Intensity of different color light & its relation to energy AND Amplitude

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SUMMARY

This discussion centers on the relationship between light intensity, amplitude, and photon energy in monochromatic light waves, specifically comparing blue and red light. Both light waves have equal intensity, meaning they deliver the same energy per second, but blue light photons possess higher energy due to their higher frequency, resulting in fewer photons compared to red light. The confusion arises from the relationship between amplitude and photon count; while amplitude is proportional to intensity squared, the number of photons is inversely related to photon energy. Thus, equal intensity does not equate to equal amplitude or photon count.

PREREQUISITES
  • Understanding of light wave properties, including frequency and wavelength
  • Knowledge of photon energy calculations using E_{photon} = \hbar \nu
  • Familiarity with the concept of intensity in physics
  • Basic grasp of amplitude and its relation to energy flux
NEXT STEPS
  • Research the relationship between light intensity and amplitude in electromagnetic waves
  • Explore the concept of photon flux and its calculation in different light sources
  • Learn about the implications of frequency on photon energy and its effects on light properties
  • Investigate the mathematical models that describe the behavior of monochromatic light waves
USEFUL FOR

Physicists, optical engineers, and students studying wave mechanics or quantum physics who are interested in the properties of light and its interactions.

mlgpawnstar
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Hi Let's stick to the classical limit.

We have 2 monochromatic light waves of same intensity. Let's say one is blue and one is red.

Now this means the individual photons of the blue light have more energy(obviously, higher frequency) and the red light photons have lower energy. But since the two light waves have equal intensity, they have the same energy arriving/second.. Right? I think this part is obvious. This then means that the red light wave has more photons and the blue light wave has fewer photons.

Well let's move on to amplitude of the light wave, now this sort of scales with the intensity squared so these two light beams should have the same Amplitude? But Amplitude can also be seen as a function of the number of photons..

Here in lies my current confusion. Sorry if this is so simple. Anyone know of a basic way to get around this and stay consistent?

Either the two light is going to have different amplitude or the same amplitude while the have the same intensity but different frequencies.
 
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There is no contradiction. The number of photons in a light wave is dependent on the amplitude. You can calculate the number per second exactly as you described, by looking at the power of the light, and using E_{photon}=\hbar \nu so red light of the same intensity has more photons.
 
Hi, light intensity or energy flux is proportional to photon flux, i.e number of photons per unit time per unit surface and to individual photon energy. Does that answer your question?
 
Hi 0xDEADBEEF if # of photons depends on Amplitude, and if Amplitude comes from Intensity, since these two beams have the same intensity, they will have the same Amplitude, and thus the same # of photons?? (Obviously this isn't the case.)

"amplitude" is what is throwing me off here.


Alkim: I understand this but how does this relate with Amplitude?
 
mlgpawnstar said:
Hi 0xDEADBEEF if # of photons depends on Amplitude, and if Amplitude comes from Intensity, since these two beams have the same intensity, they will have the same Amplitude, and thus the same # of photons?? (Obviously this isn't the case.)
How do you make that conclusion? If 'amplitude' relates to the amount of energy flux and short wavelength photons have more individual energy then either the same amplitude would imply more red photons or the same number of red photons would imply lower amplitude. (Two ways of looking at the same thing.)
 

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