Polarization by Reflection

By now you can probably guess that Polaroid sunglasses cut the glare in reflected light because that light is polarized. You can check this for yourself by holding Polaroid sunglasses in front of you and rotating them while looking at light reflected from water or glass. As you rotate the sunglasses, you will notice the light gets bright and dim, but not completely black. This implies the reflected light is partially polarized and cannot be completely blocked by a polarizing filter.

This figure illustrates what happens when unpolarized light is reflected from a surface. Vertically polarized light is preferentially refracted at the surface, so that the reflected light is left more horizontally polarized. The reasons for this phenomenon are beyond the scope of this text, but a convenient mnemonic for remembering this is to imagine the polarization direction to be like an arrow. Vertical polarization would be like an arrow perpendicular to the surface and would be more likely to stick and not be reflected. Horizontal polarization is like an arrow bouncing on its side and would be more likely to be reflected. Sunglasses with vertical axes would then block more reflected light than unpolarized light from other sources.

$The schematic shows a block of glass in air. A ray labeled unpolarized light starts at the upper left and impinges on the center of the block. Centered on this ray is a symmetric star burst pattern of double headed arrows. From this point where this ray hits the glass block there emerges a reflected ray that goes up and to the right and a refracted ray that goes down and to the right. Both of these rays are labeled partially polarized light. The reflected ray has evenly spaced large black dots on it that are labeled perpendicular to plane of paper. Centered on each black dot is a double headed arrow that is rather short and is perpendicular to the ray. The refracted ray also has evenly spaced dots, but they are much smaller. Centered on each of these small black dots are quite large doubled headed arrows that are perpendicular to the refracted ray.$

Polarization by reflection. Unpolarized light has equal amounts of vertical and horizontal polarization. After interaction with a surface, the vertical components are preferentially absorbed or refracted, leaving the reflected light more horizontally polarized. This is akin to arrows striking on their sides bouncing off, whereas arrows striking on their tips go into the surface.

Since the part of the light that is not reflected is refracted, the amount of polarization depends on the indices of refraction of the media involved. It can be shown that reflected light is completely polarized at a angle of reflection ${\theta }_{\text{b}}$, given by

$\text{tan}\phantom{\rule{0.25em}{0ex}}{\theta }_{\text{b}}=\cfrac{{n}_{2}}{{n}_{1}}\text{,}$

where ${n}_{1}$ is the medium in which the incident and reflected light travel and ${n}_{2}$ is the index of refraction of the medium that forms the interface that reflects the light. This equation is known as Brewster’s law, and ${\theta }_{\text{b}}$ is known as Brewster’s angle, named after the 19th-century Scottish physicist who discovered them.

Things Great and Small: Atomic Explanation of Polarizing Filters

Polarizing filters have a polarization axis that acts as a slit. This slit passes electromagnetic waves (often visible light) that have an electric field parallel to the axis. This is accomplished with long molecules aligned perpendicular to the axis as shown in this figure.

The schematic shows a stack of long identical horizontal molecules. A vertical axis is drawn over the molecules.

Long molecules are aligned perpendicular to the axis of a polarizing filter. The component of the electric field in an EM wave perpendicular to these molecules passes through the filter, while the component parallel to the molecules is absorbed.

This figure illustrates how the component of the electric field parallel to the long molecules is absorbed. An electromagnetic wave is composed of oscillating electric and magnetic fields. The electric field is strong compared with the magnetic field and is more effective in exerting force on charges in the molecules. The most affected charged particles are the electrons in the molecules, since electron masses are small. If the electron is forced to oscillate, it can absorb energy from the EM wave. This reduces the fields in the wave and, hence, reduces its intensity. In long molecules, electrons can more easily oscillate parallel to the molecule than in the perpendicular direction. The electrons are bound to the molecule and are more restricted in their movement perpendicular to the molecule. Thus, the electrons can absorb EM waves that have a component of their electric field parallel to the molecule. The electrons are much less responsive to electric fields perpendicular to the molecule and will allow those fields to pass. Thus the axis of the polarizing filter is perpendicular to the length of the molecule.

The figure contains two schematics. The first schematic shows a long molecule. An EM wave goes through the molecule. The ray of the EM wave is at ninety degrees to the molecular axis and the electric field of the EM wave oscillates along the molecular axis. After passing the long molecule, the magnitude of the oscillations of the EM wave are significantly reduced. The second schematic shows a similar drawing, except that the EM wave oscillates perpendicular to the axis of the long molecule. After passing the long molecule, the magnitude of the oscillation of the EM wave is unchanged.

Artist’s conception of an electron in a long molecule oscillating parallel to the molecule. The oscillation of the electron absorbs energy and reduces the intensity of the component of the EM wave that is parallel to the molecule.

Example: Calculating Polarization by Reflection

(a) At what angle will light traveling in air be completely polarized horizontally when reflected from water? (b) From glass?

Strategy

All we need to solve these problems are the indices of refraction. Air has ${n}_{1}=1.00,$ water has ${n}_{2}=1\text{.}\text{333,}$ and crown glass has ${n\prime }_{2}=1.520$. The equation $\text{tan}\phantom{\rule{0.25em}{0ex}}{\theta }_{\text{b}}=\cfrac{{n}_{2}}{{n}_{1}}$ can be directly applied to find ${\theta }_{\text{b}}$ in each case.

Solution for (a)

Putting the known quantities into the equation

$\text{tan}\phantom{\rule{0.25em}{0ex}}{\theta }_{\text{b}}=\cfrac{{n}_{2}}{{n}_{1}}$

gives

$\text{tan}\phantom{\rule{0.25em}{0ex}}{\theta }_{\text{b}}=\cfrac{{n}_{2}}{{n}_{1}}=\cfrac{1.333}{1.00}=1\text{.}\text{333.}$

Solving for the angle ${\theta }_{\text{b}}$ yields

${\theta }_{\text{b}}={\text{tan}}^{-1}\phantom{\rule{0.25em}{0ex}}1\text{.}\text{333}=\text{53}\text{.}1º.$

Solution for (b)

Similarly, for crown glass and air,

${\text{tan}\phantom{\rule{0.25em}{0ex}}\theta \prime }_{\text{b}}=\cfrac{{n\prime }_{2}}{{n}_{1}}=\cfrac{1.520}{1.00}=1\text{.}\text{52.}$

Thus,

${\theta \prime }_{\text{b}}={\text{tan}}^{-1\phantom{\rule{0.25em}{0ex}}}1\text{.}\text{52}=\text{56.7º.}$

Discussion

Light reflected at these angles could be completely blocked by a good polarizing filter held with its axis vertical. Brewster’s angle for water and air are similar to those for glass and air, so that sunglasses are equally effective for light reflected from either water or glass under similar circumstances. Light not reflected is refracted into these media. So at an incident angle equal to Brewster’s angle, the refracted light will be slightly polarized vertically. It will not be completely polarized vertically, because only a small fraction of the incident light is reflected, and so a significant amount of horizontally polarized light is refracted.

Polarization by Reflection