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Home , Crystal optics

FS 2018 - ETH Zurich with Intense Sources Prof. Manfred Fiebig

Solution of Exercise sheet #5 Discussed on 28.03.2018

Exercise 5.1: Short questions (7 points)

0 (a) The equation ij:::n = Rii0 Rjj0 :::Rnn0 i0j0:::n0 (with unitary transformation R) describing the after a coordinate transformation is not valid for the six-dimensional abbreviated nomenclature used for e.g. the elasto-optical or electro-optical tensor. This is because both coordinates in the abbreviated form must be transformed/rotated. (b) The optical (e.g. and gyrotropy) can be induced in isotropic media such as gases, liquids, glasses and cubic only if an external eld is applied which breaks the isotropy of the system and make the system uniaxial (biaxial in case of certain strains?). Magnetic elds induce optical activity rather than birefringence (due to the circular motion of the electrons). (c) The sound wave in an acusto-optical device can be assumed to be non-propagating becaue the velocity of sound is much less than light velocity in the material.

(d) The di raction lattice forms an acusto-optical modulator because the propagating sound wave leads to a density wave which leads to a periodic modulation of the . (e) Electro-optical Kerr e ect is allowed in all materials, even in isotropic media ( rst eld breaks symmetry, second eld induces changes in refractive index), whereas the Pockels e ect is only allowed in materials without inversion symmetry (like LiNbO3, KDP, BBO). (f) Mechanical shutter: normally ms or thenth of ms (s shutting times seem also to be available). The speed of an acusto-optical modulator is roughly limited by the speed of sound in the material, leading to shutting times in the order of tenth of nanoseconds. Pockels cells can have rise times in the order of hundred picoseconds, although they are also used in the nanosecond regime.

(g) !l represents the longitudinal excitation mode while !o represents the transverse excitation mode in the Kuro- sawa form of the dispersion relation. The Kurosawa form the dispersion relation is given by (see J. Phys. Soc. Jpn. 16, 1298 (1961)): ¡ c2k2 YN !2 !2 = " (!) = " (1) l i !2 (!2) !2 i=1 o i

Exercise 5.2: 4th rank tensor calculus (4 points) In the lecture an example was given for the use of the elasto-optical e ect: A model of a mechanical piece is made of amorphous material (e.g. acrylic glass) and then areas of higher mechanical workload can be made visible by observing the stress-induced birefringence. Here we want to take a closer look on the elasto-optical tensor p connecting optical indicatrix Bij and the applied force Skl: ¡Bij = pijklSkl.

(a) A cubic system is isotropic in the linear response, however it has a much lower rotational (and translational) symmetry than an amorphous material. (b) The tensor is even, polar and invariant. Together with cubic symmetry m3m one has to look up the non- vanishing components of T4: xxxx = yyyy = zzzz and xxyy(x : 3) = yyxx(y : 3) = xxzz(x : 3) = zzxx(z : 3) = yyzz(y : 3) = zzyy(z : 3). The (i : 3) means the permutation of the last three indices while the rst one remains xed. Beware: these permutations give in the rst place distinct tensor components! (c) Using xx = 1; yy = 2; zz = 3; yz = 4; xz = 5; xy = 6 one gets the three independent components 11 = 22 = 33, 12 = 13 = 21 = 23 = 31 = 32 and 44 = 55 = 66 : 0 1 0 1 p11 p12 p13 p14 p15 p16 p11 p12 p12 0 0 0 B C B C B p21 p22 p23 p24 p25 p26 C B p12 p11 p12 0 0 0 C B C B C B p31 p32 p33 p34 p35 p36 C B p12 p12 p11 0 0 0 C pijkl = p = B C = B C B p41 p42 p43 p44 p45 p46 C B 0 0 0 p44 0 0 C @ A @ A p51 p52 p53 p54 p55 p56 0 0 0 0 p44 0 p61 p62 p63 p64 p65 p66 0 0 0 0 0 p44 (d) As an amorphous structure lacks rotational symmetry, additional dependencies are expected between the two components; for example, in the lecture it was given as 2p44 = p11 p12.

Exercise 5.3: Pockels e ect (9 points) Consider the linear electro-optical e ect in potassium dihydrogen phosphate (KDP) with point group 42m (Also see by Boyd, second edn., Chapter 11):

(a) In genaral, the optical indicatrix (or the ) in its principle co-ordinate system takes the following form: x2 y2 z2 + + = 1 2 2 2 nxx nyy nzz

For a uniaxial electro-optic (EO) crystal, nxx = nyy = no and nzz = ne, the the above equation simpli es to x2 y2 z2 x2 + y2 z2 + + = + = 1 2 2 2 2 2 no no ne no ne In other co-ordinate systems, the indicatrix is given by the most general expression of an ellipsoid in the absence of any external eld can be written as:             1 1 1 1 1 1 x2 + y2 + z2 + 2 yz + 2 xz + 2 xy = 1 2 2 2 2 2 2 n 1 n 2 n 3 n 4 n 5 n 6 ¡ where 1 are the optical coecients also known as the impermeability tensor,  de ned as n2 i ij X Ei = ijDj j

2 It should be noted that ij is the matrix inverse of ij = nij. Thus, the indicatrix in terms of impermeability tensor can be written as:

2 2 2 11x + 22y + 33z + 223yz + 213xz + 212xy = 1

where ¡ ¡ ¡ 1 =  ; 1 =  ; 1 =  ; n2 1 11 n2 2 22 n2 3 33 ¡ ¡ ¡ 1 =  =  ; 1 =  =  ; 1 =  =  ; n2 4 23 32 n2 5 13 31 n2 6 12 21 Owing to the fact that the permittivity tensor in the principle axis system is: 2 3 "xx 0 0 4 5 0 "yy 0 0 0 "zz

2 2 2 This implies that 23 = 32 = 13 = 31 = 12 = 21 = 0 and 11 = nxx , 22 = nyy , 33 = nzz . THis basically gives us back the rst equation we wrote for the indicatrix. (b) In presence of an external electric eld, the above general expression of the indicatrix gets modi ed to: ¢ ¡ ¡ £ ¢ ¡ ¡ £ ¢ ¡ ¡ £ ¢ ¡ ¡ £ 1 + ¡ 1 x2 + 1 + ¡ 1 y2 + 1 + ¡ 1 z2 + 2 1 + ¡ 1 yz n2 1 n2 1 n2 2 n2 2 n2 3 n2 3 n2 4 n2 4 ¢ ¡ ¡ £ ¢ ¡ ¡ £ +2 1 + ¡ 1 xz + 2 1 + ¡ 1 xy = 1 n2 5 n2 5 n2 6 n2 6 where   1 X X  = ¡ = r E = r E ij n2 hk k ijk k h k k

Here we introduce the reduced/transformed coordinate system, such that rijk = rhk, with 8 > 1 ij = 11 > > 2 ij = 22 <> 3 ij = 33 h = > 4 ij = 23=32 > > 5 ij = 13=31 :> 6 ij = 12=21 Now, for the symmetry 42m the rijk has the tensor property: polar,odd,invariant. The Birss table gives the fol- lowing tensor components of J3 with xyz = yxz, xzy = yzx, zxy = zyx. Thus the corresponding components of the EO tensor in the reduced notation:

r xyz = r yxz = r63 |{z} |{z} |{z}12 3 |{z}21 3 6 6 r xzy = r yzx ) r52 = r41 |{z} |{z} |{z}13 2 |{z}23 1 5 4 r zxy = r zyx ) r52 = r41 |{z} |{z} |{z}31 2 |{z}32 1 5 4 ¡ P P We now evaluate all the components of  = ¡ 1 = r E = r E . We explicitely write this in ij n2 h hk k ijk k k k matrix format: 2 ¡ 3 2 3 ¡ 1 n2 ¡1 r11 r12 r13 6 ¡ 1 7 6 r r r 7 2 3 6 n2 ¡2 7 6 21 22 23 7 E 6 ¡ 1 7 6 r r r 7 x 6 n2 ¡3 7 = 6 31 32 33 7 4 E 5 6 ¡ 1 7 6 r r r 7 y 6 n2 ¡4 7 6 41 42 43 7 E 4 ¡ 1 5 4 r r r 5 z n2 ¡5 51 52 53 1 ¡ r61 r62 r63 n2 6

For 42m we have already evaluated the allowed components of rijk = rhk. Using them we nd: 2 ¡ 3 2 3 ¡ 1 n2 ¡1 0 0 0 6 ¡ 1 7 6 0 0 0 7 2 3 6 n2 ¡2 7 6 7 E 6 ¡ 1 7 6 0 0 0 7 x 6 n2 ¡3 7 = 6 7 4 E 5 6 ¡ 1 7 6 r 0 0 7 y 6 n2 ¡4 7 6 41 7 E 4 ¡ 1 5 4 0 r 0 5 z n2 ¡5 41 1 ¡ 0 0 r63 n2 6 In we now summarize, we see that: ¡ ¡ ¡ ¡ 1 = 0; ¡ 1 = 0; ¡ 1 = 0; n2 1 n2 2 n2 3 ¡ ¡ ¡ 1 = 0; 1 = 0; 1 = 0: n2 4 n2 5 n2 6 Thus the general expression for the eld-dependent indicatrix reduces to:

x2 + y2 z2 + + 2r (yzE + xzE ) + 2r xyE = 1 2 2 41 x y 63 z no ne

(c) Without eld the indicatrix along the optical axis is a circle (i.e. isotropic response), while with eld the indicatrix is elliptical with axes 45 to the crystallographic x; y axes. Therefore, a static electrical eld along z (Ez = E) induces an anisotropy in the plane perpendicular to the eld. The indicatric looks like: x2 + y2 z2 + + 2r xyE = 1 2 2 63 no ne Apply the following coordinate transformation so that the indicatrix equation is once again in the principle axis form:       x0 cos 45 sin 45 x = y0 sin 45 cos 45 y Thus,     x0 1 x y = p y0 2 x + y

This implies: x = p1 (x0 + y0), y = p1 ( x0 + y0) and z = z0. Using these in the above indicatrix equation we 2 2 obtain:     1 1 z02 x02 r E + y02 + r E + = 1 2 63 2 63 2 no no ne The eld dependent values of n1, n2 and n3 are:

  1=2 1 n1 = nx0 = 2 r63E no   1=2 1 n2 = ny0 = 2 + r63E no n3 = nz0 = ne

(d) See of solids\, by Fox, exercise (11.10), page 329 (second edition): " From above,

¡ 1 ¢ ¡£ 1 ¡ 1 2 =2 2 2 =2 2 =2 n1;2 = no § r63E = no 1 § r63noE = no 1 § r63noE

1 ¡ =2 x Using the approximation: (1 + x) = 1 2 , we get: 1 n = n ¦ r n3E 1;2 o 2 63 o

3 The change in the refractive index is de ned as: ¡n = jn1 n2j. Clearly, ¡n = r63noE. Now 50% of the unpolarized light will be eliminated after the rst . Therefore, in order to have all the rest 50% light pass through the second polarizer (in the crossed set-up), the EO-crystal (KDP) has to rotate the by 90. Assume that the beam is polarized along x after the rst polariser. In the crystal the beam splits into two beams along the major and minor axis of the indicatrix. Due to the di erent refraction indices a phase  = nd ¡ 2= is accumulated. If the phase di erence is ¡ = ¡nL2= =  then the incident light is rotated by 90, and will pass the second polariser. The length L of the crystal is irrelevant as it was asked for the voltage U and not the applied electric eld E = U=d. Now,

3 ¡n = r63noE

With ¡n = =2L, E = U=L,  = 632 nm, r63 = 11 pm/V, and no = 1:5074, we get:  632 U = = ¢ 103 V = 8:387 kV 3 3 2nor63 2 ¢ (1:5074) ¢ 11