279. PolarizationLearning Objectives
Polaroid sunglasses are familiar to most of us. They have a special ability to cut the glare of light reflected from water or glass (see Figure 1). Polaroids have this ability because of a wave characteristic of light called polarization. What is polarization? How is it produced? What are some of its uses? The answers to these questions are related to the wave character of light. Light is one type of electromagnetic (EM) wave. As noted earlier, EM waves are transverse waves consisting of varying electric and magnetic fields that oscillate perpendicular to the direction of propagation (see Figure 2). There are specific directions for the oscillations of the electric and magnetic fields. To examine this further, consider the transverse waves in the ropes shown in Figure 3. The oscillations in one rope are in a vertical plane and are said to be The Sun and many other light sources produce waves that are randomly polarized (see Figure 4). Such light is said to be Figure 6 shows the effect of two polarizing filters on originally unpolarized light. The first filter polarizes the light along its axis. When the axes of the first and second filters are aligned (parallel), then all of the polarized light passed by the first filter is also passed by the second. If the second polarizing filter is rotated, only the component of the light parallel to the second filter’s axis is passed. When the axes are perpendicular, no light is passed by the second. Only the component of the EM wave parallel to the axis of a filter is passed. Let us call the angle between the direction of polarization and the axis of a filter $\theta $. If the electric field has an amplitude $E$, then the transmitted part of the wave has an amplitude $E\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta $ (see Figure 7). Since the intensity of a wave is proportional to its amplitude squared, the intensity $I$ of the transmitted wave is related to the incident wave by $$I={I}_{0}\phantom{\rule{0.25em}{0ex}}{\text{cos}}^{2}\phantom{\rule{0.25em}{0ex}}\theta \text{,}$$where ${I}_{0}$ is the intensity of the polarized wave before passing through the filter. (The above equation is known as Malus’s law.) Example 1: Calculating Intensity Reduction by a Polarizing FilterWhat angle is needed between the direction of polarized light and the axis of a polarizing filter to reduce its intensity by $\text{90}\text{.}\mathrm{0\%}\text{}$? Strategy When the intensity is reduced by $\text{90}\text{.}\mathrm{0\%}\text{}$, it is $\text{10}\text{.}\mathrm{0\%}\text{}$ or 0.100 times its original value. That is, $I=0\text{.}\text{100}{I}_{0}$. Using this information, the equation $I={I}_{0}\phantom{\rule{0.25em}{0ex}}{\text{cos}}^{2}\phantom{\rule{0.25em}{0ex}}\theta $ can be used to solve for the needed angle. Solution Solving the equation $I={I}_{0}\phantom{\rule{0.25em}{0ex}}{\text{cos}}^{2}\phantom{\rule{0.25em}{0ex}}\theta $ for $\text{cos}\phantom{\rule{0.25em}{0ex}}\theta $ and substituting with the relationship between $I$ and ${I}_{0}$ gives $$\text{cos}\phantom{\rule{0.25em}{0ex}}\theta =\sqrt{\frac{I}{{I}_{0}}}=\sqrt{\frac{0\text{.}\text{100}{I}_{0}}{{I}_{0}}}=0\text{.}\text{3162.}$$Solving for $\theta $ yields $$\theta ={\text{cos}}^{1\phantom{\rule{0.25em}{0ex}}}0\text{.}\text{3162}=\text{71}\text{.}\mathrm{6\xba.}$$Discussion A fairly large angle between the direction of polarization and the filter axis is needed to reduce the intensity to $\text{10}\text{.}\mathrm{0\%}\text{}$ of its original value. This seems reasonable based on experimenting with polarizing films. It is interesting that, at an angle of $\text{45\xba}$, the intensity is reduced to $\text{50\%}\text{}$ of its original value (as you will show in this section’s Problems & Exercises). Note that $\text{71}\text{.}\mathrm{6\xba}$ is $\text{18}\text{.}\mathrm{4\xba}$ from reducing the intensity to zero, and that at an angle of $\text{18}\text{.}\mathrm{4\xba}$ the intensity is reduced to $\text{90}\text{.}\mathrm{0\%}\text{}$ of its original value (as you will also show in Problems & Exercises), giving evidence of symmetry. Polarization by ReflectionBy 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. Figure 8 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. 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 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 Things Great and Small: Atomic Explanation of Polarizing FiltersPolarizing 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 Figure 9. Figure 10 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. Example 2: 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 ${n}_{1}=\mathrm{1.00,}$ water has ${n}_{2}=1\text{.}\text{333,}$ and crown glass has ${n\prime}_{2}=1.520$. The equation $\text{tan}\phantom{\rule{0.25em}{0ex}}{\theta}_{\text{b}}=\frac{{n}_{2}}{{n}_{1}}$ can be directly applied to find ${\theta}_{\text{b}}$ in each case. Solution for (a) Putting the known quantities into the equation $$\text{tan}\phantom{\rule{0.25em}{0ex}}{\theta}_{\text{b}}=\frac{{n}_{2}}{{n}_{1}}$$gives $$\text{tan}\phantom{\rule{0.25em}{0ex}}{\theta}_{\text{b}}=\frac{{n}_{2}}{{n}_{1}}=\frac{1.333}{1.00}=1\text{.}\text{333.}$$Solving for the angle ${\theta}_{\text{b}}$ yields $${\theta}_{\text{b}}={\text{tan}}^{1}\phantom{\rule{0.25em}{0ex}}1\text{.}\text{333}=\text{53}\text{.}\mathrm{1\xba}.$$Solution for (b) Similarly, for crown glass and air, $${\text{tan}\phantom{\rule{0.25em}{0ex}}\theta \prime}_{\text{b}}=\frac{{n\prime}_{2}}{{n}_{1}}=\frac{1.520}{1.00}=1\text{.}\text{52.}$$Thus, $${\theta \prime}_{\text{b}}={\text{tan}}^{1\phantom{\rule{0.25em}{0ex}}}1\text{.}\text{52}=\text{56.7\xba.}$$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 ScatteringIf you hold your Polaroid sunglasses in front of you and rotate them while looking at blue sky, you will see the sky get bright and dim. This is a clear indication that light scattered by air is partially polarized. Figure 11 helps illustrate how this happens. Since light is a transverse EM wave, it vibrates the electrons of air molecules perpendicular to the direction it is traveling. The electrons then radiate like small antennae. Since they are oscillating perpendicular to the direction of the light ray, they produce EM radiation that is polarized perpendicular to the direction of the ray. When viewing the light along a line perpendicular to the original ray, as in Figure 11, there can be no polarization in the scattered light parallel to the original ray, because that would require the original ray to be a longitudinal wave. Along other directions, a component of the other polarization can be projected along the line of sight, and the scattered light will only be partially polarized. Furthermore, multiple scattering can bring light to your eyes from other directions and can contain different polarizations. Photographs of the sky can be darkened by polarizing filters, a trick used by many photographers to make clouds brighter by contrast. Scattering from other particles, such as smoke or dust, can also polarize light. Detecting polarization in scattered EM waves can be a useful analytical tool in determining the scattering source. There is a range of optical effects used in sunglasses. Besides being Polaroid, other sunglasses have colored pigments embedded in them, while others use nonreflective or even reflective coatings. A recent development is photochromic lenses, which darken in the sunlight and become clear indoors. Photochromic lenses are embedded with organic microcrystalline molecules that change their properties when exposed to UV in sunlight, but become clear in artificial lighting with no UV. TakeHome Experiment: PolarizationFind Polaroid sunglasses and rotate one while holding the other still and look at different surfaces and objects. Explain your observations. What is the difference in angle from when you see a maximum intensity to when you see a minimum intensity? Find a reflective glass surface and do the same. At what angle does the glass need to be oriented to give minimum glare? Liquid Crystals and Other Polarization Effects in MaterialsWhile you are undoubtedly aware of liquid crystal displays (LCDs) found in watches, calculators, computer screens, cellphones, flat screen televisions, and other myriad places, you may not be aware that they are based on polarization. Liquid crystals are so named because their molecules can be aligned even though they are in a liquid. Liquid crystals have the property that they can rotate the polarization of light passing through them by $\text{90\xba}$. Furthermore, this property can be turned off by the application of a voltage, as illustrated in Figure 12. It is possible to manipulate this characteristic quickly and in small welldefined regions to create the contrast patterns we see in so many LCD devices. In flat screen LCD televisions, there is a large light at the back of the TV. The light travels to the front screen through millions of tiny units called pixels (picture elements). One of these is shown in Figure 12 (a) and (b). Each unit has three cells, with red, blue, or green filters, each controlled independently. When the voltage across a liquid crystal is switched off, the liquid crystal passes the light through the particular filter. One can vary the picture contrast by varying the strength of the voltage applied to the liquid crystal. Many crystals and solutions rotate the plane of polarization of light passing through them. Such substances are said to be Glass and plastic become optically active when stressed; the greater the stress, the greater the effect. Optical stress analysis on complicated shapes can be performed by making plastic models of them and observing them through crossed filters, as seen in Figure 14. It is apparent that the effect depends on wavelength as well as stress. The wavelength dependence is sometimes also used for artistic purposes. Another interesting phenomenon associated with polarized light is the ability of some crystals to split an unpolarized beam of light into two. Such crystals are said to be Section Summary
Conceptual QuestionsExercise 1Under what circumstances is the phase of light changed by reflection? Is the phase related to polarization? Exercise 2Can a sound wave in air be polarized? Explain. Exercise 3No light passes through two perfect polarizing filters with perpendicular axes. However, if a third polarizing filter is placed between the original two, some light can pass. Why is this? Under what circumstances does most of the light pass? Exercise 4Explain what happens to the energy carried by light that it is dimmed by passing it through two crossed polarizing filters. Exercise 5When particles scattering light are much smaller than its wavelength, the amount of scattering is proportional to $1/{\lambda}^{4}$. Does this mean there is more scattering for small $\lambda $ than large $\lambda $? How does this relate to the fact that the sky is blue? Exercise 6Using the information given in the preceding question, explain why sunsets are red. Exercise 7When light is reflected at Brewster’s angle from a smooth surface, it is $\text{100\%}\text{}$ polarized parallel to the surface. Part of the light will be refracted into the surface. Describe how you would do an experiment to determine the polarization of the refracted light. What direction would you expect the polarization to have and would you expect it to be $\text{100\%}\text{}$? Problems & ExercisesExercise 1What angle is needed between the direction of polarized light and the axis of a polarizing filter to cut its intensity in half? Show/Hide Solution Solution$\text{45}\text{.}\mathrm{0\xba}$ Exercise 2The angle between the axes of two polarizing filters is $\text{45}\text{.}\mathrm{0\xba}$. By how much does the second filter reduce the intensity of the light coming through the first? Exercise 3If you have completely polarized light of intensity $\text{150 W}/{\text{m}}^{2}$, what will its intensity be after passing through a polarizing filter with its axis at an $\text{89}\text{.}\mathrm{0\xba}$ angle to the light’s polarization direction? Show/Hide Solution Solution$\text{45}\text{.}7\phantom{\rule{0.25em}{0ex}}{\text{mW/m}}^{2}$ Exercise 4What angle would the axis of a polarizing filter need to make with the direction of polarized light of intensity $1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}{\text{kW/m}}^{2}$ to reduce the intensity to $10\text{.}0\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}$? Exercise 5At the end of Example 1, it was stated that the intensity of polarized light is reduced to $\text{90}\text{.}\mathrm{0\%}\text{}$ of its original value by passing through a polarizing filter with its axis at an angle of $\text{18}\text{.}\mathrm{4\xba}$ to the direction of polarization. Verify this statement. Show/Hide Solution Solution$\text{90}\text{.}\mathrm{0\%}\text{}$ Exercise 6Show that if you have three polarizing filters, with the second at an angle of $\text{45\xba}$ to the first and the third at an angle of $\text{90}\text{.}\mathrm{0\xba}$ to the first, the intensity of light passed by the first will be reduced to $\text{25}\text{.}\mathrm{0\%}\text{}$ of its value. (This is in contrast to having only the first and third, which reduces the intensity to zero, so that placing the second between them increases the intensity of the transmitted light.) Exercise 7Prove that, if $I$ is the intensity of light transmitted by two polarizing filters with axes at an angle $\theta $ and $I\prime $ is the intensity when the axes are at an angle $\text{90.0\xba}\mathrm{\theta ,}$ then $I+I\prime ={I}_{\mathrm{0,}}$ the original intensity. (Hint: Use the trigonometric identities $\text{cos}\phantom{\rule{0.25em}{0ex}}(\mathrm{90.0\xba}\theta )=\text{sin}\phantom{\rule{0.25em}{0ex}}\theta $ and ${\text{cos}}^{2}\phantom{\rule{0.25em}{0ex}}\theta +{\text{sin}}^{2}\phantom{\rule{0.25em}{0ex}}\theta =1.$) Show/Hide Solution Solution${I}_{0}$ Exercise 8At what angle will light reflected from diamond be completely polarized? Exercise 9What is Brewster’s angle for light traveling in water that is reflected from crown glass? Show/Hide Solution Solution$\text{48}\text{.}\mathrm{8\xba}$ Exercise 10A scuba diver sees light reflected from the water’s surface. At what angle will this light be completely polarized? Exercise 11At what angle is light inside crown glass completely polarized when reflected from water, as in a fish tank? Show/Hide Solution Solution$\text{41}\text{.}\mathrm{2\xba}$ Exercise 12Light reflected at $\text{55}\text{.}\mathrm{6\xba}$ from a window is completely polarized. What is the window’s index of refraction and the likely substance of which it is made? Exercise 13(a) Light reflected at $\text{62}\text{.}\mathrm{5\xba}$ from a gemstone in a ring is completely polarized. Can the gem be a diamond? (b) At what angle would the light be completely polarized if the gem was in water? Show/Hide Solution Solution(a) 1.92, not diamond (Zircon) (b) $\text{55}\text{.}\mathrm{2\xba}$ Exercise 14If ${\theta}_{\text{b}}$ is Brewster’s angle for light reflected from the top of an interface between two substances, and ${\theta \prime}_{\text{b}}$ is Brewster’s angle for light reflected from below, prove that ${\theta}_{\text{b}}+{\theta \prime}_{\text{b}}=\text{90}\text{.}\mathrm{0\xba.}$ Exercise 15Integrated Concepts If a polarizing filter reduces the intensity of polarized light to $\text{50}\text{.}\mathrm{0\%}\text{}$ of its original value, by how much are the electric and magnetic fields reduced? Show/Hide Solution Solution${B}_{2}=0\text{.}\text{707}\phantom{\rule{0.25em}{0ex}}{B}_{1}$ Exercise 16Integrated Concepts Suppose you put on two pairs of Polaroid sunglasses with their axes at an angle of $\text{15}\text{.}\mathrm{0\xba}$. How much longer will it take the light to deposit a given amount of energy in your eye compared with a single pair of sunglasses? Assume the lenses are clear except for their polarizing characteristics. Exercise 17Integrated Concepts (a) On a day when the intensity of sunlight is $1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{kW}/{\text{m}}^{2}$, a circular lens 0.200 m in diameter focuses light onto water in a black beaker. Two polarizing sheets of plastic are placed in front of the lens with their axes at an angle of $\text{20}\text{.}\mathrm{0\xba.}$ Assuming the sunlight is unpolarized and the polarizers are $\text{100\%}\text{}$ efficient, what is the initial rate of heating of the water in $\text{\xbaC}/\text{s}$, assuming it is $\text{80}\text{.}\mathrm{0\%}\text{}$ absorbed? The aluminum beaker has a mass of 30.0 grams and contains 250 grams of water. (b) Do the polarizing filters get hot? Explain. Show/Hide Solution Solution(a) $2.07\times {10}^{2}$ °C/s (b) Yes, the polarizing filters get hot because they absorb some of the lost energy from the sunlight.
