[分享]利用MATLAB光学仿真（1） [复制链接]

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 1">]w2je:  
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1、光疏射向光密 blA]z!FU  
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clear r.;(Kx/M  
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close all `)R@\@jt  
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n1=1,n2=1.45; to*<W,I  
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theta=0:0.1:90; IRQ3>4hI  
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a=theta*pi/180; R!.HS0i.  
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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)); id.W"5+  
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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)); E5 0$y:  
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tp=2*n1*cos(a)./(n2*cos(a)+n1*sqrt(1-(n1/n2*sin(a)).^2)); 7a:mZ[Vh  
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ts=2*n1*cos(a)./(n1*cos(a)+n2*sqrt(1-(n1/n2*sin(a)).^2));  <XxFR  
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figure(1) $zBG19 [%  
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subplot(1,2,1); HSOdqjR*  
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plot(theta,rp,'-',theta,rs,'--',theta,abs(rp),':',theta,abs(rs),'-.','LineWidth',2) @.X}S"yr  
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legend('r_p','r_s','|r_p|','|r_s|') (&$|R\W.  
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xlabel('\theta_i') 1W5\
  
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ylabel('Amplitude') #={L!"3?e  
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title(['n_1=',num2str(n1),',n_2=',num2str(n2)]) <z|?C  
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axis([0 90 -1 1]) }x1mpPND  
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grid on soZw""|v  
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subplot(1,2,2); R6Cm:4m}I  
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plot(theta,tp,'-',theta,ts,'--',theta,abs(tp),':',theta,abs(ts),'-.','LineWidth',2) ]Nvtiw 6  
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legend('t_p','t_s','|t_p|','|t_s|') k[
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xlabel('\theta_i') |pp@  
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ylabel('Amplitude') Y(SgfWeK@1  
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title(['n_1=',num2str(n1),',n_2=',num2str(n2)]) N)h>Ie  
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axis([0 90 0 1]) 7Yxy2[  
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grid on Ik1,?A
 
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Rp=abs(rp).^2; ZZE  
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Rs=abs(rs).^2; RQU5T 2,  
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Rn=(Rp+Rs)/2; 2>O2#53ls0  
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Tp=1-Rp; i:kWO7aP  
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Ts=1-Rs; 0 \LkJ*i  
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Tn=(Tp+Ts)/2; G-3.-  
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figure(2)  cJ8F#t  
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subplot(1,2,1); d0 V>;Q  
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plot(theta,Rp,'-',theta,Rs,'--',theta,Rn,':','LineWidth',2) Hs=!.tZ,  
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legend('R_p','R_s','R_n') R~w(]  
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xlabel('\theta_i') YF>t{|  
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ylabel('Amplitude') faO8 &  
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title(['n_1=',num2str(n1),',n_2=',num2str(n2)]) Xe+&/J5b  
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axis([0 90 0 1]) DhZtiqL#_  
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grid on Xp} vJl   
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subplot(1,2,2); Y[~6f,?^  
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plot(theta,Tp,'-',theta,Ts,'--',theta,Tn,':','LineWidth',2) X^|oY]D  
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legend('T_p','T_s','T_n') u=E?N:I~F  
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xlabel('\theta_i')
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ylabel('Amplitude') gs8L/veP  
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title(['n_1=',num2str(n1),',n_2=',num2str(n2)]) @6j*XF  
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axis([0 90 0 1]) 7lVIN&.=  
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grid on gw*d"~A  
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2、光密射向光疏 
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clear 5{g9Wh[  
g\-3c=X  
close all V^{!d
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n1=1.45,n2=1; `e9uSF:9C  
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theta=0:0.1:90; ,\M_q">npc  
Q'a N|^w"f  
a=theta*pi/180; [;=ky<K0E  
V[I<9xaE  
rp=(n2*cos(a)-n1*sqrt(1-(n1/n2*sin(a)).^2))./(n2*cos(a)+n1*sqrt(1-(n1/n2*sin(a)).^2)); d>`(.qvxR  
=_dd4`G&<  
rs=(n1*cos(a)-n2*sqrt(1-(n1/n2*sin(a)).^2))./(n1*cos(a)+n2*sqrt(1-(n1/n2*sin(a)).^2)); 68c;V
b  
h(8;7}K  
tp=2*n1*cos(a)./(n2*cos(a)+n1*sqrt(1-(n1/n2*sin(a)).^2)); Yy)a,clZ*$  
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ts=2*n1*cos(a)./(n1*cos(a)+n2*sqrt(1-(n1/n2*sin(a)).^2)); wY.g-3  
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figure(1) 9Bk}g50$#  
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subplot(1,2,1); [~<',,tA0|  
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plot(theta,rp,'-',theta,rs,'--',theta,abs(rp),':',theta,abs(rs),'-.','LineWidth',2) NACY;XQ%  
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legend('r_p','r_s','|r_p|','|r_s|') nTU~M~gky  
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xlabel('\theta_i') V 3]p3  
#Q@6:bBzv  
ylabel('Amplitude') j_yFH#^W:  
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title(['n_1=',num2str(n1),',n_2=',num2str(n2)]) LQ@|M.$A  
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axis([0 90 -1.5 1.5]) mY8=qkZE  
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grid on N
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subplot(1,2,2); 7=.VqC^  
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plot(theta,tp,'-',theta,ts,'--',theta,abs(tp),':',theta,abs(ts),'-.','LineWidth',2) I*t)x,~3  
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legend('t_p','t_s','|t_p|','|t_s|') +Jka:]MW!  
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xlabel('\theta_i') ]GtR8w@w  
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ylabel('Amplitude') W:poUG1UR  
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title(['n_1=',num2str(n1),',n_2=',num2str(n2)]) 0bxvM  
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axis([0 90 -0.5 3]) C9g~l}=$&  
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grid on pd>a6 lI`  
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Rp=abs(rp).^2; q&$
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Rs=abs(rs).^2; .&Pe7`.BE  
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Rn=(Rp+Rs)/2; D'[P,v;Q  
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Tp=1-Rp; ,q@(L  
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Ts=1-Rs; SxW}Z_8x  
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Tn=(Tp+Ts)/2; 3zA=q[C  
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figure(2) :IT U0%;!+  
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subplot(1,2,1); v6P~XK}G  
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plot(theta,Rp,'-',theta,Rs,'--',theta,Rn,':','LineWidth',2) %M)oHX1p  
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legend('R_p','R_s','R_n') 'oiD#\t4  
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xlabel('\theta_i') Fma`Cm.  
KpbZnW}g  
ylabel('Amplitude') &8_f'+i0  
SeZT4y*=  
title(['n_1=',num2str(n1),',n_2=',num2str(n2)]) c7l!G~yx'  
(-'0g@0UA  
axis([0 90 0 1]) -m'3L7:  
#:vDBP05.m  
grid on >H1|c%w  
af?\kBm  
subplot(1,2,2); _/]:=_bf_z  
/Xu;/MMpd3  
plot(theta,Tp,'-',theta,Ts,'--',theta,Tn,':','LineWidth',2) }u3H4S<o  
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legend('T_p','T_s','T_n') kJvy<(iG  
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xlabel('\theta_i') J<($L}T*$  
sPE)m_u  
ylabel('Amplitude') %$Jqt  
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title(['n_1=',num2str(n1),',n_2=',num2str(n2)]) L=$?q/=-  
c6Yf"~TD0  
axis([0 90 0 1]) 1IF'>*  
pRiH,:\  
grid on A"BtVy[[9  
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感谢楼主分享

这个在《MATLAB在光学中的应用》这本书里都有

谢谢了楼主分享

thanks

谢谢了楼主分享

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