What Determines Light Intensity?

In summary: The more often the electron jumps down from the higher to the lower energy shell, the higher the energy of the photon. This is related to the concept of "quantum numbers" which we'll touch on later. But for now, just know that the higher the frequency of the photon, the higher the energy.
  • #1
one_raven
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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.
 
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  • #2
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
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.
 
  • #4
Just to be clear, light intensity is proportional to the square of the wave amplitude.
 
  • #5
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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  • #6
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.
 
  • #7


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 traveling 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.
 
  • #8


Originally posted by pallidin
Frank, a given photon traveling 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
Oops, I meant "blue-shift" Sorry. A photon in free travel will always decrease in frequency. Again, sorry for my mistake.
 
  • #10
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.
 
  • #11
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
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.
 
  • #13
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/hbase/bkg3k.html#c4
 
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  • #14
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?
 
  • #15
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
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
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
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
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
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
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 traveling away from you to appear red, and towards you to appear blue.


LPF
 
  • #22


Originally posted by 8LPF16
Gentlemen,...
...I submit: a photon must be traveling 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
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
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


Originally posted by 8LPF16
and it creates its' own energy as it goes along
LPF

It does? Well, maybe news to me, but...
 
  • #26
Palladin :Originally posted by 8LPF16
and it creates its' own energy as it goes along
LPF
--------------------------------------------------------------------------------



"It does? Well, maybe news to me, but..."

This is certainly a postulative statement, derived from 2 facts.
1. once created, the photon requires no more energy to move at natures fastest posted speed limit, and 2. it does so apparently in collusion with gravity, by shifting all mass to the potential side of the scale, and reading "weightless." "Recycles" would probably have been a better choice over "creates."


Jimmy,

I understand the currently held view of science on "seeing" the light. I just also think that I (we) would see a sphere of white light passing side to side or from back to front. I will describe this in a different way.

There is a definite line between light and dark, but it (the line) is not flanked by white and black, it is surrounded by gray. It is in this "grayscale" that our eyes ability to perceive movement is at its peak. So as the photon vibrated throught the darkness around me, my eyes would detect this vibration, and match it up with that of a photon, and I would "see" light. At this point, we could easily get caught up in semantics of "what is seeing", so I will just conclude with this. All experiments ever done have been judged by the perception of the human eye. In fact, it is the hallmark of scientific "proof",when I can communicate what my perception was well enough that you can recreate my experience (perception) without exception.

Perhaps with a chain of photons, the light from the last one would be reflected off of the one just in front of it, and I would see "red shifted" (the slowest, darkest color)light moving away from me.

LPF
 
  • #27
If you can indeed perceive things when they do not interact with any sensory organs then you are indeed most highly developed. The evolution of an active scanning eye will no doubt be one of the greatest evolutionary steps ever taken.

You can't perceive anything unless there's some sort of interaction to produce the correct nervous responses. Processes do not happen without cause, and what you suggest is proposterous unless your brain had some way of creating an E.M. field around your person and could detect the minute disturbances in that field caused by the oscillations of a photon and interpret it as visual sensory input.

Also photons cannot reflect off one another as their wave behaviour means they obey the superposition principle.
 
  • #28
Originally posted by Kurdt
If you can indeed perceive things when they do not interact with any sensory organs then you are indeed most highly developed. The evolution of an active scanning eye will no doubt be one of the greatest evolutionary steps ever taken.


I do not believe that LPF is suggesting that anyone can perceive things in the absence of an interaction. He specifically mentions a vibration that causes an interaction in the eye.

I need to give this some thought in order to make any reasonable queries or comments...
 
  • #29
It was a rather ambiguous statement, and in light of previous comments I interpreted it with possible inaccuracy. I apologise if that was not your point.
 
  • #30
Kurdt,

No need to apologize, what I said was challenging to the accepted parameters, you show intellectual integrity by questioning such things (as does Jimmy's pause for cause). Thanks for responding.

When I said "So as the photon vibrated throught the darkness around me, my eyes would detect this vibration, and match it up with that of a photon, and I would "see" light." , the vibration is the interaction you asked for. Let me try to clarify with two further items. 1. If the heavy mass of a planet, etc. can stretch the fabric of space-time, is it not reasonable to assume that an object of infinite energy (photon) could wiggle it just a bit? This is a vibration (signal). 2. It has been established that gravity moves at C, therefore has very similar (resonant) rate of vibration as light. Therefore, my eyes would be "fooled" into seeing light because of their inability to differentiate between frequencies so similar, and would produce the conditioned response (light). This is no different than our eyes not being able to tell the difference between, for instance, yellow & a combination of red+green.

This is all, of course, specualation. I do not own a photon machine (1 at a time version). As a weekend warrior of physics, all I can afford to do is read and do mind experiments, and then read some more. I have no lab or equipment. This forum allows much more, and your feedback drastically reduces my learning curve.

Also, when you say "unless your brain had some way of creating an E.M. field around your person and could detect the minute disturbances in that field", I thought it was accepted that we do have a small EM field, this it the thing that MRI "modulates" with to create unidirectional polarity. I do know one thing (not really related to first paragraph): if you blindfold me in a dark, cool room, I could tell when you entered the room, and follow your movement without seeing you. (limited to the mass/distance ratio) This is all about heat, or infrared light, that I am not seeing, but feeling and knowing.

LPF
 

1. What is light intensity?

Light intensity refers to the amount of light energy that is present in a given area or space. It is measured in units of watts per square meter (W/m²) or lux (lx).

2. How is light intensity determined?

Light intensity is determined by several factors, including the distance between the light source and the object, the brightness of the light source, and the properties of the medium through which the light travels (such as air or water).

3. What affects light intensity?

Light intensity can be affected by the type of light source, the distance from the light source, and the presence of any objects that may block or reflect the light. The color and surface properties of the object being illuminated can also impact light intensity.

4. How does light intensity change with distance?

As light travels away from its source, it spreads out and becomes less intense. This is known as the inverse square law, which states that the intensity of light is inversely proportional to the square of the distance from the source.

5. How is light intensity measured?

Light intensity can be measured using a light meter, which typically measures the amount of light in lux or foot-candles. These devices use sensors to detect the amount of light in a given area and display the results on a digital screen.

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