# Depolarization factors of a drude metal plasmonic spheroid

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?

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 $$L_x$$ = $$L_y$$ = $$\frac{1}{4}$$.