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**Example text**

1. Along the x-axis we have: z × Ei0 exp iβxi x + z × E˜ r0 exp iβxr x = (z × Ei0 + z × E˜ r0 ) exp iβxt x − αxt x . 61) The above equation can be satisﬁed for every x only if: βxr = βxi ; βxt = βxi ; αxt = 0 . 62) The ﬁrst two conditions in Eq. 64) and imply that θr = θi . 65) then we have: ω β t = kt n , c and the well-known Snell’s law [1] is: n sin θi = n sin θt . e. with an attenuating component in the z direction. In this case we have from Eqs. 69) c2 ω2 2αt βzt = 2α t β t 1 − sin2 (θt ) = 2 2n κ .

139) recovers the Drude model [5]. e. for wavelength much larger than the bulk unit cell. 141) where we considered an isotropic medium and deﬁne the eﬀective mass as: 1 1 d2 En,k = . 142) For intra-band transition n = n so that the matrix element in Eq. 143) due to the limit q → 0. The intra-band dielectric constant can be rewritten as: ˜ i ntra (q → 0, ω) = 1 − − 4π e2 ||q||2 V fn,k k,n 1 En,k − En,k−q − ω − iη+ 1 En,k+q − En,k − ω − iη+ . 145) where we used Eq. 141). For a simple metal with just one band we have 4π e2 N V ω2 m∗ which coincides with the result in Eq.

E. 82) where n = Re[n˜ ]. e. the light is incoming from the vacuum, then the reﬂectance and the transmittance are simply: R= 1 − n˜ 1 + n˜ T =n 2 = 2 1 + n˜ (1 − n )2 + κ 2 , (1 + n )2 + κ 2 2 = 4n . 84) Note that due to energy conservation, we have: R + T = 1. e. when ρe = 0 and Je = 0). 86) 1 ∂A . 87) The magnetic vector potential always exists as H is divergence-less (see Eq. 2), for non magnetic media) and thus it can be written as Eq. 86). The electric scalar potential also always exists: in fact from Eq.

### Advanced Light Source - Users Handbook

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