Liquid Crystal Displays. Ernst Lueder
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(4.2)
a pitch p given by
and a twist angle
at z = d. For calculations, the helix is cut into slices parallel to the x–y plane. In each slice, all molecules are assumed to be parallel. The slices are rotated from the previous slice by the angle ε. The angle ε corresponds to a thickness dε with
or
for each slice. The twist angle at z = d is from Equation (4.4) with α0 in Equation (4.3)
(4.6)
whereas the number of slices in the cell is with Equation (4.5)
The Jones vector J1 at the input is translated into the vector O1 at the output of the first slice with thickness dε. Its component J1x is parallel to the x-axis and the component J1y is parallel to the y-axis, and hence (as already known from the Fréedericksz cell),without the need to rotate the input vector, we obtain
(4.8)
or
Figure 4.2 The propagation of light from the Jones vector J1 at the input to the Jones vector Os at the output through the transmission matrices Tv and the rotation matrices Rv
with
where
The further propagation of the light through the s − 1 remaining rotated slices is depicted in Figure 4.2 with the rotation matrices
and the transmission matrices
The Jones vector Os at the output at z = d measured in the coordinates σ and τ with the angle β to the x-axis in Figure 4.1 is, with Equations (4.1) and (4.9)
(4.12)
or with Equations (4.10), (4.11), dε in Equation (4.5), s in Equation (4.7) and
(4.13)
We want to determine the lim for ε → 0 of this expression, providing an infinitesimally small thickness of the slices, and hence the exact solution. From Equation (4.14), we obtain
For an easier calculation of the lim for ε → 0, we transform
into a diagonal matrix
where