye to register an artifact (sensitivity of cones & rods, and no. of cones and rods).
How the focusing of the eye will affect the light (ie. onto what area will it focus the light).
How the ray of light disperse and attenuate with distance.
The longest wavelength of light that the eye can detect.
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william wu
wu::riddles Administrator
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Posts: 1291
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Re: Light Physics
« Reply #2 on: Jul 26th, 2002, 4:19pm » |
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Yeah, I think this is one of those questions where the more factors you can think of, the better. Similar to the classic Microsoft estimation problems (e.g. how many diapers in the usa). Another factor I thought of was the smogginess of the area where the observer resides. I was reminded of some telescope work I did when I was younger. To gather decent data, we had to travel to the mountains, where the air was purer and starlight could reach us more easily.
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[ wu ] : http://wuriddles.com / http://forums.wuriddles.com
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Kozo Morimoto
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You guys are both over complicating things. The riddle is:
"How much energy must the light contain?"
All you need is the frequency. Hertz (Hz) is a unit of energy, not an SI unit, but still a unit of energy.
You guys may be confusing it with Power or Work.
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Ryan
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Hertz is certainly NOT a unit of energy. Hertz is simply an easier way to say seconds^-1 or "per second." It's a frequency description. Energy (and work) are measured in joules or calories.
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Kozo Morimoto
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Because of the E=hf formula, Hertz, nm, J, Calories, Kelvins, evolts, ergs are all valid units of energy (when referring to em radiation) and they are completely interchangeable. Joules just happens to be the SI unit.
Look at http://www.ilpi.com/msds/ref/energyunits.html
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Ryan
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Sweet monkey! I hope you don't go around interchanging things like Hertz, joules, and Kelvin.
Joules, calories, and eV are the only units of ENERGY in that list (unless you really want to get particular about Kelvin). The other units can be used in formulas to DETERMINE energy, but they are not energy themselves. Did you even read the link you posted?
E=hf works because Planck's constant is a conversion term. It doesn't just scale the value - it changes the type of unit. Surely you don't tell people that a free-falling object has an acceleration of 20 grams.
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Kozo Morimoto
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I said it only applies to EM radiation....
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peterchen
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Ryan wrote: Sweet monkey! I hope you don't go around interchanging things like Hertz, joules, and Kelvin.
You obviously still have to see a particle physicist on a blackboard, explaining something. It's amazing! Voltage, speed, frequency, mass... it's all energy.
They typically use it to shorten their calculations and determine the required additional factors by looking at th required unit vs. the actual unit of the result.
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Anonymous Coward
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>>You are allowed to ask questions for more information.<<
Can't you just ask them for the following information:
How much energy must the light contain?
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BenderBot80
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Posts: 2
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Re: Light Physics
« Reply #10 on: Feb 3rd, 2003, 5:02pm » |
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I tried a back of the envelope type thing for this problem. (BART is wonderful to do such calculations or finishing homework at the last minute).
D = distance to sattelite (meters)
I assumed 500 km for normal orbit (5*10^5 m)
r = radius of the pupil (meters)
i am not sure if this is the right piece of eye anatomy
to measure, but it's a start. At a glance, about 3mm
(3*10^-3 m)
h = plank's constant (6.626 * 10^-34 Js )
v = frequency of red light (4.3 * 10^14 Hz)
k = min number of photons necessary for human eye to
detect a signal. This i vaguely remembered to be on the order
of 100. I checked some literature on the subject, turns out the retinal has a better-than-half chance of detecting a light signal if about 9 photons hit it. But, only 10% of the photons that arrive at the eye actually make it all the way to the retina. (Reminds me of a joke about God's bad engineering practices: putting the wires in front of the sensors  )
f = eye's framerate. (Hz)
I figured that in order for the light source in this problem to appear 'solid', it would have to emit the right amout of energy per human eye's sampling rate. For this value I figured about 30 Hz
So taking all those into account, we want to calculate the power that the light source has to generate (assuming omni-directional, not laser light)
So the minimal power going into the eye would have to be:
hvf = 6.626 * 10^-34 Js * 4.3 * 10^14 Hz * 30 Hz * 100 (min photons) =
= 0.000000000000000854754 W
But at a distance D, the eye recieves only a fraction of the power from the light source. Assuming no dissipation of the light by the atmosphere (big assumption i know, but i don't have the data for this) the fraction of the original light that hits your eye is:
(2pi(r^2)) / 4pi(D^2))
Multiplying the minimal power by the inverse of this fraction, we get:
(2 (D^2) / (r ^ 2)) * 0.000000000000000854754 =
(2 (5*10^5m)^2 / (3*10^-3m)^2)) * 0.000000000000000854754 W =
=~ 50 watts.
At first I thought that was a really low number. I mean lightbulbs are 100Watts! And they are right in the same room as we are. But this calculation assumes the absolute perfect conditions: no other light sources, no absporption of energy by the atmosphere, and a nearly perfectly tuned eye.
Then I remembered the old wisdom that
"The sensitivity of the human eye is so keen that on a clear, moonless night, a person standing on a mountain can see a match being struck as far as 50 miles away." So 50 watts doesn't seem like that far off for a number.
Any thoughts?
--VY
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Kozo Morimoto
Junior Member
Posts: 114
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Re: Light Physics
« Reply #11 on: Apr 11th, 2003, 12:20am » |
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Didn't the question ask for energy and not power? Watts is power (from memory), so you need WattHours or KWh for units of energy? I still don't see why its just not e = hv.
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DewiMorgan
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Which light? The light striking his eye must contain enough energy to trigger sufficient receptors for his brain to notice it.
The light from the satellite is a beam and may be assumed to be a laser (coherent, parallel, single-frequency).
Remember that frequency response of the eye is not a binary relation, so we need to find which point he can see it with the fewest photons at the lowest energy for the shortest time - that's going to be a 3D graph, and you want the lowest point on it.
Given that frequency, and the number of photons per second at that frequency, you have the minimum energy. This energy will be massively, massively lower than any actual energy that you could see, though because the laser won't be perfectly pointed at his eye, and won't be travelling through a perfect vacuum, and won't be perfectly coherent. The eyeball itself will absorb and reflect photons before they get to the detectors at the back of the retina.
This all assumes he sees directly, and not through any kind of light amplification system, in whcih case all he needs is a single, low-frequency photon.
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