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Unformatted text preview: Assignment 4 TAs’ Solution Problem 1 Solution: Case 1: E r perpendicular to the plane of incidence (TE mode) First, making use of the continuity of the tangential components of the E r field, we have at the boundary at any time and any point: t r i E E E = + (1) Similarly, the normal component of B r is continuous, as is the tangential component of u B H r r = . Here we assume the permeabilities of the two medium is t i u u , . Thus we get: t t t r i r i i i u B u B u B θ θ θ cos cos cos = + (2) The positive direction is that of increasing x. And as relationship between E r and B r is B v E k r r = × ˆ , we can get t t t r r r i i i v E B v E B v E B = = = , , . Since r i r i and v v θ θ = = , , Eq.(2) can be written as t t t t i r i i i E v u E E v u θ θ cos 1 cos ) ( 1 = (3) As i t t i n n v v = , the Eq.(3) can be written as t t t t i r i i i E u n E E u n θ θ cos cos ) ( = (4) Combine Eq.(4) with Eq(1), this yields t t t i i i i i i TE i t u n u n u n E E θ θ θ cos cos cos 2 + = (5) When the two mediums are dielectrics, we have u u u t i ≈ ≈ , so Eq.(5) can be written as t t i i i i TE i t n n n E E θ θ θ cos cos cos 2 + = (6) From Snell’s Law, we have i t i t i t n n n n n n θ θ θ θ 2 2 sin 1 cos sin sin , = ⇒ = = , so Eq.(6) can be written as: i i i TE i t TE n E E t...
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 Spring '11
 Prof.ShenghuiSong
 Snell's Law, Cos, refractive index, Total internal reflection

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