solution of transcedental equation

function beta1 = trial0(beta, eps1, eps2, epsm,k0,h)
S1 = sqrt(power(beta,2)-(eps1*power(k0,2)));
S2 = sqrt(power(beta,2)-(epsm*power(k0,2)));
S3 = sqrt(power(beta,2)-(eps2*power(k0,2)));
beta1 = tanh(S2*h)*((eps1*eps2*power(S2,2))+(power(epsm,2)*S1*S3)) + (S2*((eps1*S3)+(eps2*S1))*epsm);
end
eps1 = 1.5471;
eps2 = 1.5431;
eps1m = -9.894 ;
epsm2 = 1.0458;
epsm = eps1m + i*eps2m;
lambda = 633;
k0 = (2*pi)/lambda;
h = 50;

find the value of beta

[Catalytic]

@shiv gaur

You have not accepted any answers previously provided to you. That is a bit of a disincentive for people to help you.

Answer 1

[Matt J]

eps1 = 1.5471;
eps2 = 1.5431;
eps1m = -9.894 ;
eps2m = 1.0458;
epsm = eps1m + i*eps2m;
lambda = 633;
k0 = (2*pi)/lambda;
h = 50;

[b,fval]=fminsearch(@(b) abs(trial0(complex(b(1),b(2)), eps1, eps2, epsm,k0,h))  ,[1,1]);
beta=complex(b(1),b(2)), fval

beta = -0.0145 - 0.0004i

fval = 2.7735e-05

function beta1 = trial0(beta, eps1, eps2, epsm,k0,h)
S1 = sqrt(power(beta,2)-(eps1*power(k0,2)));
S2 = sqrt(power(beta,2)-(epsm*power(k0,2)));
S3 = sqrt(power(beta,2)-(eps2*power(k0,2)));
beta1 = tanh(S2*h)*((eps1*eps2*power(S2,2))+(power(epsm,2)*S1*S3)) + (S2*((eps1*S3)+(eps2*S1))*epsm);
end

[shiv gaur]

pl root of equation

[Matt J]

Your comments are too brief. Why isn't the solution acceptable? It is a root.

[b,fval]=fminsearch(@(b) abs(trial0(complex(b(1),b(2)), eps1, eps2, epsm,k0,h))  ,[1,1]);
beta=complex(b(1),b(2));

%Check if very small
trial0(beta, eps1, eps2, epsm,k0,h)

ans = 1.7858e-05 + 2.1222e-05i

[Matt J]

We can make it even smaller...

opts=optimset('TolFun',1e-16);
[b,fval]=fminsearch(@(b) abs(trial0(complex(b(1),b(2)), eps1, eps2, epsm,k0,h))  ,[1,1],opts);
beta=complex(b(1),b(2))

beta = -0.0145 - 0.0004i

%Check if very small
trial0(beta, eps1, eps2, epsm,k0,h)

ans = -1.2490e-16 - 1.2105e-16i

[shiv gaur]

Mj=[cos (kj hj) i*sin(kj hj) /kj ; i*kj*sin(kj hj) cos (kj hj) ];

kj=sqrt(nj^2*k0^2-x^2);

Take n1=1.521;n2=2.66;k0=1;h1=1.5e-6;h2=1e-6

Mj=[cos (kj hj) i*sin(kj hj) /kj ; i*kj*sin(kj hj) cos (kj hj) ];

kj=sqrt(nj^2*k0^2-x^2);

Take n1=1.521;n2=2.66;k0=1;h1=1.5e-6;h2=1e-6

We take M=M1*M2;

Where M=( m11 m12;m21 m22) as usual in

Matrix;

How to multiply this matrix for M find the multiplication of matrix and find the value of

x;

and solve the eq

ns=1.512;

nc=1.2-i7;

gs=x^2-ns^2k0^2;

gc= x*2-ns^2k0^2;

f(x)=1i*(gsm11+gcm22)-m21+gsgcm12;

and solve the eq and value of x

[Matt J]

Is this a new question?

[shiv gaur]

relates with this type of equation

[Matt J]

Well, if we've answered your first question, please Accept-click it and post the related question in its own thread.