Surface wave direction control on curved surfaces

Before our simulation work, we assume that all the surfaces considered, as well as their refractive index profile distribution, are rotationally symmetric. Figure 1 illustrates a curved surface generated by rotating the generating curve C, with axis of revolution z in a cylindrical coordinate system. We equate the radial light path L in the flat homogeneous surface with the radial light path in the rotationally symmetric curved surface C [36] as

$$\begin{equation}N(\rho ) = n\left[ {F\left( \rho \right)} \right]F\left( \rho \right)/\rho ,\end{equation} \tag{ 1 }$$

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where ρ is the distance from the axis z, N(ρ) is the refractive index of the curved surfaces which is a function only of ρ, $F\left( \rho \right) = \exp \int_1^\rho {\left[ {{s^{\textrm^^{\prime}}}\left( \rho \right){\textrm{/}}\rho } \right]} {\textrm{ }}d\rho {\textrm{ }}$ is the mapping function and s(ρ) is the arc distance measured along C, n[F(ρ)] is the refractive index of the flat surface as a function of F(ρ). By solving equation (1), we can derive the modified index profile N(ρ). Thus, the layered refractive index lens can be mapped onto an arbitrary rotationally symmetric curved surface to manipulate the surface wave on curved surfaces.

At first, we set angle ϕ as the angle between the reference line i along the incident light and the line o along the output wave propagation, as shown in figure 2(a). Here, we set angle ϕ = 0°, which describes a wave moving forward without bending and ϕ = 180° to represent a retroreflector. A planar surface with refractive index $n = {n_b}\sqrt {\frac{{\text{2}}}{r} - 1}$, which is known as an Eaton lens, can bend the surface wave with an angle ϕ = 180°, where n_b is the background refractive index. An Eaton lens is a typical gradient index lens and offers a general method to control the surface wave. Various applications based on Eaton lens' have been introduced such as an omnidirectional retroreflector [37] and waveguides [38, 39]. Here, we set the background refractive index as air (n_b = 1). This work is simulated with COMSOL, based on the finite element method. In the simulation, we set the radius of the bottom of the curved surface to 1000 nm. A perfect electric conductor boundary is used for the top and bottom of the curved surfaces and the scattering boundary conditions around the edge are used. The input source, polarizing along the z-axis, propagates from the left of the curved surfaces. Based on equation (1), we can bend the surface wave on any rotationally symmetric curved surface. We present four rotationally symmetric curved surfaces, which can bend the surface wave with an angle ϕ = 180°. The simulation, as shown in figure 2(b), is a cone surface which can be described as $s\left( \rho \right) = {k_c}\rho$, where k_c is the slope of the cone surface. The analytically solution of equation (1) is

$$\begin{equation}N\left( \rho \right) = {\rho ^{{k_c} - 1}}\sqrt {\frac{{\text{2}}}{{{\rho ^{{k_c}}}}} - 1} ,\end{equation} \tag{ 2 }$$

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and is valid for any slope k_c of the cone surface. Here, we choose k_c = 1.183 to keep the height in the same range as the other surfaces. Although the analytically solution here can perfectly match the refractive index distribution needed to bend the surface wave, the tip of the cone surface results in imperfect propagation. To achieve a better effect of controlling the surface wave, we choose a dome-like top surface for further designs. In figure 2(c), it is a hemisphere surface with arc distance $s\left( \rho \right) = {\sin ^{ - 1}}\rho$. The required index for a hemisphere surface can be calculated analytically and is given by

$$\begin{equation}N\left( \rho \right) = \frac{{{n_b}}}{{1 + \sqrt {1 - {\rho ^2}} }}\sqrt {\frac{{2\left( {1 + \sqrt {1 - {\rho ^2}} } \right)}}{\rho } - 1} .\end{equation} \tag{ 3 }$$

The generalized Rinehart surface is a special type of surface which is hard to describe in a Cartesian coordinate system but easy to express by a generated curve. The arc distance of the generalized Rinehart surface takes the form $s\left( \rho \right) = \left( {1 - {k_r}} \right)\rho + {k_r}{\sin ^{ - 1}}\rho$, where k_r is a parameter ranging within [0,1]. The generalized Rinehart surface also can be calculated analytically and the refractive index distribution can be expressed as

$$\begin{equation}N\left( \rho \right) = \frac{{{n_b}}}{{{{\left( {1 + \sqrt {1 - {\rho ^2}} } \right)}^{{k_r}}}}}\sqrt {\frac{{2{{\left( {1 + \sqrt {1 - {\rho ^2}} } \right)}^{{k_r}}}}}{\rho } - 1} .\end{equation} \tag{ 4 }$$

In the simulation shown in figure 2(d), we choose k_r = 1/2 for simple representation. The hemisphere surface is a special situation of the generalized Rinehart surface (k_r = 1). Compared with the generalized Rinehart surface, the parabolic surface with arc length $s = \frac{1}{2}\rho \sqrt {1 + 4k_p^2{\rho ^2}} + \frac{{\arcsin h\left( {2{k_p}\rho } \right)}}{{4{k_p}}}$, where k_p is the parameter of the parabolic surface, can be a more ordinary rotationally symmetric surface. The analytical solution of the parabolic surface to equation (1) is

$$\begin{equation}N\left( \rho \right) = \frac{{1 + b}}{{1 + a}}{e^{a - b}}\sqrt {\frac{{{\text{2}}\left( {1 + a} \right)}}{{\rho \left( {1 + b} \right){e^{a - b}}}} - 1}, \end{equation} \tag{ 5 }$$

where $a = \sqrt {1 + 4k_p^2{\rho ^2}}$ and $b = \sqrt {1 + 4k_p^2}$. To keep the same scale of the models, k_p = 0.63 is chosen in the simulation and the E-field profile is shown in figure 2(e). Even though the parabolic surface is more ordinary, its refractive index expression is complex.

To show the propagation direction control of the surface wave, we present simulations of four different curved surfaces which bend the surface wave at different angles ϕ. The general Eaton lens can bend the rays into different angles in the flat plane and can be expressed as [40]

$$\begin{equation}r{n^{\tfrac{{2\pi }}{\varphi }}} - 2{n^{\tfrac{{\pi - \varphi }}{\varphi }}} + r = 0,\end{equation} \tag{ 6 }$$

where r is the radius in the polar coordinate system.

For the consideration of the refractive index of the general Eaton lens in a planar surface, we can calculate the refractive index distribution of rotationally symmetric curved surfaces for all angles between 0° to 180°.

Figure 3 shows the simulations with the cone surface, hemisphere surface, Rinehart surface and the parabolic surface which control the propagation direction of the surface wave at angle ϕ = 45°, ϕ = 90° and ϕ = 135°. The incident total electric surface field $E = \sqrt {{{\left| {{E_x}} \right|}^2} + {{\left| {{E_y}} \right|}^2} + {{\left| {{E_z}} \right|}^2}}$ from the left is bent though the structure. As can be observed, all simulations with different curved surfaces can control the propagation direction of the surface wave to the specified directions accurately. It is confirmed that the same method can be used for arbitrary rotationally symmetric curved surfaces. Among the simulations above, the structures control the propagation direction of the surface waves at angle ϕ = 90° and ϕ = 180° perfectly, and the output surface waves at angle ϕ = 45° and ϕ = 135° are divergent slightly in figure 3. Based on equation (1), we have the analytical solution of the cone surface, the hemisphere surface, Rinehart surface and the parabolic surface. However, there are no analytical solution to most of the angles in equation (6), and therefore we apply a numerical approach (equation (S1)) to calculate the refractive index distribution N(ρ). The parameters used in equation (S1) are listed in table S1 (available online at stacks.iop.org/JPD/54/074003/mmedia) and the numerical solution of equation (6) and the corresponding fitting results at angle ϕ = 45° and ϕ = 135° are shown in figure S1. Due to the approximate refractive index distribution, the output surface waves at angle ϕ = 45° and ϕ = 135° are not as perfect as the ones at angle ϕ = 90° and ϕ = 180°.

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Considering all the curved surfaces mentioned above, the Rinehart surface has a horizontal tangent plane at ρ = 0, a vertical tangent plane at ρ = 1 and the Rinehart surface is concave down everywhere between ρ= 0 and ρ = 1. The characteristics of the Rinehart surface lead to good performance of direction control of surface wave. The Rinehart surface allows for the balance between the ordinary of the curved surfaces and is convenient for calculation. We take the Rinehart surface for further exploration.

The corresponding gradient refractive index distribution is hard to realize. For experimental realization of the device, the gradient refractive index distribution is discretized by dividing the Rinehart surface into layers along the z-axis and we set the background refractive index n_b = 1 and operation wavelength λ = 500 nm. Every curved layer has equal height in the layered Rinehart surface. Each curved layer in the layered Rinehart surface is given a refractive index according to the gradient refractive index distribution. Comparing the layered Rinehart surface with 10 layers, 20 layers, 30 layers and 40 layers, as shown in figure 4, the layered Rinehart surface needs 40 layers at least to achieve good performance for controlling the propagation directions of the surface waves. However, the 40-layer Rinehart surface is still complex.

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Considering the regularity of the gradient refractive index distribution, we simplify the 40-layer Rinehart surface to 14 layers. The proposed Rinehart surface with 14 layers has fewer layers at the bottom and more layers at the top. As the layers refractive index distribution is stable at the bottom and variable at the top, we combine each 10 layers at the bottom of the 40-layer Rinehart surface into one and each five layers at the middle into one. Each curved layer in the 14-layer Rinehart surface is given a refractive index according to the gradient refractive index distribution. As shown in figures 5(a)–(d), we simulated electric field propagation with the 14-layer Rinehart surface. With a proper refractive index of each layer in the 14-layer Rinehart surface, the devices control the propagation direction of the surface wave at angles ϕ = 0°, ϕ = 45°, ϕ = 90° and ϕ = 135°. The corresponding refractive index of each layer in the 14-layer Rinehart surface used in the simulation are shown in figure 5(e). Compared with the 40-layer Rinehart surface, the 14-layer Rinehart surface is sufficient to achieve the proposed functionalities.

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