Electromagnetic Wave Propagation Lecture 13: Oblique incidence II

Transcription

1 Electromagnetic Wave Propagation Lecture 13: Oblique incidence II Daniel Sjöberg Department of Electrical and Information Technology October 2016

2 Outline 1 Surface plasmons 2 Snel s law in negative-index media 3 Ray tracing 4 Conclusions 2 / 39

3 Outline 1 Surface plasmons 2 Snel s law in negative-index media 3 Ray tracing 4 Conclusions 3 / 39

4 Surface plasmons (Fig in Orfanidis) An infinite reflection coefficient corresponds to fields existing without excitation, a resonance. This is only possible for a special combination of material parameters and complex wave vectors. 4 / 39

5 Conditions for a surface plasmon A plasmon corresponds to a pole in the reflection coefficient. In order to satisfy ρ TM = k z2ɛ 1 k z1 ɛ 2 k z2 ɛ 1 + k z1 ɛ 2 = k z2 ɛ 1 + k z1 ɛ 2 = 0 we need to require (assuming real permittivities) ɛ 1 ɛ 2 < 0, ɛ 1 + ɛ 2 < 0 Writing ɛ 2 = ɛ 2r, with ɛ 2r > ɛ 1 > 0, we get ɛ1 ɛ 2r ɛ k x = k 0 1 ɛ, k z1 = jk 0, k z2 = jk 0 2r ɛ2r ɛ 1 ɛ2r ɛ 1 ɛ2r ɛ 1 5 / 39

6 Permittivity of a plasma In a gas of charged particles, a simple model for the permittivity is ) ωp ɛ(ω) = ɛ 0 (1 2 ω 2 jωγ In a good metal like copper, we have f p = ω p /2π = Hz, and γ = sec 1 (see Orfanidis Ch1). Thus, for optical frequencies f Hz, we have ɛ(ω) ɛ 0 1 ω2 p ω 2 26 < 0 A plasmon can be seen as collective oscillations of the electron gas. 6 / 39

7 Dispersion relation Inserting ɛ 1r = 1 and ɛ 2r = ωp/ω 2 2 ɛ1 ɛ 1 in k x = k 2r 0 ɛ2r ɛ 1 k 2 x = ω2 c 2 0 ω 2 p ω 2 ω 2 p 2ω 2 ω ω p = where k p = ω p /c 0. ( k x k p ) (kx k p implies ) The plasmon resonance exists for any frequency ω < ω p / 2, but requires a k x larger than the wavenumber in vacuum, k 0. It cannot be generated by a plane wave incident from vacuum. (Fig in Orfanidis) 7 / 39

8 Excitation of plasmons A wave suffering total internal reflection in a denser medium can have k x > k 0 and leak through via evanescent waves: (Fig in Orfanidis) 8 / 39

9 Other means of generation Short wavelengths can also be generated by a grating. Stefan A. Meier, Plasmonics: Fundamentals and applications, Springer, Available as ebook from 9 / 39

10 Example: silver At λ = 632 nm, silver has the relative permittivity ɛ 2 = j The wavenumbers are (k 0 = 9.94 rad/µm): k x = j rad/µm, δ x = 1 α x = 93.6 µm k z1 = j rad/µm δ z1 = 1 α z1 = 390 nm k z2 = j rad/µm δ z2 = 1 α z2 = 24 nm Thus, the plasmon is strongly confined to the surface, especially on the metal side, and can travel many wavelengths along it. 10 / 39

11 Typical field distribution 11 / 39

12 Plasmon generation Very narrow resonance, first dot corresponds to critical angle. Kretschmann-Raether setup, n a = 1.5, ɛ = 16 j0.5, d = 50 nm. 12 / 39

13 Particle plasmons Plasmon resonances occur for finite particles as well. (Fig. 5.7 in Meier, 2007) 13 / 39

14 Applications of plasmonics Plasmons are envisaged as a new means of controlling energy on a small scale. Focusing on sub-wavelength scale. Extra-ordinary transmission through small holes. Enhancement of non-linear effects. Spectroscopy and sensing. Some more material in Ch / 39

15 Outline 1 Surface plasmons 2 Snel s law in negative-index media 3 Ray tracing 4 Conclusions 15 / 39

16 Oblique incidence on negative-index media Snel s law becomes (ɛ, µ, n < 0) n sin θ = n sin θ = n sin θ = n sin( θ ) (Fig in Orfanidis) 16 / 39

17 Wave vector and power flow The general solution for a TE polarized wave is (similar for TM) [ E(r) = ŷe 0 e jkzz + ρ TE e jkzz] e jkxx H(r) = E [( 0 ˆx + k ) ( x ẑ e jkzz + ρ TE ˆx + k ) ] x ẑ e jkzz e jkxx η TE k z k z E (r) = ŷτ TE E 0 e jk zz e jkxx H (r) = τ ( TEE 0 η TE ˆx + k ) x k z ẑ e jk zz e jkxx η TE = ωµ k z, η TE = ωµ k z, ρ TE = η TE η TE η TE + η, τ TE = 1 + ρ TE TE The Poynting vector in the right medium is (µ < 0) P = 1 2 Re{E H } = 1 ( ) k 2 τ TE 2 E 0 [ẑ 2 Re z ωµ }{{ } positive ( kx +ˆx Re ωµ } {{ } negative ) ] 17 / 39

18 Typical field distribution 18 / 39

19 Experimental verification Shelby, Smith & Schultz, Science 292, p. 77 (2001). 19 / 39

20 Movie A finite pulse incident on a negative refractive index material (Wikimedia Commons). See also the youtube video below. 20 / 39

21 Outline 1 Surface plasmons 2 Snel s law in negative-index media 3 Ray tracing 4 Conclusions 21 / 39

22 Geometrical optics A plane wave propagating in direction ˆk is described by E(r) = E 0 e jnk 0ˆk r ˆk E0 = 0 H(r) = H 0 e jnk 0ˆk r H 0 = n ˆk E0 η 0 The constant-phase-planes are described by S(r) = nˆk r = const We want to generalize this, and look for solutions E(r) = E 0 (r)e jk 0S(r), H(r) = H 0 (r)e jk 0S(r) where E 0 (r) and H 0 (r) are slowly varying functions of r (compared to the wavelength λ = 2π/k 0 ). This is a high frequency approach. 22 / 39

23 Maxwell s equations Inserting E(r) = E 0 (r)e jk 0S(r) and H(r) = H 0 (r)e jk 0S(r) into Maxwell s equations implies E = e jk 0S(r) ( E 0 jk 0 S E 0 ) = jωµ 0 H 0 e jk 0S(r) H = e jk 0S(r) ( H 0 jk 0 S H 0 ) = jωn 2 ɛ 0 E 0 e jk 0S(r) Assuming E 0 k 0 S E 0, this boils down to (locally plane wave) H 0 = n η 0 ˆk E0, E 0 = η 0 n ˆk H 0, ˆk = 1 n S That ˆk is a unit vector can be written S 2 = n(r) 2 which is known as the eikonal equation. This is a nonlinear partial differential equation, which determines S(r). 23 / 39

24 Wave fronts and rays dr dl = ˆk = 1 ( n S d n dr ) = n dl dl This is a differential equation for the ray paths, r(l). (Fig in Orfanidis) 24 / 39

25 Fermat s principle (Fig in Orfanidis) S B S A = B A n dl = tb t A c 0 dt = c 0 (t B t A ) Of all possible paths connecting A and B, the geometrical optics ray path corresponds to a stationary optical path length (or travel time). Typically, this is a minimum. 25 / 39

26 Fermat s principle, example minimum path maximum path 26 / 39

27 Ray tracing Homogeneous in x, inhomogeneous in z. (Fig in Orfanidis) Rays are refracted towards higher refractive index. This can be seen from k z = (nk 0 ) 2 k 2 x: when keeping k x constant, increasing n implies increasing k z. 27 / 39

28 Mathematics of graded index refraction The ray equation is (using n(r) = n(z)) ( d n dr ) = n dl dl d dl d dl ( n dx ) = 0 dl ( n dz ) = dn dl dz The first equation implies n dx dl generalized Snel s law = n sin θ = const, or the n(z) sin θ(z) = n a sin θ a for all z The second implies (using dz = dl cos θ) cos θ d dn (n cos θ) = dz dz n cos θ d dz cos θ = sin2 θ dn dz Thus, if dn dz > 0, we have d dz cos θ > 0, meaning θ is decreasing. 28 / 39

29 Examples in spherical geometry Luneburg lens Maxwell s fish-eye n(r) = 2 ( r R )2 n(r) = n 0 1+( r R )2 Variation of refractive index can be realized by layered spheres. 29 / 39

30 FDTD simulations Regular lens Luneburg lens Fisheye lens Half fisheye 30 / 39

31 Ionospheric refraction The ionosphere consists of a plasma, where ɛ decreases with height. (Fig in Orfanidis) A radio wave is reflected if its frequency is below the plasma frequency, implying n 2 (ω) = ɛ(ω) ɛ 0 = 1 ω2 p < 0, where ω 2 ω 2 p = Ne2 ɛ 0 m. But also some higher frequencies get reflected due to refraction. 31 / 39

32 Ionospheric refraction Snel s law implies (where f 2 p(z) = sin 2 θ(z) = n2 a sin 2 θ a n 2 (z) N(z)e2 (2π) 2 ɛ 0 m ) = sin 2 θ a 1 f 2 p(z)/f 2 For propagating rays sin θ(z) 1, which implies sin 2 θ a 1 f 2 p(z) f 2 f p (z) f cos θ a If this becomes an equality, f p (z max ) = f cos θ a, then θ(z max ) = 90 and the wave is refracted downwards to earth. For a strict inequality, f p (z) < f cos θ a for all z, the wave never turns downward and eventually escapes. MUF: Maximum usable frequency, the largest frequency at which reflection occurs. LUF: Lowest usable frequency, the smallest frequency at which absorption can be ignored. 32 / 39

33 Mirages (Fig in Orfanidis) The following photos are by Mila Zinkova, released under Creative Commons / 39

34 Desert mirage 34 / 39

35 Fata morgana 35 / 39

36 Fata morgana explained The cat is viewed through three layers of different sugar concentrations, decreasing refractive index from bottom to top (photo by Shy Halatzi, released under Creative Commons 3.0). 36 / 39

37 Green flash 37 / 39

38 Outline 1 Surface plasmons 2 Snel s law in negative-index media 3 Ray tracing 4 Conclusions 38 / 39

39 Conclusions A surface plasmon can propagate on a metal surface, and can be excited by evanescent waves. Waves incident on a negative index medium are refracted the wrong way. Power still propagates away from the boundary. In geometrical optics, the ray path corresponds to a stationary optical path length, typically a minimum. Optical mirages in nature are often due to temperature variations, giving rise to change in refractive index. 39 / 39