| Date 06/09/2010 |
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| Script S4_2_8.m | |||
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%Fabry_Perot interferometer 1 %============================================== % %the wavelength 0.435 microns (the blue line of a mercury-vapor lamp) lambda=0.435; %the thickness d in microns d=1e+003; %angles of incidence in degrees and in radians tetag=0.5:0.0001:3; teta=tetag*(pi/180); %the very large number max of the angles max=length(teta) %three values of reflexivity ro=[0.3 0.6 0.9]; % %the array with 3 row and max columns is initialized %setting zero values in all positions Ir=zeros(3,max); % %============================================================== %values of Ir for different reflexivity are computed for j=1:3 ro1=ro(j); j %corresponding values od refractivity %and their squares tau1=1-ro1; ro1q=ro1^2; tau1q=tau1^2; % %maximum of Ir is always equal to one % %the minimum is given by the following formula %for the assigned value of reflectivity minIr=tau1q/((1+ro1)^2) % %corresponding phase shifts fi k=2*pi/(lambda); fi=(k*2*d)*cos(teta); %the relative intensity Ir num=tau1q; den1=1+ro1q; den2=2*(ro1*cos(fi)); den=den1-den2; %the row j is set to values of Ir for the assigned reflexivity Ir(j,:)=num./den; end %============================================================== % %plots of Ir for the considered reflexivity plot(tetag,Ir(1,:),'r-') title('Ir when ro is 0.3. Now minimum is about 0.29'),figure plot(tetag,Ir(2,:),'b-') title('Ir when ro is 0.6. Now minimum is about 0.06'),figure plot(tetag,Ir(3,:),'g-') title('Ir when ro is 0.9. Now minimum is about 0.001') %============================================== % |
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