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I read that during the process of reflection and refraction light loses some of its intensity. But the frequency of the wave remains the same. A possible explanation that I found was that some photons of the wave are absorbed while the frequency of no photon is changed, therefore the frequency of light remains the same.

Since we are associating the term energy to a photon and intensity to a wave, then the term energy of light (or any EM wave for that matter) should be incorrect.

Or the energy of a wave is equal to the sum of the energy of all the photons combined? In that case what would be the formula for energy of an EM wave?

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  • $\begingroup$ You concern is valid, but you have to watch out for context and sloppy language. Energy density is often more appropriate and correct. However, one might ask for the total energy of a light wave in an optical cavity such as a Fabry-Perot or a laser, or some other confined volume. $\endgroup$
    – garyp
    Dec 11, 2017 at 15:49
  • $\begingroup$ @Countto10 I tried to find the expression is it - 'Energy = k(A)^2' ? $\endgroup$ Dec 11, 2017 at 15:57
  • $\begingroup$ Light intensity has to do with the number of photons. It’s best if you don’t think of waves. $\endgroup$ Dec 11, 2017 at 16:34

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In most cases you would be more interested in the power or intensity of a light beam. The "energy" of the light beam would only make sense if you have observed (or absorbed) the light beam for a specified amount of time.

I read that during the process of reflection and refraction light loses some of its intensity. But the frequency of the wave remains the same. A possible explanation that I found was that some photons of the wave are absorbed while the frequency of no photon is changed, therefore the frequency of light remains the same.

This question was just asked yesterday and the answers haven't changed in the intervening 24 hours.

As some comments pointed out, reflection off an ideal mirror does actually change the frequency of the light, but the amount of change is usually negligible.

Since we are associating the term energy to a photon and intensity to a wave, then the term energy of light (or any EM wave for that matter) should be incorrect.

In a sense, yes. Energy is power accumulated over time. At one instant in time, for one surface that the beam passes through, you can only know the power or intensity of the beam. If you consider a volume of space containing the beam, or the amount passing through a surface over a period of time, then you can think about the energy contained in the volume or passed through the surface.

In that case what would be the formula for energy of an EM wave? Or the energy of a wave is equal to the sum of the energy of all the photons combined? In that case what would be the formula for energy of an EM wave?

You'd be better off forgetting about the quantum view (photons) and just think about the power in the beam in terms of waves. Quantum effects generally relate to how optical energy is generated or absorbed, or how EM radiation interacts with matter.

Wave effects generally explain how optical beams propagate in space. Even if you want to think about photons flying from place to place you have to remember that photons don't travel in straight lines like bullets --- they travel according to the same wave equation that governs classical wave optics.

You can calculate the energy passing through a surface $S$ in time $T$ as

$$E = \int_0^T \int_S \vec{S}\cdot {\rm d}\vec{n}\ {\rm d}t$$

where $\vec{S}$ is the Poynting vector derived from the E and H fields of the beam and $\vec{n}$ is the unit normal on $S$. To put it in simpler terms, you should integrate the intensity of the beam over time and over a surface that the beam is passing through.

If you insist on thinking in quantum terms, then yes, this energy is equivalent to the sum of the energies of the photons passing through the surface in the given period of time.

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