How to find the intersection values?
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GULZAR
am 17 Aug. 2023
Bearbeitet: Torsten
am 17 Aug. 2023
How to find the intersection values of line(black) and the curve(blue)
clc
close all
d1 = 0.4;d2 = 0.6;d = d1 + d2;
n1 = sqrt(12);n2 = 1;
lambda = linspace(400e-3,800e-3, 100000);
D1 = (2*pi*n1*d1)./lambda;D2 = (2*pi*n2*d2)./lambda;
RHS = cos(D1).*cos(D2) - 0.5*(n1^2+n2^2)/(n1*n2) * sin(D1) .*sin(D2);
plot(lambda,RHS)
hold on
yline(-1);
hold off
hold on
yline(1);
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Akzeptierte Antwort
Bruno Luong
am 17 Aug. 2023
d1 = 0.4;d2 = 0.6;d = d1 + d2;
n1 = sqrt(12);n2 = 1;
lambda = linspace(400e-3,800e-3, 100000);
D1 = (2*pi*n1*d1)./lambda;D2 = (2*pi*n2*d2)./lambda;
RHS = cos(D1).*cos(D2) - 0.5*(n1^2+n2^2)/(n1*n2) * sin(D1) .*sin(D2);
plot(lambda,RHS)
hold on
for yx = [-1 1]
yline(yx);
ys = RHS-yx;
i = find(ys(1:end-1).*ys(2:end) <= 0);
% linear interpolation
i1 = i; ss1 = ys(i1);
i2 = i1 + 1; ss2 = ys(i2);
w = ss2./(ss2-ss1);
xx = w.*lambda(i1) + (1-w).*lambda(i2);
plot(xx, yx+zeros(size(xx)), 'rx')
end
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Torsten
am 17 Aug. 2023
format long
d1 = 0.4;d2 = 0.6;d = d1 + d2;
n1 = sqrt(12);n2 = 1;
D1 = @(lambda)(2*pi*n1*d1)./lambda;D2 = @(lambda)(2*pi*n2*d2)./lambda;
RHS = @(lambda)cos(D1(lambda)).*cos(D2(lambda)) - 0.5*(n1^2+n2^2)/(n1*n2) * sin(D1(lambda)) .*sin(D2(lambda));
% Case -1
start = [0.42,0.45,0.56,0.59,0.72];
offset = -1;
rhs = @(lambda) RHS(lambda)-offset;
for i = 1:numel(start)
sol(i) = fsolve(rhs,start(i),optimset('Display','none'));
end
sol
% Case 1
start = [0.4,0.47,0.52,0.63,0.67];
offset = 1;
rhs = @(lambda) RHS(lambda)-offset;
for i = 1:numel(start)
sol(i) = fsolve(rhs,start(i),optimset('Display','none'));
end
sol
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