optical experiment polarization

Description This is a simulation intended to help visualize polarization. A polarizing filter has a particular transmission axis and only allows light waves aligned with that axis to pass through. In this simulation unpolarized waves pass through a vertical slit, leaving only their vertical components. This vertical transverse wave approaches a vertical slit. If the slit is rotated, only a component of the wave can pass through. If the slit is rotated 90 degrees, the wave is stopped completely. Use the check boxes to show one, two, or three slits and see what happens as the wave passes through two or three slits at various angles. Press the button at the bottom left to start or pause the animation.

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33 Polarization

33–1 The electric vector of light

In this chapter we shall consider those phenomena which depend on the fact that the electric field that describes the light is a vector. In previous chapters we have not been concerned with the direction of oscillation of the electric field, except to note that the electric vector lies in a plane perpendicular to the direction of propagation. The particular direction in this plane has not concerned us. We now consider those phenomena whose central feature is the particular direction of oscillation of the electric field.

In ideally monochromatic light, the electric field must oscillate at a definite frequency, but since the $x$-component and the $y$-component can oscillate independently at a definite frequency, we must first consider the resultant effect produced by superposing two independent oscillations at right angles to each other. What kind of electric field is made up of an $x$-component and a $y$-component which oscillate at the same frequency? If one adds to an $x$-vibration a certain amount of $y$-vibration at the same phase, the result is a vibration in a new direction in the $xy$-plane. Figure  33–1 illustrates the superposition of different amplitudes for the $x$-vibration and the $y$-vibration. But the resultants shown in Fig.  33–1 are not the only possibilities; in all of these cases we have assumed that the $x$-vibration and the $y$-vibration are in phase , but it does not have to be that way. It could be that the $x$-vibration and the $y$-vibration are out of phase.

When the $x$-vibration and the $y$-vibration are not in phase, the electric field vector moves around in an ellipse, and we can illustrate this in a familiar way. If we hang a ball from a support by a long string, so that it can swing freely in a horizontal plane, it will execute sinusoidal oscillations. If we imagine horizontal $x$- and $y$-coordinates with their origin at the rest position of the ball, the ball can swing in either the $x$- or $y$-direction with the same pendulum frequency. By selecting the proper initial displacement and initial velocity, we can set the ball in oscillation along either the $x$-axis or the $y$-axis, or along any straight line in the $xy$-plane. These motions of the ball are analogous to the oscillations of the electric field vector illustrated in Fig.  33–1 . In each instance, since the $x$-vibrations and the $y$-vibrations reach their maxima and minima at the same time, the $x$- and $y$-oscillations are in phase. But we know that the most general motion of the ball is motion in an ellipse, which corresponds to oscillations in which the $x$- and $y$-directions are not in the same phase. The superposition of $x$- and $y$-vibrations which are not in phase is illustrated in Fig.  33–2 for a variety of angles between the phase of the $x$-vibration and that of the $y$-vibration. The general result is that the electric vector moves around an ellipse. The motion in a straight line is a particular case corresponding to a phase difference of zero (or an integral multiple of $\pi$); motion in a circle corresponds to equal amplitudes with a phase difference of $90^\circ$ (or any odd integral multiple of $\pi/2$).

In Fig.  33–2 we have labeled the electric field vectors in the $x$- and $y$-directions with complex numbers, which are a convenient representation in which to express the phase difference. Do not confuse the real and imaginary components of the complex electric vector in this notation with the $x$- and $y$-coordinates of the field. The $x$- and $y$-coordinates plotted in Fig.  33–1 and Fig.  33–2 are actual electric fields that we can measure. The real and imaginary components of a complex electric field vector are only a mathematical convenience and have no physical significance.

Now for some terminology. Light is linearly polarized (sometimes called plane polarized) when the electric field oscillates on a straight line; Fig.  33–1 illustrates linear polarization. When the end of the electric field vector travels in an ellipse, the light is elliptically polarized . When the end of the electric field vector travels around a circle, we have circular polarization . If the end of the electric vector, when we look at it as the light comes straight toward us, goes around in a counterclockwise direction, we call it right-hand circular polarization. Figure  33–2 (g) illustrates right-hand circular polarization, and Fig.  33–2 (c) shows left-hand circular polarization. In both cases the light is coming out of the paper. Our convention for labeling left-hand and right-hand circular polarization is consistent with that which is used today for all the other particles in physics which exhibit polarization (e.g., electrons). However, in some books on optics the opposite conventions are used, so one must be careful.

We have considered linearly, circularly, and elliptically polarized light, which covers everything except for the case of unpolarized light. Now how can the light be unpolarized when we know that it must vibrate in one or another of these ellipses? If the light is not absolutely monochromatic, or if the $x$- and $y$-phases are not kept perfectly together, so that the electric vector first vibrates in one direction, then in another, the polarization is constantly changing. Remember that one atom emits during $10^{-8}$ sec, and if one atom emits a certain polarization, and then another atom emits light with a different polarization, the polarizations will change every $10^{-8}$ sec. If the polarization changes more rapidly than we can detect it, then we call the light unpolarized, because all the effects of the polarization average out. None of the interference effects of polarization would show up with unpolarized light. But as we see from the definition, light is unpolarized only if we are unable to find out whether the light is polarized or not.

33–2 Polarization of scattered light

The first example of the polarization effect that we have already discussed is the scattering of light. Consider a beam of light, for example from the sun, shining on the air. The electric field will produce oscillations of charges in the air, and motion of these charges will radiate light with its maximum intensity in a plane normal to the direction of vibration of the charges. The beam from the sun is unpolarized, so the direction of polarization changes constantly, and the direction of vibration of the charges in the air changes constantly. If we consider light scattered at $90^\circ$, the vibration of the charged particles radiates to the observer only when the vibration is perpendicular to the observer’s line of sight, and then light will be polarized along the direction of vibration. So scattering is an example of one means of producing polarization.

33–3 Birefringence

Another interesting effect of polarization is the fact that there are substances for which the index of refraction is different for light linearly polarized in one direction and linearly polarized in another. Suppose that we had some material which consisted of long, nonspherical molecules, longer than they are wide, and suppose that these molecules were arranged in the substance with their long axes parallel. Then what happens when the oscillating electric field passes through this substance? Suppose that because of the structure of the molecule, the electrons in the substance respond more easily to oscillations in the direction parallel to the axes of the molecules than they would respond if the electric field tries to push them at right angles to the molecular axis. In this way we expect a different response for polarization in one direction than for polarization at right angles to that direction. Let us call the direction of the axes of the molecules the optic axis . When the polarization is in the direction of the optic axis the index of refraction is different than it would be if the direction of polarization were at right angles to it. Such a substance is called birefringent . It has two refrangibilities, i.e., two indexes of refraction, depending on the direction of the polarization inside the substance. What kind of a substance can be birefringent? In a birefringent substance there must be a certain amount of lining up, for one reason or another, of unsymmetrical molecules. Certainly a cubic crystal, which has the symmetry of a cube, cannot be birefringent. But long needlelike crystals undoubtedly contain molecules that are asymmetric, and one observes this effect very easily.

Let us see what effects we would expect if we were to shine polarized light through a plate of a birefringent substance. If the polarization is parallel to the optic axis, the light will go through with one velocity; if the polarization is perpendicular to the axis, the light is transmitted with a different velocity. An interesting situation arises when, say, light is linearly polarized at $45^\circ$ to the optic axis. Now the $45^\circ$ polarization, we have already noticed, can be represented as a superposition of the $x$- and the $y$-polarizations of equal amplitude and in phase, as shown in Fig.  33–2 (a). Since the $x$- and $y$-polarizations travel with different velocities, their phases change at a different rate as the light passes through the substance. So, although at the start the $x$- and $y$-vibrations are in phase, inside the material the phase difference between $x$- and $y$-vibrations is proportional to the depth in the substance. As the light proceeds through the material the polarization changes as shown in the series of diagrams in Fig.  33–2 . If the thickness of the plate is just right to introduce a $90^\circ$ phase shift between the $x$- and $y$-polarizations, as in Fig.  33–2 (c), the light will come out circularly polarized. Such a thickness is called a quarter-wave plate, because it introduces a quarter-cycle phase difference between the $x$- and the $y$-polarizations. If linearly polarized light is sent through two quarter-wave plates, it will come out plane-polarized again, but at right angles to the original direction, as we can see from Fig.  33–2 (e).

One can easily illustrate this phenomenon with a piece of cellophane. Cellophane is made of long, fibrous molecules, and is not isotropic, since the fibers lie preferentially in a certain direction. To demonstrate birefringence we need a beam of linearly polarized light, and we can obtain this conveniently by passing unpolarized light through a sheet of polaroid. Polaroid, which we will discuss later in more detail, has the useful property that it transmits light that is linearly polarized parallel to the axis of the polaroid with very little absorption, but light polarized in a direction perpendicular to the axis of the polaroid is strongly absorbed. When we pass unpolarized light through a sheet of polaroid, only that part of the unpolarized beam which is vibrating parallel to the axis of the polaroid gets through, so that the transmitted beam is linearly polarized. This same property of polaroid is also useful in detecting the direction of polarization of a linearly polarized beam, or in determining whether a beam is linearly polarized or not. One simply passes the beam of light through the polaroid sheet and rotates the polaroid in the plane normal to the beam. If the beam is linearly polarized, it will not be transmitted through the sheet when the axis of the polaroid is normal to the direction of polarization. The transmitted beam is only slightly attenuated when the axis of the polaroid sheet is rotated through $90^\circ$. If the transmitted intensity is independent of the orientation of the polaroid, the beam is not linearly polarized.

To demonstrate the birefringence of cellophane, we use two sheets of polaroid, as shown in Fig.  33–3 . The first gives us a linearly polarized beam which we pass through the cellophane and then through the second polaroid sheet, which serves to detect any effect the cellophane may have had on the polarized light passing through it. If we first set the axes of the two polaroid sheets perpendicular to each other and remove the cellophane, no light will be transmitted through the second polaroid. If we now introduce the cellophane between the two polaroid sheets, and rotate the sheet about the beam axis, we observe that in general the cellophane makes it possible for some light to pass through the second polaroid. However, there are two orientations of the cellophane sheet, at right angles to each other, which permit no light to pass through the second polaroid. These orientations in which linearly polarized light is transmitted through the cellophane with no effect on the direction of polarization must be the directions parallel and perpendicular to the optic axis of the cellophane sheet.

We suppose that the light passes through the cellophane with two different velocities in these two different orientations, but it is transmitted without changing the direction of polarization. When the cellophane is turned halfway between these two orientations, as shown in Fig.  33–3 , we see that the light transmitted through the second polaroid is bright.

It just happens that ordinary cellophane used in commercial packaging is very close to a half-wave thickness for most of the colors in white light. Such a sheet will turn the axis of linearly polarized light through $90^\circ$ if the incident linearly polarized beam makes an angle of $45^\circ$ with the optic axis, so that the beam emerging from the cellophane is then vibrating in the right direction to pass through the second polaroid sheet.

If we use white light in our demonstration, the cellophane sheet will be of the proper half-wave thickness only for a particular component of the white light, and the transmitted beam will have the color of this component. The color transmitted depends on the thickness of the cellophane sheet, and we can vary the effective thickness of the cellophane by tilting it so that the light passes through the cellophane at an angle, consequently through a longer path in the cellophane. As the sheet is tilted the transmitted color changes. With cellophane of different thicknesses one can construct filters that will transmit different colors. These filters have the interesting property that they transmit one color when the two polaroid sheets have their axes perpendicular, and the complementary color when the axes of the two polaroid sheets are parallel.

Another interesting application of aligned molecules is quite practical. Certain plastics are composed of very long and complicated molecules all twisted together. When the plastic is solidified very carefully, the molecules are all twisted in a mass, so that there are as many aligned in one direction as another, and so the plastic is not particularly birefringent. Usually there are strains and stresses introduced when the material is solidified, so the material is not perfectly homogeneous. However, if we apply tension to a piece of this plastic material, it is as if we were pulling a whole tangle of strings, and there will be more strings preferentially aligned parallel to the tension than in any other direction. So when a stress is applied to certain plastics, they become birefringent, and one can see the effects of the birefringence by passing polarized light through the plastic. If we examine the transmitted light through a polaroid sheet, patterns of light and dark fringes will be observed (in color, if white light is used). The patterns move as stress is applied to the sample, and by counting the fringes and seeing where most of them are, one can determine what the stress is. Engineers use this phenomenon as a means of finding the stresses in odd-shaped pieces that are difficult to calculate.

Another interesting example of a way of obtaining birefringence is by means of a liquid substance. Consider a liquid composed of long asymmetric molecules which carry a plus or minus average charge near the ends of the molecule, so that the molecule is an electric dipole. In the collisions in the liquid the molecules will ordinarily be randomly oriented, with as many molecules pointed in one direction as in another. If we apply an electric field the molecules will tend to line up, and the moment they line up the liquid becomes birefringent. With two polaroid sheets and a transparent cell containing such a polar liquid, we can devise an arrangement with the property that light is transmitted only when the electric field is applied. So we have an electrical switch for light, which is called a Kerr cell . This effect, that an electric field can produce birefringence in certain liquids, is called the Kerr effect.

33–4 Polarizers

So far we have considered substances in which the refractive index is different for light polarized in different directions. Of very practical value are those crystals and other substances in which not only the index, but also the coefficient of absorption, is different for light polarized in different directions. By the same arguments which supported the idea of birefringence, it is understandable that absorption can vary with the direction in which the charges are forced to vibrate in an anisotropic substance. Tourmaline is an old, famous example and polaroid is another. Polaroid consists of a thin layer of small crystals of herapathite (a salt of iodine and quinine), all aligned with their axes parallel. These crystals absorb light when the oscillations are in one direction, and they do not absorb appreciably when the oscillations are in the other direction.

Suppose that we send light into a polaroid sheet polarized linearly at an angle $\theta$ to the passing direction. What intensity will come through? This incident light can be resolved into a component perpendicular to the pass direction which is proportional to $\sin\theta$, and a component along the pass direction which is proportional to $\cos\theta$. The amplitude which comes out of the polaroid is only the cosine $\theta$ part; the $\sin\theta$ component is absorbed. The amplitude which passes through the polaroid is smaller than the amplitude which entered, by a factor $\cos\theta$. The energy which passes through the polaroid, i.e., the intensity of the light, is proportional to the square of $\cos\theta$. $\operatorname{Cos}^2\theta$, then, is the intensity transmitted when the light enters polarized at an angle $\theta$ to the pass direction. The absorbed intensity, of course, is $\sin^2\theta$.

An interesting paradox is presented by the following situation. We know that it is not possible to send a beam of light through two polaroid sheets with their axes crossed at right angles. But if we place a third polaroid sheet between the first two, with its pass axis at $45^\circ$ to the crossed axes, some light is transmitted. We know that polaroid absorbs light, it does not create anything. Nevertheless, the addition of a third polaroid at $45^\circ$ allows more light to get through. The analysis of this phenomenon is left as an exercise for the student.

One of the most interesting examples of polarization is not in complicated crystals or difficult substances, but in one of the simplest and most familiar of situations—the reflection of light from a surface. Believe it or not, when light is reflected from a glass surface it may be polarized, and the physical explanation of this is very simple. It was discovered empirically by Brewster that light reflected from a surface is completely polarized if the reflected beam and the beam refracted into the material form a right angle. The situation is illustrated in Fig.  33–4 . If the incident beam is polarized in the plane of incidence, there will be no reflection at all. Only if the incident beam is polarized normal to the plane of incidence will it be reflected. The reason is very easy to understand. In the reflecting material the light is polarized transversely, and we know that it is the motion of the charges in the material which generates the emergent beam, which we call the reflected beam. The source of this so-called reflected light is not simply that the incident beam is reflected; our deeper understanding of this phenomenon tells us that the incident beam drives an oscillation of the charges in the material, which in turn generates the reflected beam. From Fig.  33–4 it is clear that only oscillations normal to the paper can radiate in the direction of reflection, and consequently the reflected beam will be polarized normal to the plane of incidence. If the incident beam is polarized in the plane of incidence, there will be no reflected light.

This phenomenon is readily demonstrated by reflecting a linearly polarized beam from a flat piece of glass. If the glass is turned to present different angles of incidence to the polarized beam, sharp attenuation of the reflected intensity is observed when the angle of incidence passes through Brewster’s angle. This attenuation is observed only if the plane of polarization lies in the plane of incidence. If the plane of polarization is normal to the plane of incidence, the usual reflected intensity is observed at all angles.

33–5 Optical activity

Another most remarkable effect of polarization is observed in materials composed of molecules which do not have reflection symmetry: molecules shaped something like a corkscrew, or like a gloved hand, or any shape which, if viewed through a mirror, would be reversed in the same way that a left-hand glove reflects as a right-hand glove. Suppose all of the molecules in the substance are the same, i.e., none is a mirror image of any other. Such a substance may show an interesting effect called optical activity, whereby as linearly polarized light passes through the substance, the direction of polarization rotates about the beam axis.

To understand the phenomenon of optical activity requires some calculation, but we can see qualitatively how the effect might come about, without actually carrying out the calculations. Consider an asymmetric molecule in the shape of a spiral, as shown in Fig.  33–5 . Molecules need not actually be shaped like a corkscrew in order to exhibit optical activity, but this is a simple shape which we shall take as a typical example of those that do not have reflection symmetry. When a light beam linearly polarized along the $y$-direction falls on this molecule, the electric field will drive charges up and down the helix, thereby generating a current in the $y$-direction and radiating an electric field $E_y$ polarized in the $y$-direction. However, if the electrons are constrained to move along the spiral, they must also move in the $x$-direction as they are driven up and down. When a current is flowing up the spiral, it is also flowing into the paper at $z = z_1$ and out of the paper at $z = z_1 + A$, if $A$ is the diameter of our molecular spiral. One might suppose that the current in the $x$-direction would produce no net radiation, since the currents are in opposite directions on opposite sides of the spiral. However, if we consider the $x$-components of the electric field arriving at $z = z_2$, we see that the field radiated by the current at $z = z_1 + A$ and the field radiated from $z = z_1$ arrive at $z_2$ separated in time by the amount $A/c$, and thus separated in phase by $\pi + \omega A/c$. Since the phase difference is not exactly $\pi$, the two fields do not cancel exactly, and we are left with a small $x$-component in the electric field generated by the motion of the electrons in the molecule, whereas the driving electric field had only a $y$-component. This small $x$-component, added to the large $y$-component, produces a resultant field that is tilted slightly with respect to the $y$-axis, the original direction of polarization. As the light moves through the material, the direction of polarization rotates about the beam axis. By drawing a few examples and considering the currents that will be set in motion by an incident electric field, one can convince himself that the existence of optical activity and the sign of the rotation are independent of the orientation of the molecules.

Corn syrup is a common substance which possesses optical activity. The phenomenon is easily demonstrated with a polaroid sheet to produce a linearly polarized beam, a transmission cell containing corn syrup, and a second polaroid sheet to detect the rotation of the direction of polarization as the light passes through the corn syrup.

33–6 The intensity of reflected light

Let us now consider quantitatively the reflection coefficient as a function of angle. Figure  33–6 (a) shows a beam of light striking a glass surface, where it is partly reflected and partly refracted into the glass. Let us suppose that the incident beam, of unit amplitude, is linearly polarized normal to the plane of the paper. We will call the amplitude of the reflected wave $b$, and the amplitude of the refracted wave $a$. The refracted and reflected waves will, of course, be linearly polarized, and the electric field vectors of the incident, reflected, and refracted waves are all parallel to each other. Figure  33–6 (b) shows the same situation, but now we suppose that the incident wave, of unit amplitude, is polarized in the plane of the paper. Now let us call the amplitude of the reflected and refracted wave $B$ and $A$, respectively.

We wish to calculate how strong the reflection is in the two situations illustrated in Fig. 33–6 (a) and  33–6 (b). We already know that when the angle between the reflected beam and refracted beam is a right angle, there will be no reflected wave in Fig.  33–6 (b), but let us see if we cannot get a quantitative answer—an exact formula for $B$ and $b$ as a function of the angle of incidence, $i$.

The principle that we must understand is as follows. The currents that are generated in the glass produce two waves. First, they produce the reflected wave. Moreover, we know that if there were no currents generated in the glass, the incident wave would continue straight into the glass. Remember that all the sources in the world make the net field. The source of the incident light beam produces a field of unit amplitude, which would move into the glass along the dotted line in the figure. This field is not observed, and therefore the currents generated in the glass must produce a field of amplitude $-1$, which moves along the dotted line. Using this fact, we will calculate the amplitude of the refracted waves, $a$ and $A$.

In Fig.  33–6 (a) we see that the field of amplitude $b$ is radiated by the motion of charges inside the glass which are responding to a field $a$ inside the glass, and that therefore $b$ is proportional to $a$. We might suppose that since our two figures are exactly the same, except for the direction of polarization, the ratio $B/A$ would be the same as the ratio $b/a$. This is not quite true, however, because in Fig.  33–6 (b) the polarization directions are not all parallel to each other, as they are in Fig.  33–6 (a). It is only the component of the electric field in the glass which is perpendicular to $B$, $A \cos\,(i + r)$, which is effective in producing $B$. The correct expression for the proportionality is then \begin{equation} \label{Eq:I:33:1} \frac{b}{a}=\frac{B}{A \cos\,(i + r)}. \end{equation}

Now we use a trick. We know that in both (a) and (b) of Fig.  33–6 the electric field in the glass must produce oscillations of the charges, which generate a field of amplitude $-1$, polarized parallel to the incident beam, and moving in the direction of the dotted line. But we see from part (b) of the figure that only the component of the electric field in the glass that is normal to the dashed line has the right polarization to produce this field, whereas in Fig.  33–6 (a) the full amplitude $a$ is effective, since the polarization of wave $a$ is parallel to the polarization of the wave of amplitude $-1$. Therefore we can write \begin{equation} \label{Eq:I:33:2} \frac{A \cos\,(i - r)}{a}=\frac{-1}{-1}, \end{equation} since the two amplitudes on the left side of Eq. ( 33.2 ) each produce the wave of amplitude $-1$.

Dividing Eq. ( 33.1 ) by Eq. ( 33.2 ), we obtain \begin{equation} \label{Eq:I:33:3} \frac{B}{b}=\frac{\cos\,(i+r)}{\cos\,(i-r)}, \end{equation} a result which we can check against what we already know. If we set $(i + r) = 90^\circ$, Eq. ( 33.3 ) gives $B = 0$, as Brewster says it should be, so our results so far are at least not obviously wrong.

We have assumed unit amplitudes for the incident waves, so that $\abs{B}^2/1^2$ is the reflection coefficient for waves polarized in the plane of incidence, and $\abs{b}^2/1^2$ is the reflection coefficient for waves polarized normal to the plane of incidence. The ratio of these two reflection coefficients is determined by Eq. ( 33.3 ).

Now we perform a miracle, and compute not just the ratio, but each coefficient $\abs{B}^2$ and $\abs{b}^2$ individually! We know from the conservation of energy that the energy in the refracted wave must be equal to the incident energy minus the energy in the reflected wave, $1-\abs{B}^2$ in one case, $1-\abs{b}^2$ in the other. Furthermore, the energy which passes into the glass in Fig.  33–6 (b) is to the energy which passes into the glass in Fig.  33–6 (a) as the ratio of the squares of the refracted amplitudes, $\abs{A}^2/\abs{a}^2$. One might ask whether we really know how to compute the energy inside the glass, because, after all, there are energies of motion of the atoms in addition to the energy in the electric field. But it is obvious that all of the various contributions to the total energy will be proportional to the square of the amplitude of the electric field. Therefore we can write \begin{equation} \label{Eq:I:33:4} \frac{1-\abs{B}^2}{1-\abs{b}^2}= \frac{\abs{A}^2}{\abs{a}^2}. \end{equation}

We now substitute Eq. ( 33.2 ) to eliminate $A/a$ from the expression above, and express $B$ in terms of $b$ by means of Eq. ( 33.3 ): \begin{equation} \label{Eq:I:33:5} \frac{1-\abs{b}^2\,\dfrac{\cos^2\,(i+r)}{\cos^2\,(i-r)}} {1-\abs{b}^2}=\frac{1}{\cos^2\,(i-r)}. \end{equation} This equation contains only one unknown amplitude, $b$. Solving for $\abs{b}^2$, we obtain \begin{equation} \label{Eq:I:33:6} \abs{b}^2=\frac{\sin^2\,(i-r)}{\sin^2\,(i+r)} \end{equation} and, with the aid of ( 33.3 ), \begin{equation} \label{Eq:I:33:7} \abs{B}^2=\frac{\tan^2\,(i-r)}{\tan^2\,(i+r)}. \end{equation} So we have found the reflection coefficient $\abs{b}^2$ for an incident wave polarized perpendicular to the plane of incidence, and also the reflection coefficient $\abs{B}^2$ for an incident wave polarized in the plane of incidence!

It is possible to go on with arguments of this nature and deduce that $b$ is real. To prove this, one must consider a case where light is coming from both sides of the glass surface at the same time, a situation not easy to arrange experimentally, but fun to analyze theoretically. If we analyze this general case, we can prove that $b$ must be real, and therefore, in fact, that $b = \pm\sin\,(i - r)/\sin\,(i + r)$. It is even possible to determine the sign by considering the case of a very, very thin layer in which there is reflection from the front and from the back surfaces, and calculating how much light is reflected. We know how much light should be reflected by a thin layer, because we know how much current is generated, and we have even worked out the fields produced by such currents.

One can show by these arguments that \begin{equation} \label{Eq:I:33:8} b=-\frac{\sin\,(i-r)}{\sin\,(i+r)},\quad B=\frac{\tan\,(i-r)}{\tan\,(i+r)}. \end{equation} These expressions for the reflection coefficients as a function of the angles of incidence and refraction are called Fresnel’s reflection formulas.

If we consider the limit as the angles $i$ and $r$ go to zero, we find, for the case of normal incidence, that $B^2\approx b^2\approx(i-r)^2/(i+r)^2$ for both polarizations, since the sines are practically equal to the angles, as are also the tangents. But we know that $\sin i/\sin r = n$, and when the angles are small, $i/r \approx n$. It is thus easy to show that the coefficient of reflection for normal incidence is \begin{equation*} B^2=b^2=\frac{(n-1)^2}{(n+1)^2}. \end{equation*}

It is interesting to find out how much light is reflected at normal incidence from the surface of water, for example. For water, $n$ is $4/3$, so that the reflection coefficient is $(1/7)^2 \approx 2\%$. At normal incidence, only two percent of the light is reflected from the surface of water.

33–7 Anomalous refraction

The last polarization effect we shall consider was actually one of the first to be discovered: anomalous refraction. Sailors visiting Iceland brought back to Europe crystals of Iceland spar (CaCO$_3$) which had the amusing property of making anything seen through the crystal appear doubled, i.e., as two images. This came to the attention of Huygens, and played an important role in the discovery of polarization. As is often the case, the phenomena which are discovered first are the hardest, ultimately, to explain. It is only after we understand a physical concept thoroughly that we can carefully select those phenomena which most clearly and simply demonstrate the concept.

Anomalous refraction is a particular case of the same birefringence that we considered earlier. Anomalous refraction comes about when the optic axis, the long axis of our asymmetric molecules, is not parallel to the surface of the crystal. In Fig.  33–7 are drawn two pieces of birefringent material, with the optic axis as shown. In part (a) of the figure, the incident beam falling on the material is linearly polarized in a direction perpendicular to the optic axis of the material. When this beam strikes the surface of the material, each point on the surface acts as a source of a wave which travels into the crystal with velocity $v_\perp$, the velocity of light in the crystal when the plane of polarization is normal to the optic axis. The wavefront is just the envelope or locus of all these little spherical waves, and this wavefront moves straight through the crystal and out the other side. This is just the ordinary behavior we would expect, and this ray is called the ordinary ray .

In part (b) of the figure the linearly polarized light falling on the crystal has its direction of polarization turned through $90^\circ$, so that the optic axis lies in the plane of polarization. When we now consider the little waves originating at any point on the surface of the crystal, we see that they do not spread out as spherical waves. Light travelling along the optic axis travels with velocity $v_\perp$ because the polarization is perpendicular to the optic axis, whereas the light travelling perpendicular to the optic axis travels with velocity $v_\parallel$ because the polarization is parallel to the optic axis. In a birefringent material $v_\parallel\neq v_\perp$, and in the figure $v_\parallel < v_\perp$. A more complete analysis will show that the waves spread out on the surface of an ellipsoid, with the optic axis as major axis of the ellipsoid. The envelope of all these elliptical waves is the wavefront which proceeds through the crystal in the direction shown. Again, at the back surface the beam will be deflected just as it was at the front surface, so that the light emerges parallel to the incident beam, but displaced from it. Clearly, this beam does not follow Snell’s law, but goes in an extraordinary direction. It is therefore called the extraordinary ray .

When an unpolarized beam strikes an anomalously refracting crystal, it is separated into an ordinary ray, which travels straight through in the normal manner, and an extraordinary ray which is displaced as it passes through the crystal. These two emergent rays are linearly polarized at right angles to each other. That this is true can be readily demonstrated with a sheet of polaroid to analyze the polarization of the emergent rays. We can also demonstrate that our interpretation of this phenomenon is correct by sending linearly polarized light into the crystal. By properly orienting the direction of polarization of the incident beam, we can make this light go straight through without splitting, or we can make it go through without splitting but with a displacement.

We have represented all the various polarization cases in Figs. 33–1 and  33–2 as superpositions of two special polarization cases, namely $x$ and $y$ in various amounts and phases. Other pairs could equally well have been used. Polarization along any two perpendicular axes $x'$, $y'$ inclined to $x$ and $y$ would serve as well [for example, any polarization can be made up of superpositions of cases (a) and (e) of Fig.  33–2 ]. It is interesting, however, that this idea can be extended to other cases also. For example, any linear polarization can be made up by superposing suitable amounts at suitable phases of right and left circular polarizations [cases (c) and (g) of Fig.  33–2 ], since two equal vectors rotating in opposite directions add to give a single vector oscillating in a straight line (Fig.  33–8 ). If the phase of one is shifted relative to the other, the line is inclined. Thus all the pictures of Fig.  33–1 could be labeled “the superposition of equal amounts of right and left circularly polarized light at various relative phases.” As the left slips behind the right in phase, the direction of the linear polarization changes. Therefore optically active materials are, in a sense, birefringent. Their properties can be described by saying that they have different indexes for right- and left-hand circularly polarized light. Superposition of right and left circularly polarized light of different intensities produces elliptically polarized light.

Circularly polarized light has another interesting property—it carries angular momentum (about the direction of propagation). To illustrate this, suppose that such light falls on an atom represented by a harmonic oscillator that can be displaced equally well in any direction in the plane $xy$. Then the $x$-displacement of the electron will respond to the $E_x$ component of the field, while the $y$-component responds, equally, to the equal $E_y$ component of the field but $90^\circ$ behind in phase. That is, the responding electron goes around in a circle, with angular velocity $\omega$, in response to the rotating electric field of the light (Fig.  33–9 ). Depending on the damping characteristics of the response of the oscillator, the direction of the displacement $\FLPa$ of the electron, and the direction of the force $q_e\FLPE$ on it need not be the same but they rotate around together. The $\FLPE$ may have a component at right angles to $\FLPa$, so work is done on the system and a torque $\tau$ is exerted. The work done per second is $\tau\omega$. Over a period of time $T$ the energy absorbed is $\tau\omega T$, while $\tau T$ is the angular momentum delivered to the matter absorbing the energy. We see therefore that a beam of right circularly polarized light containing a total energy $\energy$ carries an angular momentum (with vector directed along the direction of propagation) $\energy/\omega$ . For when this beam is absorbed that angular momentum is delivered to the absorber. Left-hand circular light carries angular momentum of the opposite sign, $-\energy/\omega$.

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• Published: 04 March 2021

Optical polarization analogue in free electron beams

• Hugo Lourenço-Martins 1 ,
• Davy Gérard   ORCID: orcid.org/0000-0003-4789-9888 2 &
• Mathieu Kociak   ORCID: orcid.org/0000-0001-8858-0449 3  

Nature Physics volume  17 ,  pages 598–603 ( 2021 ) Cite this article

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Spectromicroscopy techniques with fast electrons can quantitatively measure the optical response of excitations with unrivalled spatial resolution. However, owing to their inherently scalar nature, electron waves cannot access the polarization-related quantities. Despite promising attempts based on the conversion of concepts originating from singular optics (such as vortex beams), the definition of an optical polarization analogue for fast electrons has remained an open question. Here we establish such an analogue using the dipole transition vector of the electron between two well-chosen singular wave states. We show that electron energy loss spectroscopy allows the direct measurement of the polarized electromagnetic local density of states. In particular, in the case of circular polarization, it directly measures the local optical spin density. This work establishes electron energy loss spectroscopy as a quantitative technique to tackle fundamental issues in nano-optics, such as super-chirality, local polarization of dark excitations or polarization singularities at the nanoscale.

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Acknowledgements

We acknowledge J. Verbeeck for introduction to this field and thank J. Verbeeck, D. Ugarte, F. Houdellier and G. Guzzinati for insightful discussions. H.L.-M. thanks T. R. Harvey for insightful discussions. This project has received funding from the European Union’s Horizon 2020 research and innovation programme under grant agreement no. 823717-ESTEEM3 and no. 101017720-EBEAM, from the French state managed by the National Agency for Research under the programme of future investment EQUIPEX TEMPOS-CHROMATEM with the reference ANR-10-EQPX-50 and ANR-17-CE24-0039 (2D-CHIRAL).

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Hugo Lourenço-Martins

Light, nanomaterials & nanotechnologies (L2n), CNRS-ERL 7004, Université de Technologie de Troyes, Troyes, France

Davy Gérard

Université Paris-Saclay, CNRS, Laboratoire de Physique des Solides, Orsay, France

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Contributions

M.K. and H.L.-M. developed the theory. H.L.-M. wrote the simulation codes and performed the numerical simulations. M.K., H.L.-M. and D.G. developed the analogy between the Bloch and Poincaré spheres. All the authors discussed the results and contributed to the writing of the manuscript.

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Extended data

Extended data fig. 1 examples of transitions between higher order hg states possessing a non-zero opa..

Transitions a and b possess a dipole moment along the x direction and therefore measure E x (ω, qz). Transitions c and d possess a dipole moment along the y direction and therefore measure E y (ω, qz).

Extended Data Fig. 2 Illustration of the effect of the summation over the final states.

a , Wavefunction ψ i and plasmonic potential ϕ m of the four first modes m of a 100nm * 15nm silver nano-antenna. The relative position and scale of the wavefunction and potential are respected. Scales bar is 50 nm. b , pEELS spectrum as a function of the collection angle. The peaks of the four modes shown in a. are marked with arrows. Below θ ω , the second and fourth peak are suppressed due to the selection rule reminiscent of formula (184) of the SI. Above θ ω ,all the peaks are present and the resulting spectrum is essentially classical.

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Lourenço-Martins, H., Gérard, D. & Kociak, M. Optical polarization analogue in free electron beams. Nat. Phys. 17 , 598–603 (2021). https://doi.org/10.1038/s41567-021-01163-w

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Wave Optics

86 Polarization

Learning objectives.

• Discuss the meaning of polarization.
• Discuss the property of optical activity of certain materials.

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 86.1 ). 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 86.2 ). There are specific directions for the oscillations of the electric and magnetic fields. Polarization is the attribute that a wave’s oscillations have a definite direction relative to the direction of propagation of the wave. (This is not the same type of polarization as that discussed for the separation of charges.) Waves having such a direction are said to be polarized. For an EM wave, we define the direction of polarization to be the direction parallel to the electric field. Thus we can think of the electric field arrows as showing the direction of polarization, as in Figure 86.2 .

To examine this further, consider the transverse waves in the ropes shown in Figure 86.3 . The oscillations in one rope are in a vertical plane and are said to be vertically polarized . Those in the other rope are in a horizontal plane and are horizontally polarized . If a vertical slit is placed on the first rope, the waves pass through. However, a vertical slit blocks the horizontally polarized waves. For EM waves, the direction of the electric field is analogous to the disturbances on the ropes.

The Sun and many other light sources produce waves that are randomly polarized (see (Figure 86.4) ). Such light is said to be unpolarized because it is composed of many waves with all possible directions of polarization. Polaroid materials, invented by the founder of Polaroid Corporation, Edwin Land, act as a polarizing slit for light, allowing only polarization in one direction to pass through. Polarizing filters are composed of long molecules aligned in one direction. Thinking of the molecules as many slits, analogous to those for the oscillating ropes, we can understand why only light with a specific polarization can get through. The axis of a polarizing filter is the direction along which the filter passes the electric field of an EM wave (see (Figure 86.5) ).

Figure 86.6 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 [latex]\theta[/latex]. If the electric field has an amplitude [latex]E[/latex], then the transmitted part of the wave has an amplitude [latex]E\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta[/latex] (see Figure 86.7 ). Since the intensity of a wave is proportional to its amplitude squared, the intensity [latex]I[/latex] of the transmitted wave is related to the incident wave by

where [latex]{I}_{0}[/latex] is the intensity of the polarized wave before passing through the filter. (The above equation is known as Malus’s law.)

What angle is needed between the direction of polarized light and the axis of a polarizing filter to reduce its intensity by [latex]\text{90}\text{.}0%\text{}[/latex]?

When the intensity is reduced by [latex]\text{90}\text{.}0%\text{}[/latex], it is [latex]\text{10}\text{.}0%\text{}[/latex] or 0.100 times its original value. That is, [latex]I=0\text{.}\text{100}{I}_{0}[/latex]. Using this information, the equation [latex]I={I}_{0}\phantom{\rule{0.25em}{0ex}}{\text{cos}}^{2}\phantom{\rule{0.25em}{0ex}}\theta[/latex] can be used to solve for the needed angle.

Solving the equation [latex]I={I}_{0}\phantom{\rule{0.25em}{0ex}}{\text{cos}}^{2}\phantom{\rule{0.25em}{0ex}}\theta[/latex] for [latex]\text{cos}\phantom{\rule{0.25em}{0ex}}\theta[/latex] and substituting with the relationship between [latex]I[/latex] and [latex]{I}_{0}[/latex] gives

Solving for [latex]\theta[/latex] yields

A fairly large angle between the direction of polarization and the filter axis is needed to reduce the intensity to [latex]\text{10}\text{.}0%\text{}[/latex] of its original value. This seems reasonable based on experimenting with polarizing films. It is interesting that, at an angle of [latex]\text{45º}[/latex], the intensity is reduced to [latex]\text{50%}\text{}[/latex] of its original value (as you will show in this section’s Problems & Exercises). Note that [latex]\text{71}\text{.}6º[/latex] is [latex]\text{18}\text{.}4º[/latex] from reducing the intensity to zero, and that at an angle of [latex]\text{18}\text{.}4º[/latex] the intensity is reduced to [latex]\text{90}\text{.}0%\text{}[/latex] of its original value (as you will also show in Problems & Exercises), giving evidence of symmetry.

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.

Figure 86.8 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 reflected light is completely polarized at a angle of reflection [latex]{\theta }_{\text{b}}[/latex], given by

where [latex]{n}_{1}[/latex] is the medium in which the incident and reflected light travel and [latex]{n}_{2}[/latex] 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 [latex]{\theta }_{\text{b}}[/latex] is known as Brewster’s angle , named after the 19th-century Scottish physicist who discovered them.

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 Figure 86.9 .

Figure 86.10 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.

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

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

Solution for (a)

Putting the known quantities into the equation

Solving for the angle [latex]{\theta }_{\text{b}}[/latex] yields

Solution for (b)

Similarly, for crown glass and air,

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 Scattering

If 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 86.11  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 86.11 , 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 non-reflective 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.

Find 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 Materials

While 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 [latex]\text{90º}[/latex]. Furthermore, this property can be turned off by the application of a voltage, as illustrated in Figure 86.12 . It is possible to manipulate this characteristic quickly and in small well-defined 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 86.12  (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 optically active . Examples include sugar water, insulin, and collagen (see Figure 86.13 ). In addition to depending on the type of substance, the amount and direction of rotation depends on a number of factors. Among these is the concentration of the substance, the distance the light travels through it, and the wavelength of light. Optical activity is due to the asymmetric shape of molecules in the substance, such as being helical. Measurements of the rotation of polarized light passing through substances can thus be used to measure concentrations, a standard technique for sugars. It can also give information on the shapes of molecules, such as proteins, and factors that affect their shapes, such as temperature and pH.

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 86.14 . 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 birefringent (see Figure 86.15 ). Each of the separated rays has a specific polarization. One behaves normally and is called the ordinary ray, whereas the other does not obey Snell’s law and is called the extraordinary ray. Birefringent crystals can be used to produce polarized beams from unpolarized light. Some birefringent materials preferentially absorb one of the polarizations. These materials are called dichroic and can produce polarization by this preferential absorption. This is fundamentally how polarizing filters and other polarizers work. The interested reader is invited to further pursue the numerous properties of materials related to polarization.

Section Summary

• Polarization is the attribute that wave oscillations have a definite direction relative to the direction of propagation of the wave.
• EM waves are transverse waves that may be polarized.
• The direction of polarization is defined to be the direction parallel to the electric field of the EM wave.
• Unpolarized light is composed of many rays having random polarization directions.
• Light can be polarized by passing it through a polarizing filter or other polarizing material. The intensity [latex]I[/latex] of polarized light after passing through a polarizing filter is [latex]I={I}_{0}\phantom{\rule{0.25em}{0ex}}{\text{cos}}^{2}\phantom{\rule{0.25em}{0ex}}\mathrm{\theta ,}[/latex] where [latex]{I}_{0}[/latex] is the original intensity and [latex]\theta[/latex] is the angle between the direction of polarization and the axis of the filter.
• Polarization is also produced by reflection.
• Brewster’s law states that reflected light will be completely polarized at the angle of reflection [latex]{\theta }_{\text{b}}[/latex], known as Brewster’s angle, given by a statement known as Brewster’s law: [latex]\text{tan}\phantom{\rule{0.25em}{0ex}}{\theta }_{\text{b}}=\frac{{n}_{2}}{{n}_{1}}[/latex], where [latex]{n}_{1}[/latex] is the medium in which the incident and reflected light travel and [latex]{n}_{2}[/latex] is the index of refraction of the medium that forms the interface that reflects the light.
• Polarization can also be produced by scattering.
• There are a number of types of optically active substances that rotate the direction of polarization of light passing through them.

Conceptual Questions

• Under what circumstances is the phase of light changed by reflection? Is the phase related to polarization?
• Can a sound wave in air be polarized? Explain.
• No 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?
• Explain what happens to the energy carried by light that it is dimmed by passing it through two crossed polarizing filters.
• When particles scattering light are much smaller than its wavelength, the amount of scattering is proportional to [latex]1/{\lambda }^{4}[/latex]. Does this mean there is more scattering for small [latex]\lambda[/latex] than large [latex]\lambda[/latex]? How does this relate to the fact that the sky is blue?
• Using the information given in the preceding question, explain why sunsets are red.
• When light is reflected at Brewster’s angle from a smooth surface, it is [latex]\text{100%}\text{}[/latex] 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 [latex]\text{100%}\text{}[/latex]?

Problems & Exercises

• What angle is needed between the direction of polarized light and the axis of a polarizing filter to cut its intensity in half?
• The angle between the axes of two polarizing filters is [latex]\text{45}\text{.}0º[/latex]. By how much does the second filter reduce the intensity of the light coming through the first?
• If you have completely polarized light of intensity [latex]\text{150 W}/{\text{m}}^{2}[/latex], what will its intensity be after passing through a polarizing filter with its axis at an [latex]\text{89}\text{.}0º[/latex] angle to the light’s polarization direction?
• What angle would the axis of a polarizing filter need to make with the direction of polarized light of intensity [latex]1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}{\text{kW/m}}^{2}[/latex] to reduce the intensity to [latex]10\text{.}0\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}[/latex]?
• At the end of Example 86.1 , it was stated that the intensity of polarized light is reduced to [latex]\text{90}\text{.}0%\text{}[/latex] of its original value by passing through a polarizing filter with its axis at an angle of [latex]\text{18}\text{.}4º[/latex] to the direction of polarization. Verify this statement.
• Show that if you have three polarizing filters, with the second at an angle of [latex]\text{45º}[/latex] to the first and the third at an angle of [latex]\text{90}\text{.}0º[/latex] to the first, the intensity of light passed by the first will be reduced to [latex]\text{25}\text{.}0%\text{}[/latex] 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.)
• Prove that, if [latex]I[/latex] is the intensity of light transmitted by two polarizing filters with axes at an angle [latex]\theta[/latex] and [latex]I\prime[/latex] is the intensity when the axes are at an angle [latex]\text{90.0º}-\mathrm{\theta ,}[/latex] then [latex]I+I\prime ={I}_{0,}[/latex] the original intensity. (Hint: Use the trigonometric identities [latex]\text{cos}\phantom{\rule{0.25em}{0ex}}\left(90.0º-\theta \right)=\text{sin}\phantom{\rule{0.25em}{0ex}}\theta[/latex] and [latex]{\text{cos}}^{2}\phantom{\rule{0.25em}{0ex}}\theta +{\text{sin}}^{2}\phantom{\rule{0.25em}{0ex}}\theta =1.[/latex])
• At what angle will light reflected from diamond be completely polarized?
• What is Brewster’s angle for light traveling in water that is reflected from crown glass?
• A scuba diver sees light reflected from the water’s surface. At what angle will this light be completely polarized?
• At what angle is light inside crown glass completely polarized when reflected from water, as in a fish tank?
• Light reflected at [latex]\text{55}\text{.}6º[/latex] from a window is completely polarized. What is the window’s index of refraction and the likely substance of which it is made?
• (a) Light reflected at [latex]\text{62}\text{.}5º[/latex] 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?
• If [latex]{\theta }_{\text{b}}[/latex] is Brewster’s angle for light reflected from the top of an interface between two substances, and [latex]{\theta \prime }_{\text{b}}[/latex] is Brewster’s angle for light reflected from below, prove that [latex]{\theta }_{\text{b}}+{\theta \prime }_{\text{b}}=\text{90}\text{.}0º.[/latex]
• Integrated Concept If a polarizing filter reduces the intensity of polarized light to [latex]\text{50}\text{.}0%\text{}[/latex] of its original value, by how much are the electric and magnetic fields reduced?
• Integrated Concepts Suppose you put on two pairs of Polaroid sunglasses with their axes at an angle of [latex]\text{15}\text{.}0º[/latex]. 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.
• Integrated Concepts (a) On a day when the intensity of sunlight is [latex]1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{kW}/{\text{m}}^{2}[/latex], 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 [latex]\text{20}\text{.}0º.[/latex] Assuming the sunlight is unpolarized and the polarizers are [latex]\text{100%}\text{}[/latex] efficient, what is the initial rate of heating of the water in [latex]\text{ºC}/\text{s}[/latex], assuming it is [latex]\text{80}\text{.}0%\text{}[/latex] 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.

College Physics 2 Copyright © 2012 by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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