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Wide-view circular polarizer consisting of a linear polarizer and two biaxial films

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Abstract

A simple wide-view circular polarizer comprising of a linear polarizer and two biaxial films is proposed. Over the ±85° viewing cone, the produced polarization is almost circular and the light leakage from the crossed circular polarizers is calculated to be less than 8.23×10-5, provided that the air-interface surface reflections are ignored. The design tolerance within ±5% of the optimal parameters is analyzed.

©2005 Optical Society of America

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Figures (8)

Fig. 1.
Fig. 1. Configuration of a wide-view circular polarizer with a linear polarizer and two biaxial films.
Fig. 2.
Fig. 2. States of polarization inside a wide-view circular polarizer at oblique incidence θ in = 85°. Dotted lines and solid line show the polarization states when the azimuths of incident plane ϕ in are at 30° and 60°, respectively. Red and blue lines show the polarizations inside the first and second biaxial films, respectively.
Fig. 3.
Fig. 3. State of polarization emerging from a wide-view circular polarizer when θ in = 0° ~ 85° at each fixed ϕ in, where ϕ in = 0° ϕ 360° with 10° interval.
Fig. 4.
Fig. 4. Device configuration of the crossed wide-view circular polarizers.
Fig. 5.
Fig. 5. Iso-transmittance contour showing: (a) light leakage of the crossed wide-view circular polarizers, and (b) transmittance of two parallel circular polarizers. λ=550 nm.
Fig. 6.
Fig. 6. Ten-layer ideal anti-reflection film: (a) refractive indices profile, and (b) transmittance.
Fig. 7.
Fig. 7. The calculated maximum light leakage of the crossed circular polarizers at different viewing angles as a function of wavelength. The configuration of the crossed circular polarizers is in Fig. 1 and the ten-layer anti-reflection film in Fig. 6 is assumed.
Fig. 8.
Fig. 8. Design tolerance of the proposed wide-view circular polarizer. The viewing cone is ±85° and λ= 550 nm. Ten-layer anti-reflection film is assumed.

Equations (7)

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Δ P ( X ) ( RCP ) = P ( X ) P ( RCP )
= ( S 1 _ ( X ) 0 ) 2 + ( S 2 _ ( X ) 0 ) 2 + ( S 3 _ ( X ) ( 1 ) ) 2
= 2 ( S 3 _ ( X ) + 1 ) .
cos t = max { 2 ( S 3 _ ( 2 B ) + 1 ) ( θ in = 0 0 ~ 85 0 , ϕ in = 0 0 ~ 360 0 ) } ,
2 ϕ 1 4 ϕ 2 = 2 × ( 360 + 46.37 ) 4 × ( 180 + 0.68 )
= 90.02 0
90 0
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