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 05-02 电磁波在非导电介质界面上的反射和折射

经典电动力学导论 Let there be light 第五章 结论 复旦大学物理系 林志方徐建军3

Let there be light ²;>ÄåÆØ 1ÊÙµ>^ÅD § 5.2 kx = k 0 x = k 00 x , ky = k 0 y = k 00 y , ω = ω 0 = ω 00 (ص E￾ŒÆ ÔnX  Mï
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经典电动力学导论 Let there be light 第五章 结论 (1)入射、反射、折射波频率相同:u=u=u 复旦大学物理系 林志方徐建军3

Let there be light ²;>ÄåÆØ 1ÊÙµ>^ÅD § 5.2 kx = k 0 x = k 00 x , ky = k 0 y = k 00 y , ω = ω 0 = ω 00 (ص (1) \!‡!òŪǃӵω = ω 0 = ω 00 E￾ŒÆ ÔnX  Mï
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经典电动力学导论 Let there be light 第五章 a=k2=k, ky=kl=ku, w=w=w 结论 (1)入射、反射、折射波频率相同:u=u=u (2)若取入射面为xz平面(如图),则ky=0 k′=0.k=0 入射、反射、折射波矢共面,同在入射面内 复旦大学物理系 林志方徐建军3

Let there be light ²;>ÄåÆØ 1ÊÙµ>^ÅD § 5.2 kx = k 0 x = k 00 x , ky = k 0 y = k 00 y , ω = ω 0 = ω 00 (ص (1) \!‡!òŪǃӵω = ω 0 = ω 00 (2) e\¡ xz ²¡£X㤧K ky = 0 =⇒ k 0 y = 0, k00 y = 0 \!‡!òÅ¥
¡§Ó3\¡S E￾ŒÆ ÔnX  Mï
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经典电动力学导论 Let there be light 第五章 结论 (1)入射、反射、折射波频率相同:u=u (2)若取入射面为xz平面(如图),则ky=0 k′=0.k"=0 入射、反射、折射波矢共面,同在入射面内 (3)kx=k=ksin=ksin由于:k=k sine= sin e' 6=0入射角=反射角 复旦大学物理系 林志方徐建军3

Let there be light ²;>ÄåÆØ 1ÊÙµ>^ÅD § 5.2 kx = k 0 x = k 00 x , ky = k 0 y = k 00 y , ω = ω 0 = ω 00 (ص (1) \!‡!òŪǃӵω = ω 0 = ω 00 (2) e\¡ xz ²¡£X㤧K ky = 0 =⇒ k 0 y = 0, k00 y = 0 \!‡!òÅ¥
¡§Ó3\¡S (3) kx = k 0 x =⇒ k sin θ = k 0 sin θ 0 duµk = k 0 = ω √1µ1 =⇒ sin θ = sin θ 0 =⇒ θ = θ 0 \ = ‡ E￾ŒÆ ÔnX  Mï
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经典电动力学导论 Let there be light 第五章 a=k2=k, ky=kl=ku, w=w=w 结论 (1)入射、反射、折射波频率相同:u=u=u (2)若取入射面为xz平面(如图),则ky=0 k′=0.k"=0 入射、反射、折射波矢共面,同在入射面内 (3)kx=k=ksin=ksin由于:k=k sine= sin e' 6=0入射角=反射角 (4)kr= k r ksin=k"sin0”由于:k"=u"ve22 212 VE1ui sin e =√e22sin0"= Sin 6 e11 复旦大学物理系 林志方徐建军3

Let there be light ²;>ÄåÆØ 1ÊÙµ>^ÅD § 5.2 kx = k 0 x = k 00 x , ky = k 0 y = k 00 y , ω = ω 0 = ω 00 (ص (1) \!‡!òŪǃӵω = ω 0 = ω 00 (2) e\¡ xz ²¡£X㤧K ky = 0 =⇒ k 0 y = 0, k00 y = 0 \!‡!òÅ¥
¡§Ó3\¡S (3) kx = k 0 x =⇒ k sin θ = k 0 sin θ 0 duµk = k 0 = ω √1µ1 =⇒ sin θ = sin θ 0 =⇒ θ = θ 0 \ = ‡ (4) kx = k 00 x =⇒ k sin θ = k 00 sin θ 00 duµk 00 = ω 00√2µ2 =⇒ √1µ1 sin θ = √2µ2 sin θ 00 =⇒ sin θ sin θ 00 = √2µ2 √1µ1 = n2 n1 = n21 E￾ŒÆ ÔnX  Mï
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