General solutions of the theme “light propagation in optical uniaxial crystals”

Untitled TAÏP CHÍ PHAÙT TRIEÅN KH&CN, TAÄP 20, SOÁ T3–2017 Trang 79 General solutions of the theme “Light propagation in optical uniaxial crystals”  Truong Quang Nghia  Nguyen Tu Ngoc Huong Universi[.]

TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 20, SỐ T3–2017 General solutions of the theme “Light propagation in optical uniaxial crystals”  Truong Quang Nghia  Nguyen Tu Ngoc Huong University of Science, Vietnam National University-Ho Chi Minh City (Received on Deceember 13th 2016, accepted on July 26th 2017) ABSTRACT In this article, we introduce a new approach to receive general solutions which describe all of the properties of the light propagating across optical uniaxial crystals In our approach we not use the conception of refractive index ellipsoid as being done in references The solutions are given in analytical expressions so we can handly calculate or writing a small program to compute these expressions Keywords: extra-ordinary ray, light polarization, light velocity, Maxwell’s equations, optical uniaxial crystals, ordinary ray, refractive index, tensor INTRODUCTION The problem of lights propagation in optical uniaxial crystals, i.e crystals of trigonal, tetragonal and hexagonal systems, was solved by the application of Maxwell’s equations Solving the Maxwell’s equations for a plane wave light propagating in transparent non-magnetic crystals, one can derive two refractive indices of the two propagating modes of light [2, 3]: no,2e  1 11  22     11   22   4122   (1) expression (1), the light direction is taken in parallel to axis OX3 of an arbitrary coordinate axes OX i (i = 1, 2, 3) Unfortunately, in the reality it is difficult to use the general expression (1) to receive two refractive indices, because in references the components of tensor [  ij ] are often given in crystal coordinate axes OX i* (i = 1, 2, 3) where the number of independent components of this tensor is minimum, i.e 11* and * (for optical uniaxial crystals) 33 In (1),  ij (i, j = 1, 2) are the components of the dielectric impermeability tensor of crystal In X *3 m O X*2 X *1 Fig The crystal coordinate axes OX i* (i = 1, 2, 3) and the light direction m Trang 79 Science & Technology Development, Vol 3, No.T20–2017 On the other side, when the light direction varies, the components 11 ,  22 and 12 in (1) also vary in according to the light direction Therefore in references, in order to eliminate this difficulty, one can only solve this problem in crystal coordinate axes OX i* with the help of the conception of refractive index ellipsoid, but this approach can only be applied in some limited cases when light propagating in some special symmetric directions of the crystal The refractive index ellipsoid of optical uniaxial crystals is an ellipsoid of revolution It has an important property: the central section perpendicular to the light direction m   m1 , m2 , m3  is an ellipse and the refractive indices of the two waves are given by the lengths of the semi–axes of this ellipse and the directions of these semi–axes give the directions of * o  * e  oscillations of the eigen vectors D and D for each of the two modes of light By this approach, it is difficult to solve the problem when light propagating in an arbitrary m direction In order to eliminate this difficulty, in this article we introduce a new approach using the general solution (1) Here, the important query is the calculation of the components 11 ,  22 and 12 via * the components 11* and 33 given in crystal coordinate axes OX i* In order to that we have to find the transformation cosinus matrix  ik  (i, k = 1, 2, 3) of the transformation of axes from OX i* to Having found  ik  we apply the OX i transformation rule of the components of a second * rank tensor [  ij ] to derive the corresponding components  ij in an arbitrary coordinate axes OX i Replacing observed values of 11 ,  22 and 12 into general expression (1) we can solve the given problem THEORETICAL CALCULATIONS In an arbitrary coordinate axes OX i (i = 1, 2, 3) we choose the OX axis which is parallel to the light direction m, which has three components (cosinus) in crystal coordinate axes: m1 ; m2 ; m3 Thus, the unit vector u3 along axis equals to m u3  m   m1 , m2 , m3  (2) Denoting g and h two vectors (not unit vectors) prolonging axes OX , OX respectively We can write: g  1 , ,    u3 Where (1 , , 0) is the unit vector along OX 1* and  is a coefficient derived from the orthogonal condition g.u3  Applying this orthogonal  3   3  g.u  1 , ,    u  u  condition 3 We find   m1 Thus, g  1 , ,   m1u3  1  m12  ,  m1m2 ,  m1m3  (3) are the components of g in axes OX respectively OX 1* , OX 2* The vector h along the axis in the form: OX can be written h   ; ;   1 g  2 u3 Where (0 , , 0) is an unit vector along OX 2* , 1 and  are the coefficients derived from the orthogonal conditions h.g  h.u3  From these orthogonal conditions we m1m2 find: 2  m2 1   m12 mm Thus, h   ; ;   22 g  m2 u3 (4)  m1 From expressions (2), (3), (4) we can derive the components of h along the axes OX i* (i = 1, 2, 3): The transformation cosinus matrix  ik  mm   m32 mm  m12 m2 m3 m2 m , 32  h   22 1  m12   m2 m1 ,  22  m22 ,  m2 m3   0 , 2  m1  m1  m1  m1  1  m1   Trang 80 and * TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 20, SỐ T3–2017 * The orthogonal conditions of axes OX i (i =1, 2, 3) via their scalar multiplications * The determinant of matrix  ik  must be equal to The components of g and h along the OX i* are not the direction cosinus of axes OX , OX versus OX 1* , OX 2* , OX 3* , but these direction cosinus can be derived by dividing these components by their vector length, i.e g and h if the observed coordinate axes OX i form a right – handed system The components of dielectric impermeability tensors [  ij ] in coordinate axes OX i Finally, we obtain the direction cosinus matrix   of the transformation of axes from OX i* to k i OX i Applying the transformation rule of the components of a second rank tensor when the coordinate axes varies from OX i* to OX i : [1] : m1m3    m12  m2    m12   m3    m1m2    m1   ik      m1   1 m m3  m12 m2 ij  ik jk* (i, j, k, ℓ = 1, 2, 3) (5) (6) In expression (6) we used Einstein notation, i.e to take the summation of the repeated indices by running this index from to For example: We can verify the truth of this matrix by these tests:         * i1 2 i  i  (i =1, 2, 3) k * * * 11  1k 1 k*  1 1k1  1 k  1 k           1 * 11 * 12 * 13      1 * 21 * *   22  1323   13 1131*  1232*  1333*            *   11  11  11*  12  11*  13  33  2  11*   2 2 * 11 * 33 m12 m32 * 33 11*   m12 In this example we have taken into account tensor [  ij ] is diagonal for optical uniaxial crystals and *   * * and m12  m22  m32  11  22 Analogously, we can derive all the components of tensor [  ij ] in the coordinate axes  * m12 m32 *   11*  11   33  m   i j    12    13   m1m2 m3 * 33 11*   m12 m32 * m22 *  33  11  m12  m12 23   11*   1 m   m2 m3 * *      33 11   m12  11*  m32 33*  11*     m1m32  OX i as follows: * 33 (7) Trang 81 Science & Technology Development, Vol 3, No.T20–2017 Now, replacing 11 , 12 and  22 from (7) into the general expression (1) we can solve the proposed problem After a long way of calculations we derive the refractive indices for the two propagating modes of light: * * no,2e    m32 11*   m32 33   m32 33  11*  (8)        The corresponding refractive index of ordinary and extra-ordinary rays Now here, we discuss what of the sign (+ or –) in (8) of which the refractive index of ordinary ray will be taken For the convenience of discussion we rewrite expression (8) in the form: no,2e  1 * *  1  m32 11*  1  m32 33*  B   Ao,e ; no2,e  and B  1  m32 33  11  Ao,e no,e There are two cases for discussion: * * * Positive optical crystals  ne  no  or 33  11 0 In this case, because of the refractive index of ordinary ray no < ne, the quantity Ao must be greater On the * * other side, in this case 33  11   so that B  Thus no2 takes the sign (–) and therefore ne2 takes the sign (+) in expression (8) * * * Negative optical crystals  ne  no  or 33  11 0 In this case, B  and A must be smaller so no2 also takes the sign (–) and ne2 takes the sign (+) Finally, regardless of positive or negative optical crystals, the refractive index of ordinary and extra-ordinary rays have the expressions:  * *  1  m32 33  11*    no2  1  m32 11*  1  m32 33  (9)  * * 2  1  m32 33  11*    ne  1  m32 11*  1  m32 33   * For ordinary ray: 1 * * no2    m32 11*   m32 33   m32 33  11*  = 211*   11* 2 no  (10)        11* Therefore the velocity of ordinary ray propagating across the crystal: c (11)  c 11* and is independent of light direction vo  no * For extra-ordinary ray: 1 * * * *  = m3211*   m32 33 ne2    m32 11*   m32 33   m32 33  11*    2m3211*   m32 33  2 (12) ne  * * m311  1  m32 33            The velocity of extra-ordinary ray: ve  The polarization of the two rays Trang 82 c *  c m3211*  1  m32 33 ne (13) TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 20, SỐ T3–2017 Denote D o  and D e , the unit vectors of polarization of the two rays in coordinate axes light is transversal, so in OX i :    D o  D1 o  , D2 o  , and D e  D1 e , D2 e , o OX i Because the  e In order to derive D and D we have to solve the equations determined the eigen vectors of a twodimension tensor [ ij ] having known eigen values no and ne ij D j  n2 Di  (i, j = 1, 2)  ij Using the Kronecker notation  ij , we can write these above equations in the form:  n2 ij  D j  (14) * For the ordinary ray: Replacing n2  no2  11* from (10) into the equations (14) we have: 11  11*  D1 o   12 D2 o     o o *  12 D1  22  11  D2  Combining with the normalization of D o  , i.e D o   we solve the equations and derive the components of D o  as follows:  m2 D o   ,  m m2  m2  m1 m3 m m m 2 2  , 0   (15) * For extra-ordinary ray: * Replacing n2  ne2  m3211*  1  m32 33 from (12) into the equation (14) to determine D e :     m2 *  1  m2  *   D e   D e  33  12  11  11   e  * * 12 D1  22   m311  1  m3 33   D2 e      Combining with the normalized condition of D e we derive:   m1 m3 m2 D e   , , 0  m2 m2  m2  m12 m32  m22   We can verify the orthogonality of D o  and D e via their scalar product The polarization of the rays in crystal coordinate axes OX i* (16) Remember that, the light direction m   m1 , m2 , m3  was given in crystal coordinate axes to transform the polarized vectors D o and D transformation cosinus matrix is now   k i e * o  into their corresponding vectors D  , which is the inverse matrix of   OX i* , * e  and D in so we have OX i* The k i Rotating matrix from (5) around its diagonal by an angle π we have:     k  i          m12 m1m2 m3 1 m  m12 m1m3 m2 1 m  m12 2 m1 m2 m3            (17) Trang 83 Science & Technology Development, Vol 3, No.T20–2017 Applying the transformation rule of the components of a vector: Di*  ik Dk (i, k = 1, 2, 3) We derive the vectors D* o and D* e in the crystal coordinate axes  m  m1 , , 0 D* o      m2   m32    mm m2 m3 ,   m32 D* e   ,   m2  m3  OX i* : (18)     From (18) we see that the ordinary ray is always polarized in the plane perpendicular to OX 3* , i.e the optical axis of crystals (19) OX * , OX 2*  or the plane We can verify the truth of (18) and (19) by the following tests: * The orthogonality of D* o and D* e via their scalar multiplication * The orthogonalities D* o m  D* e m  * The normalized conditions of vectors D* o and D* e The lack of the coincidence between the light direction m and the direction of light energy transfer P (the Poynting vector) According to [2], [3] the angle  of the lack of coincidence between the light direction m and the direction of light energy transfer, i.e Poynting vector P is determined by the following expression: cos  E * D* E * D*  E * D* E* if vector D* is normalized Thus, in order to calculate  we have to determine the electric vector E * Because in crystal coordinate axes OX i* the dielectric impermeability tensor [  ij ] is diagonal, therefore * Ei*  ij* D*j  ii* Di* (i = 1, 2, 3) For the ordinary ray from (18): Therefore : E * o  11*  * o  m2 * o  * 11*  E1  11 D1  m    m1 m1  * *   E2* o    22  11* D2* o    22 2  m3  m3   * o  * o  *  E3  33 D3   Analogously, for the extra-ordinary ray: Trang 84 TAÏP CHÍ PHÁT TRIỂN KH&CN, TẬP 20, SỐ T3–2017  * e  m1 m3 * 11  E1   m32   m2 m3 *  11   E2* e    m32   * e  *  E3    m3 33   * Therefore : E * e  m32 11*   1  m32 33  * For the ordinary ray: cos o  E * o D* o  E * o   E1* o  D1* o   E2* o  D2*o  11*  11*    o  0o 11* (20) Thus, for the ordinary ray, there is no lack of coincidence between m and P * For the extra-ordinary ray: * m3211*   m32 33 E * e D* e cos e   * E * e m32 11*   m32 33      (21)   Before applying our results to some specific cases, we summarize all the solutions we have derived In crystal coordinate axes OX i* : light direction m   m1 , m2 , m3  * For the ordinary ray: + Refractive index : no  (22) 11* * + Light velocity : vo  c 11 where c is the light velocity in vacuum  m  m1 + Light polarization : D* o    , , 0   m2   m32   + The ordinary ray is always polarized in the plane perpendicular to the optical axe of crystals + There is a coincidence between m and P * For the extra-ordinary ray: + Refractive index : ne  * * m311  1  m32 33 * + Light velocity : ve  c m3211*  1  m32 33  mm m2 m3 + Light polarization : D* e   , ,   m32 2  1 m  m 3  + Angle  e of lack of coincidence between m and P: cos e  * m3211*  1  m32 33 * m32 11*   1  m32 33  2 (23) (24) (25) (26)     (27) (28) Trang 85 Science & Technology Development, Vol 3, No.T20–2017 APPLICATION To test the truth of our above results, in the application we use KDP crystal KDP (DihydroPhosphate-Kali: KH2PO4) is a crystal of tetragonal system Its point symmetry group is 42m It has an inverse axe A4 , two axes A2 , which are perpendicular to A4 , two mirrors M which contain A4 The Fig shows the polar projection and the crystallographic axes of KDP: * A4 OX M OX*2 A’2 A4 A’2 OX*1 A2 M’ A2 Fig A) Polar projection of point group 42m of KDP In crystallographic coordinate axes OX i* the tensor dielectric impermeability of KDP is: 0  0.43858  * ij  = 0.43858    0 0.46277  * * Because 33  0.46277  11  0.43858 , KDP is a negative optical crystal Its optical axe is the A4 axe and in this case is parallel to axis OX 3* We apply our above results in three cases: The light direction is along the optical axe of KDP In this case we have m1 = m2 = and m3 = This is the simplest case of light propagation in optical uniaxial crystals and interestingly to be discussed here In references, we know that in this case we have only one ray propagating along the optical axe of KDP This is the ordinary mode Its polarization can be taken in any direction belonging in the plane perpendicular to the optical axe Which, for our results: * For the ordinary ray: From (22), (23) we have: Trang 86 B) Crystallographic axes of KDP OX1* no  A2 ;  * 11 OX 2*  A '2 * ; OX A4 0.43858  1.51 c 300000  km / s  198675.5km/ s no 1.51 From (24) we derive the light polarization:  m   0 m1  D* o    , , 0   , ,  th 2  1 m  0   m 3   e polarization of this mode is undetermined This query will be discussed later * For the extra-ordinary ray: From (25) we have : 1 ne    no * * 11* m311  1  m3 33 vo  This means that, in this case we have only one mode propagating along optical axe of KDP It is the ordinary ray Light polarization is calculated from (27):  mm  0 m2 m3  D* e   , ,   m32    , ,  0   m2  0    m3   is also undetermined From these above results we see that the polarization of the rays is undetermined but these polarizations are certainly lying in the plane perpendicular to optical axe because 0 * o  * e   D3  D3  The ratio   will go to some 0 TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 20, SOÁ T3–2017 limitd values, which is not infinity but depends on the light polarization entering the crystal Imagine a laser beam with any polarization entering along the crystalographic axe of crystal The polarization of the laser beam can now combine two perpendicular components lying in the plane perpendicular to the optical axe of crystal Each of the components is the polarization vector for mode no or ne Although their lengths are not equal to 1, but as shown in [2] the important thing is not the eigen vector but eigen direction as all vectors of arbitrary lengths provided lying along this direction are also eigen vectors of a second rank tensor Thus, in references we frequently speak about eigen direction instead of the eigen vector In our case the laser beam will propagate across the crystal with its original polarization It is the ordinary ray The laser beam can be polarized in any direction so the plane perpendicular to optical axe of KDP is an eigen plane In conclusion of this discussion, our results are the same already known in the classical approach The light direction is along one of the two axes A2 of KDP (along OX 1* or OX 2* ) For example, m1  1, m2  m3  ne  * m3211*  1  m32 33  light direction is * For the ordinary ray: no  vo   * 11  0.43858  1.51 c 300000  km / s  198675.5km/ s no 1.51 Polarization (Fig 3) :  m D* o    ,   m2    , 1 , 0 m1  m32  , 0     90o , 180o , 90o  There is a coincidence between m and P :  o  0o * For the extra-ordinary ray: the * OX * 33  0.46277 A  1.47 A’ c 300000  km / s  204081.6 km/ s ne 1.47 Polarization (figure 3) :  mm  m2 m3 D* e   , ,   m32    m2   m32   ve  A 2 OX m   , ,  1   90o , 90o , 180o  * OX (*)o D Angle of lack of coincidence between m and P: cos e   * m3211*  1  m32 33 * m32 11*   1  m32 33  (*)e 33*    e  0o 33* There is a coincidence between m and P In this case, we can say that we have two ordinary rays propagating with different velocities along an A2 of KDP D Fig The polarization of the ordinary ray and extra-ordinary ray Trang 87 * Science & Technology Development, Vol 3, No.T20–2017  The light direction m   ,  6 ,  o o o    65.91 , 35.26 , 65.91  6 In this case it is difficult to use the refractive index ellipsoid approach to solve the problem Our results: * For the ordinary ray: 1   1.51 no  * 0.43858 11 c 300000  km / s  198675.5km/ s no 1.51 Polarization (Fig 4) :  m  m1 D* o    , , 0   m2   m32   vo  1    , , 0     25.57o , 116.57o , 90o  There is a coincidence between m and P :  o  0o * For the extra-ordinary ray: 1 ne    1.47646 * * * * m311   m3 33 11  533 c 300000 ve   km / s  203190.7 km/ s ne 1.47646 Polarization (figure 4) :  mm  m2 m3 D* e   , ,   m32  2  1 m   m3      * OX A  m * OX A’ 5  ,  30 30   ,   30 2 (*)o   79.48o , 68.58o , 155.91o  D Angle of lack of coincidence between m and P: cos e  * m3211*  1  m32 33 m    1  m   * 11  0.9998069   e  1.126o Trang 88 (*)e * 33 D A OX * Fig The polarization of the ordinary ray and extra-ordinary ray TẠP CHÍ PHÁT TRIỂN KH&CN, TẬP 20, SỐ T3–2017 CONCLUSION Based on the general expression of refractive index (1), by the transformation cosinus matrix and tensorial calculations, we have completely solved the theme “Light propagation in optical uniaxial crystals” These analytical expressions describe all the properties of light propagating across the crystal We have some remarks: the polarization of the two propagating modes depends only on the light direction whereas the light velocities and the angle of lack of coincidence between m and P of extraordinary ray depend on the crystal and light direction With the exception of the cubic system, which is an isotropic medium in optical aspect, our approach can be applied to orthorhombic and monoclinic systems Of course the calculations will be more complex and take longer time because of in these * cases 11*  22 Acknowledgments: The authors wish to give their thanks to Pr Lê Khắc Bình for his comments and helps during the preparation of this article Phương pháp giải tổng quát chủ đề “Sự truyền ánh sáng tinh thể đơn trục quang học”  Trƣơng Quang Nghĩa  Nguyễn Từ Ngọc Hƣơng Trường Đại học Khoa học Tự nhiên, Đại học Qu c gia thành ph H Ch Minh TÓM TẮT Trong báo này, giới thiệu chiết suất làm tài liệu cách thức để nhận phương pháp giải tham khảo Phương pháp đưa biểu tổng quát mơ tả tất tính chất thức đại số nên dễ dàng tính truyền ánh sáng qua tinh thể tốn viết chương trình nhỏ để tính đơn trục quang học Trong cách thức này, chúng biểu thức không sử dụng khái niệm số ellipsoid Từ khóa: tia bất thường, phân cực ánh sáng, vận tốc ánh sáng, hệ phương trình Maxwell, tinh thể đơn trục quang học, tia thường, số chiết suất, tensor REFERENCES [1] Q.H Khang, Quang học tinh thể k nh hiển vi phân cực, Nxb Đại học Trung học chuyên nghiệp, Hà Nội (1986) [2] T.Q Nghĩa, T nh chất vật lý tinh thể, Nxb, Đại học Qu c gia TP H Ch Minh (2012) [3] N.V Perelomova, M.M Tagieva, Problems in crystal physics, Mir Publishers, Moscow (1983) Trang 89 ... and the refractive indices of the two waves are given by the lengths of the semi–axes of this ellipse and the directions of these semi–axes give the directions of * o  * e  oscillations of the. .. lack of the coincidence between the light direction m and the direction of light energy transfer P (the Poynting vector) According to [2], [3] the angle  of the lack of coincidence between the. .. solved the theme “Light propagation in optical uniaxial crystals” These analytical expressions describe all the properties of light propagating across the crystal We have some remarks: the polarization