Minimum deviation angle in uniaxial prisms

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Avendaño-Alejo et al. Vol. 24, No. 8 /August 2007 /J. Opt. Soc. Am. A 2431

Minimum deviation angle in uniaxial prisms

M. Avendaño-Alejo,1,* Ivan Moreno,2 and Orestes Stavroudis3

1Universidad Nacional Autónoma de México, Centro de Ciencias Aplicadas y Desarrollo Tecnológico,Apdo. Postal 70-186, C. P. 04510, Distrito Federal, México

2Universidad Autónoma de Zacatecas, Facultad de Física, Apdo. Postal c-580, C. P. 98068, Zacatecas Zac., México3Centro de Investigaciones en Óptica, A. C., Apdo. Postal 1-948, C. P. 37150, León Guanajuato, México

*Corresponding author: maximino.avendano@ccadet.unam.mx

Received November 14, 2006; revised February 23, 2007; accepted March 3, 2007;posted March 21, 2007 (Doc. ID 77068); published July 11, 2007

Dispersion in a uniaxial prism provides an example of the application of the law of refraction for a uniaxialcrystal. Formulas for the minimum deviation angle for the extraordinary ray are given when the crystal axislies in the plane of incidence. Three particular cases for the crystal axis position are presented and are shownto have a behavior similar to that of an ordinary prism. © 2007 Optical Society of America

OCIS codes: 080.2710, 260.1180, 260.1440.

tbfewc

2AiifiYs=inifrd�

wt

c(

. INTRODUCTIONhe wide range of applications for uniaxial crystals areell known. They are used as laser harmonic generators,olarizing prisms, liquid crystal displays, and birefrin-ent filters [1–4]. When a ray of light from a monochro-atic source is incident on a uniaxial surface, two rays,

n ordinary ray and an extraordinary ray, will propagatenside the surface until they are refracted at the secondnterface of the crystal. These two rays are polarized or-hogonally; so they do not interfere. The ordinary ray sat-sfies Snell’s law, so that the normal to the wave front andhe direction of energy flux coincide. For the extraordi-ary ray the energy flux direction does not generally co-

ncide with the direction normal to the wave front.It is well known that there are several methods by

hich we can obtain the minimum deviation angle in ansotropic prism to evaluate either the index of refractionf the prism for a defined wavelength or, equivalently, itsispersion relation. A geometrical method to obtain theinimum deviation angle is given in Jenkins and White

5]. An alternative method using differential forms is ex-lained by Klein and Furtak [6], but neither of theseethods calculate the total derivative; instead they only

pply a relationship between the angles involved in theefraction process through the prism. Simon [7] obtainedhe minimum deviation angle for uniaxial prisms by us-ng an effective refractive index, which can be expressedn terms of the phase velocity.

In this paper the minimum deviation angle for the or-inary ray is obtained by two methods different fromhose explained above. In the first method we have calcu-ated the total derivative, to obtain the extremal valueshat give either the minimum or the maximum for the to-al deviation angle. The second method to obtain theinimum deviation angle for the ordinary ray is by using

1084-7529/07/082431-7/$15.00 © 2

he principle of reversibility; with this method the alge-raic solutions are reduced. By using the formulas of re-raction and applying the principle of reversibility for thextraordinary ray, we obtain its minimum deviation anglehen the crystal axis is perpendicular to the plane of in-

idence and when it lies in the plane of incidence.

. PRELIMINARIESlocal coordinate system �X� ,Y� ,Z�� is established with

ts origin located at the point of incidence of the ray at thenterface between two (isotropic-uniaxial) refracting sur-aces. The Z� axis is parallel to the normal to the refract-ng surface at the point of incidence. We assume that the�–Z� plane is the incidence plane; then the direction co-

ines for the incident ray are Si= ��i ,�i ,�i��0,sin �i , cos �i�, where �i is the angle of incidence in the

sotropic media; thus the direction cosines for the ordi-ary ray are So= ��o ,�o ,�o�= �0,sin �o1 ,cos �o1�, where �o1

s the refracted angle for the ordinary ray in the first sur-ace with respect to the axis Z�, as shown in Fig. 1. Theefraction from an isotropic medium to a birefringent me-ium for the ordinary ray is given by the usual Snell’s lawno sin �o1=ni sin �i�; thus we have

�o1 = arcsin�ni sin �i

no� , �1�

here ni and no are the incident and the ordinary refrac-ive index, respectively.

The refraction for the extraordinary ray Se when therystal axis has an arbitrary orientation is given by Eq.47) of [8] as

007 Optical Society of America

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wan

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Fnad

Fa

2432 J. Opt. Soc. Am. A/Vol. 24, No. 8 /August 2007 Avendaño-Alejo et al.

�e

�e

�e� =

1

�eg��no

2�ne2 + �no

2 − ne2��1

2 − no2�no

2 − ne2��1�1 0

− no2�no

2 − ne2��1�1 no

2�ne2 + �no

2 − ne2��1

2 0

0 0 �/�o�� �o

�o

�o� − 1�no

2 − ne2��

1 0 0

0 1 0

0 0 0��

�1

�1

1� , �2�

w

wtwajn

tcw

Ee

3OTi

wf

N

here ne is the extraordinary refractive index and we de-ne

�eg2 = ne

2�no2�2 − �no

2 − ne2��1 + no

2��1�o + �1�o�2�,

2 = ��ne2 − no

2��o2 + �o

2� + no2�no

2 − ne2���1�o + �1�o�2,

� = ne21

2 + no2��1

2 + �12�, �3�

here ��1 ,�1 ,1� are the direction cosines of the crystalxis and ��o ,�o ,�o� are the direction cosines for the ordi-ary ray given above.If we know the direction cosines of the extraordinary

ay Se1, where the subindex 1 refers to the point of con-act at the first surface, and we know the direction cosinesf the crystal axis, as shown in Fig. 1, the extraordinaryay can be traced out of the crystal by calculating a fictiverdinary ray Sof= ��of ,�of ,�of� at the point of incidence ofhe extraordinary ray and the exiting surface. The ordi-ary fictive ray from Eq. (58) of [8] is given by

ig. 1. Direction cosines for incident, ordinary, and extraordi-ary refracted rays and the optical crystal axis in both Y�–Z�nd Y�–Z� system coordinates. The fictive ordinary ray also israwn.

ig. 2. Angles involved for obtaining the total deviation angle incommon prism.

��of

�of

�of� =

1

no�e��ne

2 + �no2 − ne

2��12 �no

2 − ne2��1�1 0

�no2 − ne

2��1�1 ne2 + �no

2 − ne2��1

2 0

0 0 e/�e1��

���e1�

�e1�

�e1�� + �e1� 1�no

2 − ne2��

1 0 0

0 1 0

0 0 0��

�1

�1

1� , �4�

here we define

�e2 = ne

2 sin2 � + no2 cos2 �,

e2 = ne

2�no2 − ne

2�sin2 � + ��no2 − ne

2�1 cos � + ne2�e1� 2,

cos � = �1�e1� + �1�e1� + 1�e1� , �5�

here ��e1� ,�e1� ,�e1� � are the direction cosines of the projec-ion of the extraordinary ray at the exit surface, in whiche form a new local coordinate system �Y�–Z��; A1 and A2re parallel in a general coordinate system, but their pro-ections could have different values with respect to eachormal, as shown in Fig. 1.By using Snell’s law we refract the ordinary fictive ray

o obtain the direction of the extraordinary ray out of therystal Se2. In an isotropic media with refractive index ni,e get

��e2

�e2

�e2� =

1

ni�no 0 0

0 no 0

0 01

�of

ni2 − no

2�1 − �of2 ��� �of

�of

�of� . �6�

quation (6) gives the direction cosines for the refractedxtraordinary rays in the isotropic media.

. MINIMUM DEVIATION ANGLE FOR THERDINARY RAY

he total deviation angle in an isotropic prism, as shownn Fig. 2, is given by

o = �i + �t2 − �, � = �o1 + �i2, �7�

here � is the constant apex angle. We calculate �t2 as aunction of �i by using Snell’s law, Eq. (1),

�t2��i� = arcsin�no sin�� − �o1�

ni�

= arcsin�no sin�� − arcsin�ni sin �i

no��

ni� . �8�

ext we take the first derivative of Eq. (7) with respect to

�u

Cw

R

bt�

Bc

Tardt

wn

ipfw

Wp

4TFA=

B

w

Ft

wmftPof�

Fotgn

Avendaño-Alejo et al. Vol. 24, No. 8 /August 2007 /J. Opt. Soc. Am. A 2433

i by using Eq. (8) to obtain the extremal values that gives either minimum or maximum values for the functiono. Thus we get

d o

d�i= 1 − � ni cos �i�cos � +

ni sin � sin �i

no2 − ni

2 sin2 �i�

ni2 − �ni cos � sin �i − sin � no

2 − ni2 sin2 �i�2

= 0.

ollecting similar terms to find the roots of this equation,e get

ni2 cos2 �i�ni sin � sin �i + cos � no

2 − ni2 sin2 �i�2

= � no2 − ni

2 sin2 �i

� ni2 − �ni cos � sin �i − sin � no

2 − ni2 sin2 �i�22.

educing this formula further, we have

�ni2 − no

2�2�no4 sin2 � − 4ni

2no2 sin2 �i + 4ni

4 sin4 �isin2 � = 0,

�9�

ut ni�no and ��0; then the only solution for this equa-ion is obtained by solving the square bracketed term fori; thus we obtain four possible values, given by

�i+ = ± arcsin�no sin��� − ��/2

ni�,

�i− = ± arcsin�no sin��/2�

ni� . �10�

y substituting the four solutions into Eqs. (7) and (8) toalculate the extremal values, we get

o��i++ � = arcsin�no cos��/2�

ni� − arcsin�no cos�3�/2�

ni� − �,

o��i−− � = − arcsin�no sin��/2�

ni� + arcsin�no sin�3�/2�

ni�

− �,

o��i+− � = − �,

o��i−+ � = arcsin�no sin��/2�

ni� + arcsin�no sin��/2�

ni� − �.

�11�

he first three solutions are mathematically correct butre not physically possible, because the incident and theefracted ray have different values for the angle of inci-ence; thus the only possible solution is for o��i−

+ �, so thathe principle of reversibility is satisfied [5,6]. Thus we get

mo = 2 arcsin�no sin��/2�

ni� − �, �12�

here mo is the minimum deviation angle for the ordi-ary ray. From Eq. (12) we conclude that � =� as shown

n Fig. 3. If we know experimentally the value for mo, it isossible to obtain with high precision the value for the re-raction index of the prism, with the following expression,hich is obtained easily from Eq. (12):

no =ni sin�� mo + ��/2

sin��/2�. �13�

ith this expression it is also possible to obtain therism’s dispersion for isotropic media.

. ALTERNATIVE METHOD TO OBTAINHE MINIMUM DEVIATION ANGLEOR THE ORDINARY RAYs we know from the principle of reversibility that �i�t2, then from Eqs. (7) and (8) we have

�i = arcsin�no sin�� − arcsin�ni sin �i

no��

ni� . �14�

y solving Eq. (14) for �i, we get

ni sin �i

no= sin�� − arcsin�ni sin �i

no�� ,

hich we can reduce further to

ni2 sin2 �i − no

2 sin2��/2� = 0.

inally, solving this for �i, we obtain a solution similar tohat of �i−

+ from Eq. (10):

�imo= �i−

+ = arcsin�no sin��/2�

ni� , �15�

here �imois the angle of incidence that gives the mini-

um deviation angle for the ordinary ray. An importantact is that the number of solutions is reduced from four towo, and we take only the positive value of Eq. (15).hysically this means that if we trace a ray with an anglef incidence given by �imo

and apply Snell’s law the re-racted ray inside the prism has an angle of refraction

=� /2, and when the ordinary refracted ray inside the

ig. 3. Thicker lines give the angle of refraction for which webtain the minimum deviation angle for both ordinary and ex-raordinary rays as a function of the apex angle. Dashed linesive possible orientations of the crystal axis with respect to theormal to the refracting surface as a function of the apex angle.

o1

pafFiad

5EAApptw

oa=t

widESt=iedsr

TtE

wotph

wTomg

wpmtc

Wcotm�ia

BWd=arnitid

2434 J. Opt. Soc. Am. A/Vol. 24, No. 8 /August 2007 Avendaño-Alejo et al.

rism has this angle we obtain the minimum deviationngle. On the other hand, if the ray crosses the prism itorms an isosceles triangle inside the prism, as shown inig. 3. With this we ensure that the principle of revers-

bility is satisfied by the symmetry of the isosceles tri-ngle, which becomes a unique solution for the minimumeviation angle.

. MINIMUM DEVIATION ANGLE FOR THEXTRAORDINARY RAY. Crystal Axis Perpendicular to the Plane of Incidences we know, the direction of the extraordinary ray de-ends on the direction of the crystal axis. Throughout thisaper, a superscript on the angles involved in the rayracing process will denote the position of the crystal axisith respect to the plane of incidence.Again, if we suppose that the Y�–Z� plane is the plane

f incidence, then the conditions under which the crystalxis is perpendicular to this plane are that �1=1, �1=10, additionally �i=0, and therefore �o=0. Substituting

hese values into Eqs. (2) and (3), we obtain

�e1� = arctan� ni sin �i

ne2 − ni

2 sin2 �i� , �16�

here �e1� is the refracted angle for the extraordinary ray

nside the uniaxial prism when the crystal axis is perpen-icular to the plane of incidence. Equation (16) is equal toq. (1) with ne instead of no, or, on the other hand, it isnell’s law for an isotropic medium with index of refrac-ion ne. To calculate the exiting ray, we use the values �e�0, �1=1, �1=1=0, and �e�=sin �i2, �e�=cos �i2, where �i2

s the angle of incidence of the extraordinary ray on thexiting surface given by �i2=�−�e1

� . We calculate the or-inary fictive ray through Eqs. (4) and (5), which are sub-tituted into Eq. (6) to finally yield the extraordinary rayefracted out of the prism; in this way we have

�e2� = arctan� ne sin�� − �e1

� �

ni2 − ne

2 sin2�� − �e1� �� . �17�

he relationship for the total deviation angle for the ex-raordinary ray [7] is similar to that for the ordinary ray,q. (7), so that

e = �i + �e2 − �, �18�

here �e2 is the angle of refraction for extraordinary rayut the prism. To obtain the minimum deviation angle forhe extraordinary ray, we known that �i=�e2, or in thisarticular case, �i=�e2

� must be satisfied; thus with theelp of Eqs. (16) and (17) and by solving for �i we have

�ne4 sin2 � − 4ni

2ne2 sin2 �i + 4ni

4 sin4 �i�sin2 � = 0,

hose solutions are similar to Eq. (10) for ne instead of no.hus when the crystal axis is perpendicular to the planef incidence, the only angle of incidence that gives theinimum deviation angle for the extraordinary ray is

iven by

�ime

� = arcsin�ne sin��/2�

ni� , �19�

here the superscript � means that the crystal axis iserpendicular to the plane of incidence. Therefore theinimum deviation angle for the extraordinary ray when

he crystal axis is perpendicular to the plane of incidencean be calculated through Eq. (12) with ne instead of no:

me� = 2 arcsin�ne sin��/2�

ni� − �. �20�

hen the crystal axis is perpendicular to the plane of in-idence, both ordinary and extraordinary rays obey therdinary law of refraction with indices no and ne, respec-ively. In general, for isotropic media we can write theinimum deviation angle given by I

m�nI�=2�Im−�, forIm=arcsin��nI /ni�sin�� /2�, where ni, nI are incident andsotropic refraction indices, respectively, and � is the apexngle.

. Crystal Axis in the Plane of Incidencehen the crystal axis lies in the plane of incidence, the

irection cosines for the crystal axis are A1= ��1 ,�1 ,1��0,sin � , cos �� where � is the angle between the crystalxis and the Z� axis. The sign convention is positive to theight-hand side. The direction cosines for the extraordi-ary ray are Se= ��e ,�e ,�e�= �0,sin �e1 ,cos �e1�, where �e1

s the angle between the refracted extraordinary ray andhe Z� axis. By substituting these expressions given abovento Eqs. (2) and (3), the refracted angle for the extraor-inary ray is given by

tan �e1 =nineno sin �i + �ne

2 − no2�cos � sin � ne

2 cos2 � + no2 sin2 � − ni

2 sin2 �i

�ne2 cos2 � + no

2 sin2 �� ne2 cos2 � + no

2 sin2 � − ni2 sin2 �i

, �21�

atta

�Et

wft4s

f

B

R

�

w

Ftp

Avendaño-Alejo et al. Vol. 24, No. 8 /August 2007 /J. Opt. Soc. Am. A 2435

nd the extraordinary ray gives the direction ofhe energy flux or the Poynting vector. To tracehe extraordinary ray out of the prism, we calculate

fictive ordinary ray, with A1= �0,sin � , cos �� and w

or the extraordinary ray, Eq. (18) must be satisfied [7]. As

wwhasst

mcst�

ontg

W(rt(=a

�e� ,�e� ,�e��= �0,sin��−�e1� , cos��−�e1��; substituting intoqs. (4) and (5), we obtain the values for the ordinary fic-

ive ray, and finally, substituting these values into Eq. (6),

tan �e2 =ne

2 cos � sin � + no2 cos � sin �

ni2�no

2 cos2 � + ne2 sin2 �� − �ne

2 cos � sin � + no2 cos � sin ��2

, �22�

here �= ��−�e1−��, and �e2 gives the angle of refractionor the extraordinary ray out of the prism when the crys-al axis lies in the plane of incidence, as is shown in Fig.. When ne→no, Eq. (22) is reduced to Eq. (17) with no in-tead of ne.

In order to again obtain the minimum deviation angle

e know that the minimum deviation angle is reachedhen �i=�e2 or, equivalently, when the extraordinary rayas an angle of refraction given by �e1=� /2, as explainedbove, this means that the extraordinary ray will cross in-ide the prism, forming an isosceles triangle, and it alsoatisfies the principle of reversibility as shown in Fig. 3;

tan��

2 � =nineno sin �i + �ne

2 − no2�cos � sin � ne

2 cos2 � + no2 sin2 � − ni

2 sin2 �i

�ne2 cos2 � + no

2 sin2 �� ne2 cos2 � + no

2 sin2 � − ni2 sin2 �i

. �23�

y again solving Eq. (23) for �i, we get

�i = arcsin� �ne2 cos � sin���/2� − � + no

2 sin � cos���/2� − �� ne2 cos2 � + no

2 sin2 �

ni ne2no

2 cos2���/2� + �ne2 cos � sin���/2� − � + no

2 cos���/2� − �sin ��2� .

educing further, finally we have

ime

� = �i

= arcsin

��ne2 cos � sin���/2� − � + no

2 sin � cos���/2� − �

ni no2 cos2���/2� − � + ne

2 sin2���/2� − � � ,

�24�

here �ime

� is the angle of incidence, which gives the mini-

ig. 4. Angles and direction cosines involved for obtaining theotal deviation angle for the extraordinary ray in a uniaxialrism.

um deviation angle for the extraordinary ray when therystal axis lies in the plane of incidence. It is easy tohow from Eq. (22) that �e2=�ime

� for �e1=� /2; in this wayhe principle of reversibility is satisfied, and thereforee2=�i.

By substituting Eq. (24) into Eq. (18) for �i=�e2=�ime

� tobtain the minimum deviation angle for the extraordi-ary ray � me

� � as a function of the angle between the crys-al axis and the normal to the refracting surface ���, weet

me� = 2 arcsin

��ne2 cos � sin���/2� − � + no

2 sin � cos���/2� − �

ni no2 cos2���/2� − � + ne

2 sin2���/2� − � �− �. �25�

e now can calculate for each position of �, through Eq.25), the minimum deviation angle for the extraordinaryay. It is simple to show that if ne→no, Eq. (24) is reducedo Eq. (15) and that if no→ne, Eq. (24) is reduced to Eq.19). An example for quartz as a uniaxial crystal with ne1.5533, no=1.5442, and ni=1, for �=589.3 nm and thepex angle of �=30° is shown in Fig. 5 as a solid curve.

6Ic=dt

ABI=s

wtrpv

wcicaca=t

wnE=

BtI=t

wmt[a(fipbg

Fft=

Fwtau

Fw(dt�

2436 J. Opt. Soc. Am. A/Vol. 24, No. 8 /August 2007 Avendaño-Alejo et al.

. EXAMPLESn this section we consider a prism made from a uniaxialrystal (calcite) with ne=1.4864, no=1.65835 for �589.3 nm immersed in air �ni=1�; assume that the inci-ent light is monochromatic and that the apex angle ofhe prism is �=30°.

. Crystal Axis in the Plane of Incidence, Parallel to thease of the Uniaxial Prism „�e

¸…

n this particular case the crystal axis forms an angle �� /2 with respect to the normal as is shown in Fig. 3; byubstituting this value into Eq. (24), we get

�ime

� = arcsin�no sin��/2�

ni� = �imo

,

here �ime

� gives the angle of incidence for which we ob-ain the minimum deviation angle for the extraordinaryay when the crystal axis is parallel to the base of therism [9,10]. Is important to note that the minimum de-iation angle for the ordinary ray is reached for � ,

ig. 5. Minimum deviation angle for the extraordinary ray as aunction of the crystal axis position for quartz as a uniaxial crys-al with ne=1.5533, no=1.5442, and the apex angle given by �30°.

ig. 6. Gray curve, total deviation angle for an isotropic prismith index of refraction no and o given by Eq. (7). Dashed curve,

otal deviation angle for the extraordinary ray when the crystalxis is parallel to the base of the prism, e

� , given by Eq. (18), byse of Eqs. (21) and (22) for �=� /2.

imo

hich is equal to �ime

� for the extraordinary ray when therystal axis is parallel to base of the prism. By substitut-ng Eqs. (21) and (22) into Eq. (18) and with the numeri-al values given above, we obtain e

� as a function of thengle of incidence; thus, as �i is varied, we will find that e

�

hanges, passing through a minimum at a well-definedngle given by Eq. (15), �imo

=�ime

� =25.4176°, and mo20.8352°, given by Eq. (12), as shown in Fig. 6. To con-

inue, we have

me� = mo for �i = �imo

= �ime

� ,

e� ��i� � o��i� for �i � �imo

= �ime

� .

here e� ��i� is the total deviation angle for the extraordi-

ary ray as a function of the angle of incidence given byq. (18), where we have used Eqs. (21) and (22) for �� /2, and o is given by Eq. (7).

. Crystal Axis in the Plane of Incidence, Perpendicularo the Base of the Uniaxial Prism „�e

�…

n this particular case the crystal axis forms an angle ���−�� /2 with respect to the normal, as shown in Fig. 3;

hen, by substituting this value into Eq. (24), we get

�ime

� = arcsin�ne sin��/2�

ni� = �ime

� ,

here �ime

� is the angle of incidence for which we have theinimum deviation angle for the extraordinary ray when

he crystal axis is perpendicular to the base of the prism9,10]. Let the apex angle of the prism be �=30° and thengle of the crystal axis �=−75°. Again, substituting Eqs.21) and (22) into Eq. (18) and with the values given aboveor the calcite, we obtain e

� as a function of the angle ofncidence; thus, as �i is varied, we find that e

� changes,assing through a minimum at a well-defined angle giveny Eq. (19), �ime

� =�ime

� =22.6257°, and me� =15.2513° = me

� ,iven by Eq. (20), as shown in Fig. 7. To continue, we have

ig. 7. Solid curve, total deviation angle for an isotropic prismith index of refraction ne and e

� given by Eq. (18) by use of Eqs.16) and (17). Dashed curve, total deviation angle for the extraor-inary ray when the crystal axis is perpendicular to the base ofhe prism e

� given by Eq. (18), by use of Eqs. (21) and (22) for= ��−�� /2.

wr(−E

7Bdtnpiftot

ATN

R

1

Avendaño-Alejo et al. Vol. 24, No. 8 /August 2007 /J. Opt. Soc. Am. A 2437

me� = me

� for �i = �ime

� = �ime

� ,

e���i� � e

���i� for �i � �ime

� = �ime

� ,

here e� is the total deviation angle for the extraordinary

ay as a function of the angle of incidence given by Eq.18), where we have used Eqs. (21) and (22) for �= ���� /2, and e

� is given by Eq. (18), where we have usedqs. (16) and (17).

. CONCLUSIONSy using the formulation of Avendaño-Alejo and Stavrou-is [8] for ray tracing in a uniaxial crystal, we have ob-ained the minimum deviation angle for the extraordi-ary ray when the crystal axis lies in and when it iserpendicular to the plane of incidence. With this angle its possible to design uniaxial prisms with predeterminedeatures. Other possible applications include the charac-erization of dispersion in uniaxial devices and the devel-pment of more accurate methods to measure the refrac-

CKNOWLEDGMENThis work has been partially supported by the Consejoacional de Ciencia y Tecnología under project J43919-F.

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