12
1 Electromagnetics for Zero-Index Metamaterials
Since metals have free electrons, resonant frequency ω 0 is zero. The x-axis of the
graph has been normalized w.r.t. the resonant frequency ω 0 in Fig. 1.6a and w.r.t. the
plasma frequency ω p in Fig. 1.6b. Below the plasma frequency, η
2
− κ
2 is negative
and 2ηκ is significantly high, which accounts for the highly reflective and absorptive
nature of metals. Above the plasma frequency, η
2
− κ
2 becomes positive, thus metals
begin to acquire dielectric-like properties of wave propagation. Additionally, the
value of 2ηκ also descends to extremely low values which in turn reduces absorption
losses. Hence, for ω > ω p metals become transparent to electromagnetic waves.
1.5 Structure-Dependent Refractive Index
In the previous section, the origin of refractive index of natural materials was discussed, which was dependent on how strongly the electrons are held by their nucleus,
and hence how strongly they could respond to the electric field of the incident light.
However, there is a special class of artificial materials called the metamaterials,
for which the optical parameters like relative permittivity, relative permeability and
refractive index depend not only on their constituent materials but also on their structure [16–35]. Their peculiar structure imparts to them the exotic properties, which
are drastically different from those of their constituent materials. Presented below
is an example of one such metamaterial structure, which comprises the alternating
layers of a metal and a dielectric material, one of the easiest ways to achieve the
desired permittivity at the wavelength of interest.
Metals intrinsically exhibit negative dielectric constant below their plasma frequency due to the presence of free electrons in them. This property proves beneficial
in realization of negative refractive index where negative permittivity ( < 0) and
negative permeability (μ < 0) are required at the same wavelength. Interleaving of
dielectric layers in between metallic layers dilutes the metal in a way and allows
the liberty to reduce the permittivity to the desired magnitude [15]. The value of
effective permittivity thus obtained depends on the filling fraction of the metal and
the dielectric in a unit cell as well as on the polarization of the incident field. For the
structure shown in Fig. 1.7, let us suppose that m is the permittivity of the metal, d
is the permittivity of the dielectric, h m and h d are their respective thicknesses, and
in consequence, f m = h m /(h m + h d ) and f d = h d /(h m + h d ) = 1 − f m become the
filling fractions of the metal and the dielectric, respectively. The incident electric field
can either be polarized parallel or perpendicular to the interface [23]. In the case of
perpendicular polarization, the electric field is discontinuous across the boundary,
but the displacement vector is continuous, i.e., D m = D d = D ⊥ . On account of discontinuity, the effective electric field needs to be taken as a weighted mean and is
thus given by
E ⊥ = f m E m + f d E d
(1.46)
Hence, the effective permittivity ⊥ is given by
1 Electromagnetics for Zero-Index Metamaterials
Since metals have free electrons, resonant frequency ω 0 is zero. The x-axis of the
graph has been normalized w.r.t. the resonant frequency ω 0 in Fig. 1.6a and w.r.t. the
plasma frequency ω p in Fig. 1.6b. Below the plasma frequency, η
2
− κ
2 is negative
and 2ηκ is significantly high, which accounts for the highly reflective and absorptive
nature of metals. Above the plasma frequency, η
2
− κ
2 becomes positive, thus metals
begin to acquire dielectric-like properties of wave propagation. Additionally, the
value of 2ηκ also descends to extremely low values which in turn reduces absorption
losses. Hence, for ω > ω p metals become transparent to electromagnetic waves.
1.5 Structure-Dependent Refractive Index
In the previous section, the origin of refractive index of natural materials was discussed, which was dependent on how strongly the electrons are held by their nucleus,
and hence how strongly they could respond to the electric field of the incident light.
However, there is a special class of artificial materials called the metamaterials,
for which the optical parameters like relative permittivity, relative permeability and
refractive index depend not only on their constituent materials but also on their structure [16–35]. Their peculiar structure imparts to them the exotic properties, which
are drastically different from those of their constituent materials. Presented below
is an example of one such metamaterial structure, which comprises the alternating
layers of a metal and a dielectric material, one of the easiest ways to achieve the
desired permittivity at the wavelength of interest.
Metals intrinsically exhibit negative dielectric constant below their plasma frequency due to the presence of free electrons in them. This property proves beneficial
in realization of negative refractive index where negative permittivity ( < 0) and
negative permeability (μ < 0) are required at the same wavelength. Interleaving of
dielectric layers in between metallic layers dilutes the metal in a way and allows
the liberty to reduce the permittivity to the desired magnitude [15]. The value of
effective permittivity thus obtained depends on the filling fraction of the metal and
the dielectric in a unit cell as well as on the polarization of the incident field. For the
structure shown in Fig. 1.7, let us suppose that m is the permittivity of the metal, d
is the permittivity of the dielectric, h m and h d are their respective thicknesses, and
in consequence, f m = h m /(h m + h d ) and f d = h d /(h m + h d ) = 1 − f m become the
filling fractions of the metal and the dielectric, respectively. The incident electric field
can either be polarized parallel or perpendicular to the interface [23]. In the case of
perpendicular polarization, the electric field is discontinuous across the boundary,
but the displacement vector is continuous, i.e., D m = D d = D ⊥ . On account of discontinuity, the effective electric field needs to be taken as a weighted mean and is
thus given by
E ⊥ = f m E m + f d E d
(1.46)
Hence, the effective permittivity ⊥ is given by