General ray-tracing formulas for crystal

Wei-Quan Zhang

I develop a new method of ray tracing that can be used to analyze the refraction or reflection of rays byboth uniaxial and biaxial crystals.

Key words: Ray tracing, biaxial crystal.

1. IntroductionThe ray-tracing formulas of a light ray for uniaxialcrystal have been discussed by Simon and Echarri,1 ,2Zhang,3 and Liang.4 In this paper we discuss thepropagation of light rays in biaxial crystal, and a set ofray-tracing formulas for biaxial crystal is obtained.The propagation direction of reflected and refractedwaves can be obtained by solving this set of formulasnumerically. As for uniaxial crystal, a set of simplerformulas can be obtained. The method is based onphase-matching conditions for wave vectors. Thekey is to relate Snell's law with the expression forrefractive index by means of the transformationbetween the incident plane coordinate and the majoraxes coordinate of crystal. This set of ray-tracingformulas provides an efficient method for designingcrystal elements.

2. Direction Cosines of Refracted and Reflected Raysin Biaxial CrystalOn the boundary, the phase-matching condition ofwave vectors is

n, sin = n2 sin , (1)

where i and + are angles formed by the incident andrefracted wave vector and the surface normal, respec-tively, and n1 and n2 are the refractive indices of theincident and refractive media, respectively. The inci-dent plane coordinate axes OX2, Y1, and OZ3 arechosen as follows: The Y1 = 0 plane represents theinterface P1, and Y 1 is perpendicular to P1. Thelower half-space (Y1 < 0) is the birefringent medium.OZ3 lies in the incident plane I as shown in Fig. 1.

The author is with the Department of Physics, Zhejiang SilkEngineering Institute, Hangzhou, Zhejiang, China.

Received 29 April 1991.0003-6935/92/347328-04$05.00/0.© 1992 Optical Society of America.

In the incident plane coordinate the direction cosinesof the refracted wave vector are Sx2 = 0, Syl = -cos 4+,S,3 = sin +. The refractive index n2 of biaxial crystalis dependent on the direction of propagation and isgiven by

(1/n2)2 = (Vp/c)2= ((Vx/c)2

+ (V2/c)2

+ [(V/c)2- (VIc)2

]

x SZ2 cos2 0 - S 2 sin 2 0

± [1 - 2(SX2 sin2 0 + SZ2 cos 20)

+ (Sx2 sin2 0 - Sz2 cos2 0)2]1/2})/2, (2)

where cos2 0 = (V2 - Vz2)I(Vx2 - Vz 2); sin2 0 = (Vx2 -y2)/(V 2 - V 2); Vx = cn.; Vy = cn,; V = c/ne; nx, ny,

and n, are the main refractive indices of the crystal;Sx and S, are the direction cosines of the refractedwave vector in the major axes coordinate system XYZ;OX, OY, and OZ are three major axes of the crystal,and P3 is the YOZ plane.

In order to solve Eqs. (1) and (2) simultaneously,we must establish a relation between X2Y1Z3, andXYZ. The Eulerian angles between XYZ and X2Y1Z3are defined as follows: First, the rotation aroundOY1 with an angle y forms the X1YZ1 coordinate.Then the rotation around OX1 with an angle P formsthe X1Y1Z2 coordinate. Finally, the rotation aroundOY1 with an angle a forms the X2Y1Z3 coordinate.The transformation relation between the coordinatescan be represented by the matrix Q:

X Cos C

Y= 0 1Z - sin y 0

cos U.

x 0-sin a

O sin y 1 0

0 0 cos f

O cos y O sin

0 sina X2

1 0 Y .

0 COS ( Z3

0

-sin

cos 3

(3)

7328 APPLIED OPTICS / Vol. 31, No. 34 / 1 December 1992

Fig. 1. Transformation of coordinates in a biaxial crystal.

In Fig. 1, P2 is the plane formed by the surfacenormal 0Y1 and the major axis OY of the crystal,which is the principal section of the biaxial crystal.a is the angle between the incident plane and theprincipal section; (3 is the angle between the majoraxis OY and the surface normal 0Y1. y is the anglebetween the principal section P2 and the major axisplane YOZ (i.e., P3). Using Eq. (3), we obtain thedirection cosines of the refracted wave vector in XYZ:

S. = cos y sin a sin )

+ sin y(cos a cos ( sin - cos 4) sin Is),

Sy = -(cos 4) cos (3 + cos at sin sin 3),

S = cos y(cos ax cos sin 4-cos 4) sin (3)

- sin y sin a sin . (4)

Substituting Eqs. (1) and (4) into Eq. (2) results in anonlinear equation for sin 4). Solving this equation,we obtain sin and Vp for given incident angle i,refractive index n of the incident medium, andEulerian angles a, , y that describe the axes of thecrystal. Substituting sin 4) into Eq. (4) yields thedirection cosines of the refracted wave vector. Thedirection cosines of the refracted ray are

Vp Vk2 - Vr2

Vr Vk 2- (k = x, y, z), (5)

where the velocity Vr of the refracted ray is

= Vp2 + 1/V 2/[SX2/(V 2 - V 2)2

+S.2/ (V 2-V 2)2 + Sz2/ (V 2 - V 2)2]. (6)

Using the inverse matrix Q-1 of Q, we can transformthe direction cosines of the refracted ray in the major

axes coordinate system into those in the incidentplane coordinate system. For each set of i, ax, , andy there are two phase velocities Vp for two refractedrays. Both are extraordinary rays. Equation (1) isalso applicable to the refraction at the surface be-tween two different birefringent media.

If + is taken as the angle between the reflectedwave vector and the surface normal and n2 is therefractive index of the medium in the direction of thereflected wave vector, Eq. (1) is also applicable toreflection in crystal. In the incident plane coordi-nate X2Y 1Z3, the direction cosines of the incident andreflected wave vectors are, respectively,

SX 2 =0, S"y = -Cos i, SoZ3 =sin i,

S.2 = O Syl = COS ah' Sz'3 = sin

(7)

(8)

After a transformation of coordinates from Eqs. (7)we get

S0 = cos y sin a sin i + sin y

x (cos a cos 3 sin i - cos i sin A),

SO = cos y(cos a. cos sin i - cos i sin (3)

- sin y sin a sin i.

Substituting the above into Eq. (2), we get therefractive index of the medium in the direction of theincident wave vector, n = cVp. Similarly, fromEqs. (8) we get

S. = cos y sin a sin 4) + sin y

x (cos a cos sin 4) + cos 4) sin ),

$ = cos y(cos a cos sin 4) + cos 4) sin (3)

- sin y sin a sin 4). (9)

Substituting Eqs. (9) into Eq. (2) and using Eq. (1),we get a nonlinear equation for sin 4). Solving thisequation, we obtain sin 4) and the phase velocity Vp ofthe reflected wave vector in the biaxial crystal.Substituting sin 4) into Eqs. (9), we get the directioncosines of the reflected wave vector. From Eqs. (5)and (6), the direction cosines of the reflected ray inthe XYZ coordinate can be obtained. The directioncosines in the incident plane coordinate X2Y1Z3 can beobtained easily by the matrix Q-1 .

As an example we give the calculation of therefracted rays for an air-mica boundary: n = 1,n = 1.5601, ny = 1.5936, n, = 1.5977. Assume thaty = O. = 45°, and i = 30°, then

Sx = sin a sin , Sy = -(cos + cos a sin )/i/2,

Sz = (os a sin - cos,4)2/2. (10)

Letting a = 0 or 400 and solving Eqs. (1), (2), and (4)simultaneously, we get sin 4) and n2 = c/Vp of tworefracted wave vectors. Substituting sin 4) into Eq.(10) yields S., Sy, S. The direction cosines and Vpcombined are substituted into Eq. (6) and we get Vr.

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From Eq. (5) we get the direction cosines t, t, t oftwo refracted rays in XYZ. Finally, using the matrixQ-1 we transform these direction cosines into thedirection cosines in X2Y1Z3. The direction cosines ofeach ray, sin 4) and n2, are shown in Table 1.

3. Direction Cosines of Refracted and Reflected Raysin Uniaxial CrystalIn the major axes coordinate system XYZ of uniaxialcrystal, OZ is the optical axis. Plane P2, formed bythe surface normal OY1 and the optical axis, is theprincipal section of the crystal. For a uniaxial crys-tal the other major axis OY can be chosen arbitrarily,unlike the cases discussed in Section 2 for a biaxialcrystal. We choose the OY axis in the principalsection; thus OY, 0Y1, and OZ are coplanar, as shownin Fig. 2. Since the 0Z1 and OZ axes are the same,-y = 0 in the Eulerian angles. is the angle betweenthe optical axis and the boundary.

Letting V = Vy = V0 and V = V in Eq. (2), we get

VP = V I

2 = V 2 + (V 2 - V 2 )Sz 2 , (11)

where VO is the phase velocity of an ordinary waveand Vpe is the phase velocity of an extraordinary wave(e-ray). Settingy = 0 in Eqs. (4), we get

S. = sin a sin 4),

S = -(cos cos + cos ax sin sin 4)),

S,= cos a cos (3sin 4) - cos 4) sin (3. (12)

From Eq. (1) and Vpe we obtain the equation for sin 4:

(sin 4)/sin i)2 - n 2[l/ne2 + (1/n02 - 1/ne2)Sz 2]. (13)

Substituting S, of Eqs. (12) into Eq. (13), we can solvefor sin 4) and Vpe in terms of i, a, and . Letting V=Vy=VOV =V, and Vp = Vpe in Eq. (6), we get the rayvelocity of an e-ray:

Vr2 = Ve4 + (V04 - Ve4)Sz2]IVpe 2. (14)

Substituting Eq. (14) into Eq. (5) and letting Vk = Veor V, we get the direction cosines of an e-ray:

t = V 2S /(VrVpe)

ty = Ve2SyI(VrVpe),

= V 2S$z/(VrVpe). (15)

Table 1. Direction Cosines of Refracted Rays for theAir-Mica Boundary

a(deg) sing n2 t., tyl tz3

0 0.31311 1.5969 0 -0.9504 0.311140 0.31316 1.5966 0.00142 -0.9503 0.3114

0 0.3205 1.5601 0 -0.9420 0.335540 0.3202 1.5615 -0.0068 -0.9490 0.3152

Fig. 2. Transformation of coordinates in a uniaxial crystal.

The direction cosines in the incident plane coordinateX2Y1Z3 can be found by using matrix Q-1. Thedirection cosines of a reflected e-ray in a uniaxialcrystal can be obtained in the same manner for abiaxial crystal.

4. Conclusion

With given direction cosines of the incident ray (or inother words with given i), the main refractive indicesnx, ny, and n,, and the Eulerian angles that describethe major axes of crystal (a, (3, y), the directioncosines of each ray can be obtained by means of theabove formulas. The characteristic of this method isthe use of coordinate transformation. When crystalelements with several refractive surfaces are de-signed, we can apply the above formulas and coordi-nate transformation repeatedly to find the ray trace,which is important for the design of crystal elements.Moreover, the physical significance of all the parame-ters is quite clear.

Appendix A

1. Refracted e-Ray for the Air-Calcite Boundary

By using refractive indices n, = 1, no = 1.6584, andn = 1.4864, we can obtain the refracted e-ray for theair-calcite boundary. Assume that ( = 45 and i =450. Then

S, = sin a sin 4,

Sy = -(cos 4) + cos a sin 4))//2,

S = (cos a sin4) - cos ))v2/2.

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Table 2. Direction Cosines of the Refracted e Ray for theAir-Calcite Boundary

a(deg) sin 2 n2 ty tz3

45 0.4686 1.5092 -0.0395 -0.8524 0.521390 0.4569 1.5479 -0.0943 -0.8650 0.4928

Table 3. Direction Cosines of the Reflected e Ray in Calcite

aL

(deg) n1 sin 4) n2 tx2 ty1 tz3

45 1.4897 0.6558 1.6081 0.0987 0.7408 0.664590 1.5244 0.7073 1.5240 0.0728 0.6676 0.7409

are

From Eq. (13) we obtain the equation for sin 4):

2 sin2 4) = 0.4526-0.08901(cos a sin 4) - cos 4)2/2.

Solving this equation, we get sin 4) and n2 = c/Vpe.Then from Eqs. (14) and (15) we get the directioncosines t, t, t of an e ray refracted in XYZ. Finally,by using the matrix Q-1 these are transformed intodirection cosines in X2Y1Z3. The direction cosines ofthe e-ray, sin 4) and n2, are shown in Table 2.

2. Reflected e Ray in Calcite (i = = 450)

The direction cosine of the incident wave vector is

SOZ = (COS a - 1)/2.

From Eqs. (11) we get

(1/n1 )2 = 0.4526-0.08901(cos a - 1)2/4.

When a = 450, n = 1.4897; when a = 90°, n1 =1.5244.

The direction cosines of the reflected wave vector

S. = sin a sin 4),

Sy = (cos - sin cos a)v2/2,

S = (sin cos a + cos )/2/2.

From Eq. (13) we get

2 sin2 4) = n,2[0.4526-0.08901(sin 4) cos a + cos 4))2/2].

Solving this equation, we get the direction cosines ofthe reflected e-ray in crystal. The direction cosines,sin 4), n1, and n2, are shown in Table 3.

References1. M. C. Simon, "Ray tracing formulas for monoaxial optical

components," Appl. Opt. 22, 354-360 (1983).2. M. C. Simon and R. M. Echarri, "Ray tracing formulas for

monoaxial optical components: vectorial formulation," Appl.Opt. 25, 1935-1939 (1986).

3. Z. X. Zhang. "The trajectory of the extraordinary ray as thecrystal rotates," Acta Phys. Sin. 29, 1483-1486 (1980).

4. Q. T. Liang, "Simple ray tracing formulas for uniaxial opticalcrystals," Appl. Opt. 29, 1008-1010 (1990).

5. M. Born and E. Wolf, Principles of Optics (Pergamon, London,1959), p. 665.

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