What Determines Light Intensity?

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Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
Kurdt
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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.
 
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  • #3
Arcon
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.
 
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Just to be clear, light intensity is proportional to the square of the wave amplitude.
 
  • #5
Chi Meson
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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.
 
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Single photon energy level

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.
 
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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.
 
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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.
 
  • #9
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Oops, I meant "blue-shift" Sorry. A photon in free travel will always decrease in frequency. Again, sorry for my mistake.
 
  • #10
Kurdt
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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.
 
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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.
 
  • #12
Kurdt
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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 [Broken]

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/hbase/bkg3k.html#c4
 
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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?
 
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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.
 
  • #16
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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.
 
  • #17
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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.
 
  • #18
Kurdt
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That raises an interesting question. If the photons increase in wavelength is linked with the expansion of space-time then we could determine the rate of expansion of space time from observing the rate at which a photon increases its wavelength as it travels through space time. The energy of that photon acts like an energy density of a volume of space time linked with a particular photon, thus as that volume expands the energy density decreases which in turn means the photon frequency is lowered and the wavelength increased.

Or I may just have completely the wrong end of the stick. Still an interesting thought.
 
  • #19
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Kurdt, you certainly may have something there.
I suppose that an experiment would be virtually impossible, as the proposed effect is so dependent on such an extraordinary great distance travel of the photon that "limiting experimental models"(earth-bound) might not even reveal it, even if does occur.
Hmmm... wonder if this can be dealt with.
 
  • #20
Kurdt
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That is true. I thought about it the last couple of days and it does not seem possible to gain accurate data for the expansion of such a theory. Theres always one parameter missing (i.e such as a factor taking into account the rate of expansion or the age of the universe) which could not be taken from current estimates obviously. Also I do not have any expertise in this field of research and it may turn out that some other factor could inhibit the theory such as intersteller mediums scattering effects and so forth. Never the less it is a fun idea to ponder, and maybe some time in the future we could have developed equipment sensitive enough to have an appreciable result.
 
  • #21
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Red / Blue shift

Gentlemen,

May I suggest that since energy levels of individual frequencies do change with color (violet more than red, proportional to red's higher wavelength), and the speed of light is a constant (so in this sense can not change with frequency), and there is no drag on the photon (whatever drag there is is compensated for by a greater "thrust" - like the difference in water flow between the front and back of a catamaran hull sailboat) and it creates its' own energy as it goes along (obeying conservation of energy) then I submit: a photon must be travelling away from you to appear red, and towards you to appear blue.


LPF
 
  • #22
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Originally posted by 8LPF16
Gentlemen,...
...I submit: a photon must be travelling away from you to appear red, and towards you to appear blue.
LPF
To see the color of a photon it must enter the eye. If a photon is traveling away from you, you would not see it. It is true that if the source of light is moving away from or toward you then they will appear red or blue shifted respectively. The photons you see, however, were always moving toward you.
 
  • #23
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Jimmy,

Does this mean that a photon that shot by from your left to right would not be seen? Or one from behind you and crossing your line of sight? (of course, we can also use "light", and in this sense, photon means the last one in the "wavicle")

LPF
 
  • #24
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It's my understanding that the only photons we see are the ones that interact with our eyes. How could you see a photon that passed in front of you but did not enter your eye? If we saw every single photon all the time, I don't think we could make much sense out of all that information. We see objects because photons are reflected from the object to our eyes. An object will reflect photons in all directions but we only see the ones reflected toward us. If two people standing in two different locations are looking at the same object, they do not see the same photons. They see the same object because photons are continually being reflected in all directions but they do not see the same photons.

We are literally immersed in a see of electromagnet radiation. I'm amazed how our brains can sort out all of this information and give us a picture of our surroundings. As I type this, there are photons traveling along from my right to left and left to right in front of me. There are photons coming directly at me from my monitor. There are photons coming toward me from all directions yet my brain is able to sort out all of this information which gives me a picture of separate objects that are around me.

In a vacuum, if you were to shine a flashlight in front of someone and not into their eyes, they would not see the beam. When you see a flashlight beam passing in front of you, you do not see the photons going from left to right, but the ones that are reflected toward you by particles in the air.
 
  • #25
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Originally posted by 8LPF16
and it creates its' own energy as it goes along
LPF
It does? Well, maybe news to me, but...
 

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