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Polarization rotator using a hybrid aligned nematic liquid crystal cell

Fuzi Yang, Lizhen Ruan, S. A. Jewell, and J. R. Sambles

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Abstract

Using both direct mathematical analysis and numerical modeling based on the predictions by Jones [1] it is shown that if the director in a liquid crystal cell is in a plane which lies at 45° to the incident polarization, then, for normally incident light, the transmission signal which conserves polarization will always have a phase difference of π/2 from the transmission signal of the orthogonal polarization. This is independent of the director profile in the plane, the cell thickness, the anisotropy of the liquid crystal refractive index and the optical parameters of other isotropic layers in the cell. Based on this realization a hybrid aligned nematic liquid crystal cell has been tested as a thresholdless voltage-controlled polarization rotator. By using a quarter-wave plate to compensate for the phase difference between the two orthogonal output polarizations a simple liquid crystal spatial light modulator has been realized.

©2007 Optical Society of America

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Figures (6)
Fig. 1.
Fig. 1. The director in the lab-system, O-XYZ.

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Fig. 2.
Fig. 2. The model results for a HAN cell under an applied voltage. The liquid crystal layer has ne = 1.7400, no = 1.5200 and d = 6.00 μm. (a) The transmission intensities, Txx and Txy against voltage. (b) The transmission phases, Φx and Φy , against voltage.

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Fig. 3
Fig. 3 The model results of a HAN cell under an applied voltage. The transmission phases, Φx and Φy , against voltage when the liquid crystal layer has (a) ne = 1.7400, no = 1.5200 and d = 8.00 μm and (b) ne = 1.7400, no = 1.5800 and d = 6.00 μm.

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Fig. 4.
Fig. 4. The experimental set-up.

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Fig. 5.
Fig. 5. The polarization direction of the output beam against applied voltage for (a) a parallel aligned cell and (b) a HAN cell.

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Fig. 6.
Fig. 6. Sum of transmissivities against applied voltage for (a) a parallel cell and (b) a HAN cell.

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Equations (8)

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n e ′ = n e n o n e 2 cos 2 θ + n o 2 sin 2 θ
x 1 = 1 2 [ e i 2 π λ n e ′ ( 1 ) d 1 + e i 2 π λ n o d 1 ] = cos [ π λ ( n e ′ ( 1 ) − n o ) d 1 ] ∙ e i [ π λ ( n e ′ ( 1 ) + n o ) d 1 ] = A x 1 e iΦ x 1 ,
y 1 = 1 2 [ e i 2 π λ n e ′ ( 1 ) d 1 − e i 2 π λ n o d 1 ] = sin [ π λ ( n e ′ ( 1 ) − n o ) d 1 ] ∙ e i [ π λ ( n e ′ ( 1 ) + n o ) d 1 + π 2 ] = A y 1 e iΦ y 1
x 1 = A x 1 , y 1 = A y 1 e i π 2 and A y 1 = 1 − A x 1 2 . It follows that
x 2 = 1 2 [ ( x 1 + y 1 ) e i 2 π λ n e ′ ( 2 ) d 2 + ( x 1 − y 1 ) e i 2 π λ n o d 2 ] = A x 2 e iΦ x 2 ,
y 2 = 1 2 [ ( x 1 + y 1 ) e i 2 π λ n e ′ ( 2 ) d 2 − ( x 1 − y 1 ) e i 2 π λ n o d 2 = A y 2 e i Φ y 2 ] with A y 2 = 1 − A x 2 2 and
Φ x 2 = π λ [ n e ′ ( 2 ) + n o ] d 2 , Φ y 2 = π λ [ n e ′ ( 2 ) + n o ] d 2 + π 2 .
Φ = π λ n o [ d + n e ∫ o d dz n e 2 cos 2 θ ( z ) + n o 2 sin 2 θ ( z ) ]
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