# Why does light seem to reflect preferentially off surfaces and interfaces even when the stuff in question is non-opaque?

Glass is said to reflect about 4% of incident light. The other 96% is either absorbed or transmitted. That 4% seems to reflect from the surface of the glass. From the perspective of a 'stream of photons', what's special about the surface as opposed to all the glass below the surface? (Alternatively what's special about the 4% that reflects - though I'm guessing that's not it).

• What prior research did you do on this question? – my2cts May 30 at 8:47
• It's the presence of an abrupt change in conditions that causes reflection. – Andrew Steane May 30 at 9:56
• @AndrewSteane actually, it's the presence of a change, not necessarily abrupt: even a gradual change would in general result in reflection. There do exist reflectionless potentials, at least for Schrödinger's equation (maybe for Maxwell's too), but they are very special ones. – Ruslan May 30 at 11:34
• Feynman’s QED book talks a lot about reflection off of glass. Also, a laser seems to always reflect way more than 4%. It looks to be closer to 40% consistently. – Lambda May 30 at 15:08
• @lamba where did you get the figure of 40%? It cannot apply to optical reflection from an air to glass interface u less at grazing incidence. – my2cts May 30 at 18:49

Light is an electromagnetic wave. When light falls on glass then the charges on the glass move at the same frequency as the incoming electric field. These oscillating charges contribute to the light phenomenon in a way described by Fresnel's solutions to Maxwell's equations. The result is that 4% of the light is reflected and the rest transmitted into the glass. This is true for perpendicular incidence. At skew angles the number is different. The 4% follows from the material refractive index of about 1.5. Different materials have different RI values.

From the perspective of a 'stream of photons', what's special about the surface as opposed to all the glass below the surface?

Photons are quantum mechanical entities,i.e.are described by a wavefunction . So it is not a "stream of photons" that impinges on surfaces but a quantum mechanical superposition of zillions of photon wavefunctions which build up the classical electromagnetic fields called "light" . The way this can be derived is shown in the discussion here . It is much simpler to trust the mathematics and deal with the classical electromagnetic reflection refraction and absorption.

The special about a surface is that it is a boundary condition that defines the limits of the bulk. A single photon if there is an organized lattice of a transparent material , will be described by a scatter "Photon + lattice" and it will either scatter elastically: go through without interaction, when the overall colors are transmitted through the material, or be absorbed (transparent materials have little absorption).

Elastic scattering also happens in the backwards direction so that would be called "reflection". For images to be transmitted the phases of the zillions of photons have to be retained, as well as the energy, so there are no color changes.

Reflections from metallic surfaces involve the surface lattice configurations, as metals are not transparent to optical frequencies.

To simplify it a bit, the reflection occurs at the interface between two mediums, and the reflectivity is given by $$r=\frac{n_1-n_2}{n_1+n_2}$$ where $$n_1$$ and $$n_2$$ are the refractive indices of the two mediums. If the mediums are the same, such as a wave travelling through a medium, there is no reflections with this model as the refractive indices would be the same.

Of course anna v's answer is right, I would like to add a few things, my answer is more QM.

In this case it is very important to differentiate between visible and non-visible photons. In your case, you are talking about visible photons, so that is what I am going to discuss.

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 keeps part of its energy and changes angle

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

Now all three happen always, it is just QM, and its all about probabilities, the ratio of these three are different.

Now in the case of glass, we can talk about:

1. reflection, that is mostly 1. elastic scattering, that is why when you stand in front of a glass window, you will see an almost perfect mirror image of yourself too, those 4% of photons that you talk about will create that mirror image. It is elastic scattering, because that is the only way to keep the photons' absolute energy, phase, and relative energy and phase, and relative angle. It is important to understand that even with reflection, some of those 4% photons will inelastically scatter, or be absorbed, but those ratio in the 4% is very little

2. refraction, that is mostly 1. elastic scattering too, in the case of visible light, as you say, 96% of photons will be rafracted, that is, they will enter the glass. It is elastic scattering, this is the only way to keep the absolute energy, phase of the photons, and relative energy, phase and angle. This is the only way why you can see an almost perfect mirror image of what is inside the shop from outside, through the glass. These photons pass through the glass, and keep the mirror image. It is very important to understand that of the 96%, most of them will be elastically scattered, and keep the mirror(picture of what is on the other side of the glass) image, but some photons will be inelastically scattered or absorbed, but the ratio of these is very little.

3. absorption this happens with very little ratio of the visible light, and most of the absorption is with non-visible light, transferring energy to the vibrational energies of the molecules in the glass lattice, heats up the glass

Now you are asking why is the refraction (image through the glass) different from the reflection (mirror image). As you see, both are elastic scattering. Then what is the difference? It is the angle only. No other difference.

Why are some, 4%'s angle backwards, and 96%'s angle forwards? It is the lattice structure. Glass does have 4-96 ratio, metal has 99-0 ratio. Why? because the lattice structure is different. When the visible light photons interact with the atoms in the lattice, the photons partial waves interfere with each other, and create constructive and destructive interference. Just like in the double slit experiment.

Some photons in glass (4%) will create destructive interference in every direction, except the backwards angle. The backwards angle will be for these 4% constructive interference, and will create a mirror image.

Now some photons in glass (96%) will create destructive interference in every direction, except the forward angle. The forward angle will be for these 96% constructive and will keep the image through the glass.

It is very important to understand that the photons as per QM are going in every direction. But it is all probabilities, and each photon will create constructive interference in only one angle (with glass, visible light, it could be mostly either backwards or forwards), and destructive interference in all the other angles.