Effect of width of light source on viewing angle of one-dimensional integral imaging display

Wu, Fei; Wu, Rui; Deng, Hui; Yu, Junsheng

doi:10.1016/j.ijleo.2017.11.191

Cited by 3 publications

(3 citation statements)

References 14 publications

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“…Suppose that g is the gap between the organic light‐emitting diode and the liquid crystal panel, p is the pitch of the rectangular light source, h is the horizontal width of the rectangular light source, n is the number of the elemental images in the horizontal direction, and l is the viewing distance. Thus, the horizontal width of the viewing zone in the proposed integral imaging display H is calculated as

H = \frac{italiclp + italiclh}{g} - ((), n - 2) p .

The horizontal viewing angle of the proposed integral imaging display θ is obtained as

θ = 2 \arctan ([], \frac{p + h}{2 g} - \frac{((), n - 2) p}{2 l}) .

The horizontal viewing angle of the conventional integral imaging display θ is shown as 17

θ^{'} = 2 \arctan ([], \frac{p - h}{2 g} - \frac{((), n - 2) p}{2 l}) .

The horizontal viewing angle of the proposed integral imaging display is proportional to the horizontal width of the rectangular light source.…”

Section: Theory and Analysismentioning

confidence: 99%

Wide‐viewing integral imaging display using rectangular light sources

Xiao¹,

Wu²

2020

J Soc Info Display

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We propose a wide‐viewing integral imaging display. It consists of an organic light‐emitting diode, a barrier array, and a liquid crystal panel. The organic light‐emitting diode generates the rectangular light sources, and the liquid crystal panel displays the elemental images. The barriers are introduced to separate adjacent rectangular light sources and corresponding elemental images. Lights emitting from the rectangular light sources only illuminate the corresponding elemental images. The horizontal viewing angle and the luminance are both enhanced by increasing the horizontal width of the point light source. A prototype of the proposed integral imaging display is developed, and the experiment results agree well with the theory.

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H = \frac{italiclp + italiclh}{g} - ((), n - 2) p .

The horizontal viewing angle of the proposed integral imaging display θ is obtained as

θ = 2 \arctan ([], \frac{p + h}{2 g} - \frac{((), n - 2) p}{2 l}) .

The horizontal viewing angle of the conventional integral imaging display θ is shown as 17

θ^{'} = 2 \arctan ([], \frac{p - h}{2 g} - \frac{((), n - 2) p}{2 l}) .

The horizontal viewing angle of the proposed integral imaging display is proportional to the horizontal width of the rectangular light source.…”

Section: Theory and Analysismentioning

confidence: 99%

Wide‐viewing integral imaging display using rectangular light sources

Xiao¹,

Wu²

2020

J Soc Info Display

Self Cite

View full text Add to dashboard Cite

show abstract

“…Luminance can easily be improved by increasing the width of the light source. However, the viewing angle accordingly decreases [ 21 ]. Multiplexing was used to generate a movable pinhole array [ 22 ].…”

Section: Introductionmentioning

confidence: 99%

“…Therefore, a one-dimensional integral imaging (1DII) display based on a line light source array was proposed. The luminance is enhanced by removing the vertical parallax [ 21 ]. However, the luminance and viewing angle are mutually restricted.…”

Section: Introductionmentioning

confidence: 99%