A practice question asks for the depolarization factors $ L_{i=x,y,z} $ of a plasmonic spheroid made of a drude metal having the same resonance as the SPP resonance frequency.

The answer turns out to be $ L_x = L_y = \frac{1}{4} ; L_z = \frac{1}{2} $. I'm at a loss at how this was obtained. How can we infer those figures only knowing whats given?


1 Answer 1


Polarizability for a spheroidal particle of permittivity $\epsilon_2$ placed in a medium of permittivity $\epsilon_1$ due to electric field in direction $j$ is given by: $$ \alpha_j = 4 \pi \epsilon_0 R_x R_y R_z \cfrac{\epsilon_1 - \epsilon_2}{3 \epsilon_2 + 3 L_j (\epsilon_1 - \epsilon_2)} $$ where $R_x$, $R_y$, $R_z$ are the lengths of axes along x, y, z. Since we are interested in the resonance, we want the denominator to become zero, which gives the resonant condition: $$ 3 \epsilon_2 + 3 L_j (\epsilon_1 - \epsilon_2) = 0 $$ which leads to the relation $$ \epsilon_1 = \cfrac{L_j - 1}{L_j} \epsilon_2 $$ and the resonant frequency: $$ \omega_{0j} = \cfrac{\Omega_P}{\sqrt{\epsilon_{1} - \epsilon_2 \left(1-\cfrac{1}{L_j}\right)}} $$

Now we want it to be equal to SPP resonance which is given by: $$ \omega_{SPP} = \cfrac{\Omega_P}{\sqrt{\epsilon_{1} + \epsilon_2}} $$

Comparing the two expressions gives us: $$ -\epsilon_2 \left(1-\frac{1}{L_j}\right) = \epsilon_2 $$

and we get $ L_j = 1/2 $.

Since the answer is $L_z = \frac{1}{2}$ it means it was assumed that the incident field is along axis z. The three factors $L_x$, $L_y$, $L_z$ always sum up to 1. And since the question asks about a spheroid it means that two of the axes have the same length and the two other factors are equal, therefore $L_x$ = $L_y$ = $\frac{1}{4}$.


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