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Transverse dynamics of nematicons

Marco Peccianti, Andrea Fratalocchi, and Gaetano Assanto

Open Access Open Access

Abstract

Optical anisotropy plays a fundamental role on light propagation in nematic liquid crystals. With specific reference to nematicons, we investigate the transverse dynamics due to the interplay of nonlinear self-confinement, birefringent walk-off and a bias-dependent transverse index profile.

©2004 Optical Society of America

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Figures (7)
Fig. 1.
Fig. 1. Sketch of the NLC cell and experimental geometry.

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Fig. 2.
Fig. 2. Calculated maximum (on-axis) walk-off versus cell bias

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Fig. 3.
Fig. 3. Bias induced index profile in a cell filled with E7 and an applied voltage V=1.48V, providing maximum walk-off.

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Fig. 4.
Fig. 4. Simulated propagation of a 3mW X-polarized gaussian beam launched in a biased cell (as in Fig. 3) with k-vector parallel to Z and a) no phase front tilt ; b) a 7° tilt in order to compensate walk-off on axis.

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Fig. 5.
Fig. 5. Nematicon transverse profile in the observation plane at ϕ=45° with respect to X. a) For V 0=1.0V the small walk-off (about 2°) mediates an oscillation of modest amplitude; b) at V 0=1.6V a larger walk-off (about 7°) corresponds to a shorter period with larger elongation across X. c) By launching the input beam with a phase front tilt in order to compensate the walk-off, the nematicon at V 0=1.6V can be generated with no motion across X

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Fig. 6.
Fig. 6. Soliton trajectories for P=3.2mW versus bias V 0. The scale Δx quantifies the deviation from input position X=0.

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Fig. 7.
Fig. 7. Calculated (solid line) and measured (dashed line with dots) periodicity Λ of the nematicon transverse oscillation versus applied bias V0.

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Equations (5)

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( K 1 cos 2 Θ + K 3 sin 2 Θ ) d 2 Θ d X 2 + K 3 − K 1 2 sin 2 ξ ( d Θ d X ) 2 + 1 2 ε a ( d V d X ) 2 sin 2 Θ = 0
( ε ∥ sin 2 Θ + ε ⊥ cos 2 Θ ) d 2 V d X 2 + ε a sin 2 Θ d Θ d X d V d X = 0
δ ( Θ ) = arctan ( Δ n 2 sin ( 2 Θ ) Δ n 2 + 2 n ⊥ 2 + Δ n 2 cos ( 2 Θ ) )
j 2 k 0 n ( Θ ) ∂ E ∂ Z = ∇ ⊥ 2 E + k 0 2 ( n 2 ( θ ) − n 2 ( Θ ) ) E + j 2 k 0 n ( Θ ) tan δ ( θ ) ∂ E ∂ X
K ∇ ⊥ θ + ε 0 ( 1 2 Δ ε a ∣ d V d X ∣ 2 + 1 4 Δ n 2 ∣ E ∣ 2 ) sin 2 θ = 0
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