# How is energy conserved during simultaneous refraction and reflection of light?

My question is regarding the partial reflection and refraction of light when incident on a refracting surface.

Here, the energy of Incident Ray is E=hv of Reflected Ray is E=hv and that of Refracted Ray is also E=hv. I think so because neither refraction or reflection are accompanied by change in frequency(v) or so I have been told.

Doesn't this violate the pirinciple of conservation of energy?

As the initial energy is hv and final is 2hv.

• Besides the fact that the answers might explain this in details the picture is just a geometrical scheme. Say you have two phitons, one goes through and one gets reflected. That's way is written partial. Or whatever ratio depending on the material. – Alchimista Mar 4 '19 at 9:07

I will give you an answer on the QM level.

When a photon interacts with an atom, three things can happen:

1. elastic scattering, the photon keeps its energy, and phase, and changes angle

2. inelastic scattering, the photon gives part of its energy to the atom and changes angle

3. absorption, the photon gives all its energy to the atom

In the case of reflection, it is 1. that is elastic scattering.

In the case of refraction, it can be 1. or 2. or even partially 3., that is, the photon can be elastically scattered (glass) or inelastically scattered (UV heating up the molecules), or absorbed. Now the photon can even be partially absorbed, that is, the photon can excite multiple atoms.

You are saying that energy is E=h*f for light. Let's take twp possibilities:

1. you are talking about a single photon, in this case this photon will either reflect, or refract, it cannot do both at the same time (of course there is possibility for partial absorption)

2. you are talking about a herd of photons, that we call classically light. IN this case you mean that the herd of photons has together E=hf. In this case, some of the photons will be reflected, and some will be refracted. Now in the case of glass, when you look at a glass window, it lets some of the light through, as a mirror image, that is refraction. But the glass window can reflect too, like a mirror. So the herd of photons has E=hf. Now some of the photons will be reflected, and some refracted. Let's say that the ratio is 50%. Now half of the photons together will have E/2=hf/2, and the other half will together have E/2=hf/2. Energy is conserved. This assumes, that the reflected and refracted photons will all keep their energy, which is not true, because in the case of refraction, part of the energy will be transformed into the atoms' energy.

In reflection and refraction there is a point in time when the light has a potential energy reaction conserving energy because change in direction.

First I am going to take exception to the last sentence, "Perpendicular rays DO NOT refract (reflection only). Perpendicular rays reflect and transmit, though the transmitted rays don't change their angle (or direction). See Snell's law.

Second: It is true that the transmitted waves change their wavelength $$\lambda$$ and speed $$v$$, $$\lambda_2=\nu_1*\lambda_1/\nu_2$$, and $$v_2=\nu_1*v_1/\nu_2$$, but since frequency=speed/wavelength, $$\nu_2=\nu_1$$. (n= refraction index).

But looking on this from the point of view of $$E=h\nu$$, and getting the outcome of $$2h\nu$$ is an interesting question. That view is a photon view, and a single photon will just take one of the two paths, reflected or refracted, much like in two slit interference, or an image being formed. For the photon view to work you need to look at photon statistics. I once did a thin film coating transmission and reflection calculation doing photon statistics at each interface along with interference for each beam.

So taking many photons and the probabilities of being reflected or refracted (or?) I am going to deviate a little from some texts and say the incident power = refracted power + reflected power + absorbed power + scattered power does hold.

Another thing to consider is the intensity (power/area) of the refracted beam. Due to refraction it does not have the same cross section of the incident beam. All of this is calculated within the Fresnel equations for the electric and magnetic fields for reflection and refraction in any E&M textbook.