## What Determines Light Intensity?

I have a few pretty simple questions that I can't seem to find a clear answer to on the net...

The energy of a light wave is directly proportionate to frequency and inversely proportionate to wavelength.
This is due to the energy of a Photon being equal to hf.
"f" is a constant, so the greater the frequency, the greater the energy.
I get that.

The intensity of a light wave is based on the wave's amplitude.
Why?

My immediate intuitive response when looking at a wave diagram would be that there are more photons per waveform.
The more photons that are hitting our eyes in a given timespan, the brighter the light would seem.
Or...
It would seem that the amplitude could have an effect because the photons in a wave at X frequency would have to travel faster in a wave with greater amplitude to travel the same distance forward, therefore have a greater kinetic energy.
I know that light travels at C, but does that apply to the individual photons as well?
Which is correct?
Are there more photons, or do they travel faster?
Or is it neither?

Also:
Why does the frequency of a wave have an effect on the energy level of the individual photons?
What am I missing?

Last question...
What determines the amplitude of a wave?

Thanks for any input you might have.
I don't mind reading all this from articles myself if you don't want to go through all of it, but, like I said, I can't find the answers, so a link to a good source would be just fine, too.
 Recognitions: Gold Member Science Advisor Staff Emeritus The Intensity of the light is proportional to the square of the modulus of the electric field vector to be exact which is just in general the amplitude of the light wave. It is due to the fact that waves can interfere with each other and form constructive and destructive interference which is related to the amplitude of the wave and how they add together.

 Originally posted by one_raven The intensity of a light wave is based on the wave's amplitude. Why? My immediate intuitive response when looking at a wave diagram would be that there are more photons per waveform. The more photons that are hitting our eyes in a given timespan, the brighter the light would seem.
That is correct. The intensity of a beam of light is related to the number of photons. Higher frequency - greater energy. More photons - greater intensity.

For example: consider a laser beam from a He-Ne laser. Different He-Ne lasers have different powere levels but they all emit red light of the same frequency. Therefore all of the photons have the same energy. However if one laser is more powerfull than another laser then the more powerfull laser has a higher intensity beam. That means that comparing different beams from different power He-Ne lasers which all have beams of the same cross-sectional area the laser which has the higher intensity beam is emitting more photons per second.

And all photons travel at the same speed in an inertial frame of reference.

## What Determines Light Intensity?

Just to be clear, light intensity is proportional to the square of the wave amplitude.

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 The intensity of a light wave is based on the wave's amplitude. Why? ... What determines the amplitude of a wave?
I'm guessing you might have the following dilemma: Even though each photon has a frequency and wavelength, the photon does not have an "amplitude" per se, other than it's energy found by E=hf. Confusion arises when one thinks of the photon as the wave; it's not. The photon is a particle. A light "wave" is made of a continuous stream of photons. If this was not a point of confusion for you, never mind.
 It would seem that the amplitude could have an effect because the photons in a wave at X frequency would have to travel faster in a wave with greater amplitude to travel the same distance forward, therefore have a greater kinetic energy.
A photon does not travel "up and down" like a dolphin swimming. It just goes straight forward (through a vacuum). The "Up and down" is only the fluctuation of the strengths of the electric and magnetic fields at a point in space through which the photon travels. Since magnitudes are represented with arrows (long arrow represent stronger fields) the commonly drawn description seems to indicate a distance between the "top" and "bottom" of a photon. There is not. The only distance associated with a photon is its wavelength
 Also: Why does the frequency of a wave have an effect on the energy level of the individual photons?
IT is better to think of the photon first as being a specific quantity of energy, and the frequency is actually a function of its energy. When a (visible) photon is created, an electron "jumps" down from a high energy atomic shell to a lower energy shell. The energy difference between these shells equals the energy that the electron gives off. THis energy is released as a photon. The "color" or frequency of the photon must be equal to this energy. SO energy of a photon and frequency of a photon are essentially the same thing; Plank's constant is used to translate between the two.
 One of the questions I have is whether a single photon ever changes its energy level regardless as to how far it travels. I am assuming it doesn't hit anything. The inverse square law is only applicable to defining energy levels from a source that radiates energy spherically.

 Originally posted by FrankMak One of the questions I have is whether a single photon ever changes its energy level regardless as to how far it travels. I am assuming it doesn't hit anything. The inverse square law is only applicable to defining energy levels from a source that radiates energy spherically.
Frank, a given photon travelling unimpeded over distance will change. This change is with regards to the frequency of the photon. It's frequency will lower(red-shift) while it's speed, C, remains constant.

 Originally posted by pallidin Frank, a given photon travelling unimpeded over distance will change. This change is with regards to the frequency of the photon. It's frequency will lower(red-shift) while it's speed, C, remains constant.
Is this generally true? The frequency of a photon will red shift in a frame of reference stationary with respect to the photon? Why?

I would only expect a red shift if the photon is observed from a frame of ref. that is moving away from the photon.
 Oops, I meant "blue-shift" Sorry. A photon in free travel will always decrease in frequency. Again, sorry for my mistake.
 Recognitions: Gold Member Science Advisor Staff Emeritus Surely red shifts and blue shifts only occur when the observer and the source are moving relative to one another. If they are both stationary with respect to one another then the photon is of normal wavelength and does not lose any energy as it travels unless it hits something and is absorbed or scattered.
 Recognitions: Gold Member The microwave background radiation is an example of EM which has lost a considerable amount of energy over time as the universe has expanded. I would say that all electromagnetic waves are gradually increasing in wavelength as the universe expands.
 Recognitions: Gold Member Science Advisor Staff Emeritus But as the universe expands I would think that it corresponded to a system where the observer and the source were not stationary with respect to each other. The only reason we detect microwaves as cosmic background radiation is becasue we are moving in the expanded universe away from that radiation.

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 Kurdt: But as the universe expands I would think that it corresponded to a system where the observer and the source were not stationary with respect to each other. The only reason we detect microwaves as cosmic background radiation is becasue we are moving in the expanded universe away from that radiation.
The source of the CMB is not a point source from which we are moving away. The source is the universe itself. As the universe has expanded the wavelengths have been effectively stretched out.

http://cmb.physics.wisc.edu/tutorial/cmb.html

We are immersed in the CMB. Relative to the earth's motion through the CMB, there is a slight blue shift of the background radiation in the direction of earth's motion and a redshift in the opposite direction.

http://hyperphysics.phy-astr.gsu.edu.../bkg3k.html#c4

 Originally posted by Kurdt Surely red shifts and blue shifts only occur when the observer and the source are moving relative to one another. If they are both stationary with respect to one another then the photon is of normal wavelength and does not lose any energy as it travels unless it hits something and is absorbed or scattered.
Perhaps I am under the wrong impression and am certainly willing to be corrected. Does not a photon in free-travel over extraordinarily long distance not decrease in frequency due to its interaction of traveling through space-time itself?
 Recognitions: Gold Member In 1929, Fritz Zwicky proposed the idea tired light; Light whose wavelengths have been stretched out over long periods of time and distances to explain cosmological redshifts in support of a static universe in contrast to redshifts caused by the expansion of space. http://www.astro.ucla.edu/~wright/tiredlit.htm While I'm in no position to debate this, I do not believe this is generally accepted by the mainstream physics community.
 Thanks Jimmy. I can see from what you wrote that "tired light" is not generally accepted. I wish I would have asked the question before I posted on several previous questions, as my erroneous assumption was part of some of those responses! Dang, I really hate giving-out inaccurate information. Oh, well, now I know.
 Recognitions: Gold Member Pallidin, I wouldn't worry about that too much. I'm no cosmologist so generally I have to depend on what other (real physicists) tell me. I, like you, am perfectly willing to be corrected if I say something which is wrong. As far I really know, the idea of tired light could be correct. My guess is that it isn't correct based on current evidence in support of expansion. But, like I said, I am not a cosmologist.