Circularly and linearly polarized light in the total ...

INSTITUTE OF PHYSICS PUBLISHING

JOURNAL OF OPTICS A: PURE AND APPLIED OPTICS

J. Opt. A: Pure Appl. Opt. 3 (2001) 398–406

PII: S1464-4258(01)21365-8

Circularly and linearly polarized light in the total reflection region at isotropic medium–uniaxial crystal interfaces M C Simon and L I Perez ´ Laboratorio de Optica, Departamento de F´ısica, Facultad de Ciencias Exactas y Naturales, Universidad de Buenos Aires, (1428) Buenos Aires, Argentina E-mail: [email protected]

Received 26 January 2001, in final form 13 June 2001 Published Online at stacks.iop.org/JOptA/3 Abstract We calculate the phases of the reflected light for the parallel and perpendicular modes under total reflection conditions at interfaces formed by an isotropic medium and a uniaxial crystal considering its principal planes. It is found that for incident light that is linearly polarized we can obtain circularly polarized reflected light with only one reflection with lower refractive indices than those required for isotropic interfaces. Moreover we find that, under certain conditions, the polarization of the reflected light is the same as that of the incident light in the total reflection zone. Keywords: Birefringence, polarization, total reflection

1. Introduction When linearly polarized light is incident at an interface formed by two dielectric isotropic media and the incidence angle is greater than the limiting total reflection angle, the reflected light is usually elliptically polarized. As has been already studied for isotropic interfaces [1, 2] and for interfaces formed by an isotropic medium and a uniaxial crystal [3, 4], the eccentricity and the orientation of the polarization ellipse depend on the characteristics of both media, on the polarization direction of the incident light and on the incidence angle. If the interface is formed by two isotropic media, the eccentricity of the ellipse decreases when the value of the relative refractive index between both media increases and, for values greater than 2.41, circularly polarized light can be obtained for two different incidence angles [1]. In this paper we show that when light is incident from an isotropic medium on a uniaxial crystal, circularly polarized light can be obtained with only one reflection without the requirement of such a high refractive index. To obtain this light, the interface must verify certain conditions, which we determine performing a detailed analysis of the variation of the phases of both polarization modes (parallel and perpendicular to the incidence plane) and considering that the incidence plane coincides with one of the crystal principal planes. From this analysis we also show that, for certain incidence angles greater than the total reflection 1464-4258/01/050398+09$30.00

© 2001 IOP Publishing Ltd

angle, if the incident light is linearly polarized then the reflected light is linearly polarized in the same direction as the incident one.

2. Characteristics of the reflected light in an interface isotropic medium–uniaxial crystal Various authors during the last few decades [5–14] have studied the optical properties of an interface isotropic medium– uniaxial crystal. Here we summarize these properties, writing the equations corresponding to the particular cases, which we consider in this paper, and in a way which is adequate for the development that follows. We take into account light which is incident from an isotropic medium of refractive index n, on a uniaxial crystal of principal refractive indices no (ordinary) and ne (extraordinary). The incidence planes that we consider are the principal planes of the crystal: one of them contains the optical axis and the other is perpendicular to this axis (figure 1). This means that in both cases there is separation of modes parallel and perpendicular to the incidence plane, so

Printed in the UK

Es∗ = Rs Es

(1)

Ep∗ = Rp Ep

(2) 398

Circularly and linearly polarized light in the total reflection region at isotropic medium–uniaxial crystal interfaces

When the refracted rays propagate through the crystal, both are perpendicular to the optical axis and their polarizations are such that the electric vector Eo of the ordinary ray is normal to the optical axis, and the electric vector of the extraordinary ray Ee is parallel to the optical axis. This means that the separation in modes is such that for the perpendicular mode there is only extraordinary ray and therefore the corresponding reflection coefficient is the same as the one corresponding to an isotropic interface of refractive indices n and ne . This reflection coefficient can be written as a function of sin αeT in the following way: cos α − sin2 αeT − sin2 α Res = . (7) cos α + sin2 αeT − sin2 α

(a)

The ordinary ray, on the other hand, has the polarization corresponding to the parallel mode and the reflection coefficient is given by n2o cos α − n2 sin2 αoT − sin2 α Rop = (8) n2o cos α + n2 sin2 αoT − sin2 α When the incidence angle is greater than the corresponding angles of total reflection, these coefficients are complex and can be written as Res = eiηes with

(b) Figure 1. Coordinate systems and characteristics of the media: interface isotropic medium–uniaxial crystal. The z-axis is contained in the interface. (a) Optical axis z3 perpendicular to the incidence plane. N˘ o and N˘ e are the normals to the wavefronts and coincide with the directions of the associated rays R˘ o and R˘ e respectively. (b) Optical axis z3 in the incidence plane. N˘ is the normal to the ordinary wavefront that coincides with the direction of the associated ray R˘ , and N˘ is the normal to the extraordinary wavefront that does not coincide with the direction of the ray R˘ .

where Es and Ep are the components of the electric field of the incident wave which are parallel and perpendicular to the incidence plane, Es∗ and Ep∗ are the corresponding components to the electric field of the reflected wave, and Rs and Rp are the corresponding reflection coefficients. When the optical axis is perpendicular to the incidence plane (figure 1(a)), the refractive angles βo and βe of both refracted rays are given by Snell’s law: n sin α = no sin βo

(3)

n sin α = ne sin βe

(4)

where α is the incidence angle. When the refractive index of the isotropic medium is greater than the principal refractive indices of the crystal no and ne , there are two total reflection angles: one for the ordinary ray (αoT ) and another for the extraordinary one (αeT ) given by no n ne = . n

sin αoT =

(5)

sin αeT

(6)

sin2 α − sin2 αeT ηes = −2artan cos α

for α > αeT , and

Rop = eiηop

n2 sin2 α − sin2 αoT = −2artan n2o cos α

(9)

(10)

(11)

with ηop

(12)

for α > αoT . The other incidence plane that we consider is the plane containing the optical axis (figure 1(b)). In this case the polarization of the refracted ray is inverted with respect to the previous case. The ray that is polarized perpendicularly to the incident plane is ordinary and, therefore, corresponds to the perpendicular mode. On the other hand, the extraordinary ray corresponds to the parallel mode since the electric vector corresponding to this ray is contained in the incidence plane. The reflection coefficient of the perpendicular mode is the same as that for an isotropic interface of refractive indices n and no and can be written as a function of sin αoT in the following way: cos α − sin2 αoT − sin2 α Rss = (13) cos α + sin2 αoT − sin2 α and for α > αoT this coefficient is complex: Rss = eiηss sin2 α − sin2 αoT ηss = −2artan . cos α

(14) (15) 399

M C Simon and L I Perez

optical axis forms with the interface according to the equation n2o + n2e − n2o sin2 ϑ . (17) sin αT = n

72 αoT

70 68

For α > αT we have

66

Rpp = eiηpp

64

α′′T

with

62

56 0

15

30

45 ϑ (a)

60

75

90

78 76

αeT

74 α′′T

72

3. Circular and linear polarization of the reflected ray

70 68 αoT

66 64 0

15

30

45 ϑ (b)

60

75

90

Figure 2. Ordinary and extraordinary total reflection angles as a function of the angle ϑ between the optical axis and the interface for the case in which the optical axis is in the incidence plane for (a) isotropic medium—calcite (n = 1.7550, no = 1.6584, ne = 1.4865 for λ = 632.8 nm). (b) Isotropic medium—vaterit (n = 1.700, no = 1.550, ne = 1.650 for λ = 589 nm).

The parallel mode is more complicated since the refractive index for the extraordinary ray depends on the incidence angle [5], and the ray direction, R˘ , does not coincide with the normal to the wavefront N˘ (figure 1(b)). The reflection coefficient Rpp , which results from solving the boundary conditions [6, 7, 9], can be written as

Rpp

−no ne cos α + n2 sin2 αT − sin2 α = no ne cos α + n2 sin2 αT − sin2 α

(16)

where is the limiting angle of total reflection for the extraordinary ray [5,6], that depends on the angle ϑ which the 400

(19)

where we see that the phase npp depends on the angle between the optical axis and the interface since αT varies with ϑ. When the crystal is negative, i.e. ne < no , the limiting total reflection angle decreases when ϑ increases as can be seen in figure 2(a) and from equation (17). In contrast, for positive crystals, i.e. ne > no , αT increases with ϑ (figure 2(b)). For both types of crystals and for ϑ = 0 both total reflection angles (ordinary and extraordinary) are equal, and for ϑ = π/2 the difference between them is maximum. Moreover, this difference is the same as the one that exists between αoT and αeT when the incidence plane is perpendicular to the optical axis since αT (ϑ = π2 ) = αeT .

αeT

58

αT



ηpp

60

 n2 sin2 α − sin2 αT  = −2artan  ne no cos α

(18)

When the incident ray is linearly polarized, the reflected ray is circularly polarized if the difference between the phases of both modes is equal to ±π/2 and the amplitudes of both components are equal (Ep∗ = Es∗ ). On the other hand we obtain linearly polarized light when the difference between the phases is zero or an integer multiple of π . It is well known that when the interface is formed by two isotropic media of refractive indices n and n , the phase difference ϕp (parallel mode) and ϕs (perpendicular mode) is cos α sin2 α − sin2 αT (20) ϕp − ϕs = 2artan − sin2 α where αT is the limiting angle for total reflection given by sin αT =

n . n

(21)

In figure 3(a) we represent ϕp , ϕs and ϕp − ϕs for an isotropic interface with n = 1.6584 and n = 1.7550, and in figure 3(b), the refractive indices of the first and second media are n = 2.755 and n = 1.000. We see that only in the second case the differences between the phases can be −π/2. From equation (20) we obtain that the incidence angles αCP for which ϕp − ϕs = −π/2 are given by 2 2 2 + n2 2 − 8n2 n2 n + n ± n sin2 αCP = (22) 4 and the discriminant is greater than zero only if √ n 1 + 2. n

(23)


0

α

30

60 αT

π

φp – φs αC1

π/2

αC3

ηop– ηes

π/2

φp

0

φs

αeT 60

ηes

αoT

80

ηop

−π

(a) α 0

60

30

90 −π

αT

φp – φs π/2

Figure 4. Phases and phase differences between the perpendicular and parallel reflected fields as functions of the incidence angle α for an interface formed by an isotropic medium and a uniaxial negative crystal when the optical axis is perpendicular to the incidence plane (n = 1.7550, no = 1.6584, ne = 1.4865).

3.1. Circularly and linearly polarized reflected light when the incidence plane is perpendicular to the optical axis φs φp

−π

(b) Figure 3. Phases and phase differences between the perpendicular and parallel reflected fields as functions of the incidence angle α for interfaces formed by two isotropic media (a) n = 1.7550, n = 1.6584. (b) n = 2.7550, n = 1.0000.

This is, real values are obtained for αCP only if the relative refractive index between both media is greater than 2.41 as we said in the introduction. Furthermore, we know that for linearly polarized incident light, the reflected light in the total reflection region is not linearly polarized except for the extreme values α = αT and α = π/2 or for the case in which the incident polarization corresponds to one of the eigenmodes of the isotropic media. When the second medium is a uniaxial crystal the behaviour of the phases and of the corresponding phase differences is qualitatively different, mostly because of the existence of two different total reflection angles each corresponding to one of the modes (ordinary and extraordinary). We analyse this behaviour first for the incidence plane, which is perpendicular to the optical axis and then for the incidence plane containing the optical axis.

When the incidence plane is perpendicular to the optical axis, the phases of the components Es∗ and Ep∗ are those given in equations (10) and (12). In figures 4 and 5 we represent these phases and the difference ηop − ηes for a negative crystal (calcite: no = 1.6584, ne = 1.4865 for λ = 632.8 nm) and a positive crystal (vaterit: no = 1.550, ne = 1.650 for λ = 589 nm), respectively. In both cases two incidence angles for which the phase difference is equal to +π/2 for calcite and −π/2 for vaterit are obtained. One of these angles is between the ordinary and extraordinary total reflection angles (termed αC1 for negative crystals and αC2 for positive crystals), and the other is greater than both (termed αC3 for negative crystals and αC4 for positive crystals)1 . When the angle is between the two total reflection angles, the polarization of the incident light has to compensate for the difference appearing in the reflected amplitudes because there still is a refracted ray. When the crystal is negative, we have αoT > αeT and the relation between the reflected components is ∗ Ep n2o cos αC1 − n2 sin2 αoT − sin2 αC1 = Es∗ C1 n2o cos αC1 + n2 sin2 αoT − sin2 αC1 Ep (24) × exp(−iηes ) Es C1 so the polarization of the incident ray has to be Ep n2o cos αC1 + n2 sin2 αoT − sin2 αC1 = . Es n2 cos αC1 − n2 sin2 αoT − sin2 αC1

(25)

o

1 We will use odd subindices for negative crystals and even ones for positive crystals.

401

M C Simon and L I Perez π

π

π/2

0

αoT

αeT

60

80

αL1

0 60

ηop– ηes

– π/2

80

ηop ηes

αC2

αC4

ηes ηop

−π

−π

Figure 5. Phases and phase differences between the perpendicular and parallel reflected fields as functions of the incidence angle α for an interface formed by an isotropic medium and a uniaxial positive crystal when the optical axis is perpendicular to the incidence plane (n = 1.755, no = 1.550, ne = 1.650).

Figure 6. Phases and phase differences between the perpendicular and parallel reflected fields as functions of the incidence angle α for an interface formed by an isotropic medium and a uniaxial negative crystal when the optical axis is perpendicular to the incidence plane (n = 2.0000, no = 1.6584, ne = 1.4865).

This means that Ep > Es as was to be expected since the refracted ray that is still present is the ordinary ray and it is polarized parallel to the incidence plane. The value of the angle αC1 as a function of the refractive indices is obtained from the condition ηes = −π/2 and from equation (10) n2 − n2e sin αC1 = . (26) √ n 2 When the crystal is positive, we have αoT < αeT (figure 5) and the refracted ray that is still present is the extraordinary ray which polarization is perpendicular to the incidence plane. Then the polarization of the incident ray, which is required for the reflected light to be circularly polarized, is Ep cos αC2 − sin2 αeT − sin2 αC2 = . (27) Es C2 cos αC2 + sin2 αeT − sin2 αC2

For positive crystals, the condition for the maximum value of the refractive index of the isotropic medium results from requiring ηop = −π/2 for α = αeT . From equation (12) we obtain no n2o − n4o − 4n2e n2e − n2o n (30) 2 n2e − n2o

This is, the component parallel to the incidence plane has to be smaller than the perpendicular component. The angle αC2 results from the condition ηop = −π/2, and from equation (12) we obtain

n2o + n2 sin αC2 = no . (28) n4 + n4o On the other hand, in figures 4 and 5 we see that the difference between both total reflection angles must have a minimum value to enable the phase difference to reach the value ±π/2. When the crystal is negative, the phase ηes has to be π/2 for α = αoT . This gives rise to a condition concerning the maximum value of the refractive index of the isotropic medium that enables the existence of circular polarization. From equation (10) we find that n 2n2o − n2e . (29) 402

so it results that there is an incidence angle for which the polarization of the reflected ray is circular if the relation between the principal indices of the crystal is √ 1+ 2 n2e (31) n2o 2 so that the discriminant of equation (30) must be positive and n a real number. When the refractive index of the isotropic medium is greater than the maximum value given in equations (29) and (30), circularly polarized light is not attained because the difference between both total reflection angles is too small. In figure 6 we show the phases and phase differences for calcite, of refractive principal indices no = 1.6584 and ne = 1.4865, and an isotropic medium of refractive index n = 2, i.e. greater than 2n2o − n2e . In this figure we also see that there is an incidence angle αL1 for which both phases are equal (ηop = nes ). From equations (10) and (12)

no n4 − n2o n2e sin αL1 = . (32) n n4 − n4o For this incidence angle, the phase difference between the parallel and perpendicular reflected fields is zero and the amplitudes are equal to those of the incident light because αL1

Circularly and linearly polarized light in the total reflection region at isotropic medium–uniaxial crystal interfaces π

0

π

ϑ = 0°

α′T = α′′T α0 60

0

80 ηpp – ηss

– π/2

ηpp

ηss

ϑ = 30°

α′′T

α′T

60

80 ηpp – ηss

– π/2

ηss ηpp

−π

−π (a)

(b)

π

π ϑ = 38.7°

0

α′′T

ϑ = 90°

α′T

60

0

80

α′′T

α′T

60

80

ηpp – ηss

ηpp – ηss – π/2

– π/2 ηpp

ηss

αC8

αC7 ηss ηpp

−π

−π (c)

(d)

Figure 7. Phases and phase differences between the perpendicular and parallel reflected fields as functions of the incidence angle α for an interface formed by an isotropic medium and a uniaxial negative crystal when the optical axis is in the incidence plane (n = 1.7550, no = 1.6584, ne = 1.4865) for different directions of the optical axis: (a) ϑ = 0◦ (αB is the Brewster’s angle), (b) ϑ = 30◦ , (c) ϑ = 38.7◦ , (d) ϑ = 90◦ .

is greater than both total reflection angles (figure 6). This means that the polarization of the reflected ray is equal to the polarization of the incident ray, whatever it is. Therefore, for linearly polarized incident light, the reflected light can also be linearly polarized and with the same polarization direction for an incidence angle greater than the limiting angles of total reflection. From equation (32) it also results that αL1 is greater than αoT if the discriminant is greater than one, this is, if the crystal is negative. 3.2. Circularly and linearly polarized reflected light when the incidence plane contains the optical axis When the optical axis is in the incidence plane and forming an arbitrary angle ϑ with the interface, the phases for the perpendicular and parallel modes are those given in equations (15) and (19). We represent these equations for various directions of the optical axis in figure 7 for negative

crystals and in figure 8 for positive crystals. In the particular case that the optical axis in the interface (ϑ = 0), both total reflection angles are equal and the phase difference ηpp − ηss is very small but can be increased by increasing the refractive index of the isotropic medium and, it is ∓π/2 if n = 3.8 for calcite and n = 3.9 for vaterit. Moreover in this case, there is an additional resource. It consists in changing the direction of the optical axis and therefore changing the limiting angle of total reflection for the extraordinary ray αT (figure 2). For negative crystals αT < αoT and the phases are those shown in figures 7(a)–(d). In these figures we see that the difference between both phases increases when the angle ϑ increases, and the phase differences reach the value −π/2 if ϑ is greater than a given minimum value. From equation (19) and the condition ηpp = −π/2 for α = αoT , we obtain that for negative crystals

no ne n2 − n2o sin ϑ 2 (33) n n2o − n2e 403

M C Simon and L I Perez π

π

ϑ = 0°

ϑ = 20°

π/2

π/2

αoT = α′′T

0

0

70 ηpp – ηss

80

90

αoT α''T ηss

αL2 70

ηpp – ηss 80

90

ηpp

ηss ηpp

−π

−π (a) π

(b) π

ϑ = 60.8°

π/2

ϑ = 90°

αC6

π/2

αC8

ηpp – ηss 0

αoT

ηpp – ηss

α′′T 70

80

0 90

αoT

α′′T 70

80

90

ηpp ηss

ηpp

ηss

−π

−π (c)

(d)

Figure 8. Phases and phase differences between the perpendicular and parallel reflected fields as functions of the incidence angle α for an interface formed by an isotropic medium and a uniaxial positive crystal when the optical axis is in the incidence plane (n = 1.700, no = 1.550, ne = 1.650) for different directions of the optical axis (a) ϑ = 0◦ , (b) ϑ = 20◦ , (c) ϑ = 60.8◦ , (d) ϑ = 90◦ .

so it results that the refractive index of the isotropic medium has to be n2o ne ne − 5n2e − 4n2o n (34) 2 n2o − n2e and for the square root to be real the refractive indices of the crystal must verify n2e 4 (35) 2 no 5 and also no > ne . For positive crystals, on the other hand, αT > αoT and to obtain a phase difference of π/2 we must have ηss = −π/2 for α = αT and from equations (15) and (17) we get

n2 − n2o (36) sin ϑ 2 n2e − n2o then n 404

2n2e − n2o

(37)

for the minimum value of ϑ to be less than or equal to π/2. We also see that n is a real number for any positive crystal since no < ne . When the refractive indices and the angle ϑ verify these conditions there are two incidence angles such that the phase difference is ±π/2, as can be seen in figures 7(d) and 8(d). One of the angles is between the two total reflection angles (αC5 for negative crystal and αC6 for positive crystal) and the other (αC7 for negative crystal and αC8 for positive crystal) is greater than both. When α > αT and α > αoT , both refracted rays are evanescent and the amplitudes of both reflected components are equal to the incident ones. This means that for the two angles αC7 and αC8 the reflected light is circularly polarized if the components of the incident field are equal to each other (Ep = Es ). For the angles αC5 and αC6 , on the other hand, a refracted ray still remains and, similarly to what happens when the optical axis is perpendicular to the incidence plane, the polarization of the incident ray has to compensate for the


difference appearing in the reflected amplitudes due to the polarization of the refracted ray. For negative crystals the refracted ray that remains is the ordinary ray, and which polarization is perpendicular to the incidence plane, and therefore the relation between the component of the incident electric field must be Ep cos αC5 − sin2 αoT − sin2 αC5 = (38) Es C5 cos αC5 + sin2 αoT − sin2 αC5 where αC5 results from the condition ηpp = −π/2 and from equation (19):

n2e n2o + n4 sin2 αT sin αC5 = . (39) n2e n2o + n4 For positive crystals, on the other hand, the ray that remains is the refracted extraordinary ray and the incident polarization must be 2 n cos α + n sin2 αT − sin2 αC6 n o e C6 Ep = (40) Es C6 −n n cos α + n2 sin2 α − sin2 α o e

C6

T

C6

this is Ep > Es . The angle αC6 is obtained from the condition ηss = −π/2 and from equation (15) so n2 − n2o sin αC6 = . (41) √ n 2 From figures 7(b) and 8(b) we see that when the difference between the angles αT and αoT is small and when the crystal is positive, ηpp and ηss intersect each other because αT > αoT . That is, there is an angle αL2 for which both phases are equal. This angle is obtained from equations (15) and (19), i.e.

n4 sin2 αT − n2e n2o sin2 αT . (42) sin αL2 = n4 − n2e n2o This angle exists when the angle between the interface and the optical axis verifies

(n2e n2o − n4 )n2o sin ϑL2 (43) n4 (n2e − n2o ) which is obtained from the condition sin αL2 1 and from the expressions for sin αoT and sin αT as functions to the refractive indices. That is, when the optical axis is in the incident plane, there is an incidence angle for which the polarization of the reflected ray is equal to the polarization of the incident ray only if the crystal is positive.

4. Conclusions The possibility of obtaining circularly polarized light for only one reflection when linearly polarized light is incident on an interface formed by an isotropic dielectric medium and a uniaxial dielectric one (such that the incidence plane coincides with one of the crystal principal planes) has been analysed. Comparing the results with the well known case of

isotropic interfaces, it was obtained that there can also exist two incidence angles for which the reflected light is circularly polarized in our case. Nevertheless the physical situation is qualitatively quite different. In the case in which the second medium is anisotropic, there are two total reflection angles (ordinary and extraordinary, one for each mode of polarization). This gives rise to a great variation of the phase differences between both modes of the reflected field because one of them is null when one of the refracted rays still subsists. The absolute value of the phase difference is increased when the incidence angle is increased until the second angle of total reflection is reached and this difference begins to decrease. From this it follows that the maximum value of the phase difference increases with the difference between both angles of total reflection. However, as this last difference increases as the refractive index of the first medium decreases, it follows that we can obtain a phase difference equal or greater than π/2 for lower indices for isotropic–anisotropic interfaces while for isotropic ones the index must be increased. Furthermore, when isotropic–uniaxial interfaces are considered, one of the obtained incidence angles is greater than both the angles of total reflection and the other lies between them. For the first one it is necessary that the incident ray is polarized at 45◦ with respect to the incidence plane, while for the other angle the polarization of the incident ray has to compensate for the remaining polarization of the refracted ray. It does not happen at isotropic interfaces in which there is only one angle of total reflection and consequently the polarization of the ray must be linear at 45◦ in order to obtain circularly reflected light. As another consequence of the existence of two limiting angles of total reflection, we found a property that does not appear at isotropic interfaces: the possibility of finding incidence conditions under which the polarization of the reflected ray is the same as that of the incident ray. This occurs when the phases for both modes coincide and the absolute values of the reflection coefficients are equal to one, that is for a certain incidence angle which is greater than both total reflection angles.

Acknowledgments This work has been carried out with the support of CONICET and UBA. M C Simon and L I Perez are members of CONICET.

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M C Simon and L I Perez

[11] Zhang W Q 1992 Appl. Opt. 31 7328–31 [12] Lekner J 1993 J. Opt. Soc. Am. A 10 2059–64

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[13] Perez L I 1993 Ann. Assoc. F´ısica Argentina 5 224–8 [14] Simon M C and Far´ıas D 1994 J. Mod. Opt. 41 413–29

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