Thin metal films: beam splitters

A metal's index is mostly imaginary: silver at 550 nm has $\tilde n \approx 0.06 + 3.6i$ (Johnson and Christy 1972), so $\varepsilon = \tilde n^2 \approx -13 + 0.4i$. The negative real part is the signature of free electrons, and it is what makes the metal reflect; the small imaginary part is what makes it absorb. Both come out of the simplest model of a conductor, and both are needed to understand why a film of silver can be a beam splitter at all.

The Drude metal and the skin depth

Free electrons of density $N$ and effective mass $m$ driven by a field $E e^{-i\omega t}$ and damped at a rate $\gamma$ give the Drude permittivity

with the plasma frequency $\omega_p$ in the ultraviolet for the noble metals ($\hbar\omega_p \approx 9$ eV for silver) and $\gamma \ll \omega$ in the visible. Below $\omega_p$ the real part is large and negative: the electrons respond out of phase, screen the field, and the wave cannot propagate. Instead it decays over the skin depth

12 nm in intensity for silver in the visible ($k \approx 3.6$), 25 nm for aluminium's lower $k$ in the near ultraviolet. The interband transitions of the bound electrons (the $d$ bands) add to $\varepsilon_\infty$ and give silver its edge near 320 nm and gold its colour; the tabulated $n, k$ contain all of it, and the Drude form is the reason for their shape rather than a substitute for them. A film much thicker than $\delta$ is a mirror; a film comparable to it lets a useful fraction through. That is the whole design freedom of a metallic beam splitter: one thickness trades reflection against transmission.

Absorption is unavoidable, and how much is bounded

In a metal, the field that does get through drives currents that dissipate. However the film is chosen,

and a 50/50 silver splitter loses a few percent to absorption, more toward the blue, where silver's $k$ falls and its $n$ rises. Macleod's potential transmittance says how much a given metal layer can transmit if the dielectrics around it were chosen perfectly:

the transmittance per unit of light that enters the film. It depends on the metal layer and on the admittance presented to its exit face, not on what is in front of it, and it has a maximum over that exit admittance. For a given thickness of silver the maximum $\psi$ is the best any coating can do; design is then the choice of layers that realise it (Berning and Turner 1957). This is the principle of induced transmission: a 30 nm silver film alone transmits a few percent in the visible, but between two matching dielectric layers it transmits 60 % over a band, with the rest reflected rather than absorbed. The low-emissivity coating is this idea applied to a window.

import numpy as np
from pystratify import planar
from pystratify.docs import material  # n + ik of a tabulated page (refractiveindex.info format)

n = lambda wl: [1.0, material('docs/data/tutorials/Ag_Johnson.yml', wl), 1.52]  # incident medium, layers, substrate
d = [np.inf, 0.013, np.inf]  # thicknesses, µm
lam = (400 + 5 * np.arange(81)) * 1e-3  # wavelength, µm
th = np.radians(45)
R, T = {}, {}
for pol in ('s', 'p'):
    rows = [planar.coh_tmm(pol, n(wl), d, th, wl) for wl in lam]
    R[pol], T[pol] = np.array([r['R'] for r in rows]), np.array([r['T'] for r in rows])
R_avg, T_avg = (R['s'] + R['p']) / 2, (T['s'] + T['p']) / 2  # unpolarized

s and p at 45°

At oblique incidence the two polarizations see different tilted admittances, $n\cos\theta$ for s and $n/\cos\theta$ for p (see the transfer matrix), and a metal film therefore reflects s more strongly than p: at 45° a 50/50 silver splitter is typically 55/45 for s and 45/55 for p, mildly polarizing. Worse for interferometers, the two polarizations also pick up different phases on reflection and transmission, so a linearly polarized beam at 45° to the plane of incidence leaves slightly elliptical. Where that matters, in interferometers that recombine the two beams or in polarization-sensitive imaging, a compensating plate, a polarization-insensitive multilayer or a cube splitter with the film at a lower angle is used. The same difference is the reason the s-polarized dip of a surface-plasmon sensor is absent while the p-polarized one is deep.

Dielectric splitters

A quarter-wave stack of a few periods splits a beam with no absorption at all, at the price of a narrower band and a stronger dependence on angle and polarization: its reflectance follows the Bragg curve, 50 % only near one wavelength per design, and its s and p reflectances separate faster with angle than a metal's. Broadband dielectric splitters are therefore non-quarter-wave designs with many layers, optimised numerically, and for the wide band and modest accuracy of a visual instrument a metal film remains the simpler and cheaper part. Hybrid designs, a metal layer with dielectric enhancement on one side, sit between the two.

Measuring a metal film

The optical constants of a thin metal film are not those of the bulk: a film thinner than the electron mean free path (about 50 nm for silver) scatters its electrons at the surfaces, which raises $\gamma$ and the absorption, and below about 10 nm an evaporated film is not continuous but an island structure with its own plasmon resonance. The databases' $n, k$ are for thick, well-annealed films; a 10 nm beam splitter is measured, by ellipsometry or by $R$ and $T$ together, and the fitted thickness and index are what a design should use. The calculator's uploaded-table inputs exist for exactly this case.