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Home , Crystal optics, Plane of polarization, Refractive index

EE 485, Winter 2004, Lih Y. Lin

Chapter 6 and * - Polarization ↔ Time course of the direction of E (r,t) - Polarization affects:  Amount of reflected at material interfaces.  Absorption in some materials.  .  (thus velocity) of anisotropic materials.  Optically active materials to rotate polarization.

6.1 Polarization of Light Consider (z,t) = Re{A exp(jω(t − z )} (6.1-1) E c = + With complex A Axxˆ Ay yˆ (6.1-2) Trace the endpoint of E (z,t) at each position z as a function of time.

The polarization = ϕ = ϕ Ax ax exp( j x ), Ay ay exp( j y ) = + E (z,t) Exxˆ Ey yˆ (6.1-3) = a cos[ω(t − z )+ ϕ ] (6.1-4a) Ex x c x = a cos[ω(t − z )+ ϕ ] (6.1-4b) Ey y c y 2 E 2 EE Ex + y − ϕ x y = 2 ϕ 2 2 2cos sin (6.1-5) ax ay axay a → An ellipse. The shape of the ellipse depends on y and ϕ . The size ax a2 + a2 of the ellipse determines I = x y (η: impedance of the medium). 2η

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Rotation direction (viewed from the direction towards which the is approaching): ϕ > ϕ y x : Clockwise rotation ϕ < ϕ y x : Counter-clockwise rotation

• Linearly-polarized light = → ax 0 Ey only = → ay 0 Ex only Or ϕ = 0 or π  a = y for ϕ = 0  Ey a Ex →  x  = − ay ϕ = π Ey E for  ax = → ±  ax ay 45 polarization

• Circularly-polarized light

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= = ax ay a0 ϕ = π → Right circularly-polarized 2 ϕ = − π → Left circularly-polarized 2

B. Matrix Representation The Jones Vector * Complex envelopes for E (r,t): = ϕ = ϕ Ax ax exp( j x ), Ay ay exp( j y ) A  J ≡  x  Ay 

• Orthogonal polarizations ()= * + * = J1, J 2 A1x A2x A1y A2 y 0 (6.1-7)

• Expansion of arbitrary polarization If J1 and J2 are normalized and orthogonal to each other, then an arbitrary = α + α polarization J 1J1 2J 2 α = ()α = () 1 J,J1 , 2 J, J 2

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1 1 1 1  1  Example:   =   +   0 2  j 2 − j

Matrix representation of polarization devices A  A  Input:  1x  , Output:  2 x  A1y  A2 y        A2x T T  A1x A1x   =  11 12   ≡ T  (6.1-9) A2 y  T21 T22 A1y  A1y  T: Jones matrix ⇒ = J 2 TJ1 (6.1-10)

• Linear 1 0 T =   (for xˆ ) (6.1-11) 0 0 ⇒ = = A2x A1x , A2 y 0

• Wave retarders 1 0  T =   (6.1-12) 0 exp(− jΓ) A   A  2x = 1x    − Γ  A2 y  exp( j )A1y  → y component is delayed by a Γ. x: fast axis, y: slow axis. Examples: (1) Γ = π / 2 (quarter-wave retarder) 1  1    →   1 − j

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1 1   →    j 1 (2) Γ = π (half-wave retarder) 1  1    →   1 −1 1  1    →    j − j

• Polarization rotators cos θ − sin θ T =   (6.1-13) sin θ cos θ  cos θ  cos(θ + θ) 1 → 1  θ   θ + θ  sin 1  sin( 1 )

• Cascaded polarization devices = ⋅ ⋅ ⋅ T TM ... T2 T1

Coordinate transformation J'= R(θ)J (6.1-14)  cos θ sin θ R(θ) =   (6.1-15) − sin θ cosθ T'= R(θ)TR(−θ) (6.1-16)

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T = R(−θ)T'R(θ) (6.1-17)

Normal modes (of a polarization system) States of polarization that are not changed when the wave is transmitted through the optical system. TJ = µJ (6.1-19) µ: eigenvalue J: eigenvector If T is Hermitian, i.e., T12 = T21*, the modes are orthogonal to each other, and can be used as an expansion basis. The response to the system can be evaluated more easily if the input wave is decomposed into the two normal modes: = α + α J 1J1 2J 2 = α + α = α µ + α µ TJ T( 1J1 2J 2 ) 1 1J1 2 2J 2 Examples (Exercise 6.1-4): (a) The normal modes of the linear polarizer are linearly polarized . (b) The normal modes of the wave retarder are linearly polarized waves. (c) The normal modes of the polarization rotator are right and left circularly polarized waves.

6.2 and

E  E  E  = 1x = 2x = 3x J1  , J 2  , J3   E1y  E2 y  E3 y  = J 2 tJ1, t : 2×2 Jones matrix for transmission = J3 rJ1, r : 2×2 Jones matrix for reflection t 0  r 0  t =  x , r =  x  0 t y  0 ry 

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= = E2x tx E1x , E2 y t y E1y (6.2-2) E = r E , E = r E (6.2-3) 3x x 1x 3 y y 1y  For transverse electric (TE) polarization, E = Exˆ, H = Hy n cosθ − n cosθ r = 1 1 2 2 (6.2-4) x θ + θ n1 cos 1 n2 cos 2 t = 1+ r (6.2-5) x x  For transverse magnetic (TM) polarization, E = Ey, H = Hxˆ n cos θ − n cos θ r = 2 1 1 2 (6.2-6) y θ + θ n2 cos 1 n1 cos 2

= n1 + t y (1 ry ) (6.2-7) n2 rx, tx, ry, ty can be complex numbers → = ϕ = ϕ rx | rx | exp( j x ), ry | ry | exp( j y )

TE polarization • < External reflection (n1 n2 ) ϕ = π x n − n | r |= 2 1 at θ = 0º (θ = 0º) x + 1 2 n2 n1

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• > Internal reflection (n1 n2 ) θ θ ϕ = For 1 < c, x 0 n − n At θ = 0º (θ = 0º), r = 1 2 1 2 x + n1 n2 θ θ = At 1 = c, rx 1 θ θ = For 1 > c, total internal reflection, | rx | 1, ϕ sin 2 θ − sin 2 θ tan x = 1 c (6.2-9) θ 2 cos 1

TM polarization • < External reflection (n1 n2 ) ry is real n − n At θ = 0º (θ = 0º), r = 2 1 is positive 1 2 y + n2 n1

θ θ = −1 n2 = θ At 1 = B tan , ry 0 ( B: Brewster angle) n1 θ θ ϕ = π For 1 > B, ry becomes negative, y θ ϕ = π At 1 = 90º, ry = -1, y

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• > Internal reflection (n1 n2 ) n − n At θ = 0º (θ = 0º), r is negative, | r |= 2 1 , ϕ = π 1 2 y y + y n2 n1 θ θ θ ϕ = π For 1 < B, |ry| decreases with , y θ θ = At 1 = B, ry 0 θ θ θ ϕ = For 1 > B, ry becomes positive and increases with , y 0 θ θ = At 1 = c, ry 1 θ θ = For 1 > c, total internal reflection, | ry | 1, ϕ sin 2 θ − sin 2 θ tan x = 1 c (6.2-11) θ 2 θ 2 cos 1 sin c

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Power and transmittance R =| r 2 | (6.2-12) T = 1− R (6.2-13) At normal incidence, for both TE and TM, 2  n − n  R =  1 2   +   n1 n2  Examples: (a) (n = 1.5) and Air (n = 1) interface R = 0.04 for normal incidence (b) GaAs (n = 3.6) and Air (n = 1) interface R = 0.32 for normal incidence

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6.3 Optics of Anisotropic Media A. Refractive Indices = ε Di ∑ ij E j , i, j = 1, 2, 3 (6.3-1) j {ε }≡ ij ε: Electric permittivity tensor (3 × 3) D = εE ε ε = ε is symmetrical, ij ji → only 6 independent numbers.

Principal axes and principal refractive indices Choose a coordinate system such that ε  1 0 0 =  ε  ε  0 2 0   ε   0 0 3  = ε = ε = ε D1 1E1, D2 2 E2 , D3 3E3 (6.3-2) → This coordinate system defines the “principal axes” of the crystal. Principal refractive indices: ε ε ε n = 1 , n = 2 , n = 3 (6.3-3) 1 ε 2 ε 3 ε 0 0 0

Isotropic, uniaxial, and biaxial = = Isotropic: n1 n2 n3 = = Uniaxial: n1 n2 no (ordinary index, ordinary axes) = n3 ne (extraordinary index, extraordinary axis) > < Positive uniaxial: ne no Negative uniaxial: ne no ≠ ≠ Biaxial: n1 n2 n3

The index x2 x2 x2 1 + 2 + 3 = 2 2 2 1 n1 n2 n3 (6.3-7) Direction of D determines the refractive index from the .

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B. Propagation Along a Principal Axis Consider a plane wave propagating along z-direction. Normal modes = = ε For E E1xˆ, D 1E1xˆ ,

= → = c0 k n1k0 c n1 = = ε For E E2yˆ, D 1E2yˆ ,

= → = c0 k n2k0 c n2

Polarization along an arbitrary direction Decompose the : = + E E1xˆ E2yˆ The x- and y-components travel with different speeds. The phase retardation ϕ = − (n2 n1)k0d → Linearly-polarized wave becomes an elliptically-polarized wave. The crystal acts like a wave retarder.

C. Propagation in an Arbitrary Direction

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Determine the polarizations and refractive indices na and nb of the normal modes of a wave traveling in uˆ direction. (1) Draw a plane passing thru the origin of the index ellipsoid, normal to uˆ . The intersection of the plane with the ellipsoid is an ellipse, called the index ellipse. (2) The half-width of the major and minor axes of the index ellipse are the

refractive indices na and nb of the two normal modes. (3) The directions of the major and minor axes of the index ellipse are the

directions of the vectors Da and Db for the normal modes. These directions are orthogonal. = (4) The vectors Ea and Eb may be determined from Da and Db by D εE . k × H = −ωD (6.3-8) × = ωµ k E 0H (6.3-9)

Example: Uniaxial crystal For a wave traveling at an angle θ with the optical axis (z-axis), the index ellipse θ has half-length no and n( ) . 1 cos2 θ sin 2 θ = + (6.3-15) 2 θ 2 2 n ( ) no ne = The normal modes have refractive indices na no for ordinary wave, and = θ θ = θ θ = θ nb n( ) for extraordinary wave. n( ) no when = 0º, and n( ) ne when = 90º. For ordinary wave, D is along the principal axis, D and E are parallel. D is perpendicular to k and zˆ . For extraordinary wave, D and E are not parallel (unless D is alongzˆ ). D is in the k-z plane.

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D. Double Refraction In the crystal, the optical wave can be decomposed into the two normal-mode components, propagating in two different directions: ⊥ θ = θ For ordinary wave (TE, E optical axis), sin 1 no sin o θ = θ θ For extraordinary wave (TM), sin 1 n( e )sin e

6.4 Optical Activity and A. Optical Activity Certain materials act naturally as polarization rotators. The normal modes are circularly-polarized waves. The for the right-circularly-polarized

wave is c0 n+ , and the phase velocity of the left-circularly-polarized wave is

c0 n− .

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2π(n − n )d The material rotates a linearly-polarized wave by φ = − + after a λ 0 π(n − n ) propagation distance of d. ρ ≡ − + is called “rotatory power”. λ 0

Medium equation: = − ε ξ∇ × D εE 0 E (6.4-2) For a plane wave E(r) = Eexp(−jk ⋅r) , G ≡ ξk , = + ε × D εE j 0G E (6.4-3) → Depends on k direction. Therefore reversal of the propagation reverses the sense of rotation of the polarization plane.

B. Faraday Effect The material acts as a polarization rotator when placed in a static . The rotatory power is proportional to the component of B in k-direction. Medium equation: = + ε γ × D εE j 0 B E (6.4-8) → Independent of k, but dependent on B. Therefore reversal of the propagation does not reverse the sense of rotation of the polarization plane.

Optical isolators Acting as a “one-way valve” for the optical wave.

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6.5 Crystals

The molecules are elongated, with ne for the long axis and no for the short axis. The twist angle of the molecules: θ = αz (6.5-1) Phase retardation coefficient: β = − (ne no )k0 (6.5-2) In general, β >> α . If the incident wave at z = 0 is linearly polarized, the wave maintains its linear polarization, but the rotates in alignment with the molecular twist. → Output wave polarized at θ = αd .

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• Applications: Optical switch and display (LCD).

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