3 The Polarization Ellipse

Equations 3-3a and 3-3b are spoken of as the polarized or polarization components of the optical field.. 3.2 THE INSTANTANEOUS OPTICAL FIELD AND THE POLARIZATION ELLIPSE In previous secti

The Polarization Ellipse

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,

r2uiðr, tÞ ¼ 1

v2

@2uiðr, tÞ

where v is the velocity of propagation of the oscillation and r ¼ r(x, y, z) In a Cartesian system the components uxðr, tÞ and uyðr, tÞ are said to be the transverse components, and the component uzðr, 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

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 com-ponent (3-2c) did not exist That is, light consisted only of the transverse comcom-ponents (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

where u0xand u0yare the maximum amplitudes and xand yare 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 trans-verse 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

1015sec at optical frequencies In this chapter we find the equation that arises when the propagator is eliminated between the transverse components In order to do this

we show inFig 3-1the transverse optical field propagating in the z direction The transverse components are represented by

where  ¼ !t  z is the propagator The subscripts x and y refer to the components

in the x and y directions, E0xand E0yare the maximum amplitudes, and xand yare 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 do this (3-4a) and (3-4b) are written as

Ex

Ey

Ex

E0xsin y

Ey

Ex

E0xcos y

Ey

Squaring (3-6a) and (3-6b) and adding gives

E2x

E2

0x

þE2y

E2 0y

2Ex

E0x

Ey

E0ycos  ¼ sin

where

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 InFig 3-2the ellipse is shown inscribed within a rectangle whose sides are parallel to the coordinate axes and whose lengths are 2E0xand 2E0y

We now determine the points where the ellipse is tangent to the sides of the rectangle We write (3-7a) as

E0x2E2y ð2E0xE0yExcos ÞEyþE0y2ðE2xE0x2sin2Þ ¼0 ð3-8Þ The solution of this quadratic equation (3-8) is

Ey¼E0yExcos 

E0ysin 

E0x ðE

2

At the top and bottom of the ellipse where it is tangent to the rectangle the slope

is 0 We now differentiate (3-9), set E0y¼dEy=dEx¼0, and find that

Figure 3-1 Propagation of the transverse optical field

Substituting (3-10a) into (3-9), the corresponding values of Eyare found to be

Similarly, by considering (3-9) where the slope is E0y ¼ 1 on the sides of the rectangle, the tangent points are

Equations (3-10) and (3-11) show that the maximum length of the sides of the ellipse are Ex¼ E0x and Ey¼ E0y The ellipse is tangent to the sides of the rectangle at ðE0x, E0ycos Þ and ðE0xcos , E0yÞ We also see that (3-10) and (3-11) show that the extrema of Ex and Eyare E0xand E0y, respec-tively

InFig 3-2the ellipse is shown touching the rectangle at point A, B, C, and D, the coordinates of which are

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 Figure 3-2 An elliptically polarized wave and the polarization ellipse

the x axis the ellipse is tangent at the extrema x ¼ E0xand x ¼ þE0x The area of the ellipse above the x axis is given by

A ¼

ZþE 0x

E 0x

Substituting (3-9) into (3-13) and evaluating the integrals, we find that the area of the polarization ellipse is

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 E0xE0y, whereas for  ¼ 0 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

1 E0y¼0 In this case Ey(z, t) is zero and (3-4) becomes

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 E0x¼0 and Eyðz, tÞ 6¼ 0, then we have a linear oscillation along the

yaxis, and we speak of linear vertically polarized light

2  ¼0 or  Equation (3-7a) reduces to

E2x

E2

0x

þE2y

E2 0y

2Ex

E0x

Ey

Equation (3-16) can be written as

Ex

E0x

Ey

E0y

whence

Ey¼  E0y

E0x

Equation (3-18) is recognized as the equation of a straight line with slope ðE0y=E0xÞ and zero intercept Thus, we say that we have linearly polarized light with slope

ðE0y=E0xÞ The value  ¼ 0 yields a negative slope, and the value  ¼  a positive slope If E0x¼E0y, then we see that

The positive value is said to represent linear þ45
polarized light, and the negative value is said to represent linear 45
polarized light

3  ¼ /2 or 3/2 The polarization ellipse reduces to

E2x

E2 0x

þ E2y

E2 0y

This is the standard equation of an ellipse Note that  ¼ /2 or  ¼ 3/2 yields the identical polarization ellipse

4 E0x ¼E0y ¼E0 and  ¼ /2 or  ¼ 3/2 The polarization ellipse now reduces to

E2x

E2þE2y

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

We see that for  ¼ 0 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 E0x¼E0y¼E0 and, when  ¼ =2 or 3=2, (3-22) becomes

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 45
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

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:

E2x

E2

0x

þE2y

E2 0y

2Ex

E0x

Ey

E0ycos  ¼ sin

2

where  ¼ yx 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 Ox0and Oy0be a new set of axes along the rotated ellipse Furthermore, let ð0 Þ be the angle between Ox and the direction Ox0of the major axis The components E0x and E0y are

the equation of the ellipse in terms of Ox0and Oy0 can be written as

where  is the propagator and 0is an arbitrary phase The  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

The form of (3-25) is chosen because it is easy to see that it leads to the standard form of the ellipse, namely,

E02x

a2 þE02y

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

Ey

We then substitute (3-25) and (3-27) into (3-24), expand the terms, and write aðcos  cos 0sin  sin 0Þ ¼E0xðcos  cos xsin  sin xÞcos

þE0yðcos  cos ysin  sin yÞsin ð3-28aÞ

bðsin  cos 0þcos  sin 0Þ ¼ E0xðcos  cos xsin  sin xÞsin

þE0yðcos  cos ysin  sin yÞcos ð3-28bÞ Equating the coefficients of cos  and sin  leads to the following equations:

acos 0¼E0xcos xcos þ E0ycos ysin ð3-29aÞ

asin 0¼E0xsin xcos þ E0ysin ysin ð3-29bÞ

bcos 0¼E0xsin xsin  E0ysin ycos ð3-29cÞ

bsin 0¼E0xcos xsin  E0ycos ycos ð3-29dÞ Squaring and adding (3-29a) and (3-29b) and using  ¼ yx, we find that

a2¼E20xcos2 þ E20ysin2 þ2E0xE0ycos sin cos  ð3-30aÞ Similarly, from (3-29c) and (3-29d) we find that

b2¼E20xsin2 þ E20ycos2 2E0xE0ycos sin cos  ð3-30bÞ Hence,

Next, we multiply (3-29a) by (3-29c), (3-29b) by (3-29d), and add This gives

Further, dividing (3-29d) by (3-29a) and (3-29c) by (3-29b) leads to

ðE20xE20yÞsin 2 ¼ 2E0xE0ycos  cos 2 ð3-33aÞ or

tan 2 ¼2E0xE0ycos 

E2 0xE2 0y

ð3-33bÞ which relates the angle of rotation 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 E0xor E0yis 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

Equation (3-34) can be obtained from (3-24) by solving for Ex and Ey or, equivalently, replacing by  , Exby E0x, and Eyby E0y 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)

ellipse defined by

E0y

Then (3-33) is easily shown by using (3-34) to reduce to

tan 2 ¼2E0xE0y

E2 0xE2 0y

cos  ¼

which then yields

ð3-37Þ

We see that for  ¼ 0 or  the angle of rotation is

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 ¼b



4



We see that for linearly polarized light b ¼ 0, so ¼ 0 Similarly, for circularly polarized light b ¼ a, so ¼ =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

2ab

a2þb2¼ 2E0xE0y

E2 0xþE2 0y

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

ð3-41Þ which is the relation between the ellipticity of the polarization ellipse and the parameters E0x, E0y, and  of the polarization ellipse

We note that only for  ¼ =2 or 3=2 does (3-41) reduce to

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:



4 <



E0y

tan ¼b

We emphasize that the polarization ellipse can be described either in terms of the orientation and ellipticity angles and on the left-hand sides of (3-43a) and (3-43b) or the major and minor axes E0x and E0y and the phase shift  on the right-hand sides of (3-43a) and (3-43b)

Finally, a few words must be said on the terminology of polarization Two cases of polarization are distinguished according to the sense in which the end point of the field vector describes the ellipse It seems natural to call the polarization right-handed or left-handed according to whether the rotation of E and the direction

of propagation form a right-handed or left-handed screw The traditional terminol-ogy, however, is just the opposite and is based on the apparent behavior of E when viewed face on by the observer In this book we shall conform to the traditional, that

is, customary usage Thus, the polarization is right-handed when to an observer looking in the direction from which the light is coming, the end point of the electric vector would appear to describe the ellipse in the clockwise sense If we consider the value of (3-4) for two time instants separated by a quarter of a period, we see that

in this case sin  > 0, or by (3-43), 0 < =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 =4 < 0

REFERENCES

1 Born, M and Wolf, E., Principles of Optics, 3rd ed., Pergamon Press, New York, 1965

1988

1958

4 Jenkins, F S and White, H E., Fundamentals of Optics, McGraw-Hill, New York, 1957

by (3- 33)

ellipse defined by

E0y

Then (3- 33) is easily... polarized light b ¼ a, so ¼ =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

2ab

a2ỵb2ẳ... 2E0xE0ycos sin cos  ? ?3- 30bÞ Hence,

Next, we multiply (3- 29a) by (3- 29c), (3- 29b) by (3- 29d), and add This gives

Further, dividing (3- 29d) by (3- 29a) and (3- 29c) by (3- 29b) leads to

E20xE20yịsin