# 350lect19 - 19 Total internal reflection(TIR and evanes...

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Unformatted text preview: 19 Total internal reflection (TIR) and evanes- cent waves TE reflection: k i k r k t θ 2 θ 1 x z H i k 1 cos θ 1 k 1 sin θ 1 k 2 sin θ 2 Medium 1 Medium 2 θ 1 E yr E yi = η 2 cos θ 1- η 1 cos θ 2 η 2 cos θ 1 + η 1 cos θ 2 E yt E yi = 2 η 2 cos θ 1 η 2 cos θ 1 + η 1 cos θ 2 TM reflection: k i k r k t θ 2 θ 1 x z E i k 1 cos θ 1 k 1 sin θ 1 k 2 sin θ 2 Medium 1 Medium 2 θ 1- E r E i = η 2 cos θ 2- η 1 cos θ 1 η 2 cos θ 2 + η 1 cos θ 1 E t E i = 2 η 2 cos θ 1 η 2 cos θ 2 + η 1 cos θ 1 . • Consider a TE- or TM-polarized wave (or a superposition) incident on an interface at x = 0 surface as depicted in the margin at an incidence angle θ 1 . • Independent of the polarization of the incident wave, the angle of trans- mitted wave θ 2 can be found using Snell’s law k 1 s in θ 1 = k 2 s in θ 2 ⇒ √ μ 1 r 1 r s in θ 1 = √ μ 2 r 2 r s in θ 2 assuming lossless media on either side of the interface, where Refractive index: n = c v p = √ μ r r μ r ≡ μ μ o and r ≡ o are the relative permeability and permittivity, respectively, of the prop- agation media. Moreover, √ μ r r = √ μ √ μ o o = c v p ≡ n above can be referred to as the refractive index of the propagation medium. • Snell’s law, expressed in terms of refractive index, n 1 s in θ 1 = n 2 s in θ 2 ⇒ s in θ 2 = n 1 n 2 s in θ 1 1 shows that for a given θ 1 , the corresponding s in θ 2 can be in excess of 1 when n 1 > n 2 , that is, for propagation from a high refractive index (optically thick) material such as glass into a lower refractive index (optically thin) material such as air. – For example : if n 1 n 2 = 1 . 5 and θ 1 = 45 ◦ , then s in θ 2 = n 1 n 2 s in θ 1 = 1 . 5 s in 45 ◦ = 1 . 5 √ 2 ≈ 1 . 5 1 . 41 > 1 . But, s in θ 2 in excess of 1 cannot be solved for θ 2 as if it were a “regular” angle 1 describing the elevation of vector k t above the x-axis. • In general when n 1 > n 2 and the incidence angle Critical angle: θ c = s in- 1 n 2 n 1 θ 1 > s in- 1 n 2 n 1 = s in- 1 μ 2 2 μ 1 1 ≡ θ c we will have s in θ 2 in excess of 1 and cos θ 2 = 1- s in 2 θ 2 purely imaginary. – in such situations use s in θ 2 and cos θ 2 = 1- s in 2 θ 2 directly in the expressions for k t , Γ , and τ as illustrated below. 1 Nor if sin θ 2 is complex valued because medium 2 is lossy and we need to use 2 r = 2 o + σ 2 jω o in Snell’s law (as we already did in Lecture 18)....
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350lect19 - 19 Total internal reflection(TIR and evanes...

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