The intensity of an electromagnetic wave is only related to its amplitude $E^2$ and not its frequency. A photon has the same wavelength as the wave that's carrying it, and its energy is $h f$.

So if a laser wave is kept at the same amplitude and the wave length is reduced, why does its intensity remain the same even though its photons now carry less energy?

Why are the intensities of electromagnetic waves so different to sound waves (and other waves travelling through a medium) which are related to $f^2 E^2$?

  • $\begingroup$ In your second paragraph I think you meant to say "carry less energy"? $\endgroup$
    – twistor59
    May 5, 2012 at 8:08
  • $\begingroup$ @twistor59-If the wavelength is reduced doesn't the energy of a photon increase, because of the relation $E=hf$? So in the laser beam with reduced frequency but equal amplitude, the amount of photons decreases. $\endgroup$ Mar 7, 2017 at 16:11
  • $\begingroup$ Yes, @twistor59, if the wavelength is reduced the photons would carry more energy. $\endgroup$
    – user183966
    Sep 19, 2018 at 10:23

2 Answers 2


In order for the intensity of a light source to stay the same, while each lower frequency photon carries less energy, there must be a greater number (per time, per area) of the lower frequency photons in the beam than the original number of higher frequency photons.

As for the second part of your question, I admit that it can be confusing that the power transmitted by E&M waves depends on the amplitude of the wave, while the power transmitted by a mode of a vibrating string depends on both the amplitude and the frequency of the wave. Ultimately this comes down to fundamental differences in the physics of each wave phenomenon.

The energy in a vibrating string is reducible to the kinetic energy of the moving string elements and the potential energy from the tension felt by each element due to the position of its neighbors. So, at fixed amplitude, you can see that you get even more energy if you jiggle the rope faster.

The energy in an E&M wave is a different effect entirely: it comes from the average size of the (squared) electric field in the wave that can do work to move charged particles. At a fixed amplitude, if you increase the frequency you won't increase the average size of the field.

  • 1
    $\begingroup$ please note that lasers have practically one frequency. en.wikipedia.org/wiki/File:Helium_neon_laser_spectrum.svg A different lazing medium is necessary for a different frequency so one cannot lower the frequency while increasing the amplitude of a laser. $\endgroup$
    – anna v
    May 5, 2012 at 11:34
  • $\begingroup$ Good point, anna, +1. This was actually the first thing one should have said - and I should have, too. $\endgroup$ May 5, 2012 at 11:34
  • $\begingroup$ You're right, I'll edit the answer to replace "laser" with "light source." I was thinking about this like the thought experiment you do for the photo-electric effect. You imagine you have some frequency and amplitude knobs you can turn for some light source. $\endgroup$ May 5, 2012 at 11:40
  • $\begingroup$ A variety of technologies allow for tunable lasers: en.wikipedia.org/wiki/Tunable_laser $\endgroup$
    – tmac
    May 7, 2012 at 8:14

The frequency $f$ and the intensity or power $P=\epsilon_0 E_{\rm max}^2 Ac$ (energy per second where $A$ is area) are independent quantities so you may change each of them independently of the other. The energy of one photon is $E=hf$ so it's a simple function of the frequency; the number of photons per second is therefore $P/E=P/hf$. But note that there are at least two independent quantities here, e.g. the frequency and the power.


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