(3.60) and (3.61) indicate
as shown in Figure 3.8, where all important angles for a Fréedericksz cell are drawn.
The result for a wavelength λ0 at the output z = d of a cell without a voltage applied is linearly polarized light at an angle β = π − а, where α is the angle of the incoming linearly polarized light. If we place the analyser in the direction β = π − α the light can pass representing the normally white mode. The analyser perpendicular to β that is at an angle π/2 − α in Figure 3.8 blocks the light representing the normally black mode. We will investigate these two modes in greater detail.
We choose the angle γ for which the x′−y′ plane in Figure 3.8 is rotated from the x−y plane as γ = β = π − α. This provides, along with Equations (3.40) and (3.41),
or for the electrical field
(3.65)
and
For the wavelength λ = λ0 in Equation (3.57), we obtain at z = d
(3.67)
(3.68)
as expected, since we know already that at z = d light with wavelength λ0 is linearly polarized in the direction β = π − α. For this case, the Jones vectors provide the components of the electrical field as
(3.69)
(3.70)
Now we place the analyser perpendicular to the angle β = π − α that is in the direction with angle γ = π/2 − α in Figure 3.8. For this case Jzx′ is identical to − Jzy′ in Equation (3.64) and (3.66) and Jzy′ is identical with Jzx′ in Equations (3.63) and (3.64). Hence, we investigate Equations (3.63) through (3.66) for both cases. The intensity I′x = |Jdx′|2 for z = d is, with Equation (3.63),
(3.71)
or
For
In order to learn how to choose a, we now consider the case of a large enough voltage across the LC cell to fully orient the LC molecules apart from two thin layers on top of the orientation layer, due to Δε > 0 in parallel to the electric field. The linearly polarized light coming in at angle a no longer experiences birefringence, as it is only exposed to the refractive index n┴. It reaches the plane z = d with the phase shift 2π(n┴d/λ). Its component Ep passing an analyser with the angle (π/2) − α in Figure 3.8 is
(3.74)
whereas the component Es passing an analyser with the angle π − α in Figure 3.8 is
(3.75)
The intensities belonging to Ep and Es are
(3.76)
and
(3.77)
The bar over the cos terms means the average over time needed for calculating the intensity. The maximum of Ip occurs for α = π/4, which (according to Figure 3.8) places the polarizer and analyser in parallel. Is assumes a maximum for α = 0, for which again the polarizer and analyser pertaining to Es are in parallel.
In Figures 3.9 and 3.10, the intensities Iy′ and Ix′ in Equations (3.73) and (3.72) are plotted versus x= (πdΔn/λ) and λ = (πdΔn/x). Iy′ in Figure