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In electrodynamics, light as a wave has an intensity $I = \frac{1}{2} c \epsilon E^2,$ where $E$ is the electric field amplitude.

Meanwhile in quantum mechanics, we study $I = \frac{dn}{dt} \frac{hc}{\lambda}$ where $\lambda$ is the wavelength and $\frac{dn}{dt}$ is the number of incident photons per unit time.

So at the same time, light carries both energies and these are different right?

My question arose from when an EM wave enters a medium from vacuum as a wave. Its wavelength stays the same while the particle's wavelegnth decreases.

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  • $\begingroup$ The energy of a single photon is $hf$, and in vacuum $f=c/\lambda$ so the formula you wrote is true in vacuum but not in different media. f doesn't change when entering a medium $\endgroup$
    – Ofek Gillon
    Commented Oct 15 at 10:16
  • $\begingroup$ You need to be clearer with your questions. You really have two questions, mostly unrelated to each other, mashed into one post. For free space, there is really no problem; the classical intensity expression is easily connected to the quantum expression by changing the number of photons per unit time. As for the behaviour of electromagnetic energy inside a material, the fight between people who believe different expressions should be considered, is still ongoing, after almost a century. You should just look it up. $\endgroup$
    – naturallyInconsistent
    Commented Oct 15 at 10:18
  • $\begingroup$ An electromagnetic wave leaves various traces/wakes of its passing, but a photon leaves theoretically no trace of its passing, so the photon seems to be a more confined "particle"-form of energy transfer compared to an arbitrary electromagnetic wave-based energy transfer. $\endgroup$
    – James
    Commented Oct 15 at 10:52
  • $\begingroup$ Macron also its better to use η when talking about wave propagation because cε cannot contain phase shift between E and H while η does. $\endgroup$
    – Root Groves
    Commented Oct 15 at 13:02
  • $\begingroup$ 3blue1brown has a series of videos on physics. for an explanation of why the wavelength changes in a medium, see But why would light "slow down"? | Optics puzzles 3 $\endgroup$
    – mmesser314
    Commented Oct 15 at 15:03

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Very hard question!You need to know about EM,optics and photometrics to answer it.You are missing a intermediate step in your line of thinking

The real power of a EM wave is equal to $\frac{1}{2|\eta|}E^{2}$ where $|\eta|$ is the amplitude of the complex wave resistance of the medium.

However the total power classically can be related to the intensity of light $I=\frac{P}{A}$.And by finding the I you can find the density of photons per unit surface by:$n=\frac{I}{hf}$ where f is the frequency of the EM wave.

Hope this helps.

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