Mie scattering for inhomogeneous medium

[Haorong Wu]

Homework Statement

Suppose a particle of Ag with diameter of 100nm is partially embeded in a glass substrate so that half of the particle is surrounded by glass, while the other half is surrounded by air. The refractive index of the glass substrate is 1.5. Use Mie scattering theory to analyze the scattering and absorption spectrums between 300nm and 1000nm.

Relevant Equations

none

It is an open question. The professor asked us to find a suitable method to solve it with the help of a computer.

When I learned the Mie scattering, the equations are given for particles in homogenous medium. But now half of the particle is surrounded by glass while the other half by air.

First, I want to treat the particle as two independent particles one of which is surrounded only by glass, while the other by air only. Then I can use Mie scattering theory to calculate independently for them and then sum the results. I believe this could be an approach but it may not give a satisfying result.

Second, I think maybe I only need consider the medium in the outgoing direction. I note that the coefficients for extinction, scattering, and absorption are related to $a_n$ and $b_n$. And $a_n$ and $b_n$ are related to the scattered fields $$ \mathbf {E} _{s}=\sum _{n=1}^{\infty }E_{n}\left(ia_{n}\mathbf {N} _{e1n}^{(3)}(k,\mathbf {r} )-b_{n}\mathbf {M} _{o1n}^{(3)}(k,\mathbf {r} )\right).$$ Besides, the incident light will excite the atoms based on its frequency $\omega$ which is not related to the refractive index of the medium. Also the scattering directions are randomly distributed, so I can calculate the scattering coefficients for pure air and for pure glass, and then add them up and divide them by 2. The same method will be applied to the extinction coefficients.

Is there any more reasonable method?

 

[Paul Colby]

Mie theory relies on separation of variables in spherical coordinates of Maxwell's equations. The half space of dielectric breaks the rotational symmetry the problem so this separation of variables doesn't apply directly. I would look at using spherical modes in each half space. The angular part of these modes in each half space (vector spherical harmonics) are the same. The only difference are the radial functions which carry the only dependence on wave number(the wavelength changes between media). What this leaves you with are one dimensional mode matching problems for each of the radial functions for each angular mode. These should be all independent.

P.S. I've never attempted this solution so I'm spitballing here.

P.S.P.S One must consider a modal expansion in spherical harmonics of the incident fields. This should be doable for your case because reflection from a dielectric interface like this has a known solution in terms of plane waves in each half space. I would start with this observation.

 

[Haorong Wu]

Thanks, @Paul Colby . I have tried the easiest situation.

This is the only configuration that I can solve, because, as you mentioned, the symmetry is crucial for Mie theory, and this configuration keeps most of symmetries.

From the equation of extinction cross section, $$W_{ext}=\frac 1 2 \Re \int_0^{2 \pi} \int_0^{\pi} (E_{i\phi} H_{s\theta}^{*} -E_{i\theta} H_{s\phi}^{*} -E_{s\theta} H_{i\phi}^{*} +E_{s\phi} H_{i\theta}^{*})r^2 \sin \theta d\theta d \phi ,$$ where $E_{i \theta} = \frac {\cos \phi} {\rho} \sum_{n=1}^{\infty} E_n(\psi_n \pi_n-i \psi_n^{'} \tau_n ),$ $H_{i\theta} = \frac k {\omega \mu} \tan \phi E_{i\theta}$, etc.

We note that all variables in $Es$ and $Hs$, such as $\psi_n, \pi_n, \tau_n , \xi_n$ do not depend on $\phi$. So after integrating $\theta$ we have $$ W_{ext}\propto\Re \int d\phi (\alpha \sin^2 \phi - \beta \cos^2 \phi)$$ where $\alpha$ and $\beta$ are some constants. If the particle is surrounded by homogeneous medium, the integration will be carried over $0$ to $2\pi$, yielding $W_{ext} \propto \Re \pi (\alpha -\beta)$. However, if the system is configured as in the above picture, then the integration will be split into two part, one from $0$ to $\pi$ with the refractive index $n_1$, the other from $\pi$ to $2\pi$ with the refractive index $n_2$, giving $W_{ext} \propto \Re \frac 1 2 [\pi (\alpha_1 -\beta_1)+\pi (\alpha_2 -\beta_2)]$.

Since the extinction coefficient is proportional to its cross section, we could first calculate the extinction coefficients when the particle is surrounded by pure air and pure glass, respectively, i.e., $$ Q_{ext}^{air},~~~ Q_{ext}^{glass},$$ and then take their average $$Q_{ext}=\frac 1 2 ( Q_{ext}^{air}+ Q_{ext}^{glass}).$$

If we consider other more general situations, the integration over $\theta$ would play an important role and it would be quite difficult to figure out its analytic solution. Therefore, we will not attempt to solve them.

 

[Paul Colby]

and then take their average

But this isn't a solution of the field boundary value problem. Replace the refractive index of the half space with that of Ag. Do you still think this average yields the correct answer? Clearly it doesn't.

 

[Haorong Wu]

@Paul Colby . Thanks, I made a mistake. I forgot that the change of the medium will affect all the fields inside the particle. I will re-think your post. Well, to be honest, my professor did not taught us those details. He just gave us the conclusion. It is a good time that I will go through the theory step by step.

 

[Paul Colby]

The problem as stated isn't an easy boundary value problem and it's not clear to me that a series solution with known coefficients is attainable. At best you might obtain a series of equations for the coefficients which need to be solved. These equations could well be too complex to solve with pencil and paper. Just setting them up is a significant task in itself. I guess it depends on how badly one want's the answer. If your professor doesn't know the solution and would like it, I have very reasonable rates.

[Edit] Ah, I missed the light being along the plane separating half spaces. This could well be a special easy case that is solvable. My comments assumed arbitrary incidence angles to the plane. They still apply.