Total Internal Reflection & Metal Mirrors

Transcription

1 Phys 531 Lecture 7 15 September 2005 Total Internal Reflection & Metal Mirrors Last time, derived Fresnel relations Give amplitude of reflected, transmitted waves at boundary Focused on simple boundaries: air glass 1 Today, consider more complicated situations Total internal reflection - Evanescant waves Materials with complex index This will wrap up unit on fundamental theory Next time: ray optics 2

2 Total Internal Reflection (Hecht 4.7) So far, considered n i < n t If n i > n t, problem with Snell s Law: sin θ t = n i n t sin θ i What if (n i /n t ) sin θ i > 1? For instance, glass (n i = 1.5) air (n t = 1): Demo! if θ i > 41.8, then sin θ i > For sin θ i > sin θ c n t /n i, all light is reflected Called total internal reflection = TIR only occurs when light exits medium (high n low n) Would like to understand from Fresnel relations r = n i cos θ i n t cos θ t n i cos θ i + n t cos θ t r = n t cos θ i n i cos θ t n i cos θ t + n t cos θ i 4

3 How can we use? Don t have a θ t! Trick: use complex θ t Say θ t = a + ib Considered cos θ t in homework 1 Look at sin θ t now Use sin(a + ib) = sin(a) cos(ib) + cos(a) sin(ib) Just need sin(ib), cos(ib) 5 Euler identity gives: sin(x) = 1 2i cos(x) = 1 2 ( e ix e ix ) ( e ix e ix) So sin(ib) = 1 ( e i(ib) e i(ib) ) 2i = i ( e b e b ) 2 Hyperbolic sine: sinh(x) 1 ( e x e x ) 2 So sin(ib) = i sinh(b) 6

4 Hyperbolic sine and cosine: 7 Also cos(ib) = 1 ( e i(ib) + e i(ib) ) 2 = 1 ( e b + e b) 2 cosh(b) Hyperbolic cosine So: sin(a + ib) = sin(a) cosh(b) + i cos(a) sinh(b) For large b, sinh b, cosh b e b no problem satisfying n i sin θ i = n t sin θ t 8

5 Want n i sin θ i = n t sin θ t with complex θ t Assume n i sin θ i and n t real: Then Im [n t sin θ t ] = n t cos(a) sinh(b) = 0 Don t want b = 0, so take a = π 2 Gives sin θ t = cosh b Snell s law becomes n i sin θ i = n t cosh b Note cosh b > 1, require sin θ i > n t /n i sin θ c 9 Example: What is the complex transmission angle for light propagating from glass to air with an angle of incidence of 60? If θ i = 60 then b = cosh 1 [1.5 sin(60 )] = So θ t = π i Just need a good calculator! Question: What are the units of b? 10

6 What happens to Fresnel coefficients? ( π Need cos θ t = cos 2 + ib = i sinh(b) pure imaginary ) So r = n i cos θ i n t cos θ t n i cos θ i + n t cos θ t n i cos θ i in t sinh b n i cos θ i + in t sinh b complex! 11 Still have E r0 = re i0 complex r: phase shift between E inc and E refl Both r and r have form r = u + iv u iv r : u = n i cos θ i and v = n t sinh b r : u = n t cos θ i and v = n i sinh b If z = u + iv, then r = z/z So r = z z = 1: all light reflected 12

7 Write z = z e iφ, then r = e 2iφ Reflection phase shift: 2φ = 2 tan 1 ( v u ) To calculate numerically, just use good calculator (or computer program): Evaluate r = n i cos θ i n t cos θ t n i cos θ i + n t cos θ t for θ t = sin 1 ( ) n i n sin θ t i Let computer deal with complex math 13 Plot for glass air: Use TIR to make good mirror be aware of phase shifts 14

8 For TIR, R = r 2 = 1 so all power reflected But also have 2n t = i cos θ i n i cos θ i + n t cos θ t 2n t = i cos θ t n i cos θ t + n t cos θ i do not equal zero!? Have a transmitted field (t 0) but doesn t carry any energy (R = 1) Question: How might this contradiction be resolved? 15 To see: Transmitted wave: E t = E t0 e i(k t r ωt) with k t = k t (sin θ tˆx + cos θ t ẑ) For TIR, θ t = π 2 + b sin θ t cosh b cos θ t i sinh b So k t k t (cosh bˆx + i sinh bẑ) n i n t θ i θ t y z x 16

9 Transmitted field E t E t0 e i[k t(x cosh b+iz sinh b) ωt] = E t0 e k tz sinh b e i(k t x cosh b ωt) Wave propagates in x direction Decays exponentially in z direction Carries no energy away from surface Called evanenscent wave (Not same as exponential decay from absorption!) 17 Evanescent wave can be observed Single surface: z n i n t Introduce second surface: Transmitted wave appears! n i n t n i 18

10 Get transmission from tail of evanescent wave For gap d, amplitude of transmitted wave e k td sinh b reflection 0 smoothly as d 0 Called frustrated total internal reflection Completely analogous to tunneling in QM 19 Still hope T = 0 for plain TIR Should have defined T = Re [ nt cos θ t n i cos θ i ] t 2 In TIR, cos θ t = i sinh b real part = 0 So T = 0, as needed Raises question: what if n i or n t is complex? 20

11 Reflection from metals (Hecht 4.8) Saw previously that in absorbing medium n n + i α 2k 0 α = absorption coefficient Get E = E 0 e i(nk 0 r ωt) E 0 e αˆk r/2 e i(nk r ωt) and I E 0 2 e αˆk r Wave attenuates as it propagates 21 Question: How could you distinguish an evanescent wave from TIR and a plane wave exponentially decaying due to absorption? (Supposing no knowledge about the media.) Normally don t want optical material to absorb Important exception: mirrors light doesn t penetrate medium not much loss How to apply Fresnel relations? Typically n i, θ i real n t complex 22

12 Same equations apply Snell s law: sin θ t = n i n t sin θ i (so θ t complex) Use cos θ t = 1 sin 2 θ t as before Plug into equations for r, r Get complex result Hard to do by hand; easy on computer 23 Consider a very good absorber α Then n i sin θ i = means θ t 0 ( n t + i α 2k 0 ) sin θ t and r = n i cos θ i n t cos θ t n i cos θ i + n t cos θ t n i cos θ i n t iα/2k 0 n i cos θ i + n t + iα/2k

13 Similarly r 1 So perfect absorber = perfect reflector To make a true absorber: need moderate α + porous surface reflected waves bounce many times Example: soot ( = carbon dust) Use high-α material for mirror 25 Highest α: metals Homework problem 5 from assignment 1: In conductive medium get current J = σe σ = conductivity Got k = ɛµ 0 ω 2 + iωµ 0 σ Rewrite k = k 0 ɛ ɛ 0 + i σ ɛ 0 ω Good conductor: silver σ (Ω m) 1 (at dc) 26

14 If λ = 500 nm and ɛ ɛ 0 σ ɛ 0 ω 2000 So k k i 45k i 2 = 30k 0 (1 + i) Then expect n α/2k 0 30 Actually, not that good at optical freqs find n = 0.3 and α/2k 0 = 4 Still get R 0.95 across visible 27 Practical notes on mirrors Typical metals: Silver: R 0.95 in visible, NIR - oxidizes quickly in air Gold: R 0.95 in NIR - doesn t oxidize Aluminum: R 0.85 in visible, NIR - oxidizes but easy to protect (SiO) Metals don t have Brewster angle typically dip in R, but not to zero Wave penetrates fraction of λ typical nm 28

15 Get better mirrors using dielectric layers - discuss theory later - can get R = 0.99 easily, with effort - more expensive than metal Could use TIR: Drawbacks: - Reflection losses from first surface - Beam displacement inconvenient - Limited range of θ i Usually use when displacement desired 29 Summary Get TIR with internal incidence, θ i > θ c Perfect reflection, with phase shift Evanescant wave at surface For TIR or absorbing media, Fresnel equations are complex Highly absorbing medium good mirror 30