Foundations of Optics

Lecture 1: Foundations of Optics

Outline

1 Electromagnetic Waves 2 Material Properties 3 Electromagnetic Waves Across Interfaces 4 Fresnel Equations

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 1 Electromagnetic Waves

Electromagnetic Waves in Matter Maxwell’s equations ⇒ electromagnetic waves optics: interaction of electromagnetic waves with matter as described by material equations polarization of electromagnetic waves are integral part of optics

Maxwell’s Equations in Matter Symbols D~ electric displacement ∇ · D~ = 4πρ ρ electric charge density ~ 1 ∂D~ 4π H magnetic ﬁeld ∇ × H~ − = ~j c ∂t c c speed of light in vacuum ~j electric current density 1 ∂B~ ∇ × E~ + = 0 E~ electric ﬁeld c ∂t B~ magnetic induction ∇ · B~ = 0 t time

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 2 Linear Material Equations Symbols dielectric constant D~ = E~ µ magnetic permeability B~ = µH~ σ electrical conductivity ~j = σE~

Isotropic and Anisotropic Media isotropic media: and µ are scalars anisotropic media: and µ are tensors of rank 2 isotropy of medium broken by anisotropy of material itself (e.g. crystals) external ﬁelds (e.g. Kerr effect) mechanical deformation (e.g. stress birefringence)

assumption of isotropy may depend on wavelength (e.g. CaF2 for semiconductor manufacturing)

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 3 Wave Equation in Matter static, homogeneous medium with no net charges: ρ = 0 for most materials: µ = 1 combine Maxwell, material equations ⇒ differential equations for damped (vector) wave

µ ∂2E~ 4πµσ ∂E~ ∇2E~ − − = 0 c2 ∂t2 c2 ∂t

µ ∂2H~ 4πµσ ∂H~ ∇2H~ − − = 0 c2 ∂t2 c2 ∂t damping controlled by conductivity σ E~ and H~ are equivalent ⇒ sufﬁcient to consider E~ interaction with matter almost always through E~ but: at interfaces, boundary conditions for H~ are crucial

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 4 Plane-Wave Solutions ~ ~ i(~k·~x−ωt) Plane Vector Wave ansatz E = E0e ~k spatially and temporally constant wave vector ~k normal to surfaces of constant phase |~k| wave number ~x spatial location ω angular frequency (2π× frequency ν) t time ~ E0 (generally complex) vector independent of time and space ~ ~ −i(~k·~x−ωt) could also use E = E0e damping if ~k is complex ~ E0 describes the polarization and the absolute phase real electric ﬁeld vector given by real part of E~ sum of solutions is also a solution

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 5 Complex Index of Refraction temporal derivatives ⇒ Helmholtz equation

ω2µ 4πσ ∇2E~ + + i E~ = 0 c2 ω spatial derivatives ⇒ dispersion relation between ~k and ω

ω2µ 4πσ ~k · ~k = + i c2 ω complex index of refraction

4πσ ω2 n˜2 = µ + i , ~k · ~k = n˜2 ω c2 split into real (n: index of refraction) and imaginary parts (k: extinction coefﬁcient) n˜ = n + ik

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 6 Transverse Waves

plane-wave solution must also fulﬁll Maxwell’s equations

~ ~ ~ ~ ~ ~ n˜ k ~ E0 · k = 0, H0 · k = 0, H0 = × E0 µ |~k|

isotropic media: electric, magnetic ﬁeld vectors normal to wave vector ⇒ transverse waves ~ ~ ~ E0, H0, and k orthogonal to each other, right-handed vector-triple ~ ~ conductive medium ⇒ complex n˜, E0 and H0 out of phase ~ ~ ~ E0 and H0 have constant relationship ⇒ consider only E

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 7 Energy Propagation in Isotropic Media time-averaged Poynting vector D E c S~ = Re E~ × H~ ∗ 8π 0 0

Re real part of complex expression ∗ complex conjugate h.i time average energy ﬂow parallel to wave vector in isotropic media

~ D~ E c |n˜| 2 k S = |E0| 8π µ |~k|

energy ﬂow is proprtional to index of refraction! in anisotropic materials (e.g. crystals), energy propagation and wave vector are not parallel!

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 8 Quasi-Monochromatic Light monochromatic light: purely theoretical concept monochromatic light wave always fully polarized real life: light includes range of wavelengths ⇒ quasi-monochromatic light quasi-monochromatic: superposition of mutually incoherent monochromatic light beams whose wavelengths vary in narrow range δλ around central wavelength λ0 δλ 1 λ measurement of quasi-monochromatic light: integral over measurement time tm amplitude, phase (slow) functions of time for given spatial location slow: variations occur on time scales much longer than the mean period of the wave

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 9 Polychromatic Light or White Light δλ wavelength range comparable wavelength ( λ ∼ 1) incoherent sum of quasi-monochromatic beams that have large variations in wavelength cannot write electric ﬁeld vector in a plane-wave form must take into account frequency-dependent material characteristics intensity of polychromatic light is given by sum of intensities of constituting quasi-monochromatic beams

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 10 Material Properties

Index of Refraction complex index of refraction

4πσ ω2 n˜2 = µ + i , ~k · ~k = n˜2 ω c2

no electrical conductivity ⇒ real index of refraction transparent, dielectric materials: real index of refraction conducting materials (metal): complex index of refraction also transparent, conducting materials, e.g. ITO index of refraction depends on wavelength (dispersion) index of refraction depends on temperature index of refraction roughly proportional to density

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 11 Glass Dispersion

en.wikipedia.org/wiki/File:Dispersion-curve.png

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 12 Wavelength Dependence of Index of Refraction tabulated by glass manufacturer various approximations to express wavelength dependence with a few parameters typically index increases with decreasing wavelength Abbé number: nd − 1 νd = nF − nC

nd : index of refraction at Fraunhofer d line (587.6 nm)

nF : index of refraction at Fraunhofer F line (486.1 nm)

nC: index of refraction at Fraunhofer C line (656.3 nm)

low dispersion materials have high values of νd

Abbe diagram: νd vs nd

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 13 Glasses

ν d 95 90 85 80 75 70 65 60 55 50 45 40 35 30 25 20 2.05 2.05

68 nd 2.00 2.00 nd ν Abbe-Diagram nd – d Description of Symbols 1.95 1.95 N- or P-glass Lead containing glass 66 46A 67 1.90 N-glass or lead containing glass 1.90 Glass suitable for Precision Molding 31A LASF Fused Silica

1.85 9 57 1.85 41 40 SF 51 50 44 47 45 6 1.80 43 1.80 21 33 56A 11 34 14 33A 37 LAF 4 1.75 7 1.75 35 2 10 34 69 10 KZFS 8 1 8 LAK 64 1.70 14 15 1.70 35 9 BASF 8 12 10 5 2 7 51 KZFS5 1.65 22 5 2 1.65 21 SSK KZFS11 BAF 2 2 16 60 8 KZFS4 53 A 4 2 52 14 4 5 1.60 1.60 5 SK F 58A 57 4 5 PSK 1 4 11 BALF KZFS2 LF 3 1.55 2 BAK 5 1 1.55 LLF 51 KF 5 PK 7 7 9 K 10 BK ZK7 1.50 52 A 10 1.50

51A 5 FK FS 1.45 1.45 August 2010 95 90 85 80 75 70 65 60 55 50 45 40 35 30 25 20 ν d

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 14 Empirical Models of Index of Refraction tabulated, measured index as function of wavelength approximations to express smooth wavelength dependence most common: Sellmeier equation and coefﬁcients:

B λ2 B λ2 B λ2 2(λ) = + 1 + 2 + 3 n 1 2 2 2 λ − C1 λ − C2 λ − C3 reﬂects resonances that drive index variation with wavelength only applicable over a limited wavelength range

BK7: B1 =1.03961212, C1 =6.00069867e-3, B2 =2.31792344e-1, C2 =2.00179144e-2,B3 =1.01046945, C3 =1.03560653e2

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 15 Transparent Materials

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 16 Internal Transmission

internal transmission per cm typically strong absorption in the blue and UV almost all glass absorbs above 2 µm

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 17 Metal Reﬂectivity

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 18 Electromagnetic Waves Across Interfaces

Introduction classical optics due to interfaces between 2 different media from Maxwell’s equations in integral form at interface from medium 1 to medium 2 ~ ~ D2 − D1 · ~n = 4πΣ ~ ~ B2 − B1 · ~n = 0 ~ ~ E2 − E1 × ~n = 0 4π H~ − H~ × ~n = − K~ 2 1 c

~n normal on interface, points from medium 1 to medium 2 Σ surface charge density on interface K~ surface current density on interface

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 19 Fields at Interfaces Σ = 0 in general, K~ = 0 for dielectrics complex index of refraction includes effects of currents ⇒ K~ = 0 requirements at interface between media 1 and 2

~ ~ D2 − D1 · ~n = 0 ~ ~ B2 − B1 · ~n = 0 ~ ~ E2 − E1 × ~n = 0 ~ ~ H2 − H1 × ~n = 0

normal components of D~ and B~ are continuous across interface tangential components of E~ and H~ are continuous across interface

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 20 Plane of Incidence

plane wave onto interface incident (i ), reﬂected (r ), and transmitted (t ) waves

~ i,r,t ~ i,r,t i(~k i,r,t ·~x−ωt) E = E0 e c H~ i,r,t = ~k i,r,t × E~ i,r,t µω

interface normal ~n k z-axis

spatial, temporal behavior at interface the same for all 3 waves

~ i ~ r ~ t (k · ~x)z=0 = (k · ~x)z=0 = (k · ~x)z=0

valid for all ~x in interface ⇒ all 3 wave vectors in one plane, plane of incidence

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 21 Snell’s Law

spatial, temporal behavior the same for all three waves

~ i ~ r ~ t (k ·~x)z=0 = (k ·~x)z=0 = (k ·~x)z=0

~ = ω ˜ k c n ω, c the same for all 3 waves Snell’s law

n˜1 sin θi = n˜1 sin θr = n˜2 sin θt

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 22 Monochromatic Wave at Interface c c H~ i,r,t = ~k i,r,t × E~ i,r,t , B~ i,r,t = ~k i,r,t × E~ i,r,t 0 ωµ 0 0 ω 0 boundary conditions for monochromatic plane wave:

2 ~ i 2 ~ r 2 ~ t n˜1E0 + n˜1E0 − n˜2E0 · ~n = 0 ~ i ~ i ~ r ~ r ~ t ~ t k × E0 + k × E0 − k × E0 · ~n = 0 ~ i ~ r ~ t E0 + E0 − E0 × ~n = 0 1 ~ i ~ i 1 ~ r ~ r 1 ~ t ~ t k × E0 + k × E0 − k × E0 × ~n = 0 µ1 µ1 µ2 4 equations are not independent only need to consider last two equations (tangential components ~ ~ of E0 and H0 are continuous)

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 23 Two Special (Polarization) Cases

TM, p TE, s

1 electric field parallel to plane of incidence ⇒ magnetic field is transverse to plane of incidence (TM) 2 electric field particular (German: senkrecht) or transverse to plane of incidence (TE)

general solution as (coherent) superposition of two cases choose direction of magnetic ﬁeld vector such that Poynting vector parallel, same direction as corresponding wave vector

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 24 Electric Field Perpendicular to Plane of Incidence

E E r i H H i r

E t H t

electric ﬁeld also perpendicular to interface normal ~n ~ i ~ r ~ t ~ i,r,t ~ i,r,t E0 + E0 − E0 × n = 0 becomes (with E0 instead of E0 )

i r t E0 + E0 − E0 = 0 ~ i ~ i ~ r ~ r ~ t ~ t ~ k × E0 + k × E0 − k × E0 × n = 0 becomes

i r t n˜1E0 cos θi − n˜1E0 cos θr − n˜2E0 cos θt = 0

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 25 Electric Field Perpendicular to Plane of Incidence

E E r i H H i r

E t H t

from previous slide:

i r t n˜1E0 cos θi − n˜1E0 cos θr − n˜2E0 cos θt = 0

~ i,r,t ~ i,r,t ~ i,r,t k × E0 in direction of H0 Poynting vector in same direction as wave vector ⇒ flip sign of tangential component of magnetic field vector of reflected wave reason for minus sign for reflected component in above equation ~ i,r,t ~ i,r,t cos θi,r,t terms from projecting k × E0 onto interface plane

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 26 Electric Field Perpendicular to Plane of Incidence

E E r i H H i r

E t H t

θr = θi ratios of reﬂected and transmitted to incident wave amplitudes

Er n˜ cos θ − n˜ cos θ r = 0 = 1 i 2 t s Ei 0 n˜1 cos θi + n˜2 cos θt

Et 2n˜ cos θ t = 0 = 1 i s Ei 0 n˜1 cos θi + n˜2 cos θt

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 27 Electric Field in Plane of Incidence

E i E r H i H r

E t

H t

~ i ~ r ~ t ~ E0 + E0 − E0 × n = 0 becomes

i r t E0 cos θi − E0 cos θr − E0 cos θt = 0 .

ﬂip tangential component of electric ﬁeld vector to align Poynting and wave vectors ~ i,r,t ~ cos θi,r,t terms are due to the cross products E0 × n

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 28 Electric Field in Plane of Incidence

E i E r H i H r

E t

H t

~ i ~ i ~ r ~ r ~ t ~ t ~ k × E0 + k × E0 − k × E0 × n = 0 becomes

i r t n˜1E0 + n˜1E0 − n˜2E0 = 0

~ i,r,t ~ i,r,t ~ i,r,t k × E0 is in direction of H0

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 29 Electric Field in Plane of Incidence

E i E r H i H r

E t

H t

ratios of reﬂected and transmitted to incident wave amplitudes

Er n˜ cos θ − n˜ cos θ r = 0 = 2 i 1 t p Ei 0 n˜2 cos θi + n˜1 cos θt

Et 2n˜ cos θ t = 0 = 1 i p Ei 0 n˜2 cos θi + n˜1 cos θt

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 30 Summary of Fresnel Equations q ˜ ˜2 ˜2 2 eliminate θt with Snell’s law n2 cos θt = n2 − n1 sin θi

µ1/µ2 ≈ 1 for most materials

electric ﬁeld amplitude transmission ts,p, reﬂection rs,p

2n˜1 cos θi ts = q ˜ ˜2 ˜2 2 n1 cos θi + n2 − n1 sin θi

2n˜1n˜2 cos θi tp = q ˜2 ˜ ˜2 ˜2 2 n2 cos θi + n1 n2 − n1 sin θi q ˜ ˜2 ˜2 2 n1 cos θi − n2 − n1 sin θi rs = q ˜ ˜2 ˜2 2 n1 cos θi + n2 − n1 sin θi q ˜2 ˜ ˜2 ˜2 2 n2 cos θi − n1 n2 − n1 sin θi rp = q ˜2 ˜ ˜2 ˜2 2 n2 cos θi + n1 n2 − n1 sin θi

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 31 Consequences of Fresnel Equations

complex index of refraction ⇒ ts, tp, rs, rp (generally) complex real indices ⇒ argument of square root in Snell’s law q ˜ ˜2 ˜2 2 n2 cos θt = n2 − n1 sin θi can still be negative ⇒ complex ts, tp, rs, rp real indices, arguments of square roots positive (e.g. dielectric without total internal reﬂection)

therefore ts,p ≥ 0, real ⇒ incident and transmitted waves will have same phase therefore rs,p real, but become negative when n2 > n1 ⇒ negative ratios indicate phase change by 180◦ on reﬂection by medium with larger index of refraction

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 32 Other Form of Fresnel Equations using trigonometric identities

2 sin θt cos θi ts = sin(θi +θt ) 2 sin θt cos θi tp = sin(θi +θt ) cos(θi −θt ) sin(θi −θt ) rs = − sin(θi +θt ) tan(θi −θt ) rp = tan(θi +θt )

refractive indices “hidden” in angle of transmitted wave, θt can always rework Fresnel equations such that only ratio of refractive indices appears ⇒ Fresnel equations do not depend on absolute values of indices can arbitrarily set index of air to 1; then only use indices of media measured relative to air

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 33 Reflectivity Fresnel equations apply to electric field amplitude need to determine equations for intensity of waves D E ~ time-averaged Poynting vector S~ = c |n˜| |E |2 k 8π µ 0 |~k| absolute value of complex index of refraction enters energy along wave vector and not along interface normal each wave propagates in different direction ⇒ consider energy of each wave passing through unit surface area on interface does not matter for reflected wave ⇒ ratio of reflected and incident intensities is independent of these two effects relative intensity of reflected wave (reflectivity)

2 Er R = 0 i 2 E0

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 34 Transmissivity transmitted intensity: multiplying amplitude squared ratios with ratios of indices of refraction (different speeds of light) projected area on interface relative intensity of transmitted wave (transmissivity)

t 2 |n˜2| cos θt E T = 0 ˜ i 2 |n1| cos θi E0

energy conservation ⇒ R + T = 1 for transparent materials (not for absorbing materials)

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 35 Arbitrary Polarization ~ i i electric ﬁeld vector of incident wave E0, length E0, at angle α to plane of incidence decompose into 2 components: parallel and perpendicular to interface i i i i E0,p = E0 cos α , E0,s = E0 sin α use Fresnel equations to obtain corresponding (complex) amplitudes of reﬂected and transmitted waves

r,t i r,t i E0,p = (rp, tp) E0 cos α , E0,s = (rs, ts) E0 sin α reﬂectivity R and transmissivity T

2 2 2 2 R = |rp| cos α + |rs| sin α

|n˜2| cos θt 2 2 2 2 T = |tp| cos α + |ts| sin α |n˜1| cos θi

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 36 Reﬂection and Transmission

Christoph U. Keller, Leiden University, [email protected] Lecture 1: Foundations of Optics 37