1. Introduction

Stratified optical media are stacks of micron or sub-micron layers of optical materials. [1–4] A stratified medium usually consists of a stack of two alternating optical layers of different refractive indices. The two principal optical axes of the layers are usually orthogonal to each other, which has been intensively studied and used in many applications, such as flat panel displays, data storage, optical coatings, optical filters, and medical devices. [5–15] In this paper, we report a study of the reflection of stratified media where the principal optical axis is stepwise twisted by an arbitrary angle. The stepwise twisted stratified media exhibit some very interesting and important reflection features.

As an example, anisotropic uniaxial optical films can be obtained by stretching polymer films. [16–19] The uniaxial axis, which is also the principal optical axis, is usually along the stretching direction. The refractive index for light linearly polarized along the principal axis is ${n_e}$, known as the extraordinary refractive index, which is larger than ${n_o}$, known as the ordinary refractive index, for light linearly polarized in the perpendicular direction. Because the optical properties of uniaxially stretched polymer films is similar to those of (nematic) liquid crystals which consist of rod-like elongated organic molecules, [20–23] and furthermore, many methods and terminologies used to study liquid crystals, can be used to analyze the stretched films, in the following discussions, we will call the optical medium liquid crystal.

The orientation of liquid crystal molecules is specified by the liquid crystal director, which is the average direction of the long molecular axis, and is denoted by a unit vector $\vec{n}$. They exhibit anisotropic physical properties due to the orientational order of the molecules. For example, they exhibit birefringence: an ordinary refractive index ${n_o}$ for light with polarization perpendicular to the liquid crystal director and an extraordinary refractive index ${n_e}$ for light with polarization parallel to the liquid crystal director. If some or all of the constituent molecules are chiral, the liquid crystal will possess a continuously twisted helical structure, [20,21,23–25] and the mesophase is known as cholesteric liquid crystal (CLC, also called chiral nematic liquid crystal). The distance over which for the liquid crystal director to twist 360° is known as the pitch and is denoted by $P$. Because the molecules are not polar, $\vec{n}$ and $- \vec{n}$ are equivalent, namely the physical properties are the same in the $\vec{n}$ and $- \vec{n}$ directions. Therefore the periodicity of the CLC is $P/2$. Due to its periodic helical structure, the CLC exhibits a selective Bragg reflection band. The central wavelength and bandwidth of the reflection band are given by ${\lambda _o} = [({n_e} + {n_o})/2]P$ and $\Delta \lambda = ({n_e} - {n_o})P$, respectively. [20,23,24] The reflected light is circularly polarized with the same handedness as the helical structure of the CLC. An unpolarized incident light can be decomposed into two circularly polarized components with opposite handedness; the component of the incident light with the same handedness as the CLC is reflected, and the other component is transmitted. Therefore for an unpolarized incident light, the maximum reflectance of a single layer CLC is 50%. The reflection of CLCs has been studied intensively and made use of in many applications. [26–37]

Typically the diameter and the length of a liquid crystal molecule are about 0.5 nm and 2 nm, respectively. For a CLC reflecting visible light, the pitch is about 400 nm. The twist angle between two neighboring molecules is $(0.5\,nm/400\,nm) \times {360^o} = {0.45^o}$, which is small, and the twisting of the liquid crystal director can be considered as continuous. An interesting question arises that if the twist is not continuous but stepwise, how the reflection will change? For example, dual brightness enhancement film (DBEF), [5,38–40] where twist angle between two neighboring homogeneously aligned polymer layers is 90°, reflects a linearly polarized incident light and transmits the orthogonal linearly polarized light. In this paper we present a study of the reflection of stepwise twisted liquid crystals. The results show that stepwise twisted liquid crystals can simultaneously possess both lefthanded and right-handed helical structures. The reflectance can be increased and bandwidth can be broadened when proper step twist angles are used. This result is in agreement with other previous theoretical studies. [41,42]

2. Coexistence of right-handed and left-handed helical components

The stepwise twisted liquid crystal is schematically shown in Fig. 1(a). The liquid crystal director is stepwise twisted by the angle $\Delta \phi$ from one layer to the top neighboring layer counterclockwise (looking down from top). The layer thickness is $\Delta d$. For the purpose of simplicity, we first consider the cases where $\Delta \phi$ and $({180^o} - \Delta \phi )$ commensurate with 360°. It is a right-handed helix, as shown in Fig. 1(b), and has the pitch (the distance over which the liquid crystal director twists 360°)

The periodicity P of the liquid crystal director is the distance over which the director rotates 360°, and therefore is equal to ${P_R}$. The dielectric tensor of the liquid crystal has a right-handed helical component of the periodicity ${P_R}$. Because the liquid crystal is apolar, $\vec{n}$ and $- \vec{n}$ are equivalent, and therefore a rotation of $\Delta \phi$ counterclockwise is equivalent to a rotation of ${180^o} - \Delta \phi$ clockwise. Hence the helical structure can also be considered as left-handed, as shown in Fig. 1(c), and has the pitch

 

Fig. 1. Schematic diagram of the stepwise twisted liquid crystal with the step twist angle of Δϕ.

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The dielectric tensor of the liquid crystal also has a left-handed helical component of the periodicity ${P_L}$. For a regular (say right-handed) cholesteric liquid crystal, both $\Delta \phi \quad (\Delta \phi < < {180^o})$ and $\Delta d$ are small, the period ${P_R}$ of the right-handed helical component is finite (say $\bar{n}{P_R}$ is in visible light region, where $\bar{n}$ is the average refractive index). ${180^o} - \Delta \phi$ is close to ${180^o}$. The period ${P_L}$ of the left-handed helical component is much smaller than ${P_R}$. Therefore we only have to consider the reflection produced by the right-handed component. For example, when $\Delta \phi = {60^o}$, the period of the right-handed helical component is ${P_R} = ({360^o}/{60^o})\Delta d = 6\Delta d$, and the period of the left-handed helical component is ${P_L} = [{360^o}/({180^o} - {60^o})]\Delta d = 3\Delta d = {P_R}/2$. We may have to consider the reflection produced by both the right-handed and left-handed components. For another example, when $\Delta \phi = {90^o}$, the period of the right-handed helical component is ${P_R} = ({360^o}/{90^o})\Delta d = 4\Delta d$, and the period of the left-handed helical pitch is ${P_L} = [{360^o}/({180^o} - {90^o})]\Delta d = 4\Delta d = {P_R}$. We certainly have to consider the reflection produced by both the right-handed and left-handed components.

3. Fourier transformation and reflection spectrum

3.1 Fourier transformation

For a non-uniform optical medium, whose refractive index n varies in space, light reflection by it can be analyzed by considering Fourier transformation of the dielectric constant $\varepsilon$. We first consider an isotropic optical medium, whose dielectric constant $\varepsilon$ only varies in the z direction and is described by $\varepsilon (z)$. The refractive index n is related to the dielectric constant by $n(z) = \sqrt {\varepsilon (z)}$. The ordinary Fourier transformation to the wavevector space is given by

The CLC is uniaxial and anisotropic, and its dielectric property is described by the dielectric tensor

The twist angle (the angle between the liquid crystal director and the x axis)$\phi$ is a function of z, the liquid crystal director is given by

The dielectric tensor is

For a regular cholesteric liquid crystal of a right-handed helix with pitch $P\;( > 0)$, the helical wavevector is ${q_o}\; = 2\pi /P$, and the liquid crystal director is given by

The tensor product of the liquid crystal director of right-handed helix is

We consider the case where the incident light propagates in the $+ z$ direction and the reflected light propagates in the $- z$ direction. The difference between the wavevectors of the incident light and reflected light is positive. In the calculation of the reflection, only the Fourier components of positive wavevector q needs to be considered. The Fourier component of the dielectric tensor of the right-handed helix is

Therefore the light is reflected. When the incident light is left-handed circularly polarized, the polarization vector is ${\vec{J}_{\textrm{Incident}}} = {\vec{J}_{I/L}} = \left( {\begin{array}{c} 1\\ { - i} \end{array}} \right)$, the reflection amplitude is

Therefore the light is not reflected. For a regular cholesteric liquid crystal of a left-handed helix with the helical wavevector ${q_o}$, the liquid crystal director is given by

The dielectric tensor is

The Fourier component of the dielectric tensor of the left-handed helix is

The Fourier component of the dielectric tensor is peaked at $q = 2{q_o}$. The wavelength of the reflected light is given by ${\lambda _o} = \frac{{2\pi \bar{n}}}{{{q_o}}} = \frac{{2\pi \bar{n}}}{{(2\pi /P)}} = \bar{n}P$. The reflection amplitude is proportional to

Therefore the light is reflected. When the incident light is right-handed circularly polarized, the polarization vector is ${\vec{J}_{\textrm{Incident}}} = {\vec{J}_{I/R}} = \left( {\begin{array}{c} 1\\ i \end{array}} \right)$, the reflection amplitude is

Therefore the light is not reflected.

We numerically calculated Fourier components of the dielectric tensor of the stepwise twisted liquid crystal. It can be seen from Eq. (6) that $\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\leftrightarrow$}} \over \varepsilon } (q) = \left( {\begin{array}{cc} {{\varepsilon_{11}}(q)}&{{\varepsilon_{12}}(q)}\\ {{\varepsilon_{12}}(q)}&{ - {\varepsilon_{11}}(q)} \end{array}} \right)$, therefore we only have to calculate ${\varepsilon _{11}}(q)$ and ${\varepsilon _{12}}(q)$. In the calculation, the following equations was used.

In the calculation the cell thickness $h = 20P$ and birefringence $\Delta n = 0.2$ were used, with which saturated reflection could be obtained.

3.2 Examples of stratified films with various twist angles

3.2.1 72° twist angle

As an example, consider a case where the stepwise twist angle is $\Delta \phi = {360^o}/5 = {72^o}$ and the layer thickness is $\Delta d = P/5$. From Eq. (1) it can be obtained that the periodicity of the right-handed component is ${P_R} = \frac{{{{360}^o}}}{{{{72}^o}}}\frac{P}{5} = P$. The periodicity of the liquid crystal director is P and the corresponding wavevector is ${q_o} = 2\pi /P$. From Eq. (2) it can be obtained that the periodicity of the left-handed component is ${P_L} = \frac{{{{360}^o}}}{{{{180}^o} - {{72}^o}}}\frac{P}{5} = \frac{{2P}}{3}$. The numerically calculated Fourier component of the dielectric tensor is shown in Fig. 2. In the wavevector region from 0 to $9{q_o}$, Fourier component of the dielectric tensor has 4 peaks. There are more peaks beyond $9{q_o}$ due to multiple reflections. For the purpose of simplicity, in this paper we only consider Fourier components up to $9{q_o}$.

 

Fig. 2. Fourier components of the dielectric tensor of the stepwise twisted liquid crystal with the step twist angle of 72° and layer thickness of 0.2P.

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It is well-known that the reflection of stratified anisotropic Media can be numerically calculated by using Berreman 4 × 4 matrix method. [45–49] We used this method to numerically calculate the reflection of the stepwise twisted liquid crystals. In our calculation, the following parameters are used: The period P is 400 nm. The ordinary refractive index is 1.5 and the extraordinary refractive index is 1.7. The total thickness of the liquid crystal film is 20 periods. The liquid crystal is sandwiched between two parallel glass plates with a refractive index of 1.5. The product of average refractive index and period is $\bar{n}P = [(1.7 + 1.5)/2] \times 400\textrm{nm} = \textrm{640}\;\textrm{nm}$. The helical wavevector ${q_o} = 2\pi /P = (2\pi /400)\;\textrm{n}{\textrm{m}^{\textrm{ - 1}}}$. In the analysis of Fourier component of the dielectric tensor, the maximum wavevector considered is $9{q_o}$, which corresponds to the wavelength $0.21\bar{n}P$. The reflection spectrum in the wavelength region from $1.2\bar{n}P$ to $0.2\bar{n}P$ is calculated.

When the stepwise twist angle is $\Delta \phi = {360^o}/5 = {72^o}$, corresponding to the layer thickness $\Delta d = P/5$, the numerically calculated reflection spectra under various polarizations of the stepwise twisted liquid crystal with the stepwise twist angle of 72° are shown in Fig. 3. When the incident light is right-handed circular polarized and the right-handed circular polarized component of the reflected light is detected, the result is shown in Fig. 3(a). It has two reflection peaks. The first reflection peak is at the wavelength ${\lambda _1} = \bar{n}P = {\lambda _c}$ . The second peak is at the wavelength ${\lambda _2} = 2\bar{n}P/7 = 2{\lambda _c}/7$. When the incident light is left-handed circular polarized and the left-handed circular polarized component of the reflected light is detected, the result is shown in Fig. 3(b). It also has two reflection peaks. The first reflection peak is at the wavelength ${\lambda _1} = 2\bar{n}P/3 = 2{\lambda _c}/3$ . The second peak is at the wavelength ${\lambda _2} = \bar{n}P/4 = {\lambda _c}/4$. The wavelength and polarization state of the reflected light all agree with those obtained in the Fourier analysis. The peak reflection is slightly less than 1 mainly due to the reflection from the reflection from the air-glass interface. When a right-handed circular polarized incident is reflected from the interface, its polarization is changed to left-handed circular polarization. If the incident light is linearly polarized, there will be a reflection peak $0.4{\lambda _c}$, corresponding to the wavevector ${q_{layer}} = 5{q_o}$. This reflection is produced by the constructive interference of the light reflected from the interfaces between the layers (more discussion on this reflection will be presented later in this paper).

 

Fig. 3. Reflection spectrum of the stepwise twisted liquid crystal with the step twist angle of 72° and layer thickness of 0.2P.

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3.2.1 90° twist angle

Next we consider the case where the stepwise twist angle is $\Delta \phi = {360^o}/4 = {90^o}$ and the layer thickness is $\Delta d = P/4$. The structure and optical properties of this twisted liquid crystal are similar, but not exactly the same, to a dielectric mirror. A dielectric mirror consists of two alternating layers of materials of different refractive indices. Both layers are isotropic. While for the stack of LC layers, each layer is the same and anisotropic, and only the uniaxial axis is rotated. From Eq. (1) it can be obtained that the periodicity of the right-handed component is ${P_R} = \frac{{{{360}^o}}}{{{{90}^o}}}\frac{P}{4} = P$. From Eq. (2) it can be obtained that the periodicity of the left-handed component is ${P_L} = \frac{{{{360}^o}}}{{{{180}^o} - {{90}^o}}}\frac{P}{4} = P$, the same as that of the right-handed component. The calculated Fourier components of the dielectric tensor are shown in Fig. 4. The Fourier component has two peaks. The first peak is at $2{q_o}$, where ${\varepsilon _{11}}(q = 2{q_o}) = 0.318\Delta \varepsilon$ and ${\varepsilon _{12}}(q = 2{q_o}) = 0$. The Fourier component of the dielectric tensor is $\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\leftrightarrow$}} \over \varepsilon } (q = 2{q_o}) = 0.318\Delta \varepsilon \left( {\begin{array}{cc} 1&0\\ 0&{ - 1} \end{array}} \right)$, which can be decomposed into $0.159\Delta \varepsilon \left( {\begin{array}{cc} 1&{ - i}\\ { - i}&{ - 1} \end{array}} \right) + 0.159\Delta \varepsilon \left( {\begin{array}{cc} 1&i\\ i&{ - 1} \end{array}} \right)$. Therefore this peak can be considered as a superposition of right-handed and left-handed components of the same wavevector. The periodicities of the liquid crystal director $\vec{n}$ in the right-handed and left-handed helices are both P. The period of both components of the dielectric tensor is $P/2$, and thus the corresponding wavevector is $\frac{{2\pi }}{{P/2}} = 2\frac{{2\pi }}{P} = 2{q_o}$, agreeing with the wavevector value of this peak. This Fourier component will result in a reflection of both right-handed and left-handed circularly polarized light at the same wavelength of $4\pi \bar{n}/2{q_o} = 4\pi \bar{n}/[2(2\pi /P)] = \bar{n}P$. The second peak is at $6{q_o}$, where ${\varepsilon _{11}}(q = 6{q_o}) ={-} 0.106\Delta \varepsilon$ and ${\varepsilon _{12}}(q = 6{q_o}) = 0$. The Fourier component of the dielectric tensor is $\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\leftrightarrow$}} \over \varepsilon } (q = 6{q_o}) ={-} 0.106\Delta \varepsilon \left( {\begin{array}{cc} 1&0\\ 0&{ - 1} \end{array}} \right)$, which can be decomposed into $- 0.053\Delta \varepsilon \left( {\begin{array}{cc} 1&{ - i}\\ { - i}&{ - 1} \end{array}} \right) - 0.053\Delta \varepsilon \left( {\begin{array}{cc} 1&i\\ i&{ - 1} \end{array}} \right)$. Therefore this peak can also be considered as a superposition of right-handed and left-handed components of the same wavevector. This peak will result in a reflection of both right-handed and left-handed circularly polarized light at the same wavelength of $4\pi \bar{n}/6{q_o} = 4\pi \bar{n}/[6(2\pi /P)] = \bar{n}P/3$. Here the wavevector ${q_{layer}}$ for the wavelength equaling the layer thickness is $4{q_o}$. The wavevector for the 2^nd reflection peak is the sum of the wavevector for the 1^st reflection peak and ${q_{layer}}$

 

Fig. 4. Fourier components of the dielectric tensor of the stepwise twisted liquid crystal with the step twist angle of 90° and layer thickness of 0.25P.

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The numerically calculated reflection spectra under various polarizations of the stepwise twisted liquid crystal with the stepwise twist angle of 90° are shown in Fig. 5. When the incident light is right-handed circular polarized and the right-handed circular polarized component of the reflected light is detected, the result is shown in Fig. 5(a). It has two reflection peaks. The first reflection peak is at the wavelength ${\lambda _1} = \bar{n}P = {\lambda _c}$ . The second peak is at the wavelength ${\lambda _2} = \bar{n}P/3 = {\lambda _c}/3$. When the incident light is left-handed circular polarized and the left-handed circular polarized component of the reflected light is detected, the result is shown in Fig. 5(b). It also two reflection peaks. The first reflection peak is at the wavelength ${\lambda _1} = \bar{n}P$ . The second peak is at the wavelength ${\lambda _2} = \bar{n}P/3 = {\lambda _c}/3$. The wavelength and polarization state of the reflected light all agree with those obtained in the Fourier analysis. This stepwise twisted liquid crystal can be used to achieve high reflection of unpolarized incident light. For an unpolarized incident light with the wavelength equal to $\bar{n}P$, it can be decomposed into a right-handed circular polarized component and a left-handed circular polarized component. Both components will be reflected, and thus reflectance of 100% is achieved. When the incident light is unpolarized and reflected light of any polarization is measured, the reflection spectrum is shown in Fig. 5(c). Indeed the reflectance at the wavelength of $\bar{n}P$ is almost 100%. Here all the reflected light is measured, independent of polarization. The light reflected from the glass-air interface is included, and therefore the reflectance is nearly 100%. Note that there is additional peak at the wavelength $0.5{\lambda _c}$, which is caused by the constructive interference of the light reflected from the interfaces between the layers. [50–52] This reflection is not related to helical Fourier components of the dielectric tensor, and thus the reflected light is not circularly polarized,

 

Fig. 5. Reflection spectrum of the stepwise twisted liquid crystal with the step twist angle of 90° and layer thickness of 0.25P.

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3.2.3 85.5° twist angle

As mentioned in the introduction section, the reflection band exhibited by a cholesteric liquid crystal is at the wavelength ${\lambda _c} = [({n_e} + {n_o})/2]P$ and has the bandwidth $\Delta \lambda = ({n_e} - {n_o})P$. If $P = 400\;\textrm{nm}$ and $\Delta n = 0.2$, the bandwidth will be 80 nm, which is not wide enough for some applications. The stepwide twisted liquid crystal can be used to enlarge the bandwidth. The stepwise twist angle can be chosen in the following way. According Eqs. (1) and (2), the first reflection peak of the right-handed helical component is at $\lambda _{R/1} = \bar{n}P_R = \bar{n}\displaystyle{{{360}^o} \over {\Delta \phi }}\left( {\displaystyle{{\Delta \phi } \over {{360}^o}}P} \right) = \bar{n}P$, and the first reflection peak of the left-handed helical component is at ${\lambda _{L/1}} = \bar{n}{P_L} = \bar{n}\frac{{{{360}^o}}}{{{{180}^o} - \Delta \phi }}\left( {\frac{{\Delta \phi P}}{{{{360}^o}}}} \right) = \frac{{\Delta \phi (\bar{n}P)}}{{{{180}^o} - \Delta \phi }}$. If ${\lambda _{R/1}} - {\lambda _{L/1}} = \frac{1}{2}({n_e} - {n_o}){P_R} + \frac{1}{2}({n_e} - {n_o}){P_L}$, the two reflection peaks will be adjacent to each other to result in a doubled reflection bandwidth. Therefore the equation to determine $\Delta \phi$ is

For example, ${n_e} = 1.7$ and ${n_o} = 1.5$, the stepwise twist angle $\Delta \phi$ should be 85°. In order to assure the two reflection peaks are close enough, the stepwise twist angle $\Delta \phi$ is chosen to be 85.5°. Note that although 85.5° does not commensurate with 360°, the analysis is still correct (detailed consideration is presented later). The layer thickness is $\Delta d = \frac{{{{85.5}^o}}}{{{{360}^o}}}P = 0.2375P$. The periodicity of the right-handed component is ${P_R} = P$. From Eq. (2) it can be obtained that the periodicity of the left-handed component is ${P_L} = \frac{{{{360}^o}}}{{{{180}^o} - {{85.5}^o}}}0.2375P = 0.9047P$. The calculated Fourier components of the dielectric tensor are shown in Fig. 6. The first peak is at $2{q_o}$, where ${\varepsilon _{11}}(q = 2{q_o}) = 0.173\Delta \varepsilon$ and ${\varepsilon _{12}}(q = 2{q_o}) ={-} (0.173\Delta \varepsilon )i$. The Fourier component of the dielectric tensor is $\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\leftrightarrow$}} \over \varepsilon } (q = 2{q_o}) = 0.173\Delta \varepsilon \left( {\begin{array}{cc} 1&{ - i}\\ { - i}&{ - 1} \end{array}} \right)$, Therefore this peak is a right-handed helical component of the dielectric tensor. This Fourier component will result in a reflection of right-handed circularly polarized light at the wavelength of $4\pi \bar{n}/2{q_o} = 4\pi \bar{n}/[2(2\pi /P)] = \bar{n}P$. The second peak is at $2.210{q_o}$, which equals to $2\pi /({P_L}/2)$, where ${\varepsilon _{11}}(q = 2.210{q_o}) = 0.154\Delta \varepsilon$ and ${\varepsilon _{12}}(q = 2.210{q_o}) = (0.145\Delta \varepsilon )i$. The slight difference between the amplitudes of ${\varepsilon _{11}}$ and ${\varepsilon _{12}}$ is probably caused by numerical error. The Fourier component of the dielectric tensor is $\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\leftrightarrow$}} \over \varepsilon } (q = 2.210{q_o}) = 0.154\Delta \varepsilon \left( {\begin{array}{cc} 1&i\\ i&{ - 1} \end{array}} \right)$. Therefore this is a left-handed helical component of the dielectric tensor. This Fourier component will result in a reflection of left-handed circularly polarized light at the wavelength of $4\pi \bar{n}/2.210{q_o} = 4\pi \bar{n}/[2.210(2\pi /P)] = 0.905\bar{n}P$. The third peak is at $6.21{q_o}$, where ${\varepsilon _{11}}(q = 6.28{q_o}) ={-} 0.054\Delta \varepsilon$ and ${\varepsilon _{12}}(q = 6.21{q_o}) = (0.051\Delta \varepsilon )i$. The Fourier component of the dielectric tensor is $\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\leftrightarrow$}} \over \varepsilon } (q = 6.21{q_o}) ={-} 0.054\Delta \varepsilon \left( {\begin{array}{cc} 1&{ - i}\\ { - i}&{ - 1} \end{array}} \right)$. Therefore this peak is generated by the right-handed helical component of the dielectric tensor, which will also result in a reflection of right-handed circularly polarized light at the wavelength of $4\pi \bar{n}/6.21{q_o} = 4\pi \bar{n}/[6.21(2\pi /P)] = 0.322\bar{n}P$. The wavevector ${q_{layer}}$ for the wavelength equaling the layer thickness is $4.21{q_o}$. The wavevector for the 3^rd reflection peak is the sum of the wavevector for the 1^st reflection peak and ${q_{layer}}$. The fourth peak is at $6.42{q_o}$, where ${\varepsilon _{11}}(q = 6.42{q_o}) ={-} 0.051\Delta \varepsilon$ and ${\varepsilon _{12}}(q = 6.42{q_o}) = ( - 0.051\Delta \varepsilon )i$. The Fourier component of the dielectric tensor is $\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\leftrightarrow$}} \over \varepsilon } (q = 6.42{q_o}) ={-} 0.051\Delta \varepsilon \left( {\begin{array}{cc} 1&i\\ i&{ - 1} \end{array}} \right)$. Therefore this peak corresponds to the left-handed helical component of the dielectric tensor, which will also result in a reflection of left-handed circularly polarized light at the wavelength of $4\pi \bar{n}/6.42{q_o} = 4\pi \bar{n}/[6.42(2\pi /P)] = 0.311\bar{n}P)$. The wavevector for the 3^rd reflection peak is the sum of the wavevector for the 2^nd reflection peak and ${q_{layer}}$.

 

Fig. 6. Fourier components of the dielectric tensor of the stepwise twisted liquid crystal with the step twist angle of 85.5° and layer thickness of 0.2375P.

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The numerically calculated reflection spectra under various polarizations of the stepwise twisted liquid crystal with the stepwise twist angle of 85.5° are shown in Fig. 7. When the incident light is right-handed circular polarized and the right-handed circular polarized component of the reflected light is detected, the result is shown in Fig. 7(a). There are two reflection peaks. The first reflection peak is at the wavelength ${\lambda _1} = \bar{n}P = {\lambda _c}$ . The second peak is at the wavelength ${\lambda _2} = 2\bar{n}P/6.21 = 0.322{\lambda _c}$. When the incident light is left-handed circular polarized and the left-handed circular polarized component of the reflected light is detected, the result is shown in Fig. 7(b). It has multiple reflection peaks. The first reflection peak is at the wavelength ${\lambda _1} = 2\bar{n}P/2.21 = 0.905{\lambda _c}$. The peak reflectance is higher than 50% but less than 60%. For an unpolarized incident light, 8% light is reflected from the glass-air interface. The rest of the incident light can be decomposed into two components: one with right-handed circular polarization and another one with left-handed circular polarization. In the high wavelength half of the reflection peak, only the component of the right-handed circular polarization is reflected. Therefore the reflectance is higher than 50% but less than 60%. The same argument can made on the low wavelength half of the reflection peak, where only the component of the left-handed circular polarization is reflected. If the reflectance of 100% of unpolarized incident light in broadband is required, it can be achieved by stacking two stepwise twisted stratified film: one film with the twist angle of 85.5° and the other film with the twist angle of 94.5°. The second peak is at the wavelength ${\lambda _2} = \bar{n}P/6.42 = 0.311{\lambda _c}$. The wavelength and polarization state of the reflected light all agree with those obtained in the Fourier analysis. When the incident light is unpolarized and reflected light of any polarization is measured, the reflection spectrum is shown in Fig. 7(c), where the reflection bandwidth is doubled. Note that there is additional peak at the wavelength of $0.475{\lambda _c}$, which corresponds to the wavevector of the ${q_{layer}} = 4.21{q_o}$. It is produced by the constructive interference of the light reflected from the interfaces between the layers, as discussed before.

 

Fig. 7. Reflection spectrum of the stepwise twisted liquid crystal with the step twist angle of 85.5° and layer thickness of 0.233P.

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Although the stepwise twist angle $\Delta \phi$ of 85.5° does not commensurate with 360°, the LC film still exhibits strong reflection, which can be qualitatively explained below. Light is reflected from the interface between two neighboring LC layers, where the refractive index changes discontinuously. For an incident with the wavelength $\lambda = \bar{n}P$, the phase difference of the light rays reflected from between two adjacent interfaces is

If the two light rays reflected from between two adjacent interfaces is exactly in phase, the electric field of light reflected from 21 LC layers is proportional to ${E_o} = \sum\limits_{k = 0}^{20} {{e^{i({{360}^o})k}}} = 21$. Therefore the light reflected by one period is still significant. The LC film thickness is $20P = 20\frac{{{{360}^o}}}{{{{85.5}^o}}}\Delta d = 84.2\Delta d$, and thus it has 4 periods. The light rays reflected by the 4 periods are in phase and interference constructively. Therefore the reflection of the LC film is still strong.

4. Conclusion

We studied the reflection of stepwise twisted liquid crystals. In reality, stepwise twisted liquid crystals do not exist, except cholesteric liquid crystals where the stepwise twist angle is small. However, the presented analysis and results are valid for stepwise twisted stratified optical media, which can be fabricated by stacking stretched polymer films. The techniques to fabricate the stepwise twisted stratified optical films already exist. For example 3M has been using the technique to produce DBEF [5], where the stepwise twist angle is 90°. By using Fourier transform of the dielectric tensor, we showed that right-handed helix and left-handed helix can coexist in stepwise twisted media. We also used the Berreman 4 × 4 matrix method to simulate the reflection spectrum of stepwise twisted media. The Fourier transform analysis successfully explained the multiple reflection peaks and the circular polarization of the reflected light. At this moment we are not able to compare our theoretical results with experimental data, because a lack of experimental data. However, our results agree with other theoretical predictions obtained by solving Maxwell equation. [42,52] In the future, the theoretical approach will be used to study stepwise twisted stratified films with non-constant twist angles. The theoretical predictions will be compared experimental results. The results reported here should be very useful in the development of optical films with superior optical properties. By using stepwise twist angle equal to 90°, high reflectance near 100% can be achieved for unpolarized incident light. By using stepwise twist angle near 85°, broad reflection bandwidth can be accomplished. Furthermore, by utilizing high order helical Fourier components, reflections at wavelengthes much shorter than the stratified layer thickness can be realized. The stepwise twisted stratified optical films can be used in flat panel displays for light efficiency enhancement, in optical fiber telecommunication for wavelength selection and in laser hazard protection of human eye and instruments.