Optics and physics

Resources: Powerpoint lecture material, lecture notes, and the two codes provided.

Note: For Questions 1-5, use the Rayleigh Code. For the remaining questions, use the Mie Code.

Optics and physics Question 1:

Use the Rayleigh Code. Carefully review the input and output parameters. Make a list of these parameters, and name the optical/physical quantities they correspond to.

Optics and physics Question 2:

What is the forward-to-backward scattering ratio of a 5 nm particle?A 20 nm particle?

Question 3:

For equal concentrations, what are the scattering cross-sections of a 5 nm and a 20 nm particle? How many times larger is the scattering cross-section of the 20 than the 5 nm particle?

Optics and physics Question 4:

Construct graphs showing the power-law dependence of scattered intensity (use the average, or pick an angle) dependence on wavelength and particle radius. What are the power-law exponents?How good is the fit, for a wavelength from 350-700 nm and a size parameter from 0-1?

Question 5:

Can you make the scattered intensity of a collection of 5 and 10 nm particles to be the same?What parameter values for each collection of scattering objects permits this?

Question 6:

Next, run the Mie Code and as before, make a list of any extra variables that represent optical parameters not found in the Rayleigh code. What is the physical/optical meaning of these new parameters?

Question 7:

Examine Figures 4-6. Explain these graphs in words. Specifically address why Mu_S is smaller than Mu. For the values in the original code, what is the size parameter in which the anisotropy factor is approximately 0.5? What does this mean for light scattering?Hint: Look at the Jacques paper for basic definitions and examples of the anisotropy factor, g. Similarly, what is the physical meaning of the reduced scattering coefficient?

Optics and physics Question 8 Extra Credit:

Code in the expression for Mie Theory scattered light intensity from a sphere found in the class notes. Plot the scattered power spectrum for a size parameter of x=0.1, x=1, and x=3.

% Mie theory
% Use SI units
MUS = [];
MUSP = [];
GMx = [];
QS = [];
for z = 10:10:500
diameter = z*1e-9; %input(‘Diameter of sphere in nm(e.g., 579):’)*1e-9;
radius = diameter/2;
lambda = 400*1e-9; %input(‘Wavelength in nm(e.g., 400):’)*1e-9;
n_s = 1.57; %input(‘Refractive index of sphere (e.g., 1.57):’);
n_b = 1.33; %input(‘Refractive index of background (e.g., 1.33):’);
w_s = 1.05*1e3; %input(‘Specific weight of sphere(e.g., 1.05 g/cc):’)*1e3;
w_b = 1.0*1e3; %input(‘Specific weight of background(e.g., 1.0 g/cc):’)*1e3;
concentration = 0.002; %input(‘Concentration by weight (e.g., 0.002):’);

k = 2*pi*n_b/lambda
x = k*radius %size parameter
n_rel = n_s/n_b
y = n_rel*x %another size parameter

% Calculate the summations
err = 1e-8;
Qs = 0;
gQs = 0;
for n = 1:200000
Snx = sqrt(pi*x/2)*besselj(n+0.5,x);
Sny = sqrt(pi*y/2)*besselj(n+0.5,y);
Cnx = -sqrt(pi*x/2)*bessely(n+0.5,x);
Zetax = Snx+i*Cnx;

% Calculate the first-order derivatives
Snx_prime = – (n/x)*Snx+sqrt(pi*x/2)*besselj(n-0.5,x);
Sny_prime = – (n/y)*Sny+sqrt(pi*y/2)*besselj(n-0.5,y);
Cnx_prime = – (n/x)*Cnx-sqrt(pi*x/2)*bessely(n-0.5,x);
Zetax_prime = Snx_prime + i*Cnx_prime;

an_num = Sny_prime*Snx-n_rel*Sny*Snx_prime;
an_den = Sny_prime*Zetax-n_rel*Sny*Zetax_prime;
an = an_num/an_den;

bn_num = n_rel*Sny_prime*Snx-Sny*Snx_prime;
bn_den = n_rel*Sny_prime*Zetax-Sny*Zetax_prime;
bn = bn_num/bn_den;

Qs1 = (2*n+1)*(abs(an)^2+abs(bn)^2);
Qs = Qs+Qs1;

if n > 1
gQs1 = (n-1)*(n+1)/n*real(an_1*conj(an)+bn_1*conj(bn))…
+(2*n-1)/((n-1)*n)*real(an_1*conj(bn_1));
gQs = gQs+gQs1;
end

an_1 = an;
bn_1 = bn;

if abs(Qs1)<(err*Qs) & abs(gQs1)<(err*gQs)
break;
end
end

Qs = (2/x^2)*Qs; %scattering efficiency
gQs = (4/x^2)*gQs;
g = gQs/Qs; %anisotropy factor

vol_s = 4*pi/3*radius^3
N_s = concentration*w_b/(vol_s*w_s)
sigma_s = Qs*pi*radius^2; %scattering cross-section
mu_s = N_s*sigma_s %scattering coefficient

mu_s_prime = mu_s*(1-g); %reduced scattering coefficient

% Output results
{‘wavelength[nm]’,’Qs [-]’,’g [-]’,’mus [/cm]’,’mus_prime[/cm]’;…
lambda*1e9,Qs,g,mu_s*1e-2,mu_s_prime*1e-2}
MUS = [MUS; mu_s];
MUSP = [MUSP; mu_s_prime];
GMx = [GMx; g];
QS = [QS; Qs];
end
size_param_vector = [10:10:500]*(k)*(1e-9)/2;
size(size_param_vector)
figure(4), plot(size_param_vector, MUS)
hold on
figure(4), plot(size_param_vector, MUSP)
figure(5), plot(size_param_vector, GMx)
figure(6), plot(size_param_vector, QS)

% Rayleigh scattering
% Use SI units
I_lam=[];
lambdaVec=[];
QsVec=[];
for z = 400:10:750
diameter=20*1e-9; %input(‘Diameter of sphere in nm(e.g., 20):’)*
radius=diameter/2;
lambda=z*1e-9; %input(‘Wavelength in nm(e.g., 400):’)
n_sphere=1.57; % input(‘Refractive index of sphere (e.g., 1.57):’);
n_background=1.33% input(‘Refractive index of background (e.g., 1.33):’);
w_sphere=1.05*1e-3/(1e-2)^3;%input(‘Specific weight of sphere(e.g., 1.05 g/cc):’)
w_background=1.0*1e-3/(1e-2)^3;%input(‘Specific weight of background(e.g., 1.0 g/cc):’)
concentration=1e-5; %input(‘Concentration by weight (e.g., 1e-5):’);

k=2*pi*n_background/lambda
x=k*radius
n_rel=n_sphere/n_background

Qs = 8*x^4/3*abs((n_rel^2 – 1)/(n_rel^2 + 2))^2
sigma_s=Qs*pi*radius^2

vol_sphere = 4*pi/3*radius^3
N_s=concentration*w_background/(vol_sphere*w_sphere)
mu_s=N_s*sigma_s

% Output results
{‘wavelength[nm]’,’Qs [-]’,’mus [/cm]’; lambda*1e9, Qs, mu_s/1e2}

% calculate intensity vs. angle at fixed r=1;
nrel = n_sphere/n_background;
alpha = (((nrel*nrel)-1)/((nrel*nrel)+2))*(diameter^3)/8;
anglerange=[0:pi()/200:pi()];
for i = 1:length(anglerange)
I(i)=(1+cos(anglerange(i))*cos(anglerange(i)))*k*k*k*k*alpha*alpha/2;
end
I_lam = [I_lam; mean(mean(I))];
lambdaVec=[lambdaVec; lambda];
QsVec=[QsVec; Qs];
end

figure(1), plot(lambdaVec, QsVec)

I_lam=[];
size_param=[];
for z = 0.1:0.1:2
diameter=20*1e-9*z; %input(‘Diameter of sphere in nm(e.g., 20):’)*
radius=diameter/2;
lambda=550*1e-9; %input(‘Wavelength in nm(e.g., 400):’)
n_sphere=1.57; % input(‘Refractive index of sphere (e.g., 1.57):’);
n_background=1.33% input(‘Refractive index of background (e.g., 1.33):’);
w_sphere=1.05*1e-3/(1e-2)^3;%input(‘Specific weight of sphere(e.g., 1.05 g/cc):’)
w_background=1.0*1e-3/(1e-2)^3;%input(‘Specific weight of background(e.g., 1.0 g/cc):’)
concentration=1e-5; %input(‘Concentration by weight (e.g., 1e-5):’);

k=2*pi*n_background/lambda
x=k*radius
n_rel=n_sphere/n_background

Qs = 8*x^4/3*abs((n_rel^2 – 1)/(n_rel^2 + 2))^2
sigma_s=Qs*pi*radius^2

vol_sphere = 4*pi/3*radius^3
N_s=concentration*w_background/(vol_sphere*w_sphere)
mu_s=N_s*sigma_s

% Output results
{‘wavelength[nm]’,’Qs [-]’,’mus [/cm]’; lambda*1e9, Qs, mu_s/1e2}

% calculate intensity vs. angle at fixed r=1;
nrel = n_sphere/n_background;
alpha = (((nrel*nrel)-1)/((nrel*nrel)+2))*(diameter^3)/8;
anglerange=[0:pi()/200:pi()];
for i = 1:length(anglerange)
I(i)=(1+cos(anglerange(i))*cos(anglerange(i)))*k*k*k*k*alpha*alpha/2; %set r=1
end
I_lam = [I_lam; mean(mean(I))];
size_param=[size_param; x];
end

figure(2), plot(size_param, I_lam)

%scattered far field from a hertzian dipole (Rayleigh approximation)
for j = 1:1000
for k = 1:315
r=j/10;
gamma=(k-1)/100;
E_sc(j,k)=k*k*1000*sin(gamma)*exp(-1*sqrt(-1)*k*r)/r;
end
end

figure(3), plot([0.1:0.1:100],real(E_sc(:,20)))
figure(4), plot([0:0.01:3.14],real(E_sc(1,:)))

Last Updated on February 11, 2019 by Essay Pro