So I was reading this thread and I got to thinking.
look out! -
Now as I understand it, your given light packet (photon) has a certain energy associated with it. To our eyes visible light is at an energy that is not harmful. If you increase the number of photons in a given area you increase the brightness, but the energy is still the same. So if you have a typical light bulb that is emitting white light (everything from visible red to visible violet), and it is at a typical reading distance, it is not hurting you. So why then if you got your eye closer to the light (increase brightness, number of photons per unit area), or if you shined a white light laser into your eye it hurts and can/will cause permanent eye damage?
But how is having more photons damaging? It cannot be increasing the energy. How is it “burning” you if the energy is the same no matter how many photons fall on your eye?
Sorry, that looks like it is reitrating something you already realised; I don’t actually understand your problem; deliver too many photons to a given area of your retina in too short a space of time and the energy they are transporting won’t be dissipated quickly enough - the affected area will heat up and this might cause damage. In practice, your brain is programmed to trigger a pain response before this happens, at least in the case of visible light.
The energy drops off over distance because it is spreading out spherically through space; if you built a set of different-sized spherical shells and put an identical light source in the centre of each one, the sum of energy captured across the entire inner surface would be the same in all of them, but in the smaller ones, the shells would be warmer because any given area is receiving a larger quantity of photons in a given time.
I don’t think that energy is the same no matter how many photons fall on a particular area of the eye, or anything else. The energy of one photon is proportional to the frequency of the light. In fact it is equal to Planck’s Constant, h, multiplied by the frequency.
If one photon/sec falls on an sq. cm. then hf energy has been dissipated somehow in that sq. cm. If 1000 photons/sec falls on the same sq. cm. then 1000h*f energy has been dissapated there. If 10[sup]10[/sup]/sec photons fall … then 10[sup]10[/sup]hf etcetera, etcetera, etcetera.
I know the energy is the same no matter how many photons you got. Different types of light have a specific frequency which translates to a specific energy, and incresing the intensity does not change a frequency. Perhaps it does have something to do with dissipating the energy, and if there are too many…
You can shine white light on your arm all day long. You shine some ultraviolet light on your arm for a couple hours you’re gonna burn. You shine some gamma rays on your arm for a couple seconds you’re gonna burn.
Microwaves resonate with water moelcules in your arm, that’s the damage you get. But you point some longer wavelength radio waves at your arm - nothing.
I’m not so sure about the dissipating energy thing. Anyone else can help or back that up?
I think you may be mixing up what should be a wavelength equation with a quantum equation. Number of photons corresponds to the amplitude of the wave (which has nothing to do with it’s energy), and should not involve Planck’s Constant.
OK, let’s look at the thing as an electromagnetic wave. The energy contained in a wave is certainly proportional to the magnitude of its electric and magnetic field vectors.
The power (energy/unit time) of an EM wave is the cross product of E and H, the electric and magnetic vectors. The magnitude is the power is simply the product of the magnitudes of E and H.
UV burns because the UV photon is so energetic it can penetrate and damage the skin cells. But if you increase the number of UV photons you increase the total energy and increase the damage. Visible light doesn’t penetrate, or at least not so far, so isn’t as damage.
I am finding in some places on the web where they are mixing “intensity” with “energy”. Here’s one
When I say energy I mean frequency. How energetic is the wave?
Perhaps I don’t understand watts and such and so “energy” is getting mixed up with “power”.
I suppose if no one out there understands what I’m talking about I’m simply insane. I will keep researching it and try to post if I can better explain and cite.
This is your problem. The enegy of a single photon is determined by the frequency. The energy disapated in your eye depends on the freuency of the photons and on the number of photons.
The area of physics with which you are dealing is called Radiometry. The quantitative methods are well established but as you are discovering they turn out to be not quite as easy to understand as one first expects.
Here is another good link, designed specifically for computer graphics geeks.
The site doesn’t so much mix them as define intensity as average energy. I think light intensity is better defined as energy per unit of area.
This statement illustrates a confusion I think. Energy is not frequency. Energy is equal to the amount of work that is done on something or the capacity of that “something” to do work on something else. Frequency is how many times something happens per unit of time. Like the frequency of midnight is one/day. The frequency of the minute hand passing the number 12 on a watch face is one/hour. The frequency of an EM electric vector is the number of times per second that the vector reaches some value, like its maximum, in a specified direction.
Power is nothing but energy per unit of time. The unit of energy in the unit system of meter-kilogram-second is one joule. That is the work done in exerting a force of one Newton through a distance of one meter. One Newton is the force needed to give a mass of one kilogram an acceleration of one meter/second.
Since power is energy per unit of time one Watt is equal to one joule of energy expended or gained per second.
You seem to have grabbed on to the idea that the energy of one photon of a particular frequency of light is constant and assumed that means that the total energy is constant regardless of the number of photons that arrive. That’s not true. Each photon delivers a quantum of energy and if more photons arrive then more energy is delivered. As I said before, the energy of one photon is Planck’s constant multiplied by the frequency of the wave.
As to damage to the eye. Anything that delivers too much energy to the retina can cause damage. Visible light that is too intense, i. e. too much energy per unit of area of the retina, can cause damage. Ultra Violet light can cause damage even faster because each photon is more energetic than that of visisble and whether or not the eye can see it, the retina can be seriously damaged, and so can the skin. I once watched an arc welder working on a project of mine for a half hour or so. I had a welder’s mask but the collar of my shirt was unbuttoned and there was a gap in the shirt below the mask. I wound up with quite a sunburn from the UV because I have fair skin and am susceptible to sunburn. That’s why when you see a welder he has on a long sleeved shirt, all buttoned up and big gloves besides.