Why is the light reflected from the metal surface not polarized light?

Visible light source (sun, incandescent lamp, etc.). ) is non-polarized in daily life. Unpolarized light is randomly composed of polarized light in different directions. A simple example of polarized light is monochromatic plane wave, which we can simply imagine as sine wave. The principle of polarizer is to filter or absorb most unpolarized light, and only let the light needed by photographer pass through polarizer. Figure 1 directly shows the principle of polarizer: unpolarized light (2) emitted by incandescent lamp (1) is filtered by polarizer (3) and becomes polarized light (4) oscillating in one direction.

Figure 1: principle of polarizer (source: polarization)

There are two kinds of polarizers for photography, one is linear polarizer and the other is CPL (circular? Polarization? Light)-circular polarizer. The principle of these two polarizers can be explained by figure 1 Because this paper focuses on the metal reflection light, the difference between the two polarizers will not be expanded, just a passage from Wikipedia:

There are two kinds of polarizers: linear polarizer (PL) and circular polarizer (CPL filter). Old-fashioned cameras usually use linear polarizers, but they can't be used with SLR cameras because of mirrors and optical/focal spectrometers. A circular polarizer can be used in any kind of camera. (Source: Ticket Dealer)

Now let's talk about the difference between metal reflection and "general plane" (glass, water, wood) reflection. Figure 2 is a reflection diagram of light that everyone has seen in middle school. But this picture adds something: incident light? Polarized light (unpolarized light) in any direction can be decomposed into a part that oscillates parallel to the reflecting surface (gray) and a part that oscillates perpendicular to the reflecting surface.

Figure 2: reflection of light (source: see note)

Whether a beam of unpolarized light is reflected on a metal surface or a "general plane" can be described by Figure 2. Having said that, there is still a difference. In Figure 3, we might as well classify "general plane" as insulator more strictly. r？ Is the reflectivity of light on the surface of an object, t? Is the incident angle of light. p？ Is the polarized light parallel to the reflecting surface of light (i.e. in Figure 2)? s？ It is polarized light perpendicular to the reflective surface of light (i.e., the reflective surface of fig. 2). How is Figure 3 drawn? ——? Fresnel equation? (Fresnel equation). What the equation looks like is not very important for understanding the problem, so skip it.

Figure 3: Relationship between light reflectivity and incident angle (source: see note)

The problem is solved here: the P part reflected by the insulator is very few, and the photographer only needs to filter out the S part with a polarizer to get a satisfactory photo. As for the metal, no matter whether the S part or the P part is filtered, the other part of the light will still be reflected into the lens, so no matter how the polarizer rotates, the reflection of the metal will not be filtered out.

Students who want to find out why the metal reflectivity is so high can continue to look down here and see the supplementary question: Why is the light reflected by metal not polarized light? I want to say that the light reflected by insulators is not polarized light in most cases. Figure 3 illustrates everything. In the figure, only the light reflected by the insulator is polarized light of about 50 ~ 60, because only the S part is left in this angle range. This angle is called. Brewster pt? (Bruce features/? Deflection angle). The Chinese name declination angle vividly describes the characteristics of this angle. Another phenomenon is that when the incident angle is equal to the deflection angle, the included angle between the reflected light and the refracted light is 90, which is also the reason for the deflection angle. Students who want to explore further can refer to the keyword "Hertz dipole". So why isn't the light reflected by metal polarized? This question can still be answered by what I said above: the P part reflected by the insulator is very small, and the photographer only needs to filter out the S part with a polarizer to get a satisfactory photo.

Go on to explain why the metal reflectivity is so high.

(Why do you want to explain? Figure 3 is the key. If the P part of the metal can be moved down, then filtering the S part with a polarizer can filter out most of the metal reflection, just like treating an insulator. Therefore, the polarizer can not filter out the reflection of metal, and the key is the high reflectivity of S and P parts of metal. )

Druid model? (Druid model) tells us that the electrons in the metal don't pay around the atom like the electrons in the insulator, but run around. Suppose an independent electron is put into a beam of electromagnetic waves (light), then the electron will follow the electromagnetic field to make regular oscillation motion, and the energy of the electron itself will remain unchanged. However, if the electrons in the metal are irradiated by electromagnetic waves (light), the electrons will collide with the surrounding atoms or ions when they oscillate. Every time they collide, the electrons gain more energy and move in different directions.

Review insulator: When the incident angle is equal to the polarization angle, the P part will not be reflected at all. So where is part p? -Concentrate. So absorption is the key. The incident light will not only be absorbed, but also refracted through the object. )

It is difficult for metals to absorb the energy of light, because electrons in metals can only absorb energy when they collide with atoms or ions. If electrons want to absorb more energy, there will be more collisions, which requires a high collision frequency. Fig. 4 shows the relationship between the reflectivity of light with an incident angle of 0.d egree. (vertical incidence) and the frequency on the aluminum surface. The dotted line is the numerical value calculated by the Druid model. It can be seen that the frequency is about equal to? At Hz, the reflectivity suddenly drops to 0. This frequency is called the plasma frequency. Its size is proportional to the electron density of the metal. So only high-frequency light can be absorbed by metal. The wavelength range of visible light of human eyes is about 400 to 800 nanometers (nanometers), which is converted into hertz. This frequency band is far from being absorbed by metals.

Fig. 4: Relationship between light reflectivity r of aluminum surface and frequency (incidence angle is 0) (source: see note).

* Note: Image source: Anwendungen der LaserTechnik 2. Vorlesung, Professor Reinhart Poprawe, LLT Institute of Laser Technology, RWT Aachen.