Fresnel Equations - Trinity College Dublin, the … Equations Consider reflection and transmission...
Transcript of Fresnel Equations - Trinity College Dublin, the … Equations Consider reflection and transmission...
Fresnel Equations Consider reflection and transmission of light at dielectric/dielectric boundary
Calculate reflection and transmission coefficients, R and T, as a function of incident light polarisation and angle of incidence using EM boundary conditions s-polarisation E perpendicular to plane of incidence p-polarisation E parallel to plane of incidence
n1 = √ε1 µ1 = 1
n2 = √ε2 µ2 = 1
θt
θi θr
s-polarisation p-polarisation
Fresnel Equations Snell’s Law
Boundary conditions apply across the entire, flat interface (z = 0) Incident, reflected and transmitted waves are like
EI = (ey cosθi + ez sinθi) EoI ei(ωt - kI.r)
ER = (-ey cosθr + ez sinθr) EoR ei(ωt - kR.r)
ET = (ey cosθt + ez sinθt) EoT ei(ωt - kT.r)
To satisfy BC (kI . r)z=0 = (kR . r) z=0 = (kT . r) z=0 (1) wave vectors lie in single plane (2) projection of wave vectors on xy plane is same From (1) From (2) kI sin θi = kR sinθr = kT sinθt kI = kR = ωc µ1ε1 kT = ωc µ2ε2
kI sin θi = kT sinθt becomes sin θi / sinθt = µ2ε2 / µ1ε1
n1 = √ε1 µ1 = 1
n2 = √ε2 µ2 = 1
θt
θi θr kI
kR
kT z
y
θi = θr
Boundary conditions on E E fields at matter/vacuum interface
Boundary conditions on E from Faraday’s Law ∮ E.dℓ = −𝐶ddt∫ B.dS𝑆
∮ E.dℓ = 𝐶 EL.dℓL + ER.dℓR (as ∆t → 0)
∫ B.dS𝑆 → 0 (as ∆t → 0) -EL sinθLdℓL + ER sinθR dℓR = 0
EL sinθL = ER sinθR E||L = E||R E|| continuous
EL
ER
θL θR dℓL
dℓR
∆t
Boundary conditions on H H fields at matter/vacuum interface Boundary conditions on H from Ampère’s Law ∇ x H = jfree + ∂D/∂t
∇ x H .dS = jfree +∂D∂t
.dS =H.dℓ
D, ∂D/∂t are everywhere finite, so as as ∆t → 0, ∫ ∂D∂t .dS → 0
For materials of finite conductivity, jfree is finite, so ∫ jfree.dS → 0 as ∆t → 0 For materials of infinite conductivity, jfree is infinite, so ∫ jfree.dS → jfree,surface
dℓ as ∆t → 0 jfree,surface is surface current per unit length
HL
HR
θL θR dℓL
dℓR
∆t
Boundary conditions on H H.dℓ = jfree,surface
dℓ
-HL sinθLdℓL + HR sinθR dℓR = jfree,surface dℓ
HR sinθR dℓ = HL sinθL dℓ + jfree,surface dℓ
H||R = H||L + jfree,surface Infinite conductivity at interface H||R = H||L Finite conductivity at interface
HL
HR
θL θR dℓL
dℓR
∆t
Boundary conditions on B
⊥⊥ =⇒=−
=⇒=∇ ∫
21
2211
BB0dS cos BdS cos B
0.d0 .
θθ
S
SBB
1
2
B1 B2
θ2
θ1
dS1,2
B field at matter/vacuum interface
Boundary conditions on D D field at matter/vacuum interface ∇.D = ρfree
∫D.dS = ∫ρfree dv ∫ D.dS = 0 No free charges at interface ∫ D.dS = ∫ ∇.D dv = σfree dS Free charge density σfree at interface
∫ D1. dS1 + ∫D2.dS2 = ∫ρfree dv
D 1 dS - D 2 dS = σfree dS dS1 = dS2 = dS
D1 = D2 No interface free charges D1 - D2 = σfree Interface free charges
1
2
D1 D2
θ2
θ1
dS1,2
Boundary conditions summary E||L = E||R E|| continuous B1 = B2 B continuous D1 = D2 D continuous No interface free charges D1 - D2 = σfree Interface free charges
H||R = H||L H|| continuous Finite conductivity at interface H||R = H||L + jfree,surface Infinite conductivity at interface
Fresnel Equations Reflection coefficient R and Transmission coefficient T ∇2E - µoεoε ∂2E/∂t2 = 0
E(r, t) = Eo ex Reei(ωt - k.r) k = ω√(µoµεoε)
∇ x E = -i k x E take curl of plane wave E ∇ x E = - ∂B/∂t Faraday’s law - ∂B/∂t = -iω B time harmonic, plane wave B -iω B = -i k x E B = k x E / ω = k ek x E / ω = ω√(µoµεoε) ek x E / ω = √(µε) ek x E / c
Fresnel Equations B = k x E / ω = k ek x E / ω = ω√(µoµεoε) ek x E / ω = √(µε) ek x E / c N = E x H = E x B / µoµ = E x (√(µoµεoε) ek x E) / µoµ N = E2 √(εoε /µoµ) R = reflected energy / incident energy = ER
2 √(εoε1 /µoµ1) / EI2 √(εoε1 /µoµ1)
= ER2 / EI
2 T = transmitted energy / incident energy = ET
2 √(εoε2 /µoµ2) / EI2 √(εoε1 /µoµ1)
= ET2 / EI
2 n2 / n1 (if µ1 = µ2 = 1) R = ER
2 / EI2
T = ET2 / EI
2 n2 / n1
Fresnel Equations Normal Incidence
x
z
y
ER EI ET
BT BI
BR kI
kR kT
Fields EI = ex EoI ei(ωt - k1z)
BI = ey BoI ei(ωt - k1z)
ER = ex EoR ei(ωt + k1z)
BR = -ey BoR ei(ωt + k1z)
ET = ex EoT ei(ωt - k2z)
BT = ey BoT ei(ωt - k2z) Boundary conditions
E||1 = E||2 EoI + EoR = EoT B = D = 0 (normal incidence) H||1 = H||2 (BoI - BoR) / µ1µo = BoT / µ2µo B = µµo H µ1 = µ2 = 1
n1 = √ε1 µ1 = 1 kI = kR = k1
n2 = √ε2 µ2 = 1 kT = k2
Fresnel Equations BoI = n1 EoI / c BoR = n1 EoR / c BoT = n2 EoT / c n1 (EoI - EoR) = n2 EoT from BoI - BoR = BoT when µ1 = µ2 = 1 EoI + EoR = EoT
EoT = EoI + EoR = n1(EoI - EoR) / n2 Eliminate EoT
EoR (n1 + n2) = EoI (n1 - n2) EoR / EoI = (n1 - n2) / (n1 + n2) EoR / EoI < 0 if n1 < n2) => π change of phase
Fresnel Equations n1 (EoI - EoR) = n2 EoT Eliminate EoR
EoI + EoR = EoT
EoR = EoT - EoI = EoI - n2 EoT / n1
EoT (n1 + n2) = 2n1 EoI EoT / EoI = 2n1 / (n1 + n2) (EoR / EoI)2 + (EoT / EoI)2
= (n1 - n2)2 / (n1 + n2)2 + 4n12 / (n1 + n2)2 ≠ 1!
Fresnel Equations Reflectivity R = (EoR / EoI)2 = (n1 - n2)2 / (n1 + n2)2
Transmittivity T = (EoT / EoI)2 √(µ2ε2) / √(µ1ε1) = 4n1
2 / (n1 + n2)2 (n2 / n1) = 4n1n2 / (n1 + n2)2
Energy conservation R + T = (n1 - n2)2 / (n1 + n2)2 + 4n1n2 / (n1 + n2)2 = 1
Fresnel Equations Off-normal incidence, s-polarisation
Fields EI = ex EoI ei(ωt - k1.r)
BI = (ey cosθi + ez sinθi) BoI ei(ωt - k1.r)
ER = ex EoR ei(ωt + k1.r)
BR = (-ey cosθr + ez sinθr) BoR ei(ωt + k1.r)
ET = ex EoT ei(ωt - k2.r)
BT = (ey cosθt + ez sinθt) BoT ei(ωt - k2.r) Boundary conditions
E||1 = E||2 EoI + EoR = EoT H||1 = H||2 (BoI - BoR) cosθi / µ1µo = BoT cosθt / µ2µo µ1 = µ2 = 1
θt
θi θr
EI kI BI ER kR
BR
BT kT ET
z
y
n1 = √ε1 µ1 = 1 kI = kR = k1
n2 = √ε2 µ2 = 1 kT = k2
Fresnel Equations B = √(µε) k x E / ck = n / (ck) k x E in uniform dielectric BoI = n1 EoI / c BoR = n1 EoR / c BoT = n2 EoT / c n1 (EoI - EoR) cosθi = n2 EoT cosθt from (BoI - BoR) cosθi / µ1µo = BoT cosθt / µ2µo with µ1 = µ2 = 1 EoI + EoR = EoT Eliminate EoT
EoT = EoI + EoR = n1(EoI - EoR) cosθi / (n2 cosθt )
EoR (n1 cosθi + n2 cosθt) = EoI (n1 cosθi - n2 cosθt) EoR / EoI = (n1 cosθi - n2 cosθt) / (n1 cosθi + n2 cosθt)
Fresnel Equations n1 cosθi (EoI - EoR) = n2 cosθt EoT Eliminate EoR EoI + EoR = EoT
EoR = EoT - EoI = EoI - n2 cosθt EoT / (n1 cosθi)
EoT (n1 cosθi + n2 cosθt) = 2n1 cosθi EoI EoT / EoI = 2n1 cosθi / (n1 cosθi + n2 cosθt) Reflectivity RS = (EoR / EoI)2 = (n1 cosθi - n2 cosθt)2 / (n1 cosθi + n2 cosθt)2
Fresnel Equations Transmittivity TS = (EoT / EoI)2 √(µ2ε2) cosθt / √(µ1ε1) cosθi = 4n1
2 cos2θi / (n1cosθi + n2cosθt) 2 (n2cosθt / n1cosθi) = 4n1n2 cosθi cosθt / (n1cosθi + n2cosθt) 2 Energy conservation R+T =(n1cosθi - n2cosθt)2 /(n1cosθi + n2cosθt)2 + 4n1n2cos2θi /(n1cosθi + n2cosθt) 2
= (n12cos2θi - 2n1n2cosθi cosθt+ n2
2cos2θt + 4n1n2cosθi cosθt) /(n1cosθi + n2cosθt) 2
= 1
Fresnel Equations Off-normal incidence, p-polarisation
Fields
EI = (ey cosθi + ez sinθi) EoI ei(ωt - k1.r)
BI = -ex BoI ei(ωt - k1.r)
ER = (-ey cosθr + ez sinθr) EoR ei(ωt + k1.r)
BR = -ex BoR ei(ωt + k1.r)
ET = (ey cosθt + ez sinθt) EoT ei(ωt - k2.r)
BT = -ex BoT ei(ωt - k2.r) Boundary conditions
E||1 = E||2 (EoI - EoR) cosθi = EoT cosθt H||1 = H||2 (BoI + BoR) / µ1µo = BoT / µ2µo µ1 = µ2 = 1
n1 = √ε1 µ1 = 1
n2 = √ε2 µ2 = 1
θt
θi θr
BI kI EI BR kR
ER
ET kT BT
z
y
X
X
X
Fresnel Equations B = √(µε) k x E / ck = n / (ck) k x E in uniform dielectric BoI = n1 EoI / c BoR = n1 EoR / c BoT = n2 EoT / c n1 (EoI + EoR) = n2 EoT from (BoI + BoR) / µ1µo = BoT / µ2µo with µ1 = µ2 = 1 (EoI - EoR) cosθi = EoT cosθt EoT = (EoI + EoR) n1 / n2 = (EoI - EoR) cosθi / cosθt Eliminate EoT
EoR (n1 / n2 + cosθi / cosθt) = EoI (- n1 / n2 + cosθi / cosθt) EoR / EoI = (- n1 / n2 + cosθi / cosθt) / (n1 / n2 + cosθi / cosθt) = (n2cosθi - n1cosθt) / (n2cosθi + n1cosθt)
Fresnel Equations Reflectivity RP = (EoR / EoI)2 = (n2cosθi - n1cosθt)2 / (n2cosθi + n1cosθt)2 EoR = EoT n2 / n1 - EoI = EoI - EoT cosθt / cosθi Eliminate EoR EoT (n2 / n1 + cosθt / cosθi) = 2EoI
EoT / EoI = 2EoI / (n2 / n1 + cosθt / cosθi) EoT / EoI = 2 / (n2 / n1 + cosθt / cosθi) = 2n1cosθi / (n1cosθt +n2cosθi)
Fresnel Equations Transmittivity TP = (EoT / EoI)2√(µ2ε2) cosθt / √(µ1ε1) cosθi = 4n1
2cos2θi n2cosθt / (n1cosθt + n2cosθi) 2 n1cosθi = 4n1cosθi n2cosθt / (n1cosθt + n2cosθi) 2 Energy conservation RP + TP = ((n2cosθi - n1cosθt)2 + 4n1cosθi n2cosθt ) / (n2cosθi + n1cosθt)2 = 1
Fresnel Equations Normal incidence R = (n1 - n2)2 / (n1 + n2)2 T = 4 n1n2 / (n1 + n2)2
S-polarisation RS = (n1cosθi - n2cosθt)2 / (n1cosθi + n2cosθt)2
TS = 4n1n2 cosθi cosθt / (n1cosθi + n2cosθt) 2 P-polarisation RP = (n2cosθi - n1cosθt)2 / (n2cosθi + n1cosθt)2 TP = 4n1n2 cosθi cosθt / (n1cosθt + n2cosθi) 2 Energy conservation R + T = 1 in each case
Fresnel Equations Light polarisation by reflection - the Brewster angle RS = (n1cosθi - n2cosθt)2 / (n1cosθi + n2cosθt)2
RP = (n2cosθi - n1cosθt)2 / (n2cosθi + n1cosθt)2 If n1 < n2 (e.g. n1 = 1, n2 > 1), θi > θt then n1cosθi < n2cosθt Consequently RS ≠ 0 for any θi If n1 < n2, n2cosθi = n1cosθt then RP = 0 for θi = θB Brewster angle
0 20 40 60 80
0.10.20.30.40.50.60.7
Ref
lect
ivity
Angle of Incidence
n1 = 1, n2 = 1.5 RS
RP θB
θB θB
EI unpolarised
ER s-polarised
ET partly polarised
θt = π/2 - θB
Fresnel Equations Normal Incidence, metal-vacuum interfaces
x
z
y
ER EI ET
BT BI
BR kI
kR kT
Fields EI = ex EoI ei(ωt - k1z)
BI = ey BoI ei(ωt - k1z)
ER = ex EoR ei(ωt + k1z)
BR = -ey BoR ei(ωt + k1z)
ET = ex EoT ei(ωt - αz) e -αz
BT = ey BoT ei(ωt - αz) e -αz Boundary conditions
E||1 = E||2 EoI + EoR = EoT B = D = 0 (normal incidence) H||1 = H||2 (BoI - BoR) / µ1µo = BoT / µ2µo B = µµo H µ1 = µ2 = 1
n1 = 1 µ1 = 1
α = √(ωσµo/2) µ2 = 1
Fresnel Equations B ≠ √(µε) k x E / ck in lossy matter, use Faraday’s law instead ∇ x ET = − ∂BT
∂t
∇ x ET = − ey EoT ei(ωt - αz) e -αz α(1 + i)
− ∂BT∂t = −iω ey BoT ei(ωt - αz) e -αz
EoT α(1 + i) = iω BoT
BoI = n1 EoI / c BoR = n1 EoR / c BoT = α(1 − i) EoT / ω H||1 = H||2 becomes n1 (EoI - EoR) / c = α(1 − i) EoT / ω α = √(ωσµo/2)
E||1 = E||2 becomes EoI + EoR = EoT set n1 = µ1 = µ2 = 1
Fresnel Equations (EoI - EoR) / c = α(1 − i) EoT / ω EoI + EoR = EoT Eliminate EoT
EoT
= EoI +
EoR = (EoI - EoR) / a(1 − i) a = ω / αc = √(σ/2εoω)
EoR (a(1 − i) + 1) = EoI (1 - a(1 − i)) EoR / EoI = (1 - a(1 − i)) / (1 + a(1 − i)) R = |EoR / EoI|2 ≈ 1 - 2 / a = 1 - 2 √(2εoω/σ) For Cu metal, σ = 6.7 x 107 (Ωm)-1 For ω = 7 x 1014 R ≈ 1 - 2.8 x 10-2 = 0.97