Essentials of Electromagnetic Field Theory Maxwell s equations serve as a fundamental tool in photonics Updated: 19:3 1
Light is both an Electromagnetic Wave and a Particle Electromagnetic waves are described by Maxwell s Equations The solid state behavior is described using quantum mechanics which is minimally considered in this course. In this course we concentrate on the wave nature of light and we aim to minimize their use of Maxwell s Equations, but they are very necessary to understand what s happening and we will use them to establish fundamental results.
Maxwell s Equations (as used for optics) Differential Integral Form Form E = B t H = D t + J i D = ρ i B = 0 # E i d l = t C B i d a # H i d l = D i d a t C S $ D i d a = ρ dv S V B i d a $ = 0 S S + S =Q enclosed J i d a 3
Field Quantities Symbol Meaning Units E Electric Field Vector Volts/meter ( V/m) H Magnetic Field Vector Amps/meter ( A/m) D Electric Flux Density Coulombs/meter C/m ( ) B Magnetic Flux Density Webers/meter ( Wb/m ) J Current Density Amps/meter ( A/m ) ρ Volume Charge Density Coulombs/meter 3 ( C/m 3 ) Q Charge Coulombs ( C) 4
Time and frequency domains: Time domain quantities: E, H, D, B, J,Q,... Frequency domain (phasor) quantities: E, H, D, B, J, Q,... 5
Constitutive Relations: Recall: D = ε E B = µ H More realistically: D r,ω ( ) = ε ( ω ) E ( r,ω ) B ( r,ω ) = µ ( ω ) H ( r,ω ) 6
Constitutive Relations: D r,t ( ) = "ε t ( )* E r,t ( ) D r,ω ( ) = ε ω ( ) E r,ω ( ) B r,t ( ) = "µ t ( )* H r,t ( ) B r,ω ( ) = µ ω ( ) H r,ω ( ) Ohm s Law: J r,t ( ) = " σ t ( )* E r,t ( ) J r,ω ( ) = σ ω ( ) E r,ω ( ) * designates convolution ε ( ω ) dielectric permittivity Farads/meter µ ( ω ) dielectric permeability Henrys/meter 7
Constitutive Relations: The permittivity and permeability can be tensors, i.e., their Values depend on the orientation of the field in the material. 8
Photonic Crystals Photonic Routing Invisibility Hollow Core Fiber 9
Constitutive Relations: The permittivity and permeability can be tensors, i.e., their Values depend on the orientation of the field in the material. In this case the permittivity and permeability are matrices, e.g., D r,ω ( ) = ε xx ε yx ε zx ( ω ) ε ( xy ω ) ε ( xz ω ) ( ω ) ε ( yy ω ) ε ( yz ω ) ( ω ) ε ( zy ω ) ε ( zz ω ) E ( x r,ω ) E ( y r,ω ) E ( z r,ω ) The permittivity and permeability can also be functions of position. 10
Constitutive Relations: Dielectric Constant: ε ε o = ε r Index of Refraction: n i = µε µ o ε o Usually for optics, µ = µ o n i = ε r Free Space Permittivity: ε o = 10 9 = 8.854 10 1 36π (Farads/meter) Free Space Permeability: µ o = 4π 10 7 (Henrys/meter) 11
Complex Permittivity In the frequency domain: ( ) = ε ( ω ) j ε ( ω ) ε ω The real and imaginary parts are not arbitrary. 1
Kramers-Kronig Relations (aka Hilbert Transforms) Recall the Fourier transform and several of its properties: where ( ) = f t F ω F ( ω ) = R( ω ) + jx ω ( ) ( )e jωt dt The Fourier transform of an even function is real: ( ) = f even t R ω ( )e jωt dt, f even t ( ) = f ( even t) The Fourier transform of an odd function is imaginary: jx ω ( ) = f odd t ( )e jωt dt, f odd t ( ) = f ( odd t) 13
Kramers-Kronig Relations (aka Hilbert Transforms) Any function can be written as a sum of an even plus an odd term: where: f ( t) = f ( even t) + f ( odd t) f even ( t) = 1 f t ( ) + f ( t) and: f odd ( t) = 1 f t ( ) f ( t) 14
Kramers-Kronig Relations (aka Hilbert Transforms) Reality Postulate: The response of a real system to a real excitation must be real. Causality Postulate: The response of any real system must be non-predictive; known as a causal system. Corollary: For a linear causal system, the impulse response must be a causal signal. i.e., for an impulse applied at t = 0, the impulse response f(t) = 0 for t < 0. 15
Kramers-Kronig Relations (aka Hilbert Transforms) For a causal signal: f odd f odd ( t) = f ( even t), t > 0 ( t) = f ( even t), t < 0 Recall: sgn( t) = 1 t > 0 1 t < 0 Thus: f odd f even ( t) = sgn( t) f ( even t) ( t) = sgn( t) f ( odd t) 16
Kramers-Kronig Relations (aka Hilbert Transforms) f odd f odd ( t) = f even ( t), t > 0 t ( ), t < 0 ( ) = f even t 17
Kramers-Kronig Relations (aka Hilbert Transforms) Putting it all together: f even Hence: ( t) R( ω ), f ( odd t) jx ( ω ), sgn( t) f odd f even δ ( t) 1, 1 πδ ( ω ) u( t) 1 jω + πδ ( ω ) ( t) = sgn( t) f ( even t) jx ( ω ) = 1 π ( t) = sgn( t) f ( odd t) R( ω ) = 1 π sgn t ( ) = u t ( ) 1 jω + πδ ω ( ) πδ ω ( ) = jω jω jω * R ω ( ) jω * jx ω ( ) 18
Kramers-Kronig Relations (aka Hilbert Transforms) Putting it all together: Similarly: jx ( ω ) = 1 π R( ω ) = 1 π jω * R ω ( ) X ω ( ) = 1 π jω * jx ω ( ) R ω ( ) = 1 π ( ) R ω ω ω d ω ( ) X ω ω ω d ω 19
Kramers-Kronig Relations (aka Hilbert Transforms) An alternate form: X ( ω ) = 1 π = 1 π Similarly: R ω 0 ( ) R ω ω ω d ω ( ) R ω ω + ω d ω = 1 π R ω ( ) = π 0 0 ( ) ( ) ω X ω ω ω d ω ( ) = 1 0 R ω π ω ω d ω 1 π # #" ### $ 1 π 0 ω ω ( ) R ω ω ω d ω = 1 π 1 ω + ω + 1 ω ω d ω = ω π 0 0 ( ) R ω ω ω d ω ( ) R ω ω + ω d ω 0 ( ) 1 π R ω ω ω d ω 0 ( ) R ω ω ω d ω 0
Kramers-Kronig Relations (aka Hilbert Transforms) Yet another form integrate by parts: X ( ω ) = 1 π u = R ω X ( ω ) = uv ( ) R ω ω ω d ω ( ) du = dr ( ω ) v du = 1 π ln ω ω d ω, dv = d ω ω ω v = ln ω ω ( ) dr ( ω ) ( ) d ω 0 1 π = 1 ln ω ω dr ω π d ω d ω = 1 π ln ω + ω ω ω ( ) This form is useful for numerical approximations. 0 ( ) dr ( ω ) ln ω ω ( ) dr ω d ω d ω d ω d ω 1
Kramers-Kronig Relations ( ) = ε ( ω ) j ε ( ω ) ε c ω ε ε ( ω ) = 1 π ( ω ) = 1 π The real and imaginary parts of the complex permittivity cannot be selected independently. We ll use these later in the course ε ( ω ) ω ω d ω ε ( ω ) ω ω d ω
Kramers-Kronig Relations Homework: 1. Using the Kramers-Kronig Relations show that given R( ω ) = Find X(ω). a a + b ω, F ω ( ) = R ω ( ) + jx ω ( ). Given that R(ω) is a rectangular pulse of width ω o with height R o and zero elsewhere, find X(ω). R o R( ω ) = R o ω o ω ω o 0 otherwise ω o ω o ω Collected, due in one week. 3
For the sake of completeness: F ( ω ) = R( ω ) + jx ( ω ) R( ω ) = 1 π ( ) X ω ω ω d ω, X ( ω ) = 1 π ( ) R ω ω ω d ω F ( ω ) = A( ω )e jφ ( ω ) ln F ( ω ) = ln A( ω ) + jφ( ω ) ln A( ω ) = 1 π Enough ( ) Φ ω ω ω d ω, Φ( ω ) = 1 π ( ) ln A ω ω ω d ω An ideal low-pass cannot be realized. 4
The Wave Equation 5
The Wave Equation For linear media and a source free region of space: ρ 0, J 0 E = B t, H = D t E = B µ H = t t E = µ D = µε E t t t = µ t ( H ) This assumes that the permittivity and permeability and not functions of time. 6
Identity: A = ( i A ) A E = i E i D = ρ = 0, ( ) E = µε E D = ε E, ε = ε r ( ) t i D = 0, i( ε E ) = ε i E + ε i E = 0 i E = E i ε ε E µε E t = E i ε ε " $ # $$ % = 0 ( ) ε ε r 7
Single Mode Optical Fiber Core Cladding Multi Mode Graded Index Optical Fiber Core ε 0 ε Buffer Air ε = 0 ε Cladding Air ε = 0 r r Not to scale 8
The Wave Equation For Source-Free, Linear, (Piecewise) Homogeneous Media: Then: ε = 0 E µε E t = 0 H µε H = 0 t 9
In Rectangular Coordinates: E µε E t = 0 E = ˆxE x + ŷe y + ẑe z E = ˆx E x + ŷ E y + ẑ E z = x + y + z 30
Solution to Wave Equation Three (very similar) equations: E µε E t = 0 x-component: E x µε E x t = 0 y-component: z-component: E y µε E y t = 0 E z µε E z t = 0 31
Solution to Wave Equation Note that a differential equation of the form: f ( x,t) 1 f ( x,t) a = 0 x t is satisfied with any functions of the form: Since: f ( x,t) = f ( x ± at) f ( x,t) = f ( x ± at) x f ( x,t) = a f ( x ± at) t 3
Solution to Wave Equation Let us solve the first of these for the x-component: The method used is known as Separation of Variables. Assume that the equation is separable, i.e.,: E x ( x, y,z) = X ( x)y ( y)z ( z) Also assume sinusoidal steady-state time variation: E x µε E x t = 0 E x + ω µε E x = 0 t = jω 33
Solution to Wave Equation In the frequency domain: E x + ω µεe x = 0, E x ( x, y, z) = X ( x)y ( y)z ( z), = x + y + z E x + ω µεe x = 0 x X ( x )Y ( y)z ( z) + y X ( x )Y ( y)z ( z) + z X ( x )Y ( y)z ( z) +ω µε X ( x)y ( y)z z ( ) = 0 34
Solution to Wave Equation Y ( y)z z ( ) X ( x) x Y ( y)z ( z) X ( x)y ( y)z ( z) + X ( x)z z X ( x) + x ( ) Y ( y) y X ( x)z ( z) X ( x)y ( y)z z + X ( x)y y ( ) Y ( y) + y ( ) Z ( z) + ω µε X ( x)y ( y)z ( z) = 0 z X ( x)y ( y) X ( x)y ( y)z z ( ) Z ( z) + ω µε = 0 z 1 X x ( ) d dx X x ( ) + 1 d Y ( y) dy Y y ( ) + 1 d Z ( z) dz Z z ( ) = ω µε 35
Solution to Wave Equation 1 d X ( x) dx X ( x ) + 1 d Y ( y) dy Y y ( ) + 1 d Z ( z) dz Z z ( ) = ω µε 1 ( ) d X x 1 ( ) d Y y 1 ( ) d Z z dx X x dy Y y dz Z z ( ) = k x ( ) = k y ( ) = k z k x, k y, and k z are known as Separation Constants 36
Solution to Wave Equation 1 ( ) d X x 1 ( ) d Y y 1 dx X x ( ) d Z z dy Y y dz Z z ( ) = k x d ( ) = k y d ( ) = k z d dx X x dy Y y dz Z z ( ) + k x X ( x) = 0 X ( x) = X ox e ± jk x x ( ) + k y Y ( y) = 0 Y ( y) = Y ox e ± jk y y ( ) + k z Z ( z) = 0 Z ( z) = Z ox e ± jk z z E ( x x, y,z) = X ( x)y ( y)z z ( ) = E ox e ± j k x x+k y y+k z z ( ) = X ox Y ox Z ox e ± jk x x e ± jk y y e ± jk z z 37
Solution to Wave Equation Note from: 1 d X ( x) dx X ( x ) + 1 d Y ( y) dy Y ( y ) + 1 d Z ( z) dz Z ( z ) k x k y k z = ω µε k x + k y + k z = ω µε = k = n i k o k o = ω k = ω µ o ε o = ω c o = π f c o µ o ε r ε o = ω c = n i = π λ o ω π f = n c i o c o = n i π λ o 38
The Solutions are Termed Plane Waves Let: and: then: with: E x k = ˆxk x + ŷk y + ẑk z r = ˆxx + ŷy + ẑz ( x, y,z) = E ox e ± j ( k x x+k y y+k z z ) E x ( r choose the sign for now ) = E ox e jki r k = k x + k y + k z = k = ω µε = k k = k 39
For the Remaining Components: E(x, y,z) = ˆxE x (x, y,z) = ˆxE ox e j(k x x+k y y+k z z) ŷe y (x, y,z) = ŷe oy e j(k x x+k y y+k z z) ẑe z (x, y,z) = ẑe oz e j(k x x+k y y+k z z) = E o e j(k x x+k y y+k z z) E( r ) = ˆxE x ( r ) = ˆxE ox e j kir ŷe y ( r ) = ŷe oy e j kir ẑe z ( r ) = ẑe oz e j kir = E o e j ki r 40
Properties of Plane Waves For r = 0: From Maxwell s equations: Vector Identity: E( r = 0) = E o i E = 0 = i i( fa ) = Ai f + f i A, i E = i E o e j ki r ( ) = E o i e j ki r + e j ki r i E o =0 = j k i E o e j ki r = j k i E = 0 k i E = 0 k E E o e j ki r ( ) A = E o, f = e j ki r 41
Properties of Plane Waves k = â x k x + â y k y + â z k z k x + k y + k z = k = n i k o k = n i k k = ˆn k refractive index k = ˆnk, ˆn = unit normal k ε E( r) = E wavenumber: k = ω µε = ω µ o e jk ˆni r o ε o ε o k = n i k o k o free-space wavenumber: k o = ω µ o ε o n i refractive index: n i = ε ε o 4
Properties of Plane Waves From Maxwell s equations: Vector Identity: H = j ωµ E = j ωµ E o e jkˆni r H = j ωµ E ( fa ) = f A + f A ( ) = j ( ( ) E o + e jkˆnir E ) o ωµ e jkˆni r = j ( ωµ jk )e jkˆnir ˆn E o = k ωµ ˆn E o e jkˆni r = 1 η ˆn E o e jkˆnir = 1 η ˆn E Intrinsic Impedance (Ohms): η = ωµ k H = 1 η ˆn E H ˆn and H E 43
Properties of Plane Waves E( r) = E o e jk ˆni r, H = 1 η ˆn E, η = ωµ k E ˆn, H ˆn, H E, k = ˆnk = ˆnn i k o 44
Why call it a plane wave? E( r) = E θ o e jk ˆni r θ o = kˆn i r o = k i r o = k x x o + k y y o + k z z o θ = k ˆn i r = k i r = k x x + k y y + k z z For a constant phase: θ = θ o θ θ o = 0 = k x ( x x ) o + k ( y y y ) o + k ( z z z ) o This is the equation of a plane containing the point (x o, y o, z o ) k x k, k y k, k z k are the direction cosines of for the plane Hence k ( x x x ) o + k ( y y y ) o + k z z z o describes a plane of constant phase. ( ) = 0 45
Traveling Waves Consider a plane wave is traveling along the z-direction with an electric field polarized in the x-direction: E x = Re E xo exp( jk z z)exp jω t ( ) ( ) = E cos ω t k x o z z # " $# θ z To keep θ constant as t increases z must increase k x = k y = 0 k z = k 46
Traveling Waves Consider a plane wave is traveling along the z-direction and with an electric field polarized in the x-direction: E x = Re E xo exp ( + jk z z)exp jω t ( ) ( ) = E cos ω t + k x o z z # " $# θ z To keep θ constant as t increases r must decrease 47
Traveling Wave What is the velocity of propagation? θ = ω t kz = constant dθ dt = 0 = ω k dz dt = ω kv p v p = ω k Phase Velocity v p = ω k = ω ω µε = 1 µε Index of Refraction: n = c v p, c = 1 µ o ε o n = µε µ o ε o = µ=µo ε ε o 48
Wavelength z o z o + λ z ( ) = exp jk ( z o + λ) exp jkz o kλ = π k = π λ ( ) 6 rad For λ = 1500nm, k = 4. 10 m 49
Two Important Theorems: Poynting s Theorem & The Reciprocity Theorem 50
Poynting s Theorem Maxwell s Equations: E = B t Perform the following operations: and apply the identity: Define the Poynting Vector: H = D t + J H i E = B t E i H = D t + J H i E E i H = i S = E H E H ( ) 51
Poynting s Theorem Put it together: For B = µ H, S = E H = H i E E i H = H i B D = ε E t E i D t E i J then: i S = H i µ H t E i ε E t E i J The electric and magnetic energy densities are: w e = ε E i E, w m = µ H i H Finally: i S = t ( w e + w m ) E i J 5
Poynting s Theorem Integrating over a volume V enclosed by the surface S we obtain Poynting s Theorem: V i S dv = V ( w t e + w m )dv V E i J dv Applying the Divergence Theorem of Gauss: S d a " + S V t ( w e + w m )dv = V E i J dv Poynting s Theorem 53
Reciprocity Theorem Consider two time-harmonic sources producing two sets of fields in the same medium: and E 1 = jω B 1 H 1 = D 1 + J 1 E = jω B H = jω D + J Perform the following operation: H i E 1 = jω B 1 ( ) E 1 i( H = jω D + J ) H 1 i( E = jω B ) E i( H 1 = jω D 1 + J 1 ) 54
Reciprocity Theorem i Adv = V Ai d s " = S= V " S= V Ai ˆnds Simplifying, and using the Divergence Theorem: ( H i E 1 E 1 i H ) " $$$$ # $$$$ i E 1 H % ( H 1 i E E i H 1 ) " $$$$ # $$$$ i E H % 1 i ( ) ( ) = H 1 i jω B H i jω B 1 + E i jω D 1 E 1 i jω D E 1 i J + E i J 1 E 1 H E H 1 ( ) = jωµ ( H 1 i H H i H 1 ) ( ) " $$$ # $$$ % + jωε E i E 1 E 1 i E " $$ # $$ % E 1 i J + E i J 1 = 0 = 0 hence: i E 1 H E H 1 ( ) = E 1 i J + E i J 1 55
Reciprocity Theorem Simplifying, and using the Divergence Theorem: i( E 1 H E H 1 )dv = V ( E 1 H E H 1 )i d a It is generally assumed here that the integral on the left goes to zero as the outer boundary of the surface foes to infinity. Then: V ( E i J 1 E 1 i J )dv = 0 V 1 E i J 1 dv = " = S V V V ( E i J 1 E 1 i J )dv ( E i J 1 E 1 i J )dv E 1 i J dv The sources are assumed finite and the integration extends only over the volume of the source. 56
Reciprocity Theorem The electric fields are those of the other occupying volumes. This is the reciprocity theorem. E i J 1 dv = V 1 V E 1 i J dv 57
Reciprocity Theorem We can easily generalize our earlier result for fields in two different materials ε 1 ε, but still µ 1 = µ. i E 1 H E H 1 ( ) = jωµ ( H 1 i H H i H 1 ) = jω ε r ( ) " $$$ # $$$ % + jω E i ε E1 1 E 1 i ε E " $$$ # $$$ % E 1 i J + E i J 1 = 0 0 ( ( ) ε 1 ( r )) E 1 i E E 1 i J + E i J 1 If we again integrate over an infinite volume we obtain a similar result as before only with the extra terms due to the difference between the two dielectric functions: i E 1 H E H 1 ( ) = jω ε ( r ) ε 1 ( r ) ( ) E 1 i E E 1 i J + E i J 1 58
Reciprocity Theorem However, dielectric waveguide structures are translationally invariant in the z-direction. We can then integrate only over the volume of a disk with thickness Δz which tens to zero and radius R the tends to infinity: z Δz z + Δz x z y R Dielectric Waveguides ẑ S V S S 1 ẑ 59
V Reciprocity Theorem Integrating: i( E 1 H E H 1 )dv = = = " S ( E 1 H E H 1 )i d a + S # 0 ( E 1 H E H 1 )i ẑ dx dy + S 1 S z+δz ( E 1 H E H 1 )i d a + = z z+δz ( E 1 H E H 1 )i ẑ dx dy z ( E 1 H E H 1 )i ẑ Δz ( )dx dy 60
Reciprocity Theorem Since the current densities are both zero in the dielectric waveguides: z ( E 1 H E H 1 )i ẑ dx dy ( ) ε 1 ( x, y) = jω ( ε x, y ) E 1 i E dx dy This will be the reciprocity relation used for coupled dielectric waveguides. 61
Plane Waves or Transverse Electromagnetic Waves 6
Again consider a plane wave is traveling in the z-direction with an electric field polarized along the x-axis, E r,t ( ) ( ) = Re ˆx E o e j ω t kz H r,t ( ) =? 63
Time-Harmonic Electromagnetic Fields Time-Harmonic: All time variation of the form exp( jωt) t = jω Assume a source-free region of space. E = jωµ H E z y E y z = jωµ o H x E x z E z x = jωµ H o y E y x E x y = jωµ H o z H = jωε E H z y H y z = jωεe x H x z H z x = jωεe y H y x H x y = jωεe z 64
Consider a plane wave polarized in the x-direction and traveling in the z-direction: E( r,t) = Re ˆx E o e j ( ω t kz ) E = jωµ H E z y E y z E x z E z x = jωµ o H x = 0 = jωµ o H y E y x E x y = jωµ o H z = 0 E x z = jωµ H H = j E x o y y ωµ o z 65
H y = H y = j ωµ o E x z, H = ŷ Re E = Re E e j ( ω t kz) x o j ωµ o E x z H y = Re = ŷ Re = ŷ k E ωµ o e j ( ω t kz) o ω µ o ε E ωµ o e j ω t kz o ε Re E µ o e j ω t kz o j jk ωµ o ( ) ( ) ( ) E o e j ω t kz ( ) = ŷ 1 η Re E e j ( ω t kz ) o 66
Transverse Electromagnetic Waves E = ˆx Re E o e j ( ω t kz ), 1 H = ŷ η Re E e j ( ω t kz) o, η = µ o ε Characteristic Impedance or Intrinsic Impedance of the material η o µ o = = 10π 377 Ω ε o For Plane or Transverse Electromagnetic (TEM) waves, E and H are perpendicular to each other and to the direction of propagation. 67
The Poynting Vector Useful for measuring power flow S E H The (instantaneous) Poynting vector (units of watts/meter ) is a measure of the power flowing through a surface and points in the direction of power flow. For time harmonic fields, the time-averaged Poynting vector is: S = 1 Re E H * ( ) 68
Boundary Conditions for Dielectric Interfaces: Reflection and Refraction Boundary conditions connect the solutions at the interface between two dissimilar regions Region 1 Region 69
Electromagnetic Boundary Conditions: Source free region J,ρ = 0 ˆn ( E E ) 1 = 0 tangential components of E are continuous ( ) = 0 tangential components of H are continuous ˆn H H 1 ˆn i D D 1 ˆn i ( ) = 0 normal components of D are continuous ( B B ) 1 = 0 normal components of B are continuous Note that E = ˆnE n + E t ˆn E = ˆn ( ˆnE n + E ) t = ˆn ˆnE n + ˆn E t = ˆn E t = E t 70
Example: An incident wave strikes the surface. Determine the reflected and transmitted waves. ε 1,µ 1,n 1 ε,µ,n Reflected wave H r E r k r k t Transmitted wave E t θ r θ i x ˆn Unit normal θ t H t z Incident wave E i H i k i y 71
Note that there are two possible orientations (polarizations) for the electric field with respect to the interface. The field can be either perpendicular of parallel to the plane of incidence. Plane of Incidence: Plane formed by the normal vector n to the reflecting surface and the k vector. The State of Polarization (SOP) is determined by the electric field. The electric field is decomposed into perpendicular and parallel components relative to the plane of incidence. When the electric field is perpendicular to the plane of incidence, it is known as a transverse electric wave (TE). 7
Incident Fields: r = ˆxx + ŷy + ẑz k = ˆxk x + ŷk y + ẑk z k i E i θ i z y H i k i = k ( 1 ŷsinθ i + ẑcosθ ) i k i i r = k 1 ysinθ i + zcosθ i ( ) E i = ˆxE i e j k i i r = ˆxE i e jk 1( ysinθ i +zcosθ i ) H i = ( ŷcosθ i + ẑsinθ ) i H i e j k i ir = ( ŷcosθ i + ẑsinθ ) E i i e jk 1 ( ysinθ i+zcosθ i ) η 1 73
Reflected Fields: k r H r E r k r = k ( 1 ŷsinθ r ẑcosθ ) r k r i r = k 1 ysinθ r zcosθ r ( ) θ r y z E r = ˆxE r e j k r i r = ˆxE r e jk 1( ysinθ r zcosθ r ) H r = ( ŷcosθ r + ẑsinθ ) r H r e j k r ir = ( ŷcosθ r + ẑsinθ ) E r r e jk 1 ( ysinθ r zcosθ r ) η 1 74
Transmitted Fields: y E t H t z k t θ t k t = k k t i r = k ( ŷsinθ t + ẑcosθ ) t ( ysinθ t + zcosθ ) t E t = ˆxE t e j k t i r = ˆxE t e jk ( ysinθ t +zcosθ t ) H t = ( ŷcosθ t + ẑsinθ ) t H t e j k ir = ( ŷcosθ t + ẑsinθ ) E t t e jk ( ysinθ t +zcosθ t ) η 75
Boundary Conditions at z = 0: ( ) = 0 ˆn E = ˆn E 1 ( ) = ˆn E t ˆn E E 1 ˆn E i + E r ( ) = 0 ˆn H = ˆn H 1 ( ) = ˆn H t ˆn H H 1 ˆn H i + H r 76
Boundary Conditions at z = 0: ˆn ( E i + E ) r = ˆn E t ( ) = ẑ ˆxE t e jk ysinθ t +zcosθ t ẑ ˆx E i e jk 1( ysinθ i +zcosθ i ) + Er e jk 1( ysinθ r zcosθ r ) ( ) z=0 ˆn ẑ ( H i + H ) r = ˆn H t ( ) E i ŷcosθ i + ẑsinθ i e jk 1 ( ysinθ i+zcosθ i ) η 1 ( ) + ( ŷcosθ r + ẑsinθ ) E r r e jk 1 ysinθ r zcosθ r η 1 = ẑ ( ŷcosθ t + ẑsinθ ) E t t e jk ( ysinθ t +zcosθ t ) η z=0 77
Boundary Conditions at z = 0: Ee + Ee = Ee jk y sinθ jk y sinθ i r t 1 i 1 r jk y sinθ t Ei jk y sinθ E i r zˆ ( yˆcosθ ˆsin ) ( ˆ i + z θi e + ycosθr + zˆsin θr) e η1 η1 Et jk y sinθt = zˆ ( yˆ cosθt + zˆsin θt) e η 1 1 jk y sinθ r zˆ zˆ 0 78
Boundary Conditions at z = 0: Ee + Ee = Ee jk y sinθ jk y sinθ i r t 1 i 1 r jk y sinθ t E E E e e e η η η i jk y sinθ r jk y sinθ t cosθi cosθr = cosθt 1 1 1 i 1 r jk y sinθ t These equations can be satisfied for all y only if the phase factors are equal, k 1 sinθ i = k 1 sinθ = k sinθ r t θ i = θ r E i + E r = E t cosθ i E i η 1 cosθ r E r η 1 = cosθ t E t η 79
k 1 = ω µ o ε 1 = ω ε 1 ε o µ o ε o = n 1 k o k = n k o θ i = θ r n 1 sinθ i = n sinθ t Snell's Law for reflection for refraction 80
Reflection and Transmission Coefficients (TE Waves) E + E = E i r t E E E cosθ cosθ = cosθ η η η i r t i i t 1 1 Er ηcosθi η1cosθt = E η cosθ + η cosθ E E t = i i 1 t η cosθi η cosθ + η cosθ i i 1 t 81
Reflection and Transmission Coefficients (TE Waves) Note : η = µ o ε = ε o ε µ o ε o = η o n E r E i = n 1 cosθ i n cosθ t n 1 cosθ i + n cosθ t E t E i = n 1 cosθ i n 1 cosθ i + n cosθ t 8
Reflection and Transmission Coefficients Normal Incidence: θ = 0, n sinθ = n sinθ θ = 0 i 1 i t t E n cosθ n cosθ n n = = E n cosθ + n cosθ n + n r 1 i t 1 i 1 i t 1 Et n1cosθ i n1 = = E n cosθ + n cosθ n + n i 1 i t 1 83
Homework: For the TM polarized wave shown below, show that: E i ε 1, µ 1,n1 + H r E r k r θ r θ i ki x ε, µ,n kt ˆn θ t E t H t E n cosθ n = E n cosθ + n r 1 t i i 1 t i i i 1 t z cosθ cosθ Et n1 cosθi = E n cosθ + n cosθ H i y 84
Total Internal Reflection (TE or TM) Set Er 1 E = i Er n1cosθi ncosθt = = 1 E n cosθ + n cosθ i 1 i t Snell s Law: n n n sinθ = n 1 i sin θ = sinθ n sin 1 i ( θ ) t sin θ = n 1 cos 1 i n cosθt = 1 sin n 1 t θ t θ i 85
Total Internal Reflection (TE or TM) E n cosθ n cosθ n 1, cos 1 sin E n n n r 1 i t 1 = = θt = i 1cosθi + cosθt θ i E E r i n n cos 1 sin 1 1 θi n n = = 1 n n cos 1 sin θ 1 1 θi + n n θ i i a jb n = 1 inθi a + jb 1 1 sinθi s n n n 1 86
Total Internal Reflection (TE or TM) sin i n θ n < n1 n1 Possible only when going from a more optically dense media to a lessdense media. Critical angle: n θ c = sin 1 n 1 For total internal reflection n1 n θ < θ i c θ θ n1 n θ = θ i c i c n1 n θ > θ 1 i c n > n 1 n n 87
What is the transmission angle when θ i = θ c? Snell s Law: Known as a surface wave. 1 1 sinθt = sinθi = = 1 n θ n n1 i = θc θ = t n π n n Total internal reflection and surface waves are key concepts for understanding optical waveguides. n < n 1 n 1 88
What is the transmission amplitude when θ i = θ c? θ t = π cosθ t = 0 E r E i = n 1 cosθ i n cosθ t n 1 cosθ i + n cosθ t = 1 E t E i = n 1 cosθ i n 1 cosθ i + n cosθ t = What does this mean? ( ) 89
What are the transmitted fields when θ i = θ c? E t = ˆxE i e jk y sinθ =1 t +z cosθt H t = ( ŷ cosθ t + ẑsinθ ) E i t η = ˆxEi e jk y e jk ( ysinθt +z cosθt ) = ẑ E i η e jk y This waves travels in the negative y-direction, parallel to the surface. It is known as a surface wave. 90
What is the power density delivered to region () when θ i = θ c? S t = 1 Re E t H * t θi =θ c = 1 * Re ˆx ẑ E e E jk y i i η E = ŷ i η No power is transferred in the z-direction. e jk y 91
cosθ t θ i >θ c = 1 sin θ t θ i >θ c What happens when θ i > θ c? We have from Snell s law that sinθ t = n 1 sinθ Recall: n i sinθ θ i >θ c c = n n 1 = 1 n 1 n sin θ i >1 θ i >θ c = ± j n 1 n sin θ i 1θi>θc = ± jα 9
Examine the transmitted fields for θ i > θ c cosθ t θ i >θ c = ± j n 1 n sin θ i 1θi>θc = ± jα sinθ t = n 1 n sinθ i ( Snell) E t = ˆxE t e jk ( ysinθ t +zcosθ t ) = ˆxEt e jk y n 1 H t = sinθ n i ± jα z ( ŷcosθ t + ẑsinθ t ) E t e jk ( ysinθ t +zcosθ t ) η = jα ŷ + ẑ n E 1 sinθ t n i e kα z+ jky η n 1 n sinθ i = ˆxEt e ± k α z+ jk y n 1 n sinθ i use (-) sign (Why?) 93
Examine the transmitted fields for θ i > θ c ε 1,µ 1,n 1 ε,µ,n Direction of Propagation n > n 1 θ r y E t = ˆxE t e k α z+ jk y n 1 n sinθ i θ i Exponential decay or an evanescent field in z-direction a surface wave 94
Examine the transmitted fields for θ i > θ c n > ε 1,µ 1,n 1 n 1 θ r θ i y ε,µ,n Velocity of Propagation in n v p = ω k y = ω k n 1 n sinθ i E t = ˆxE t e k α z+ jk y Exponential decay or an evanescent field in z-direction a surface wave n 1 n sinθ i 95
Velocity of Propagation in Region : v p = ω k y = v p v = ω k 0 n 1 sinθ i ω k 0 n ω k n 1 n sinθ i = = n n 1 sinθ i = Snell ω n k 0 n 1 n sinθ i = n n sinθ t = 1 ω k 0 n 1 sinθ i sinθ t >1 < 1 A slow wave 96
Reflection and Transmission Coefficients 97
Reflection and Transmission Coefficients 98
Reflection and Transmission Coefficients 99
Example: A Nonlinear Optical Switch Consider the interface between a nonlinear dielectric of high index and a linear dielectric of lower index. The nonlinear media has an index that depends on the intensity of the electromagnetic field (known as the optical Kerr effect). The Kerr effect modifies the index of refraction in the following way: where δn is a small constant. n( I ) = n 1 + I iδ n δ n = 10 10 cm / W Input Laser Assume that (unusually large). Non-Linear Crystal Detector Also assume that n( I ) = 1.5 for low intensities. 100
Example: A Nonlinear Optical Switch a) For low intensities (i.e., neglect δn), what is the critical angle for a ray incident from n? b) If the angle of incidence is 0.5 o smaller than θ c, the wave will have a finite transmission into the lower index media for low intensities. What is the power of the transmitted field? c) If the angle of incidence is 0.5 o smaller than θ c, at what intensity will the interface cease to transmit light? General approach (model): n( I ) Low = n 1 n( I ) High = n 1 + I iδ n θ < θ θi > θc i c n = n o = 1 Low Intensity Signal on detector High Intensity TIR occurs, no signal on detector n 101
Example: A Nonlinear Optical Switch a) Low Intensity no TIR. What is θ c? sinθ c = n n( 0) = 1 1.5 θ = 41.81 c b) Set θ i = 41.81 0.5 = 41.31 θ t = n n( 0) sinθ = 1 sin41.31 = 6.11 i 1.5 10
Example: A Nonlinear Optical Switch b) Continued T = E t E i = = n 1 cosθ i n 1 cosθ i + n cosθ t ( ) ( ) +1 cos( 6.11 ) = 1.113 1.5 cos 41.31 1.5 cos 41.31 103
Example: A Nonlinear Optical Switch b) Continued S t = 1 E t H t * E t = ˆxE t e jk ( ysinθ t +zcosθ t ), Ht = ( ŷcosθ t + ẑsinθ ) E t t e jk ( ysinθ t +zcosθ t ) η S t = 1 E t H t * = ˆx ŷcosθ t + ˆx ẑsinθ t ( ) E t η S t = n T E i = 1.0 η o 377 1.11 E i = 3.7 10 3 E i 104
Example: A Nonlinear Optical Switch b) Continued Intensity of incident wave: S i = 1 E i H * i = E i η 1 = n 1 E i η o = 3.979 10 3 E i Normalized intensity: S t S i = 3.7 10 3 E i 3.979 10 3 E i = 0.81 8% transmission 105
Example: A Nonlinear Optical Switch c) For total internal reflection the index must increase to: sin( 41.31 ) = n n n high = high 1 ( ) = 1.515 sin 41.31 The required intensity is: ( ) = n 1 + Iiδn I = n ( I ) n 1 n I δn = 1.515 1.5 10 10 cm / W = W 1.5 108 cm This is an intense beam Homework: For a 1mm beam diameter, how much laser power would be required to achieve this intensity? 106
Example: Fields for Total Internal Reflection n 1 = 3.4 n = 1 E x 45 45 H A TE wave is incident from GaAs with an index of 3.4 onto the GaAs-Air interface at an incidence angle of 45 degrees. Describe the electric field in the air region. Assume that the light has a wavelength of 1 micron. y z 107
Example: Fields for Total Internal Reflection Critical angle: Since θ > θ c sinθ c = n = 1 n 1 3.4 θ = 17.1 c total internal reflection will occur. The electric field in the air region is: E t = ˆxE t e k α z+ jk y n 1 n sinθ i = ˆxE t e n k 0 α z+ jn 1 k 0 ysinθ i α = n 1 n sin θ i 1 θ i >θ c E t = ΤE i, Τ = E t E i = n 1 cosθ i n 1 cosθ i + n cosθ t 108
Example: Fields for Total Internal Reflection cosθ t = ± 1 n 1 Τ = α = n n 1 cosθ i n 1 cosθ i + n cosθ t = = 1.539exp( 4.8 ) n 1 n sin θ i = 1 3.4 1 sin θ i 1 =.186 π k = n k o = n λ = π rad 106 m sin ( 45 ) = j.186 3.4cos( 45 ) 3.4 cos45 j.186 = 4.808.404 j.186 109
Example: Fields for Total Internal Reflection E t ( y,z,t) = ˆx Re E t e n k 0 α z+ jn 1 k 0 ysinθ i + j = 1.539 Re exp j4.8 πc λ t ( ) π 10 6.186 z+ j3.4 π 10 6 ysin 45 ˆxE i e = 1.539Re ˆxE i e 13.74 106 z+ j15.11 10 6 y+ j1.885 10 8 t+ j4.8 ( )+ j ( ) = ˆx1.539 E i e 13.74 106 z cos 1.885 10 8 t +15.11 10 6 y + 4.8 π 3 108 10 6 t Assuming that E i is real. 110
One Final Comment Brewster Angle Reflection Coefficient Transmission Coefficient TE n 1 cosθ i n cosθ t n 1 cosθ i + n cosθ t n 1 cosθ i n 1 cosθ i + n cosθ t TM n 1 cosθ t n cosθ i n 1 cosθ t + n cosθ i n 1 cosθ i n cosθ i + n 1 cosθ t Note the difference. 111
Brewster Angle: For TM Polarization Only n 1 > n Brewster Angle 11
Brewster Angle: For TM Polarization Only n 1 < n Brewster Angle 113
Brewster Angle: For TM Polarization Only (Polarized) Laser Brewster Window 114
Brewster Angle: For TM Polarization Only R TM = 0 = n cosθ n cosθ 1 t B n n 1 cosθ t + n cosθ cos θ B = n 1 cos θ t = n ( 1 1 sin θ ) t B sin θ t = n 1 n sin θ B n cos θ B = n 1 n n 1 via Snell ( 1 sin θ ) t = n 1 ( 1 sin θ ) B = 1 n 1 n sin θ B 1 n 1 n sin θ B 115
Brewster Angle: For TM Polarization Only n n 1 ( 1 sin θ ) B = 1 n 1 n sin θ B sinθ B = n n 1 + n θ B = arcsin n n 1 + n 116
Brewster Angle: For TE Polarization? No R TE = 0 n 1 cos θ B = n 1 cos θ t = n n 1 n ( 1 sin θ ) B = 1 n 1 n sin θ B n 1 n sin θ B n 1 n sin θ B via Snell ( 1 sin θ ) t = n 1 n 1 n sin θ B n 1 n = 1 Only the trivial solution exists. 117
Phase Shift of Reflected Wave We found, for TIR (for TE Waves): E r E i = n cosθ n cosθ 1 i t = 1, cosθ n 1 cosθ i + n cosθ t = 1 n 1 t n sin θ i E r E i = n n 1 cosθ i jn 1 n sin θ i 1 n n 1 cosθ i + jn 1 n sin θ i 1 = 1 a jb a + jb = 1 What about the phase? 118
Phase Shift of Reflected Wave What about the phase? Φ TE = arg E r E i = arctan n n 1 n 1 cosθ i n sin θ i 1,θ > θ i c n 1 = 3.4 n = 1 119
Phase Shift of Reflected Wave Why is this important? Laser cavity round-trip phase shift of nπ Corner Reflector via TIR 10
An Alternate View The Goos-Hänchen Shift θ i = arctan β κ z s θ z s Reflected Wave n c Incident Beam (Ray Bundle) n f ( ) = e A x, z ( ) B ( x, z ) A x, z For the Incident Wave: x ( j κ x ( β+δβ )z j κ x ( ) β Δβ )z + e A( 0, z) = ( e jδβz + e jδβz )e jβz = cos( Δβz)e jβz z 11
An Alternate View The Goos-Hänchen Shift θ i = arctan β κ z s θ z s Reflected Wave n c Incident Beam n f B( x, z) = e ( ) B ( x, z ) A x, z For the Reflected Wave: j κ x ( β+δβ )z e jφ TE θ i β+δβ x z ( ( )) j κ x β Δβ + e ( ( )) ( )z e jφ TE θ i β Δβ 1
An Alternate View The Goos-Hänchen Shift For the Reflected Wave: B( 0, z) = e j β+δβ = e j z+ Φ TE ( )z e jφ TE θ i( β+δβ ) dφ TE dβ ( β ± Δβ ) = Φ ( TE β ) ± dφ TE dβ ( ) + e j β Δβ Δβ + e + j z+ dφ TE dβ Δβ ( ) ( )z e jφ TE θ i( β Δβ ) Δβ e jφte β = cos z + dφ TE dβ Δβ e jφ TE( β ) e jβz ( ) e jβz 13
An Alternate View The Goos-Hänchen Shift Recall: Φ TE = arctan n n 1 n 1 cosθ i n sin θ i 1 Also: θ i = arctan β κ, sinθ = i β = β, cosθ β +κ n 1 k i = o κ β +κ = κ n 1 k o Thus: Φ TE = arctan β n k o n 1 k o β 14
An Alternate View The Goos-Hänchen Shift Differentiating: dφ TE dβ = d dβ arctan β n k o n 1 k o β β = ( β n k )( o n 1 k o β ) = β κ β n k o The units are in meters, hence: z s = β κ β n k o 15
An Alternate View The Goos-Hänchen Shift Thus: B 0, z ( )Δβ ( ) = cos z z s e jφ TE β ( ) e jβz Where z s = β κ β n k o 16
An Alternate View The Goos-Hänchen Shift θ i = arctan β κ z s θ z s Reflected Wave n c Incident Beam n f ( ) B ( x, z ) A x, z x z z s = β κ = tanθ i β n k o β n k o β n 1 k o tanθ i k o n 1 n 17
An Alternate View The Goos-Hänchen Shift Consider an acrylic (n = 1.49)-air interface: θ c = arcsin n n 1 = arcsin 1 1.49 = 4.15 Green light (510 nm): k o = π λ = 1.31997 106 z s = tanθ i k o n 1 n = 1.3 10 6 tanθ i 1.105 = 147 10 9 tanθ i for θ i = 45, z s = 0.94 microns 18
An Alternate View The Goos-Hänchen Shift Substrate n n 1 Beam Waveguide Ray TIR Corner Reflector Portion of a Photonic Integrated Circuit 19
An Alternate View The Goos-Hänchen Shift Substrate The lateral displacement of the beam really does occur. n n 1 If not compensated for it results in additional loss due to reduced coupling efficiency for the input to the output waveguide. Waveguide x z To compensate, the reflecting facet must be moved toward the inner corner by an amount: x = z s tanθ i = Since beam quality also suffers, such compensation is crucial for laser design. β n k o = γ 130
ASIDE: A Ring Cavity Laser Gain Section Input Coupler Pump (Input) Output Coupler Frequency Selective Grating Filter Output Goos-Hänchen Shift must be taken into account at each of the corner reflectors 131
Next Planar Optical Waveguides 13