3
$\begingroup$

Why photon reflects and refracts through glass? Some photons pass through glass and some reflects.I know this is due to energy levels of electrons of glass, an incoming photon is unable to excite the electron to a different energy orbit.

But I want to know, if an incoming photon is unable to excite an electron, then why not all the photons pass through glass? i.e. photons should not reflect off glass, all the photons should pass through glass.

(Not sure! Just a thought)Is this due to atoms of different substances like water,glass or wood etc curve spacetime differently and thus it influences how photons interact with matter? i.e. some photons reflect and some refract.

$\endgroup$
3
$\begingroup$

In classical electrodynamics, the process of how much light refracts, passing through the glass, and how much light reflects, is determined by the Huygens-Fresnel principle.

This principle, named after Christiaan Huygens and Augustin-Jean Fresnel, is a method of analyzing the wave propagation patterns of light, especially in diffraction and refraction. It states that every unobstructed point on a wave-front emanates secondary spherical waves in all directions. Hence, the net light amplitude at a given point is the vector sum of all wave amplitudes at that point. This principle makes it very useful in visualizing what happens during light diffraction.

Although, as Alex says in his answer, you can use the QFT approach, I would like to provide an alternative answer, using classical, (that is not quantum based) reasoning. It's just easier, for me anyway, to understand :), and, hopefully, to answer.

From Wikipedia: Fresnel Equations

In classical electrodynamics, light is considered as an electromagnetic wave, which is described by Maxwell's equations. Light waves incident on a material induce small oscillations of polarisation in the individual atoms (or oscillation of electrons, in metals), causing each particle to radiate a small secondary wave in all directions, like a dipole antenna. All these waves add up to give specular reflection and refraction, according to the Huygens–Fresnel principle.

In the case of dielectrics such as glass, the electric field of the light acts on the electrons in the material, and the moving electrons generate fields and become new radiators. The refracted light in the glass is the combination of the forward radiation of the electrons and the incident light. The reflected light is the combination of the backward radiation of all of the electrons

When light moves from a medium of a given refractive index n1 into a second medium with refractive index n2, both reflection and refraction of the light may occur. The Fresnel equations describe what fraction of the light is reflected and what fraction is refracted (i.e., transmitted). They also describe the phase shift of the reflected light.

The incident light is polarized with its electric field perpendicular to the plane containing the incident, reflected, and refracted rays. This plane is called the plane of incidence; it is the plane of the diagram below. The light is said to be s-polarized. The incident light is polarized with its electric field parallel to the plane of incidence. Such light is described as p-polarized.

Snell's Law

An incident light ray IO strikes the interface between two media of refractive indices n1 and n2 at point O. Part of the ray is reflected as ray OR and part refracted as ray OT. The angles that the incident, reflected and refracted rays make to the normal of the interface are given as θi, θr and θt, respectively.

The relationship between these angles is given by the law of reflection and Snell's law:

The fraction of the incident power that is reflected from the interface is given by the reflectance or reflectivity R and the fraction that is refracted is given by the transmittance or transmissivity T (unrelated to the transmission through a medium).

If you can follow the math, mostly just trigonometry, you can get the proportion of light passing through the glass, and the proportion that reflects, here:

Refraction and Reflection Coefficients

It's not easy , for me at least, to immediately find an answer to your question based on QFT, as most of the QFT explanations seem to deal with mirrors and how they reflect light, rather than explain how some goes through the glass and some reflects, (as in your particular question), but a good explanation, which is basically a copy of Feymann's book, can be found here:

Light Reflection

Just a thought. Is this due to atoms of different substances like water,glass or wood etc curve spacetime differently and thus it influences how photons interact with matter? i.e. some photons reflect and some refract.

I would say no to that reasoning, spacetime is not curved enough in a plate of glass to have a significant effect.

$\endgroup$
1
$\begingroup$

But I want to know, if an incoming photon is unable to excite an electron, then why not all the photons pass through glass? i.e. photons should not reflect off glass, all the photons should pass through glass.

While I don't know the exact details of reflection off glass (it is related to solid state physics from what I recall, with plasmon and such pseudoparticles), this I know.

While individual atoms do only accept particular energy levels, this does not apply in general, due to several reasons :

  • The most basic is the time-energy uncertainty, $\Delta t \Delta x \geq \hbar/2$. All energy levels will accept with some probability very close energies to their own.
  • The Doppler effect applies at temperatures $> 0 K$, as atoms will move around. From their perspective, the energy of the incoming photon will be redshifted or blue shifted, hence from the outside perspective, their energy levels will shift. As this effect is a distribution of probabilities, this further increase the range of energies that the material can absorb
  • Atoms interacting with each other will create new energy levels. The most basic example is that two hydrogen atoms bound together will double each energy levels, the new levels being respectively slightly superior and slightly inferior to the free atom's levels. In large solids, this can, in conjunction with the previous effects, lead to the creation of energy bands which are continuous.
  • Probably not the case in glass, but in some solids, the outermost electrons can be treated as free, and hence do not really have energy levels.
$\endgroup$
0
$\begingroup$

Instead of thinking about it deterministically think about it terms of the probability that a photon will be reflected. These probabilities can be calculated using Quantum Electrodynamics. Check out this lecture series by Richard Feynman for a detailed explanation. https://www.youtube.com/watch?v=eLQ2atfqk2c

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.