} LIBRA R.Y OF THE U N IVERSITY OF 1LLI NOIS 510.84 K6r no. 301-307 cop. 2 The person charging this material is re- sponsible for its return on or before the Latest Date stamped below. Theft, mutilation, ond underlining of books are reasons for disciplinary action and may result in dismissal from the University. UN ,V E RS,TY OF ^UO^im^^^^^^L DEC t 3 W71 i APR 6»Efctl FEB 1 Am L161— O-1096 Digitized by the Internet Archive in 2013 http://archive.org/details/theoreticalanaly303sand 10. * v ^ REPORT NO. 303 ^u^d, C00-1U69-0108 A THEORETICAL ANALYSIS OF THE MODULATION CHARACTERISTICS OF AN ELECTRO- OPTIC LIGHT VALVE by DOUGLAS STUART SAND JANUARY, 1969 AN-l* Report No. 303 A THEORETICAL ANALYSIS OF THE MODULATION CHARACTERISTICS OF AN ELECTRO-OPTIC LIGHT VALVE* by DOUGLAS STUART SAND January, 1969 Department of Computer Science University of Illinois Urbana, Illinois 6l801 * Submitted in partial fulfillment for the degree of Master of Science in Electrical Engineering, at the University of Illinois, January, 1969. iii ACKNOWLEDGEMENT The author is very grateful to his advisor, Professor W. J. Poppelbaum, for suggesting this problem and for much guidance and encouragement . He is also indebted to Professor Michael Faiman for his suggestions and proofreading. He would also like to thank Miss Carla Donaldson for typing the manuscript. IV TABLE OF CONTENTS Page I. INTRODUCTION 1 II. ANISOTROPIC STATIC FIELD SOLUTION 8 III. FIRST - ORDER OPTICAL SOLUTION 23 SUMMARY 58 LIST OF REFERENCES 59 LIST OF FIGURES Figure Page 1. Electro - optic Light Modulator 3 2. Crystal Configuration for Static Fields 9 3. Crystal Configuration for Optical Fields 32 k. Zero - Order Optical Solution 35 I . INTRODUCTION The electro-optic effect is an interaction between high-frequency electromagnetic fields (such as light) and low- frequency electric fields (such as static fields or microwaves). The electro-optic effect is an example of a large class of phenomena (including optical second-harmonic generation, Stoke 's radiation, and the stimulated Raman effect) generally called "nonlinear optics." ' Certain materials, such as the liquid nitrobenzene and the crystals quartz and KH PO, (KDP) , exhibit a strong (measurable) electro-optic effect, as well as other nonlinear optical effects. For analysis, the electro-optic effect is well approximated by expressing the optical dielectric properties of the material as a function of the local strength of the applied static field. Thus, for example, the local phase velocity (or index of refraction) becomes a function of the static field, so that the relative phase retardation of light passing through the material may be controlled by externally applied static or radio- frequency fields . In application the electro-optic effect is the basis for a variety of devices called "electro-optic light modulators." Perhaps the most familiar of these devices is the Kerr cell (or "light shutter"), used since the 1920's for high-speed photography. In recent years the electro-optic effect has been used in more sophisticated devices to produce modulation of phase, polar of light, '• 3 ^' W, [5], [6] ization, and amplitude of light, ' and to control the direction of light beam propagation. Most electro-optic light modulators are designed for time modulation of light, similar to modulation of a radio-frequency carrier. The "electro- optic light valve" (sometimes called an "Ardenne tube") is a spatial modulation .ce; that is, the light valve imposes a modulation pattern 'across the surface area of a wavefront, analogous to the intensity-modulated wavefront of light which has passed through a photographic transparency. The light valve is potentially capable of reproducing a television image in the form of a "real - time photographic transparency. " This potential has resulted in investigations in the use of the electro-optic light valve as part of a [71 television projection system and as the image source for real-time optical data processing. The configuration of a typical electro-optic light valve is shown in Figure 1. The light valve consists of a vacuum chamber containing an electron gun and a large flat thin electro-optic crystal, such as KDP. One face of the crystal is covered by a transparent conducting coating which is electrically grounded. The electron gun deposits a charge pattern on the other face of the crystal. The charge pattern induces a static electric field inside the crystal, so that light passing through is phase-modulated by the electro-optic effect. If the electron beam is scanned over the crystal face in a raster pattern and the beam current is controlled by a television video signal, the resulting charge pattern is a representation of a television picture. If the crystal is thin enough, the local charge density controls the strength of the local electric field, which in turn controls the local phase retardation of the emerging wavefront of light. Thus the emerging light is spatially modulated into a pattern resembling the original television image. . sible (intensity modulated) image may be obtained if the light is polarized before entering the crystal, and passes through an analyzer after emerging from the crystal (as in Figure 1). i o 3 T3 ■p ho ■H o •H ■P ? O •P o V H w H •H H Z til G 2 Various versions of the electro-optic light valve have been under development for about ten years. The physical con- struction and maintenance of these devices have presented many difficulties, but most have been solved by developing suitable technologies . A more serious problem has yet to be resolved: the modulation characteristics have been :locre at best, so that - in particular - the maximum pattern resolution currently feasible is less than 25% of standard television resolution. As will be seen, the pattern resolution is primarily determined by the thickness of the crystal relative to the size of its faces. This problem [9] was recognized quite early, and various attempts have been made to develop a theoretical analysis of the light valve modulation characteristics in order to provide a quantitative explanation of resolution. We now briefly consider two of these analyses . The simplest (and perhaps earliest) analysis considers a disk of diamater D located a distance T from a grounded conducting plane. The disk is charged to a potential V, and in the plane of the disk the surrounding potential falls to approximately V/2 at a diameter D + O.M+T. Assuming that the intensity (power) of the transmitted light is proportional to the potential in the plane of the spot, the half -power diameter of the light spot is equal to the V/2 diameter above. In general the electron beam spot size (~ D) is smaller than T, so the half-power diameter is approximately T/2. If we define the minimum resolvable separation of two light spots as approximately twice the half-power diameter, then resolution spacing is roughly T, the thickness the crystal. Standard television resolution is 500 line-pairs, correspon- ding to 500 spots in a line, so that this resolution would require that the face of the crystal be at least 500 times the crystal thickness. Thus a 1" diameter crystal must be polished to less than 2 mils in thickness. [12] [131 The second analysis ' is considerably' more detailed and explicit than the first, but follows similar arguments. Again we have a disk at potential V located a distance T from the grounded conducting plane, but a quantitative expression is derived for the potential at all points in the plane of the disk. Further, it is assumed that the intensity of transmitted light is proportional to sin [a V(p) + 0], -where V(p) is the disk-plane potential at a distance p from the spot center and where a and p are suitable 2 constants (this sin dependence is found in typical electro-optic light [3] modulators which use an analyzer to convert polarization modulation to intensity modulation) . The resulting expressions for the intensity distri- bution of the light spot are analyzed by a computer program to obtain reso- lution characteristics for a set of charge spot diameters and crystal thickness [13] Another approach is used to obtain the normalized Fourier decomposition of the disk-surface potential V(x). This Fourier decomposition represents the spatial frequency response of the surface potential in terms of the charge pattern and can be used to obtain an equivalent "pass-band" for the device. This (normalized) Fourier decomposition V(N)/V(0) is given by V(N) 1 - e " 2ltNT r , _x * nN -2jtNT,-1 vM = n K m [(K + 1) + (K - 1)e ] where N is the resolution in television lines (half -cycles) per millimeter, T is the crystal thickness in millimeters, and k is the relative dielectric constant of the crystal. This spatial response analysis gives resolution characteristics similar to those of the light-spot analysis. The analysis concludes with a statement predicting that for a crystal of thickness T = O.36 mm and a resolution of 500 television lines, the crystal face must be approximately 118 mm square. By linear interpolation we conclude that a 1" diameter crystal must have a thickness of less than 3 mils. The quantitative results of these two analyses are similar: a 1" crystal must be polished to a thickness of 2-3 mils. Electro-optic crystals of useable quality (e.g., uniformity) are currently available in 1" - 2" sizes, but crystals of this size have not yet been polished to a thickness of less than about 10 mils. Thus tolerable resolution from an electro-optic light valve is in essence a technological problem, at least until crystals are available in the dimensions indicated by the preceding two analyses. In considering these two resolution analyses, one finds they share two assumptions: First, the static fields are determined for an isotropic crystal; and second, the transmitted light intensity is considered a function of the surface potential only. As will be seen, the first assumption — an isotropic crystal -- is reasonable since even though the electro-optic crystals are in general anisotropic , the static field equations and solutions have the same functional form in either an isotropic or an anisotropic dielectric. However, the resulting quantitative descriptions are somewhat different. Tne second assumption essentially implies that each element of the crystal face is considered as a local intensity modulator, with a uniform static field in the direction of light propagation determined by the local surface potential (as in a parallel plate capacitor). This local modulator is analyzed as a "standard" longitudinal (r^-) linear electro-optic polari- r 3i zation modulator. That is, the crystal is normally uniaxial but becomes biaxial in the presence of a static field along the optic (x ) axis. Light propagating through the crystal in the x direction splits into two orthogonal components, each traveling at a different phase velocity controlled by the value of the static field. Leaving the crystal the two components recombine with a new polarization determined by the differential phase modulation and thus by the value of the static field. As a first-order approximation for spatial as well as polarization modulation this analysis is only partially valid, Spatial modulation requires a local variation in the direction of propagation, so the light no longer travels along the optic axis. Since the two biaxial phase velocities are a function of direction of propagation, we might expect that pattern direction (in addition to static-field characteristics) will determine the spatial resolution of the light modulator. As will be seen in the first-order analysis that follows, the resolution characteristics are a function of direction, at least in theory. The present work is an attempt to develop a more rigorous theoretical analysis of the modulation characteristics of the electro-optic light valve. In Section II the complete static field distribution is obtained for an anisotropic crystal, and for the spatial Fourier decomposition of an arbitrary charge pattern. This results in the anisotropic equivalent of the resolution characteristics V(N)/V(0) discussed previously. In Section III the static field solution is combined with the linear electro-optic dielectric tensor and Maxwell's equations, and a first-order solution of the resulting wave equation is obtained by a. perturbation analysis. This first order solution is in the form of a far-field radiation pattern, giving a direct expression of the first-order optical modulation characteristics of the electro-optic light valve . The static field solution and the first order optical solution are obtained by a generalized method for boundary conditions based on classical image sources (essentially an application of Green's theorem). These solutions and the method for their development are, to the best of the author ' s knowledge , original . II. ANISOTROPIC STATIC FIELD SOLUTION The static field equations for anisotropic media are very similar to those for ordinary isotropic media, since here the material is assumed to be homogeneous. However, the anisotropic parameters (the dielectric tensor c) complicate the analysis somewhat. The physical configuration of the light valve is such that the general static field equations and boundary conditions can be simplified considerably, so that we will first define the configuration to be analyzed. The configuration which describes the static field conditions appropriate for the light valve is shown in Figure 2. As discussed in Section I, the static field is established by a surface charge density located on one face of a flat thin crystal. This crystal has been oriented and cut so that the surfaces of the broad flat parallel faces are normal to one of the principal axes of the crystal. Let the coordinate system of the fields coincide with the principal axes of the crystal with, say, the x - axis normal to the crystal faces. Let the x - thickness of the crystal be T and choose the origin and direction of the x axis so that the grounded con- ducting surface is on the right, at x =0, and the charged surface on the left, at x = -T. Denote the surface charge distribution on the left by o(x, , x ) (this charge layer has been deposited by the electron beam). Since the size of the charge distributions to be considered is small compared with that of the face of the crystal, we can ignore edge effects and extend the crystal to infinity in the x, , x directions. Finally, we will refer to the left half-space, x < -T, as Region 1 (free space), and to the crystal region, 0, as Region 2. Region 1 (Free Space €.,_ = € 2 = e 3 =1 a(x 1? x 2 ) / / / Region 2 ( Crystal} £ l r £ 2 ' € 3 y Grounded Conducting Plane ( (2.3) For a linear anisotropic material the electric material equation is in general a second-order tensor relation between D and E. Here we assume the crystal is homogeneous so the elements of the relative dielectric tensor e are constants. In the principal-axes coordinate system e is diagonal, and the electric material equation reduces to D. = e e.E., i = 1, 2, 3 (2.4) l o 1 i' ' ' Using (2.4), the divergence equation (2.2) becomes I >1 E 1 + € 2 a 2 E 2 + € 3 a 3 E 3 = p / € o (2 * 5) where the operator d. denotes d/dx. . In terms of the potential $ the above equation (2-5) is U l\ + e 2^2 2 + € 3 a 3 2) $ = " p / € o (2 * 6) 11 This is the anisotropic Poisson's equation, similar to the isotropic Poisson's equation except for the coefficients e. . In fact, if we use the coordinate transformation, x. = u./FT, then (2.6) in the u. - coordinate system is '111' 1 an isotropic Poisson's equation. By use of the inverse of the above trans- formation, any known isotropic solution for Poisson's equation can be con- verted into the equivalent anisotropic solution, if one is careful to observe that this coordinate transformation is not measure - preserving (that is, one must follow the general tensor rules for a coordinate transformation). Let us denote the potential inside the crystal (Region 2) by § , and that in free space (Region l) by $ . The $ must satisfy (2.6) with the L K e. determined by the particular crystal in use. However, $ satisfies the free-space (isotropic) Poisson's equation, which is (2.6) .with e = e = e = 1, In addition, $ and $ must satisfy the boundary conditions at the inter- L R faces x_ = and x = -T. The general anisotropic boundary conditions are Ilk] rather complicated, but in this case the surface normals coincide with the principal axes of the crystal. Then at the left-hand interface, con- tinuity of the normal component of D requires a 3 L - e 3 3 3 $ R = c/e at x 3 = -T (2.7) where a is the surface charge distribution described above. Continuity of the transverse component of E requires, as usual, °L = °R at X 3 = " T ( 2 * 8 ) 12 Since the right-hand interface is grounded we must have 3> R = at x = (2.9) Finally, since this grounded plane is conducting, the transverse component of E vanishes, requiring \ $ R = ^2°R = ° at x 3 = ° ( 2 - 10 ) These boundary conditions (2.7) through (2.10), together with the appropriate forms of Poisson's equation (2.6), are necessary and sufficient to completely define the static fields in terms of arbitrary source charge distributions and p. Actually, we will assume that p vanishes (since the resistivity of the crystal is very large), but it will prove convenient to retain p in (2.6) until later. We will obtain the static field solutions by using Fourier trans- forms. This approach is advantageous for two reasons: first, to obtain the optical solutions we will use only the Fourier transform of the static field solutions; and second, in this particular case the Fourier transform solution is quite simple. We assume most functions of interest have three-dimension Fourier transforms, so that some scalar function f(x) and its Fourier trans- form f (k) are related by f(k) = /*" d 3 x f (x) e _tj - ' - (2.11) ^ -00 f(x) -f CO J^j3 f (k) e*~ " 2 (2.12) 13 where k • x = k^ + k g x 2 + k x 3 By differentiating (2.12) with respect to x. , it is easy to show that the functions d.f(x), Vf(x), and V^x) 1 — have as their respective Fourier transforms jk.f(k), jkf(k), and -k 2 f(k) where , 2 ,2 ,2 .2 1 + k 2 3 Thus the anisotropic Poisson's equation (2.6), which is in x - space representation, transforms into the k - space representation (e l k l 2 + € 2 k 2 2 + e 3 k 3 2) 0(k) = P&Ve o ( 2 ' 1 3) while the k - space isotropic Poisson's equation is simply k 2 $(k) = p(k)/e (2.U+) 11+ The Fourier transforms of the boundary conditions (two - dimension trans- forms, since x is fixed) will be considered later. Let us for the moment consider the isotropic Poisson's equation in k - space, (2.l4). For a unit point - charge located at x = a, p(x) = o(x - a) = 6(x 1 - a x )o(x 2 - a 2 )o(x - a^) By (2.11) we evidently have /. >. -jk • a p(k) = e °- - so that $(k) is simply, from (2.l4), 4>(k) = e"^ ' ^/e o k 2 It can easily be shown that the inverse transform of $ above is given by $(x) = 1/kne r where r = |x - a| This $(x) is obviously the familiar Coulomb potential of a unit point - charge. However, for our purposes we will transform $(k) back along only one direction, say x : 15 (DC^, k 2> x 3 ) = J 2T * (k)e 3 * 3 * J 2a€ o k/ + k/ + k 3 2 If we treat this integral as a contour integral, there are two poles in the complex k - plane : k 3 = + jTk^Tk^ = + jk r , say (2.15) This is an ordinary Fourier integral so that for convergence of a half - plane contour there are two cases of interest: for x > a we must use an upper half - plane contour (lm(k ) > 0), and for x < a we must use a ~j -J lower half - plane contour. Then $(k , k , x ) = + 2njZ (residues) -k |x - a = e - j(k i a i + k 2 a 2 } -$■ 2k e r o This result is not remarkable in itself, but it suggests that a "mixed - space" representation (that is, (k, , k , x ) - space) might be particularly useful if the boundary conditions apply to x only, since <3>(k , k , x ) for a point charge is a simple exponential. We will make frequent use of this mixed - space representation in -what follows. Next we consider the method of images. Recall the elementary image - source distribution for a grounded conducting plane: the fields on 16 one side of the plane may be determined by replacing the conducting plane and opposite half - space with a negative mirror image of the charge dis- tribution in the half - space of interest. In our case the conducting plane is defined by x = 0. Using the isotropic Poisson's equation as an example, the potential $(x) then satisfies -V^Cx) = — [pU l5 x , x ) - p(x , x , - x )] (2.16) o ^ where p(x) is assumed to vanish for x > 0. It is easily shown that $(x) determined by (2.16) completely satisfies the required boundary conditions at the grounded conducting plane ((2.9) and (2.10)) and satisfies the original Poisson's equation using only p(x). By the uniqueness theorem this is the solution (in the half - space x < in this example). For use later, the k - space representation of (2.16) is e o k 2 X 3 ) = J4 k 2 + k 2 &'& 3 r 2 2 2 where k = k + k " as in (2.15). We are interested in Region 1, x„ < -T, 3 r ' ' 3 and by definition the image source p must vanish everywhere in Region 1, so that singularities of p (k) do not contribute to the contour integral of Li (2.19) evaluated for x < -T. Then the only singularity of interest is the lower - half k - plane pole k = -jk . We then have, from (2.19), k x_ r 3 d> L (k 1 , k 2 , x ) = p L (k 1 , k 2 ) 1^^- for x < -T (2.20) o r where P L (\, ^ 2 ) = P L (k_ L , k 2 , k^ = -jk^) To obtain the general solution for <£>_ in Region 2, we impose two K conditions on the image source p D . First, the image condition requires that K p vanish in Region 2, for -T < x < 0. For convenience we will use the jk T form p-.(x + T), so that in k - space we have p (k)e as the operational A J — K ~~ 19 form of the image source. Second, we will use the mirror image condition, as in (2.17), to satisfy the boundary condition at x» = 0. Then in k - space, from (2.13), we have i k T - i k T (e l k l 2 + £ 2 k 2 2 + € 3 k 3 2) °R = p R ( - )e 3 + p R (k l' k 2' " k 3 )e ' (2 * 21) Let us define k s ■ Ve lV + ^2 d l^l (2 - 22) Then because of the image condition the k - plane singularities of interest are at k = +jk . As before we evaluate an ordinary Fourier integral to 3 s obtain the mixed - space representation of $ in Region 2: R -k T P R ( k V k ?) e V k l' V x 3 } = FT^ Sinh k s x 3 (2 ' 23) 03s where p (k , k ) = p (k , k , k = jk ) . It is easily shown that the i\±idri±djS R given by (2.23) satisfies the boundary conditions at x =0. Now we apply the boundary conditions. It is clear that (2.7) and (2.8) transform directly into mixed - space representation, so that at x = -T we have S 3°L (k l' k 2' " T) " e 3 a 3 R (k l' k 2' " T) = a(k l' V^o (2 ' 24) and L (k l9 k 2 , x 3 = -T) = ^(k^ k 2 , x 3 = -T) (2.25) 20 From (2.20) and (2.23) we obtain the required quantities for the mixed - space boundary conditions above, and, upon substitution, (2.24) and (2.25) reduce to and -k T -k T p T e + 2p D e s cosh k T = 2a L K S -k T -k T e 3 k s p L e = 2k r P R 6 Sinh k s T From these two equations we obtain expressions for p and p which give the R L desired solutions : k (x + T) c(k k ) e r 5 sinh k T $ L (* l9 k 2 , x ) = - k sinh k T + e k cosh k T (2 ' 26) o r s 3 s s a(k l5 k 2 ) sinh kx $ R (k 1 , k 2 , x ) = - k sinh R T cQsh k T (2.27) o r s 3 s s Since the potential $ expressed as (2.26) in Region 1 and (2.27) in Region 2 satisfies the appropriate boundary conditions and differential equations, by the uniqueness theorem this is the solution for the static fields in the light - valve configuration. We will not attempt to evaluate the x - space representation of $ except for the simple charge pattern consisting of a constant, say o(x 1 ,x 2 ) = A. Then o(k 1? k 2 ) = (2n) 2 6(k 1 ) 6(k 2 )A 21 so that we may obtain $(x) by determining the limit of 0(k , k , x ) as k , k vanish. If this is done, we find that V- } = " X 3 A / £ o € 3 (2 * 28) and L (x) = TA/e o £ 3 (2.29) As might be expected, this potential is identical to that of an infinite parallel - plate capacitor filled with a material of relative dielectric constant e . To conclude this section we will comment on some of the immediate implications of the potential $ (k , k , x ) expressed by (2.27). Recall R 1 2 3 that the relative frequency response of a system is given by the Fourier transform of the system response to a unit impulse function. The unit impulse function for a two - dimension spatial pattern is &(x ) &(x ), which is also a(x n , x ) for a unit point charge. For a unit point charge, a(k , k ) = 1, so $ (k , k , x ) with a = 1 is the relative spatial frequency J. c. K 1 2 3 response of the potential function in any plane x = constant (in Region 2) . For example, at x = -T we have 3 sinh k T s o^ v 1' 2' 3 " k sinh k T + eJt cosh k T r s 3 s s This is a two - dimension spatial frequency response. The frequency response to a pattern of lines is given by (k , k ) evaluated along a line 22 in the k , k plane. If, in general, e ^ e then the value of k = Ve k d : k ^/TiT (and, in turn, the relative line response) will S J. X d. d. j depend on the direction of the k , k line (which is normal to the direction of the x , x lines). However, if we have a uniaxial crystal (such as KDP) , then £ = £ and the frequency response to a line pattern is given by (setting k = 0, say) *(k r 0, -T) = -± sinh k T 1 e k n sinh k T' + VeTT,, cosh k T' o 1 1 13 1 where T' = T/e, /e . If this $(k ) is normalized (i.e., $(k )/$(k = 0)) and the hyperbolic functions changed into exponentials, the resulting expression will be very similar to the function V(N)/V(0) discussed in Section I, if we identify k I 1 in $(k )/$(0) with the exponent jtNT in V(N)/V(0). The critical difference is that the equivalent crystal thickness I" = TveT/eT is a function of the ratio of the relative dielectric constants e-. and e^. For ,C3] -•1 ~~ ^3 example, using published values of €-, and e o for some typical light - valve crystals we have KH P0, (KDP) T' ~ l.UT 2 4 KD P0, (KD*P) T' ~ 1.1T NH, H P0, (ADP) T* ~ 1.9T Thus, for a given crystal thickness, the normalized frequency response of the potential function should be best for KD*P, for which the effective thickness is nearly equal to the real thickness. However, this does not imply that of the three crystals KD*P is best for use in a light valve, since we have not considered the comparative effect of the electro-optic co- efficients of each crystal. 23 III. FIRST - ORDER OPTICAL SOLUTION • Materials such as crystals which are electrically anisotropic for static and low - frequency fields are also anisotropic for optical - frequency fields. Thus the optical electric material equation is also a tensor relation, although the optical dielectric constants may differ considerably from the low - frequency dielectric constants, since dielectric fpl properties are frequency - dependent. Further, it can he shown that the electro-optic effect can be treated by introducing additional terms into the optical dielectric tensor. These electro-optic terms are small compared with the "natural" dielectric coefficients, so that they may be treated as a perturbation effect. Using this approach we will obtain a first - order perturbation solution for the propagation of light through the electro - optic crystal in the light valve. The linear electric material equation is given by the tensor relation D = e o e • E (3-1) where e is the relative dielectric tensor. In the absence of the electro - optic effect the elements of e are constants (the "natural" or unperturbed dielectric coefficients). To express the electro - optic effect additional terms (functions of the static field present) are added to the natural di- electric coefficients. The functional expression of these electro - optical r 21 terms depends on the symmetry class of the crystal being considered. We will consider only the symmetry class ^+2m (which includes KDP and its isomorphs), for which e has the following form in the principal -axes 24 coordinate syst (3.2) The constants n and n are respectively the ordinary and extraordinary indices of refraction (n, = n so that ^+2m crystals are uniaxial ) . The quantities a. are the linear electro - optic terms and are related to the static field E and the linear electro - optic constants r. , , r^ by- 2 -! 22 a i = - n i n 3 r 4l E sl a 2 = -n x 2 n 3 2 r Ul E s2 (3-3) k 3 1 63 s3 In the light valve the static field is in general a function of position so that the a. in (3-3) are not simple constants. In particular, the static field is determined by the potential given by (2.27): $(1^, k 2 , x ) = p(k l5 k g ) sinh k g x (3-*0 for -T < x_ < and where we have defined p as - a (k , k )/e aflr k ) II L d O (^.5) P^j ^ 2 ) - k sinh k T + k cosh k T VJ •; r s 3 s s 25 By definition E = - v§> so that in mixed - space representation (where the — s operator o\ is unchanged but o\ -* jk , d -» jk ) we have 2 2 a 1 (k 1 , k 2 , x ) = jk^ n 3 r^ p sinh k^ 2 2 a 2 (k 1? k g , x ) = jk^ n 3 r^ p sinh k g x (3-6) a 3 (k 1 , k 2 , x 3 ) = k^ r 63 p cosh k^ As mentioned Section II, we will not need the x - space representation of the functions a. above, since the analysis to follow uses the a. in mixed - and k - space only. It is convenient to separate the dielectric tensor e into two terms, one containing the constant coefficients n. and the other containing the functions a. , so that (3-2) becomes | = | n + l (3-7) where I n =[0 n/ | (3.8; and 26" a 3 a 1 -I a 3 ° a ] a 2 &1 L (3.9) To obtain the appropriate electromagnetic field equations we must start with Maxwell's equations. We assume all field and source quantities are time - harmonic, so that, for example, the electric field vector E is given by E(x, t) = E(x)e _Ja3t (3.10) Then Maxwell's equations, relating the usual field vectors E(x) ,' D(x) , H(x), and B(x) to the source current density J(x) and the source charge density p(x), are given by V x H = J -joD (3-11) V X E = jouB (3-12) V • B = (3-13) V • D = p (3-1*0 By combining the divergence of (3. 11) with (3.1*0 we obtain another useful equation, the equation of source continuity: V • J = joop (3.15) 27 In addition to Maxwell's equations, we must use the material equations to relate D to E and B to H. The electric material equation is given by (3-1)? and the magnetic material equation for most crystals (in particular, KDP and its isomorphs) is simply the free - space permeability equation B = h-oS (3.16) The above equations together with appropriate boundary conditions are sufficient to determine any of the field vectors. Further, if any one field is known, say E, the other three are given directly by the first two of Maxwell's equations and the two material equations. This known field is the solution of a wave equation that relates the single vector to the sources. In our case the most convenient wave equation involves E. One such wave equation for E is obtained by an appropriate combination of (3. 11), the curl of (3. 12), and the material equations (3*1) an d (3-l6): V X V X E -k '" 1 • E = J041 J (3.17) 2 2 where k = co \i e . However, this wave equation is difficult to analyze, because of the term V x V x E. By use of the vector identity V X V X E he V(V • E) - S^E we obtain 28 V E + k € • E = -joun J + V(V • E) an appropriate combination of (3-1) , (3.1*0 3 (3-15) and the expression (3.7) f° r § w e have \ * ■ £ ■ l^f - V • 1 • £ - (n 3 2 - n ± S ) 9 3 E 3 When this expression for V • E is substituted into the wave equation we obtain 2 ^ + k o 2 ln • e + (^ - 1) va 3 E 3 = " [ 4 + 2 2 W] ' [(jC4i o- + k o\ ' - ] (3 ' l8) n. k 1 o where I is the identity tensor. For our purposes the wave equation (3 .18) is more convenient than (3.17). Also, in free space n = n = n = 1 and £ vanishes, so that the free - space wave equation corresponding to (3.18) is ^ E + k Q 2 E = -jo^ o [I + -^ W] • J = M, say, (3-19) k o where the "modified" source current M is defined to be M = -jcm [I + -^= W] -J (3-20) k o Equation (3.19) represents an uncoupled partial differential nation. That is, (3 .19) is three separate wave equations, one for each 29 component of E. However, the crystal wave equation (3.18) is coupled because of the terms involving ^ • E and VcLE . Further, since the coefficients of y are functions of position. (3.18) is a linear differential equation with non - constant coefficients, so that the solutions must be nonlinear functionals of the elements of £. In the light valve this means that the visible picture is a nonlinear function of the applied charge pattern. As will be seen, however, the first - order perturbation solution is linear in form. We have stated that the elements of the effective electro - optic tensor y are small relative to those of the dielectric tensor € • This is ^ =n true if the magnitude of the static field is reasonable. For example, a typical light - valve crystal has e ~ 25, n ~ n ~ 1.5, and r^- ~ r, ~ 10 m/y. Then for a typical crystal surface potential ~5kV and crystal thickness ~0.2mm, the magnitude of the static electric field is ~10 v/m, so -5 that an element of y is ~10 relative to an element of e . Thus x is indeed ^ =n = small compared with e , and a first order perturbation solution should give a reasonable approximation to the exact solution. T151 Recall that the perturbation approach may be applied to a two - operator equation of the form H • E - N + \G - E which relates the solution vector E to the source vector N. The operator H represents the "unperturbed" system (for which, presumably, all solutions are known or can be obtained), and the operator G represents the system 30 perturbation. The parameter X is a small constant representing the "strength" of the perturbation (the perturbation is "turned on" as \ is increased from to 1). The first order perturbation solution for X = 1 is then given by E ~ So + h where EL is the solution for the unperturbed system and E, satisfies H.^-H H • E_ x = G • EL We will refer to EL as the zero - order term and E, as the first - order term of the complete first - order solution. Note that the source N and the uniqueness theorem preclude the possibility that the unperturbed solution EL is degenerate. In the absence of the electro - optic effect (when the static field vanishes so that ^ = 0) the crystal wave equation (3«l8) becomes 2 ^E + k 2 e n ' E + (-2- - 1) ?c)E - o =n - 2 3 3 = -W [ I + ~Y~2 W] ' ^ (3-21) n k^ 1 = N, say, 31 where the modified crystal source current N is defined to be N = -ja*i [I + -~2 W] * ^ (3 ' 22) n. k o Evidently, the unperturbed system operator H is the partial differential operator on the left of (3.21), so that the zero - order term E satisfies (3.21). Further, the perturbation operator G is clearly -[k 2 I + -\ W] • x n l so that the first - order term E satisfies (3.21) if the "zero - order source current" jcou J is replaced by the "first - order source current" The uniaxial wave equation (3.21) only partially describes the unperturbed system, since it governs only the fields inside the crystal. All fields and sources outside the crystal must satisfy the free - space wave equation (3.19) ■> and the dielectric discontinuity at the crystal surfaces introduces boundary conditions. The optical configuration of the crystal (see Figure 3) is similar to that for the static field: the geometry is identical but the right - hand interface is not a conducting plane (the con- ducting film --such as vacuum - deposited gold — is essentially transparent to light ) . We will use the same coordinate system as before and again assume the crystal extends to infinity in the x x plane. Regions 1 and 2 are defined as before, and Region 3 denotes the right half - space x > 0. How- ever, we use subscripts "L" and "R" to denote variables for Regions 1 and 3, respectively, and the subscript "C" for Region 2 variables. 32 / / / Sy Region 1 (Free Space) Region 2 jS (Crystal) s In' I '/ Region 3 (Free Space) x 3 . -T ^/ X 3 = ° %' Figure 3. Crystal Configuration for Optical Fields. 33 Accordingly, the boundary conditions are as follows: continuity of the normal component of D and the transverse component of E requires n 3 E C3 = E f3 E C1 = E fl' E C2 = E f2 3.23) at x = or x = -T (with the subscript "f" taken as "R" or "L"). Con- tinuity of H at the interfaces requires V X E^ = V X E f (3-24) at each interface. The Zero - Order Term . The zero - order term E can easily be obtained by conventional methods, since the "source" is an x - directed plane wave in Region 1, so we have normal incidence at the crystal interface x = -T. Recall that for an isotropic dielectric sheet we assume that the solution vectors for a plane wave at normal incidence have no x-3 - components Assuming then, that E _ has no j wave equation (3.21) reduces to Assuming then, that E__ has no x - component, we find that the uniaxial UL 3 VE M + n, \ "1 "o -0C ° where N = since the source is confined to Region 1. This is identical to the isotropic wave equation so that the analysis and solution for the isotropic plane sheet are valid here. This solution 1 - is indicated in 3^ Figure k and is as follows . In Region 1 we have the incident plane wave E. = E e '3, where E is a constant vector and E • x„ = 0. The -mc -a -a -a -3 reflected wave E „, is of the form e 3 and is related to the incident — refl wave by a (reflection) coefficient r, , where jL , 2 ,v n -j2n n k T n iPk T (n l - 1) [1 " e d 1 o ] r = e 3 dK o L ± (3.25) L / , ,2 -j2n,k T , n x2 K3"&J (n 1 + 1 ) e 1 o - {n ± - 1) so that Sol " 5a C^* 3 + r L e " Jk ° X3 l <3.26) The transmitted wave E. is of the form e 3 in Region 3 and is — tran determined by the (transmission) coefficient r D , Fr = („ x + I) 2 e"^ 1 - ( ni - I) 2 e jn l k o T (3,27> so that £or " Sa r R edk ° X3 (3.28) In the crystal, Region 2, we have a positive - directed and a negative - directed plane wave, with wave number n.k . The solution for E-- is then l o — uu given by ^0C = E Q ^ c [e Jn l k o x 3 + T!e- dn l k o x 3] ( 3 .2 9 ) 35 Region 1 E. — mc / Region 2 / >/ >^ (crystal) >^ ^ E, — tran F -refl x 3 = -T /// X 3 = ° % Figure 4. Zero - Order Optical Solution. 36 where (3.30) and 2(n + 1) e" Jk o T " C = (n x + I) 2 e-^l k o T - (n] _ - if e^lV (3>31) It is not difficult to show that the solution for the zero - order term given above satisfies all the conditions required. The First - Order Term . The solution for the zero - order term was easily obtained with an assumption suggested by the equivalent isotropic configuration. No such analogy is evident when the source is inside the crystal, as in the case of the first - order term. However, a solution can be found by a more general approach. For example, the method of images is appropriate here because of the form of the boundary conditions. We will first obtain the general form of the solution in the free ■ space Regions 1 and 3- In k - space the wave equation (3. 19) is (k 2 - k Q 2 ) E(k) = - M(k) (3.32) is an image source, the mixed - space representation of E is determined by the singularity at k = +k , where k f = (k o 2 - k r 2)l/2 ' k r 2 = k l 2 + k 2 2 (3 ' 33) 37 Since both k singularities are on the path of integration, we invoke the radiation condition (which requires that a source emit waves) to select the desired singularity. Accordingly, k = -k represents waves propagating in the -x direction, so that the general solution for E in Region 1 is ~3 h given by -jk f Xo E L (k 1 , k 2 , x 3 ) - ^ -—- (3.3k) f and in Region 3 we have V k l' k 2' X 3 } =^W (3 - 35: f where «l " 4 (k i' k 2> k 3 = "V 4 = V k l> k 2> k 3 " V !3.36) We are ultimately interested in an expression for E , the first - order term for the transmitted wave. The general forms of E and E and the boundary conditions can be — R L used to obtain a set of conditions which the crystal field EL, must satisfy — u at the interfaces. Since there are no sources in Regions 1 and 3? the free - space fields must satisfy V • E = everywhere in the appropriate regions. 38 Thus, evaluation of 7 • E = (in mixed space) at the interfaces, followed by- appropriate substitutions from the boundary conditions (3-23) yields k l E Cl + k 2 E C2 " n 3 2 k f E C3 = ° at X 3 = ' T (3-37) k l E Cl + k 2 E C2 + n 3 Vc 3 =0at V° If we evaluate the free - space components of the boundary condition (3.24) and make appropriate substitutions from (3-23) we obtain four more indepen- dent relations for the interface values of E^, of which two contain E : jk x (n 3 2 - 1)E C3 = -(jk f + B 3 )E C1 at x 3 = -T (3-38) jk x (n 3 2 - 1)E C3 = (jk f - ^ 3 )E C1 at x 3 = The other two equations are similar and relate E and E . An appropriate combination of these equations (3-38) yields (k f - jd 3 )(k 2 E cl - k ± E c2 ) =0 at x 3 = -T (3-39) (k f + j£> 3 )(k 2 E cl - k^) =0 at x 3 = Another combination, of the equations in (3«38) and those in (3.37)? results in (n 3 2 k Q 2 - k/ - jn 3 2 k f a 3 )(k lEci + k 2 E C2 ) =0 at x 3 = -T (3^0) (n 3 2 k Q 2 - k r 2 + jn 3 2 k f a 3 )(k 1 E cl + k^) =0 at x 3 = To obtain the crystal field EL, we need the k - space uniaxial wave equation which, from (3.21), is 2 k 2 ^ - k 2 e • E_ + (-^- - 1) k k E -C o =n -C K 2 ~ 3 C3 n, J 1 (3.^1) ^o 2 2 = "I = 2 2 Cn i o - _ - ( ^ ' ^ )] n n k 1 o It is possible to manipulate (3. 1+1) such that three equations are obtained, one for each component of EL. With these equations one may easily obtain two useful expressions: A l S k 2 E Cl " R 1 E C2 = t A 0-te) k 3 " k o U k • K A 2 = k l E Cl + k 2 E C2 " 7 2 \ 2 --2—2 (3 ' U3) k 3 " k e k 3 " k c where for convenience we have defined / 2. 2 . 2x1/2 k = (n k - k ) ' c 1 o r ' k . - 1/2 (3. mo The definitions of A and A are appropriate because of the form of the 1+0 interface conditions (3«39) and (3-^0) and because these expressions are useful in the analysis to follow. The image source solution requires two image sources, one for each of the image regions x < -T and x > 0. Until the boundary conditions are applied, the two image sources must be treated as independent functions and have the following characteristics: the left - hand image source N vanishes for x > -T and produces positive - directed waves in Region 2, and the right - hand image source N vanishes for x < and produces negative - - K 3 directed waves in Region 2. These conditions on the image sources are required since Maxwell's equations are linear, and, in general, conditions at one boundary may be independent of those at the other. For each of A and A we have A. = A^ + A , where A . is determined by the true source, and A by the image sources. If we define k c P L " k 2 N Ll " k l N L2 and k c P R " k 2 N Rl " k A 2 then A , obtained from (3.1+2) has the mixed - space representation 2jA I1 = P L e JkcX 3 + P R e" Jk c x 3 (3.I+5) Expressions for P and P in terms of the source function A may be in obtained from the interface conditions (3 • 39) 3 by substitution and evaluation of A Tn as given above. However, we only need A,... at x =0, for which we II II 3 have, from (3.U0) and (3-1+5) p +p _ 2 . ^1^ + C l ( °^ k c - k f) eJk ° T + K + V 6 '^ (3.46) L R (k + kj 2 e" Jk c T - (k - kj 2 e jkcT where C-^-T) = (k f - jd ) A^ at x 3 = -T and ^(0) s (k f + j^ 3 ) A N1 at x = The function A, = A + A is, then, at x = 0: A l ■ \L + 5J (P L + P R ) (3.W where (P + P_.) is obtained as above. L K To determine the appropriate expression for P^ from (3-1+3) , we note that by definition in (3. 1+1), the quantity k • N reduces to -J w M- o 2 2s k • N = — g (n 1 k Q " - k")(k • J) n_, k 1 o 1+2 so that an alternate expression for (3.^+3) is k N jujx A 2 = 727^--2-2- ' " (3.W) k 3 " k e n l k o For the image function A-r 9 , we require that the image sources vanish in Region 2, so the term in k • J does not affect the mixed - space representation of A.,.- in Region 2. Accordingly, we have -2JA Ip =N L3 e 3.^3 e 3 so that evaluation of the interface conditions (3*^0) yields N L3 ■ N R3 (3 ' U9) 2n 1 2 k f C 2 (-T) + C (0)[(k + n i 2 k f )e e - (k - n \ )e e ] = 2j (k e + n 1 \) 2 e-^ T - (k e - n;L 2 k f ) 2 e Jk e- where C 2 ( - T) s r K 2 - J-^A' V at x„ = -T and C 2 (0) = i (k e 2 + jn 1 2 k f B 3 )A N2 at x 3 = e Then we have A p = A^ p + A evaluated at x = 0: 1*3 A 2 " V " Ti < N L3 - V (3 ' 50) where (N T - N ) is given above. To evaluate the iL . we need the first - order current source, determined by the dot - product of the electro - optic tensor (3«9) and the zero - order crystal field (3-29). With the following definitions s i s r c pn i%%i g 3 s r V- T > + ' k c + VV°> e c (3-59) " c (k + k f ) 2 e" , ' kcT - (k - k f ) 2 e dk c T ^7 where V- T > E V k i>>v x 3 ■ " T) with a similar expression for A^, (0), at x = 0. The exact expression for A at x = is then obtained by- substituting into ( 3- 59) the interface values of G for those of A^ and evaluating the k ' convolution as indicated by the definition of 3 G and (3-58) • The result is the following expression: A 1 (k 1 ,k 2 ,x^ = 0) = g 3 k s k o 2(k lV k 2 F l> 2 2 2 2 " Jk c T 2 ^ k c T V*r^\ + (r-i)k r ][(V k f) e - (VV e - ik T x {[k c ( 7 -l)(l + n) + n 1 k o ( 7+ l)( n -i)](k c+ k f )e ' ik T - [k c ( 7 -l)(l +1 ) - ni k o ( 7+ l)(ri-l)](k c -k f )e c -jn k Q T Jn k Q T - 2k coshk T[k ( 7 -l)(e + n e ) -jn n k T jn n k Tx n i/ ,\/ 1 o ° 1 o ) ] " n 1 k ^ +1 )( e " n e jn x k o T jn^T - 2o( 7 /k s )sinhk s T[2 ni k o k f (e - rje ) (0n 2 V 2 .2, _.; " Jn lV Jn l k o T )]} + (2n 1 k Q + k r ( 7 -l))(e + n e J2S 1+8 where y = £-,/€-,. As can be seen, this exact expression is quite complicated. However, a reasonable approximation results in considerable simplification. We are discussing an optical solution, so that the free-space wave-number k = 2n/\ is very large. Thus for light of o ^ ' o ~ 0.6um, we have k n ~ 10 m . This value is to be compared with typical values of k and k . From (3. 51) and (3*5) we find that the above expression for A, (0) contains the factor a(k,,k^) -- the Fourier compos i . f the crystal charge pattern -- so that the meaningful range of k, and k^. is determined by the maximum frequency content of s(k, ,k ). For standard television we have a maximum resolution of 500 lines (half -cycles) in any direction, which, for a typical k -1 crystal size of 25mm square, gives ~ 10 m for a maximum value of 2 2 2 2 k or k . Then for, say, k = k -k -k , the approximation k ~ k J. c. 1 O -L <— IO is very good for all values of k, ,k of interest. Also, from (3.UU), we have k ~ n,k . These approximations reduce the expression for A, (0) to only one term of magnitude k relative to the remaining terms Retaining only this k term, we obtain A 1 (k ] _,k 2 ,x 3 = 0) = k 2 E cl -k lEc2 Z k_F n - k-P . 2 J 2n i k T (jg_k sinhk T) 2 X . 1 2 - 1 + H e . . _ (3-60) Vde> 3 o s ' n 4l 2 j2n 1 k Q T 1 - q e where, as before, rj = (n -l)/(n ,-»l). To obtain an expression fc using the first - order source current ( 3 • 55) : To obtain an expression for A at x = 0, we begin with (3.48), 1*9 k N \ 2 = 2" P "— R*[k- L (k-)] k 3 " k e n l By substitution for N , from (3. 1+1), this becomes n A 2 (k L + k L ) - n A 2 (k • L) A = R* 3 ° Li ?_2 1 r (3.61) n 3 (k 3 - k e ) When this expression is evaluated, the term obtained from the (k,L, + k L ) component of (3.61) will be of order k relative to that obtained from the k • L component. This is reasonable since the k • L term, when evaluated, obtains a factor of order k (from k L„) while » ' o 3 3 2 the term for (k, L + k L ) already possesses the factor k , as in (3.6I), Since the approximation for A,, (3.6O), contains only the terms of highest order in k , we may then approximate A^ by ^2 ~ K R * \\ V (3.62) k- - k 3 e Note that this approximation implies that only r^ modulation is retained in this first-order evaluation, since the factor r, occurs only in the term k • L. The form of (3.62) above is similar to that for A^ in (3.58), and the analysis to follow parallels that for A,. If we define a G (k, k ? ) as in the definition for G above (but with k substituted e ~~ 3 c e for k ), the interface expressions for G and cLG have the same form c" e 3 e as those for G and S G , so that equations (3-^9) and (3'50) defining A reduce to 50 A 2 (k 1 ,k 2 ,x 3 = 0) 2n x k f 2, < k e + n l\> ° 6 V°) - < k e - "A* V' 1 * (3.63) p 2 ~^ k e T 2 2 ^ k e T (k e + n x K f ) e - (k g - n ± k f ) e where A^ (-T) = A^ (k ,k x = -T) with a similar definition for A^(0) at x_ = 0. before, we substitute G for A in (3*63) and evaluate the convolution integral. The result is A 2 (k 1 ,k 2 ,x 3 - 0) a g 3 n i 2k f k s (k o 2 ' vk r 2 )(k 1 F 1 + k 2 F 2 ) 2 2 2 "^ k e T 2 2 ^ k e T ^ 7n l k o + ^" v ^ k r ^ n l k f + k e ) e " W k f " k e^ e ^ - ik T X C[(l+i|)(7-v) - (l- n )( 7 +v)(n 1 k o /k e )](k e+ n 1 2 k f )e ' ik T + [(1+t,)(7-v) + (l- n )( 7 + v)(n 1 k o /k e )](k e -n 1 2 k f )e e -n k T jn k T - 2n 1 coshk s T[n 1 k f ( 7 -v)(e X ° + n e )] -Jn^k T jn^T -k Q (7+v)(e - tie )] -jn k T jn k T - 2j( 7 /k s )sinhk s T[2n 1 \ o k f (e 1 ° - r\e X ° ) o o o "jn k T J n n k T + (Sn^ 2 + k r 2 ( 7 -v))(e l0 + n e 1 ° )]} 51 2 2 where v = n /n . As before, we use the appropriate approximations, k ~ n, k and k„ ~ k , to reduce the above expression to only one term e 1 o f o' of relative order k . Discarding all but this highest order term, the o result is A 2 (k r k 2 ,x 3 = 0) = k lEcl + k 2 E c2 S k F + k F (-jg 3 k o sinh k s T) 1 I I s (3.64) Finally, by comparison of (3 .37) and (3-64) it is evident that _ at x_ c3 3 to order k_ we may neglect the terms involving E_ n at x = 0, so that our solution involves only E , and E ^. cl c2 The approximations for E , and E at x = are easily obtained by appropriate combinations of (3.60) and (3.6^). For example, we have E cl (k l' k 2' X 3 = 0) = (k 2 A l + k l A 2V k r 2 -jg_k sinh k T 3 o s 2 2Jn l k o T (n 1+ l)(l-^e x u ) 2F k k k 2 - k 2 2jn k T X ( 2 \ 2 + F [-+ r- 2 - - n 2 e lo ]} (3.65) k k r r For E the expression is similar to (3.65) differing only by the inter- change of F, and F^ as well as that of k and k . 12 12 52 We can obtain a more compact expression by a change in coordinate representation, to k ?k polar coordinates. Let so that R = tan" 1 k 2 /k 1 (3.66) k.. = k cose, and k^ = k sine, Irk 2 r k Then the factor in braces in (3*65) reduces to j2n k T F, cos26, + F sin20, - F n n e 1 k 2 k 1 ' Further, recall (3.52) relating F and F to E^, which by (3-26) defines the magnitude and polarization of the incident plane wave. Let define the angle of polarization, so that F, = E _ = E sine l cc2 a a F^ = E , = E cose 2 al a a (3.6?) (Note that elliptical polarization can be described by a complex -valued angle. J ) This definition for 6 , together with certain obvious triginometric identities, allow a further reduction, so that (3 .65) becomes 53 -jg k sinh k T E S ^_o s cl 2jn k T (n x + 1)(1 - n e ± ° ) 2 in k T x E a [sin(2e k + e a ) - n 2 e ' 1 ° sinej (3.68) Now, from (3«^) and (3. 51) we have g n sinh k T = n, iv^r $(-T) & 3 s 1 63 c ' where 3» ( " T)r c E a E c2 (k l5 k 2 ,x 3 = 0) = g 2Jn k T) (n 1 + 1)(1 - r) e 2 2 J n i k o T X [cos(20 1 + e ) + n e cos 9 ] (3.69) (3.70) 5U Finally, from (3o7) it is evident that the approximation for = is not of order k and so may be neglected. o We have now obtained the approximation for EL, at x = 0, which, together with the boundary conditions (3«23) and the expression (3.35) for E D , determine the first - order term of the transmitted wave in — K Region 3. The approximation for the first - order solution in Region 3 is given by the sum of E D as in (3 • 35 ) and the zero - order transmitted — K wave EL (3»28). However, since the resulting expression would be rather complicated (as in (3-70) ) we will consider a special case. If the light - valve is used to obtain visible pictures (intensity modulation, then the incident light is polarized, and the transmitted light passed through an analyzer oriented perpendicular to the polarization of the incident light. In particular, if the incident light is polarized in the x, - direction, say, we would be interested in the amplitude (or intensity) of the E component of the 2 transmitted wave relative to that of the E component. If we obtain the ratio of E Q2 , given by (3-70), to (EL^) at x = 0, given by (3-28), we should have a good estimate of the amplitude modulation characteristics for this case. For E.. - polarization we have =0 and F., = E _ = 0, 1 * a 1 a2 ' so from (3.27) through (3-31), e^(x =0) = x lEa r R = Wc _^l (3.71) n., + 1 55 Then, from (3-70) and (3-71), jn A. k cJ>(-T) (V /F ^ I - b ^ ° v c2' 0Rl y| x =0 2jn_k T 3 2(1 - n 2 e X ° ) 2 2Jn l k o T X [cos 28, + t] e ] (3-72) where $(-T) = $(k ,k ,x = -T) as given by (2.27). Now the far - field ri fii diffraction pattern can be shown to be related to the Fourier decom- position of the source pattern, which in our case (k ~ k ) is simply the ratio given above (the ratio is used since it is independent of the amplitude and phase of the incident light). If we recall the discussion in Section II concerning the static field resolution at x = -T, the function <|>(-T) determines the pattern resolution characteristics of the static field. From the ratio given above, $(-T) is also quite important in determining the pattern resolution characteristics of the transmitted light. However, the presence of the factor 2 2Jn l k o T cos 20, + r\ e indicates that the pattern resolution is a function of direction in the pattern, and that the response should be minimum along the direction determined by = n/h. This directional dependence is not unreasonable, since we have observed that the quantity k E + k E is associated with the wave - number k , which in turn is related to the extraordinary wave number n^k , and similarily, kJ5„, - k,E_^ is related to the 3o' ■" 2 CI 1 C2 ordinary wave number n k . The two components then travel at different 56 velocities inside the crystal, and reach the interface x = out of 2 phase. The result is an amplitude which depends on the values of k 2 and k . Thus the pattern resolution characteristics of the light valve are, to this first - order approximation, closely related to those of the static - field surface potential, but resolution is non-uniform as a function of k ,k direction. Finally, we compare the result (3-72) to that obtained by considering i element of the crystal to be a local longitudinal polarization modulator. In this approximation we have E o ~? E n R i sin( *&/*.) where 7 is a factor related to r , $ = $(-T) is the crystal surface potential, and $ is the full-wave retardation voltage which for r^ modulation is given by \ = X / n l' r 63 = 2 */ n l 3r 63 k o Substitution for $ yields E 2 ~ ? E 0R1 Sin K 3r 63 k o* ( - T) Then, using the Taylor expansion x 3 sin x = x - — + 57 we have, for small , V E 0R1 ~ I n l 3r 63 k o* ( " T) Clearly, the first-order term above is almost identical to the form of (3.72), except for the absence of the dependence. It would be of interest to obtain an experimental evaluation of the resolution as a function of Q . If the predicted 9 dependence is not observed, then the first-order approximation is not adequate, so that the perturbation series must then include higher-order terms. 58 SUMMARY Our aim in the present work has been to develop a theoretical analysis of the modulation characteristics of an electro - optic light valve. The development has been based on Maxwell's equations and a physical configuration which is representative of that of a typical light valve. We have attempted to use a minimal number of assumptions and to justify approximations by order - of - magnitude estimates. The fundamental assump- tion is that the linear electro - optic effect may be expressed by an r 21 appropriate optical dielectric tensor. In Section II we obtained the exact expression for the static field present in the anisotropic light - valve crystal, in terms of the Fourier composition of an arbitrary surface charge distribution. In Section III we used the linear electro - optic tensor to express the crystal wave equation as a function of the static field. A first - order perturbation treatment was used to obtain an approximation for the optical fields transmitted through the crystal. This approximation was then used to obtain an expression for the resolution characteristics of an intensity - modulation light valve. The development for the solutions of both the static fields and the optical fields was based on a generalized method of images. 59 LIST OF REFERENCES 1. N. Bloembergen, "Nonlinear Optics," Benjamin, New York, 1965- 2. P. N. Butcher, "Nonlinear Optical Phenomena," Bulletin 200, Engineering Experiment Station, Ohio State University, 1965. 3. I. P. Kaminow and E. H. Turner, Proc. IEEE %k, 137^ (1966). k. G. K. Ujhelyi, "Novel Electro - Optical Light Modulators," Report No. 170, Digital Computer Laboratory of the University of Illinois, 196k. 5. W. Kulke, et al., Proc. IEEE ^k, 1^19 (1966). 6. V. J". Fowler and J. Schlafer, Proc. IEEE 5k, 1^37 (1966). 7. E. J. Calucci, "Solid State Light Valve Study," 5th National Symposium, Society for Information Display, February 1965- 8. W. J. Poppelbaum, "Adaptive On - Line Fourier Transform," in "Pictorial Pattern Recognition," Thompson, Washington, D.C., 1968. 9. R. Peterson, "Development of a Light Valve Cathode Ray Tube," ADIU9OOI, Wiley Electronics Co., Phoenix, Arizona, 1957- Final Report on Signal Corps 'Contract No. DA 36-039 SC-72399- 10. E. Lindberg, J. Hatchett, and T. Cole, "Solid State Beam Controlled Light Modulator," ADU13U03, Motorola, Inc., Chicago, 1963. Rome Air Development Center Contract No. AF 30(602)-26 J +5. 11. R. S. Stites, W. E. Meyer, and C. L. Buddecke, "Electro - Optic Pro- jection Study," AD6I7087, Autonetics, Anaheim, Calif., 1965. Rome Air Development Center Technical Report No. RADC-TR-65-25. 12. W. E. Stoney, "Electro - Optic Projector Study," C6-1256/3^, Autonetics, Anaheim, Calif., 1966. Final Report on Rome Air Development Center Contract No. AF30(602)-3720. 13. T. H. Moore and E. A. Reitz, "Design Criteria for Electro - Optic Light Valves," T6-3077/U01, Autonetics, Anaheim, Calif., 1966. Ik. J. A. Stratten, "Electromagnetic Theory, " McGraw-Hill, New York, 19^1. 15. P. M. Morse and ft. Feshbach, "Methods of Theoretical Physics," McGraw- Hill, New York, 1953. 16. P. E. Mayes, "Electromagnetics for Engineers," Edwards Brothers, Ann Arbor, Michigan, 1965. F*m AEC-427 (4-46) aECPR 9-4 U.S. ATOMIC ENERGY COMMISSION CONTRACTOR'S RECOMMENDATION FOR DISPOSITION OF R! SEARCH DOCUMENT (See Instructions on Reverse Side) 1 TITU Of DOCUMENT A THEORETICAL ANALYSIS OF THE MODULATION CHARACTERISTICS OF AN ELECTRO-OPTIC LIGHT VALVE 2 PERSONAL AUTHOR! S) DOUGLAS STUART SAM) J CONTRACTOR S NAME AND ADDRESS UNIVERSITY OF ILLINOIS, URBANA, ILLINOIS 6l801 4 CONTRACT NUM»ER AEC AT (11-1) 1469 5. DATE Of DOCUMENT JANUARY, 1969 6. REPORT NUMBER coo- 1469- 0108 Complete only one item below, either 7, or 8, or 9. ] 7. DOCUMENT HAS BEEN OR WILL BE SUBMITTED FOR JOURNAL PUBLICA- TION. a NAME OF JOURNAL b. ESTIMATED DATE OF PUBLICATION a. DATE OF MEETING b. LOCATION c SPONSORING ORGANIZATION (Check one: d or e) Z} 8. DOCUMENT HAS BEEN OR WILL BE PRESENTED ORALLY. d. [ ] AEC's standard announcement and distribution procedures may prevail after date of meeting. e. [ ] Recommend that there be no public announcement or distribution because: I] 9. DOCUMENT IS A PROG- RESS REPORT NOT IN- TENDED FOR JOURNAL PUBLICATION OR ORAL PRESENTATION. (Check one.- a or b) a. ](3] AEC's standard announcement and distribution procedures may prevail. b. Q Recommend that there be no public announcement or distribution. SUBMITTED BY (flute print *r type) •KSNATURE DOUGLAS STUART SAND ORGANIZATION DEPARTMENT OF COMPUTER SCIENCE, UNIVERSITY OF ILLINOIS TITLE RESEARCH ASSISTANT DATE JAN. 8, 1969 t_y Q-Ux-r. L*-< --^-^^sX^