复旦大学：《电动力学》课程电子讲义（PDF演示版）第五章 电磁波的传播 05-05 表面等离激元

经典电动力学导论 Let there be light 第五章:电磁波的传播§5.5 cnee C nw MXe程:VxE=0 OD i0,× +3=-iwEoE +o 0 V·B 复旦大学物理系 林志方徐建军2

Let there be light ²;>ÄåÆØ 1ÊÙµ>^ÅD § 5.5 ~j = inee 2 mω E~ =⇒ σc = inee 2 mω Maxwell §µ ∇ × E~ = − ∂B~ ∂t = iωµ0H~ , ∇ × H~ = ∂D~ ∂t + j = −iω0E~ + σcE~ ∇ · D~ = 0∇ · E~ = 0, ∇ · B~ = µ0∇ · H~ = 0 E￾ŒÆ ÔnX  Mï
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经典电动力学导论 Let there be light 第五章:电磁波的传播§5.5 C nw 0 OD Maxwell|方程 V×E i0,× +3=-iwEoE +o 0 → Helmholtz方程:V2+u2epE=0, 复旦大学物理系 林志方徐建军2

Let there be light ²;>ÄåÆØ 1ÊÙµ>^ÅD § 5.5 ~j = inee 2 mω E~ =⇒ σc = inee 2 mω Maxwell §µ ∇ × E~ = − ∂B~ ∂t = iωµ0H~ , ∇ × H~ = ∂D~ ∂t + j = −iω0E~ + σcE~ ∇ · D~ = 0∇ · E~ = 0, ∇ · B~ = µ0∇ · H~ = 0 =⇒ Helmholtz §µ∇2E~ + ω 2  0 µ0E~ = 0,  0 = 0 + iσc ω = 0 1 − ω 2 p ω2 ! E￾ŒÆ ÔnX  Mï
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经典电动力学导论 Let there be light 第五章:电磁波的传播§5.5 cnee C nw 0 OD Maxwell|方程 V×E i0,× +3=-iwEoE+ocE 0 → Helmholtz方程:V2+u2epE=0, T。e 等离子体频率 m∈ 复旦大学物理系 林志方徐建军2

Let there be light ²;>ÄåÆØ 1ÊÙµ>^ÅD § 5.5 ~j = inee 2 mω E~ =⇒ σc = inee 2 mω Maxwell §µ ∇ × E~ = − ∂B~ ∂t = iωµ0H~ , ∇ × H~ = ∂D~ ∂t + j = −iω0E~ + σcE~ ∇ · D~ = 0∇ · E~ = 0, ∇ · B~ = µ0∇ · H~ = 0 =⇒ Helmholtz §µ∇2E~ + ω 2  0 µ0E~ = 0,  0 = 0 + iσc ω = 0 1 − ω 2 p ω2 ! ωp = s nee 2 m0 lfNªÇ E￾ŒÆ ÔnX  Mï
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经典电动力学导论 Let there be light 第五章:电磁波的传播§5.5 cnee C nw MxMe方程:VxE=0B OD i0,× +3=-iwEoE+ocE 0 → Helmholtz方程:V2e+u2e10E=0, T。e 等离子体频率 m∈ 平面波解:E(中)= Teik 复旦大学物理系 林志方徐建军2

Let there be light ²;>ÄåÆØ 1ÊÙµ>^ÅD § 5.5 ~j = inee 2 mω E~ =⇒ σc = inee 2 mω Maxwell §µ ∇ × E~ = − ∂B~ ∂t = iωµ0H~ , ∇ × H~ = ∂D~ ∂t + j = −iω0E~ + σcE~ ∇ · D~ = 0∇ · E~ = 0, ∇ · B~ = µ0∇ · H~ = 0 =⇒ Helmholtz §µ∇2E~ + ω 2  0 µ0E~ = 0,  0 = 0 + iσc ω = 0 1 − ω 2 p ω2 ! ωp = s nee 2 m0 lfNªÇ ²¡Å)µ E~ (r~) = E~ 0e i~k·r~ E￾ŒÆ ÔnX  Mï
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经典电动力学导论 Let there be light 第五章:电磁波的传播§5.5 cnee C nw MxMe方程:VxE=0B OD i0,× +3=-iwEoE+ocE 0 V·B →Hh程是:+m=0,4=60+10=6(1-2 T。e 等离子体频率 m∈ 平面波解:E(中)= Teik 其中波矢应满足:k·k=u2∈A 复旦大学物理系 林志方徐建军2

Let there be light ²;>ÄåÆØ 1ÊÙµ>^ÅD § 5.5 ~j = inee 2 mω E~ =⇒ σc = inee 2 mω Maxwell §µ ∇ × E~ = − ∂B~ ∂t = iωµ0H~ , ∇ × H~ = ∂D~ ∂t + j = −iω0E~ + σcE~ ∇ · D~ = 0∇ · E~ = 0, ∇ · B~ = µ0∇ · H~ = 0 =⇒ Helmholtz §µ∇2E~ + ω 2  0 µ0E~ = 0,  0 = 0 + iσc ω = 0 1 − ω 2 p ω2 ! ωp = s nee 2 m0 lfNªÇ ²¡Å)µ E~ (r~) = E~ 0e i~k·r~ ٥ť ~k A÷vµ~k · ~k = ω 2  0 µ0 =⇒ k 2 = ω 2 c 2 1 − ω 2 p ω2 ! =⇒ k = ω c s 1 − ω 2 p ω2 E￾ŒÆ ÔnX  Mï
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