利用菲涅尔公式计算光波在两种介质表面折反射率及折反射能流密度 *=QWx[K|
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clear j&?@:Zg v
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close all ~<_PjV
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n1=1,n2=1.45; 6$+F5T
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theta=0:0.1:90; K30{Fcb< h
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a=theta*pi/180; ADZU?7)
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rp=(n2*cos(a)-n1*sqrt(1-(n1/n2*sin(a)).^2))./(n2*cos(a)+n1*sqrt(1-(n1/n2*sin(a)).^2)); Y&b JKX
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rs=(n1*cos(a)-n2*sqrt(1-(n1/n2*sin(a)).^2))./(n1*cos(a)+n2*sqrt(1-(n1/n2*sin(a)).^2)); !vSj1w
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tp=2*n1*cos(a)./(n2*cos(a)+n1*sqrt(1-(n1/n2*sin(a)).^2)); #F >R5 D
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ts=2*n1*cos(a)./(n1*cos(a)+n2*sqrt(1-(n1/n2*sin(a)).^2)); :'gX//b):
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figure(1) uHO>FM,
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subplot(1,2,1); &Sp2['a!
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plot(theta,rp,'-',theta,rs,'--',theta,abs(rp),':',theta,abs(rs),'-.','LineWidth',2) $~.'Tnk)
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legend('r_p','r_s','|r_p|','|r_s|') q6>%1~?
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xlabel('\theta_i') LZ*ZXFIg
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ylabel('Amplitude') |"EQyV
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title(['n_1=',num2str(n1),',n_2=',num2str(n2)]) QPpC_pZh
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axis([0 90 -1 1]) L(qQ,1VY
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grid on ZncJ
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subplot(1,2,2); Bgj^n{9x
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plot(theta,tp,'-',theta,ts,'--',theta,abs(tp),':',theta,abs(ts),'-.','LineWidth',2) e47N 9&4
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legend('t_p','t_s','|t_p|','|t_s|') 4XXuj
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xlabel('\theta_i') nhP ua&
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ylabel('Amplitude') z#elwL6
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title(['n_1=',num2str(n1),',n_2=',num2str(n2)]) KI{B<S3*Z
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axis([0 90 0 1]) u'+;/8
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grid on
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Rp=abs(rp).^2; y d4\%%]
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Rs=abs(rs).^2; 3=~0m
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Rn=(Rp+Rs)/2; iSg^np
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Tp=1-Rp; <*55d2
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Ts=1-Rs; TY6Q;BTU
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Tn=(Tp+Ts)/2; %^CoWbU
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figure(2) 9(
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subplot(1,2,1); f,}9~r#
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plot(theta,Rp,'-',theta,Rs,'--',theta,Rn,':','LineWidth',2) &@h(6
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legend('R_p','R_s','R_n') (!B1}5"
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xlabel('\theta_i') ?(d<n
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ylabel('Amplitude') 1Vy8TV3D
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title(['n_1=',num2str(n1),',n_2=',num2str(n2)]) QpZCU]
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axis([0 90 0 1]) 2=/,9ka~
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grid on `8sC>)lrwu
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subplot(1,2,2); $"[5]{'J
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plot(theta,Tp,'-',theta,Ts,'--',theta,Tn,':','LineWidth',2) g ONybz6]
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legend('T_p','T_s','T_n') -L&%,%
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xlabel('\theta_i') A5[iFT>
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ylabel('Amplitude') L.tW]43K
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title(['n_1=',num2str(n1),',n_2=',num2str(n2)]) C_ 4(-OWq
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axis([0 90 0 1]) f\U(7)2
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grid on 3JJEj1O
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