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On Total Internal Reflection? ECEECE 318318 FundamentalsFundamentals ofof OpticsOptics LectureLecture 1616 && 1717 Light as Waves VI • Total Internal Reflection * Topics, questions, or homework problems with an asterisk (*) are at a difficulty level or having a scope beyond the expectation of this course. Only qualitative understanding is required. TotalTotal InternalInternal ReflectionReflection So far we have considered the cases where there always exists a propagating transmitted wave. Snell’s law is satisfied with a transmitted angle θt between 0 and π/2. Now we are going to consider a special case called Total Internal Reflection (TIR). Two conditions must be satisfied for TIR to occur: (1) ni>nt, that is, the incident light wave is in the denser medium (2) θi >θc , that is, the incident angle is greater than the critical angle In what follows, we’ll show that, upon TIR, both TE and TM polarizations are 100% reflected, however, they will experience different phase shifts. Q16-1: Next time you go for a swim, think about how you could verify first hand the zero transmission phenomenon when total internal reflection occurs at the water-air boundary. 2 © Li Qian Lecture 16 &17 CriticalCritical AngleAngle Critical angle refers to the incident angle when the refracted angle is exactly π/2. This is a special case that occurs only for internal reflections. Snell’s Law: (1) ni sinθi = nt sinθt k θc r ki Therefore: n i π k n t ni sinθc = nt sin (2) t 2 or n −1 t θc ≡ sin (3) ni Above: Solving for θc graphically using k-matching at the boundary. Example: For a glass-air interface, the critical angle occurs at 41.8o, for a water-air interface, this occurs at 48.8o. 3 © Li Qian Lecture 16 &17 WhatWhat happenshappens toto Snell’sSnell’s LawLaw beyondbeyond thethe criticalcritical angle?angle? It can be shown that (see Homework #1), even in the case of total internal reflection, Snell’s law still holds, except that the sine of the transmitted angle becomes greater than 1, and the cosine of the transmitted angle becomes imaginary. Though there is no physical interpretation of the “transmitted angle”, it is still useful as a mathematical tool. The Fresnel reflection formulae (Lecture 14&15, Eq (24) & (26)) still apply if we continue to use the cosθt term, despite its imaginary value. ni sinθi sinθt = >1 (4) nt 2 n sinθ i i cosθt = i −1 nt (5) Q16-2: What’s the origin of Snell’s Law from the Wave Optics point of view? 4 © Li Qian Lecture 16 &17 “Transmitted”“Transmitted” wavewave vectorvector isis complexcomplex Let’s analyze the wave vector kt of the “transmitted wave”, assuming it exists. In order to satisfy the wave equations (Lecture 9, Eq (27) & (28)), we must have: y ω k = k 2 + k 2 = n (6) t tx ty c t And in order to satisfy the boundary phase- ki θi θr kr ni matching condition, we must have ω x (7) n ktx = kix = ki sinθi = ni sinθi t kt=? c And in order to satisfy both (6) and (7), and noting that nisinθi > nt, kty must become Above: Boundary conditions cannot be imaginary! satisfied with a real kt . 2 ω ω n sinθ 2 2 2 2 2 2 i i kty = kt − ktx = kt − ()ki sinθi = nt − ()ni sinθi = ±i nt −1 c c nt (8) ≡ ±iβ 5 © Li Qian Lecture 16 &17 “Transmitted”“Transmitted” wavewave isis evanescentevanescent What does an imaginary kty signify and what is the consequence? If we substitute (7) and (8) into our usual wave expression for the transmitted wave, we will have: βy ω Et = Eto exp[]i()ktx x + kty y −ωt +φ = Etoe expi ni sinθi x −ωt +φ (9) c Amplitude exponentially A wave propagating in the x decays with –y. direction with vφ=c/(nisinθi) Note 1. In deriving (9), only kty=–iβ is taken, since kty=+iβ leads to an unphysical result. Note 2. For finite incident beam size, the x-direction propagation of the evanescent wave is sustained only within the extent of the incident beam. The wave expressed by (9) is a peculiar wave: it differs from the regular plane wave in that its amplitude exponentially decays in the directions orthogonal to its propagation direction. It is called an evanescent wave. Q16-3: Does this evanescent wave have planar wavefronts? Q16-4: In an absorbing medium, the E field also decays exponentially (see Lecture 12&13, Eq(15)). How is the wave propagating in an absorbing medium different from the evanescent wave described here? 6 © Li Qian Lecture 16 &17 PenetrationPenetration DepthsDepths From Eq (9), we can draw three conclusions regarding the amplitude of the transmitted E field: (1) E field is finite in the nt medium, that is, there is field penetration into the transmitted medium. The field does not stop abruptly at the interface. (2) E field is constant in the planes parallel to the interface, i.e., for y = constant. (3) E field decays exponentially with the distance away from the interface. We can define two penetration depths: (1) The distance from the interface at which the E field decays to its 1/e value Field penetration depth = 1/β (10) (2) The distance from the interface at which the intensity decays to its 1/e value Intensity penetration depth = 1/2β (11) Q16-5: Penetration depth is a function of incident angle. At what angle is penetration depth the greatest? Q16-6: For optical waves, what’s the approximate length scale of the penetration depths for large incident angles? 7 © Li Qian Lecture 16 &17 “Transmitted”“Transmitted” wavewave isis non-transverse*non-transverse* It can be shown that the evanescent wave is non-transverse, that is, it has an E or an H component in the direction of propagation (x-direction). For example, if the incident polarization is TE, then the transmitted wave can be written as: βy (12) Et = Etoe exp[i(kix x −ωt +φ)]zˆ Now if we calculate the transmitted H field from the E field using one of the Maxwell’s Equations: ∂B ∂H ∇ ×E = − = −µ ∂t ∂t (13) We obtain: ∂E ∂H β t = −µ x ⇒ βE = iµωH ⇒ H = −i E eβy exp[]i()k −ωt +φ (14) ∂y ∂t t x x µω to tx ∂E ∂H k − t = −µ y ⇒ − ik E = iµωH ⇒ H = − tx E eβy exp[]i()k −ωt +φ (15) ∂x ∂t tx t y y µω to tx Eq (14) indicates that the H field has a finite x-component. Note that Hx is π/2 out-of-phase with Ez . This is another peculiar feature of the evanescent wave. (Recall that, for a “normal” propagating wave, E and H are in phase.) 8 © Li Qian Lecture 16 &17 “Transmitted”“Transmitted” wavewave isis notnot transmittingtransmitting energyenergy acrossacross thethe boundary*boundary* We can also show that there is no net energy transport across the boundary. Again, take TE polarization as an example, from (12), (14) & (15), we can write the E and H fields using real representations: βy Et = Etoe cos(ktx x −ωt +φ)zˆ (16) βy β ktx Ht = Etoe sin()ktx x −ωt +φ xˆ − cos()ktx x −ωt +φ yˆ (17) µω µω Poynting vector of the “transmitted” wave becomes: β k E ×H = E 2 e2βy cos()k x −ωt +φ sin()k x −ωt +φ yˆ + tx E 2 e2βy cos2 ()k x −ωt +φ xˆ (18) t t µω to tx tx µω to tx The time-average of the y-component of the Poynting vector is zero, as shown below. This signifies that, on average, no net energy flow in the y-direction. β E × H ⋅yˆ = E 2 e2βy cos()k x −ωt +φ sin()k x −ωt +φ t t ωµ to tx tx β = E 2 e2βy sin()2k x − 2ωt + 2φ = 0 2ωµ to tx (19) Q16-7: How do you reconcile the fact that there is finite field penetration into the nt medium, but no net energy transport? 9 © Li Qian Lecture 16 &17 ComplexComplex FresnelFresnel ReflectionReflection CoefficientsCoefficients As mentioned before, the Fresnel reflection coefficients can still be applied to the TIR case, as long as we incorporate the imaginary cosθt term, expressed by Eq (5). TE-case: TM-case: n cosθ − in ()n sinθ n 2 −1 − n cosθ + in ()n sinθ n 2 −1 r = i i t i i t (20) r = t i i i i t (23) TE 2 TM 2 ni cosθi + int ()ni sinθi nt −1 nt cosθi + ini ()ni sinθi nt −1 iφTE iφTM ∴ rTE = e (21) ∴ rTM = e (24) n ()n sinθ n 2 −1 n ()n sinθ n 2 −1 φ = −2tan −1 t i i t (22) φ = −2 tan−1 i i i t +π (25) TE n cosθ TM n cosθ i i t i Important conclusions: (1) The wave is 100% reflected; (2) The reflected wave is not in phase with the incident wave; (3) The additional phase incurred in the reflected wave is different from TE and TM polarized waves. 10 © Li Qian Lecture 16 &17 PhasePhase ShiftsShifts uponupon TotalTotal InternalInternal ReflectionReflection The figure below shows φTE, φTM, and the relative phase difference (φTM – φTE) as a function of incident angle for a glass-to-air interface. π ∆φ=φΤΜ−φΤΕ π/2 φΤΜ 0 θi Phase 0 θc π/2 −π/2 φΤΕ −π Q16-8: Do the wavefronts of the incident, reflected and transmitted wave match at the boundary upon total internal reflection? 11 © Li Qian Lecture 16 &17 ReflectanceReflectance andand TransmittanceTransmittance From (21) and (24), rTE and rTM both have a magnitude of unity in the region of Total Internal Reflection (left plot).
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