3 The Polarization Ellipse 3.1 INTRODUCTION Christian Huygens was the first to suggest that light was not a scalar quantity based on his work on the propagation of light through crystals; it appeared that light had ‘‘sides’’ in the words of Newton This vectorial nature of light is called polarization If we follow mechanics and equate an optical medium to an isotropic elastic medium, it should be capable of supporting three independent oscillations (optical disturbances): ux(r, t), uy(r, t), and uz(r, t) Correspondingly, three independent wave equations are then required to describe the propagation of the optical disturbance, namely, @2 ui ðr, tÞ i ¼ x, y, z ð3-1Þ v2 @t2 where v is the velocity of propagation of the oscillation and r ¼ r(x, y, z) In a Cartesian system the components ux ðr, tÞ and uy ðr, tÞ are said to be the transverse components, and the component uz ðr, tÞ is said to be the longitudinal component when the propagation is in the z direction Thus, according to (3-1) the optical field components should be ð3-2aÞ ux ðr, tÞ ¼ u0x cosð!t À k Á r þ x Þ r2 ui ðr, tÞ ¼ uy ðr, tÞ ¼ u0y cosð!t À k Á r þ y Þ ð3-2bÞ uz ðr, tÞ ¼ u0z cosð!t À k Á r þ z Þ ð3-2cÞ In 1818 Fresnel and Arago carried out a series of fundamental investigations on Young’s interference experiment using polarized light After a considerable amount of experimentation they were forced to conclude that the longitudinal component (3-2c) did not exist That is, light consisted only of the transverse components (3-2a) and (3-2b) If we take the direction of propagation to be in the z direction, then the optical field in free space must be described only by ux ðz, tÞ ¼ u0x cosð!t À kz þ x Þ ð3-3aÞ uy ðz, tÞ ¼ u0y cosð!t À kz þ y Þ Copyright © 2003 by Marcel Dekker, Inc All Rights Reserved ð3-3bÞ where u0x and u0y are the maximum amplitudes and x and y are arbitrary phases There is no reason, a priori, for the existence of only transverse components on the basis of an elastic medium (the ‘‘ether’’ in optics) It was considered to be a defect in Fresnel’s theory Nevertheless, in spite of this (3-3a) and (3-3b) were found to describe satisfactorily the phenomenon of interference using polarized light The ‘‘defect’’ in Fresnel’s theory was overcome by the development of a new theory, which we now call Maxwell’s electrodynamic theory and his equations One of the immediate results of solving his equations was that in free space only transverse components arose; there was no longitudinal component This was one of the first triumphs of Maxwell’s theory Nevertheless, Maxwell’s theory took nearly 40 years to be accepted in optics due, in large part, to the fact that up to the end of the nineteenth century it led to practically nothing that could not be explained or understood by Fresnel’s theory Equations (3-3a) and (3-3b) are spoken of as the polarized or polarization components of the optical field In this chapter we consider the consequences of these equations The results are very interesting and lead to a surprising number of revelations about the nature of light 3.2 THE INSTANTANEOUS OPTICAL FIELD AND THE POLARIZATION ELLIPSE In previous sections we pointed out that the experiments of Fresnel and Arago led to the discovery that light consisted only of two transverse components The components were perpendicular to each other and could be chosen for convenience to be propagating in the z direction The waves are said to be ‘‘instantaneous’’ in the sense that the time duration for the wave to go through one complete cycle is only 10À15 sec at optical frequencies In this chapter we find the equation that arises when the propagator is eliminated between the transverse components In order to this we show in Fig 3-1 the transverse optical field propagating in the z direction The transverse components are represented by Ex ðz, tÞ ¼ E0x cosð þ x Þ ð3-4aÞ Ey ðz, tÞ ¼ E0y cosð þ y Þ ð3-4bÞ where ¼ !t À z is the propagator The subscripts x and y refer to the components in the x and y directions, E0x and E0y are the maximum amplitudes, and x and y are the phases, respectively As the field propagates, Ex(z, t) and Ey(z, t) give rise to a resultant vector This vector describes a locus of points in space, and the curve generated by those points will now be derived In order to this (3-4a) and (3-4b) are written as Ex ¼ cos cos x À sin sin x E0x ð3-5aÞ Ey ¼ cos cos y À sin sin y E0y ð3-5bÞ Copyright © 2003 by Marcel Dekker, Inc All Rights Reserved Figure 3-1 Propagation of the transverse optical field Hence, Ey Ex sin y À sin x ¼ cos sinðy À x Þ E0x E0y ð3-6aÞ Ey Ex cos y À cos x ¼ sin sinðy À x Þ E0x E0y ð3-6bÞ Squaring (3-6a) and (3-6b) and adding gives E2y E2x E Ey þ À2 x cos ¼ sin2 E0x E0y E20x E20y ð3-7aÞ where ¼ y À x ð3-7bÞ Equation (3-7a) is recognized as the equation of an ellipse and shows that at any instant of time the locus of points described by the optical field as it propagates is an ellipse This behavior is spoken of as optical polarization, and (3-7a) is called the polarization ellipse In Fig 3-2 the ellipse is shown inscribed within a rectangle whose sides are parallel to the coordinate axes and whose lengths are 2E0x and 2E0y We now determine the points where the ellipse is tangent to the sides of the rectangle We write (3-7a) as E0x2 E2y À ð2E0x E0y Ex cos ÞEy þ E0y2 ðE2x À E0x2 sin2 Þ ¼ ð3-8Þ The solution of this quadratic equation (3-8) is Ey ¼ E0y Ex cos E0y sin Æ ðE0x2 À E2x Þ1=2 E0x E0x ð3-9Þ At the top and bottom of the ellipse where it is tangent to the rectangle the slope is We now differentiate (3-9), set E 0y ¼ dEy =dEx ¼ 0, and find that Ex ¼ ÆE0x cos Copyright © 2003 by Marcel Dekker, Inc All Rights Reserved ð3-10aÞ Figure 3-2 An elliptically polarized wave and the polarization ellipse Substituting (3-10a) into (3-9), the corresponding values of Ey are found to be Ey ¼ ÆE0y ð3-10bÞ Similarly, by considering (3-9) where the slope is E 0y ¼ on the sides of the rectangle, the tangent points are Ex ¼ ÆE0x ð3-11aÞ Ey ¼ ÆE0y cos ð3-11bÞ Equations (3-10) and (3-11) show that the maximum length of the sides of the ellipse are Ex ¼ ÆE0x and Ey ¼ ÆE0y The ellipse is tangent to the sides of the rectangle at ðÆE0x , ÆE0y cos Þ and ðÆE0x cos , ÆE0y Þ We also see that (3-10) and (3-11) show that the extrema of Ex and Ey are ÆE0x and ÆE0y, respectively In Fig 3-2 the ellipse is shown touching the rectangle at point A, B, C, and D, the coordinates of which are A: þE0x cos , þ E0y ð3-12aÞ B: þE0x , þ E0y cos ð3-12bÞ C: ÀE0x cos , À E0y ð3-12cÞ D: ÀE0x , À E0y cos ð3-12dÞ The presence of the ‘‘cross term’’ in (3-7a) shows that the polarization ellipse is, in general, rotated, and this behavior is shown in Fig 3-2 where the ellipse is shown rotated through an angle More will be said about this later It is also of interest to determine the maximum and minimum areas of the polarization ellipse which can be inscribed within the rectangle We see that along Copyright © 2003 by Marcel Dekker, Inc All Rights Reserved the x axis the ellipse is tangent at the extrema x ¼ ÀE0x and x ¼ þE0x The area of the ellipse above the x axis is given by Z þE0x A¼ Ey dx ð3-13Þ ÀE0x Substituting (3-9) into (3-13) and evaluating the integrals, we find that the area of the polarization ellipse is A ¼ E0x E0y sin ð3-14Þ Thus, the area of the polarization ellipse depends on the lengths of the major and minor axes, E0x and E0y, and the phase shift between the orthogonal transverse components We see that for ¼ /2 the area is E0x E0y , whereas for ¼ the area is zero The significance of these results will soon become apparent In general, completely polarized light is elliptically polarized However, there are certain degenerate forms of the polarization ellipse which are continually encountered in the study of polarized light Because of the importance of these special degenerate forms we now discuss them as special cases in the following section These are the cases where either E0x or E0y is zero or E0x and E0y are equal and/or where ¼ 0, /2, or radians 3.3 SPECIALIZED (DEGENERATE) FORMS OF THE POLARIZATION ELLIPSE The polarization ellipse (3-7a) degenerates to special forms for certain values of E0x, E0y, and We now consider these special forms E0y ¼ In this case Ey(z, t) is zero and (3-4) becomes Ex ðz, tÞ ¼ E0x cosð þ x Þ ð3-15aÞ Ey ðz, tÞ ¼ ð3-15bÞ In this case there is an oscillation only in the x direction The light is then said to be linearly polarized in the x direction, and we call this linear horizontally polarized light Similarly, if E0x ¼ and Ey ðz, tÞ 6¼ 0, then we have a linear oscillation along the y axis, and we speak of linear vertically polarized light ¼ or Equation (3-7a) reduces to E2y E2x E Ey þ Æ2 x ¼0 2 E E0x E0y 0x E0y Equation (3-16) can be written as Ey Ex Æ ¼0 E0x E0y ð3-16Þ ð3-17Þ whence E0y Ey ¼ Æ E E0x x Copyright © 2003 by Marcel Dekker, Inc All Rights Reserved ð3-18Þ Equation (3-18) is recognized as the equation of a straight line with slope ÆðE0y =E0x Þ and zero intercept Thus, we say that we have linearly polarized light with slope ÆðE0y =E0x Þ The value ¼ yields a negative slope, and the value ¼ a positive slope If E0x ¼ E0y, then we see that Ey ¼ ÆEx ð3-19Þ The positive value is said to represent linear þ45 polarized light, and the negative value is said to represent linear À45 polarized light ¼ /2 or 3/2 The polarization ellipse reduces to E2y E2x þ ¼1 E20x E20y ð3-20Þ This is the standard equation of an ellipse Note that ¼ /2 or ¼ 3/2 yields the identical polarization ellipse E0x ¼ E0y ¼ E0 and ¼ /2 or ¼ 3/2 The polarization ellipse now reduces to E2x Ey þ ¼1 E20 E20 ð3-21Þ Equation (3-21) describes the equation of a circle Thus, for this condition the light is said to be right or left circularly polarized ( ¼ /2 and 3/2, respectively) Again, we note that (3-21) shows that it alone cannot determine if the value of is /2 or 3/2 Finally, in the previous section we showed that the area of the polarization ellipse was A ¼ E0x E0y sin ð3-22Þ We see that for ¼ or the area of the polarization ellipse is zero, which is to be expected for linearly polarized light For ¼ /2 or 3/2 the area of the ellipse is a maximum; that is, E0xE0y It is important to note that even if the phase shift between the orthogonal components is /2 or 3/2, the light is, in general, elliptically polarized Furthermore, the polarization ellipse shows that it is in the standard form as given by (3-20) For the more restrictive condition where the orthogonal amplitudes are equal so that E0x ¼ E0y ¼ E0 and, when ¼ =2 or 3=2, (3-22) becomes A ¼ E20 ð3-23Þ which is, of course, the area of a circle The previous special forms of the polarization ellipse are spoken of as being degenerate states We can summarize these results by saying that the degenerate states of the polarization ellipse are (1) linear horizontally or vertically polarized light, (2) linear þ45 or À45 polarized light, and (3) right or left circularly polarized light Aside from the fact that these degenerate states appear quite naturally as special cases of the polarization ellipse, there is a fundamental reason for their importance: they are relatively easy to create in an optical laboratory and can be Copyright © 2003 by Marcel Dekker, Inc All Rights Reserved used to create ‘‘null-intensity’’ conditions Polarization instruments, which may be based on null-intensity conditions, enable very accurate measurements to be made 3.4 ELLIPTICAL PARAMETERS OF THE POLARIZATION ELLIPSE The polarization ellipse has the form: E2y E2x E Ey þ À2 x cos ¼ sin2 2 E0x E0y E0x E0y ð3-7aÞ where ¼ y À x In general, the axes of the ellipse are not in the Ox and Oy directions In (3-7a) the presence of the ‘‘product’’ term ExEy shows that it is actually a rotated ellipse; in the standard form of an ellipse the product term is not present In this section we find the mathematical relations between the parameters of the polarization ellipse, E0x , E0y , and and the angle of rotation , and another important parameter, , the ellipticity angle In Fig 3-3 we show the rotated ellipse Let Ox and Oy be the initial, unrotated, axes, and let Ox0 and Oy0 be a new set of axes along the rotated ellipse Furthermore, let ð0 Þ be the angle between Ox and the direction Ox0 of the major axis The components E 0x and E 0y are E 0x ¼ Ex cos E 0y ¼ ÀEx sin þ Ey sin þ Ey cos ð3-24aÞ ð3-24bÞ If 2a and 2b (a ! b) are the lengths of the major and minor axes, respectively, then the equation of the ellipse in terms of Ox0 and Oy0 can be written as E 0x ¼ a cosð þ Þ ð3-25aÞ E 0y ¼ Æb sinð þ Þ ð3-25bÞ where is the propagator and 0 is an arbitrary phase The Æ sign describes the two possible senses in which the end point of the field vector can describe the ellipse Figure 3-3 The rotated polarization ellipse Copyright © 2003 by Marcel Dekker, Inc All Rights Reserved The form of (3-25) is chosen because it is easy to see that it leads to the standard form of the ellipse, namely, 02 Ey E 02 x þ ¼1 a b ð3-26Þ We can relate a and b in (3-25) to the parameters E0x and E0y in (3-7a) by recalling that the original equations for the optical field are Ex ¼ cosð þ x Þ E0x ð3-27aÞ Ey ¼ cosð þ y Þ E0y ð3-27bÞ We then substitute (3-25) and (3-27) into (3-24), expand the terms, and write aðcos cos À sin sin Þ ¼ E0x ðcos cos x À sin sin x Þ cos þ E0y ðcos cos y À sin sin y Þ sin ð3-28aÞ Æbðsin cos þ cos sin Þ ¼ ÀE0x ðcos cos x À sin sin x Þ sin þE0y ðcos cos y À sin sin y Þ cos ð3-28bÞ Equating the coefficients of cos and sin leads to the following equations: a cos ¼ E0x cos x cos þ E0y cos y sin ð3-29aÞ þ E0y sin y sin ð3-29bÞ Æb cos ¼ E0x sin x sin À E0y sin y cos ð3-29cÞ Æb sin ¼ E0x cos x sin À E0y cos y cos ð3-29dÞ a sin ¼ E0x sin x cos Squaring and adding (3-29a) and (3-29b) and using ¼ y À x , we find that a2 ¼ E20x cos2 þ E20y sin2 þ 2E0x E0y cos sin cos ð3-30aÞ sin cos ð3-30bÞ Similarly, from (3-29c) and (3-29d) we find that b2 ¼ E20x sin2 þ E20y cos2 À 2E0x E0y cos Hence, a2 þ b2 ¼ E20x þ E20y ð3-31Þ Next, we multiply (3-29a) by (3-29c), (3-29b) by (3-29d), and add This gives Æab ¼ E0x E0y sin ð3-32Þ Further, dividing (3-29d) by (3-29a) and (3-29c) by (3-29b) leads to ðE20x À E20y Þ sin ¼ 2E0x E0y cos cos ð3-33aÞ or tan ¼ 2E0x E0y cos E20x À E20y which relates the angle of rotation Copyright © 2003 by Marcel Dekker, Inc All Rights Reserved ð3-33bÞ to E0x, E0y, and We note that, in terms of the phase , is equal to zero only for ¼ 90 or 270 Similarly, in terms of amplitude, only if E0x or E0y is equal to zero is equal to zero An alternative method for determining is to transform (3-7a) directly to (3-26) To show this we write (3-24a) and (3-24b) as Ex ¼ E 0x cos À E 0y sin ð3-34aÞ Ey ¼ E 0x sin þ E 0y cos ð3-34bÞ Equation (3-34) can be obtained from (3-24) by solving for Ex and Ey or, equivalently, replacing by À , Ex by E 0x , and Ey by E 0y On substituting (3-34a) and (3-34b) into (3-7a), the cross term is seen to vanish only for the condition given by (3-33) It is useful to introduce an auxiliary angle ð0 =2Þ for the polarization ellipse defined by tan ¼ E0y E0x ð3-35Þ Then (3-33) is easily shown by using (3-34) to reduce to tan ¼ 2E0x E0y tan cos ¼ cos E20x E20y À tan2 ð3-36Þ which then yields tan ¼ ðtan 2 Þ cos ð3-37Þ We see that for ¼ or the angle of rotation is ¼ Æ ð3-38Þ For ¼ =2 or 3=2 we have ¼ 0, so the angle of rotation is also zero Another important parameter of interest is the angle of ellipticity, This is defined by tan ¼ Æb a À ð3-39Þ We see that for linearly polarized light b ¼ 0, so ¼ Similarly, for circularly polarized light b ¼ a, so ¼ Æ=4 Thus, (3-39) describes the extremes of the ellipticity of the polarization ellipse Using (3-31), (3-32), and (3-35), we easily find that 2E0x E0y Æ2ab ¼ sin ¼ ðsin 2 Þ sin 2 E0x þ E20y a þb ð3-40Þ Next, using (3-39) we easily see that the left-hand side of (3-40) reduces to sin 2 , so we can write (3-40) as sin 2 ¼ ðsin 2 Þ sin ð3-41Þ which is the relation between the ellipticity of the polarization ellipse and the parameters E0x , E0y , and of the polarization ellipse Copyright © 2003 by Marcel Dekker, Inc All Rights Reserved We note that only for ¼ =2 or 3=2 does (3-41) reduce to ¼ Æ ð3-42Þ which is to be expected The results that we have obtained here will be used again, so it is useful to summarize them The elliptical parameters E0x, E0y, and of the polarization ellipse are related to the orientation angle and ellipticity angle by the following equations: tan ¼ ðtan 2 Þ cos ð3-43aÞ ð3-43bÞ sin 2 ¼ ðsin 2 Þ sin À 0, or by (3-43), < =4 For left-handed polarization the opposite is the case; i.e., to an observer looking in the direction from which the light is propagated, the electric vector would appear to describe the ellipse counterclockwise; in this case sin < 0, so that À=4 < REFERENCES Born, M and Wolf, E., Principles of Optics, 3rd ed., Pergamon Press, New York, 1965 Wood, R W., Physical Optics, 3rd ed., Optical Society of America, Washington, D.C., 1988 Strong, J., Concepts of Classical Optics, W H Freeman and Company, San Francisco, 1958 Jenkins, F S and White, H E., Fundamentals of Optics, McGraw-Hill, New York, 1957 Stone, J M., Radiation and Optics, McGraw-Hill, New York, 1963 Copyright © 2003 by Marcel Dekker, Inc All Rights Reserved