| Date 08/09/2010 |
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| Script S4_2_11.m | |||
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%==============================================
%Fabry-Perot 4 %============================================== % %the sodium line 0.58900 microns lambda=0.58900; %the thickness t in microns t=2.0e+003; %angles on incidence between 0° and 1.35° %in degrees tetag=eps:0.001:1.35 %and in radians teta=tetag*(pi/180); %reflectivity, refractivity and their squares %e i loro quadrati ro=0.60; tau=1-ro; roq=ro^2; tauq=tau^2; %values of k and of fi k=2*pi/(lambda); fi=(k*2*t)*cos(teta); % %============================================== %orders m1MAX and m2MAX m1MAX=floor(k*t/pi) m2MAX=m1MAX-1 % %preliminary calculi num3=tauq; dena=1+roq; %============================================== %Calculus of Ir %============================================== denb1=2*(ro.*cos(fi)); denb11=dena-denb1; Ir=num3./denb11 % %============================================== %for the first maximum % %only a part of the array of Ir is used %to find values and positions for the maximum and minimum Ir1p=Ir(201:601); teta1=tetag(201:601); [Ir1max,p1max]=max(Ir1p) ang1max=teta1(p1max) Ir11p=Ir(601:1001); teta11=tetag(601:1001); [Ir1min,p1min]=min(Ir11p) ang1min=teta11(p1min) %countercheck (see previous two problems) fi1max=m1MAX*2*pi; Ir1maxprova=num3/(dena-2*ro*cos(fi1max)) %============================================== %for the second maximum %only a part of the array of Ir is used %to find values and positions for the maximum and minimum Ir2p=Ir(1001:1201); teta2=tetag(1001:1201); [Ir2max,p2max]=max(Ir2p) ang2max=teta2(p2max) %countercheck (see previous two problems) fi2max=m2MAX*2*pi; Ir2maxprova=num3/(dena-2*ro*cos(fi2max)) %============================================== % %plot of the first two fringes plot(tetag,Ir,'r-') axis([0 1.35 0 1]) title('Ir of the first two fringes when tetag varies from 0° to 1.35°') grid on %============================================== % %angles to the right of a maximum %corresponding to Ir = 0.5 %the array of Ir is observed finding the values %0.5 in the position 498 for the first maximum %hence d1d=tetag(498) %and in the position 1103 for the second maximum %hence d2d=tetag(1103) %============================================== % |
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