专

Organic solvent type coating compsn. used as primer for plastics - contg. epoxide adduct of graft polymer of olefin! polymer and maleic acid or anhydride.

126200803982A

发明人

MATOBA T, IWASE O, NEZU T, SUGIURA S H, SUGIURA S

受让人

KANSAI PAINT CO LTD (KAPA-C)

申请人

LEE JIN SEOK

申请号

1020197009089

申请日

2020-06-02

公开(公告)号

126200803982A

公开(公告)日

2003-09-12

IPC分类号

C07D501/28C07D499/64C12D009/14C07D499/46C07B029/02C07D501/60C12D013/00C12D009/04A61K031/395C07D501/20C12D001/02C12D009/06C12D009/00A61K031/43A61K031/545C07D499/44C12P035/04C12P037/04

CPC分类号

优先权号

0350231

优先权日

2003-06-17

摘要

NOVELTY - A ferritic heat-resistant steel (1) comprises a base material (2), and an oxide layer (A) provided on the surface of the base material. The base material comprises a chemical composition containing 0.01-0.3 %mass carbon, 7.0-14.0 %mass chromium, 0.005-0.15 %mass nitrogen, 0.001-0.3 %mass soluble aluminum. The oxide layer (A) comprises a chemical composition comprising 0-5.0 %mass copper, 0-5.0 %mass nickel, 0-0.1 %mass calcium, 0-0.1 %mass magnesium, 0-0.1 %mass zirconium, 0-0.1 %mass boron, 20-45 %mass chromium and manganese, and 0-0.1 %mass rare earth elements, and remaining iron and impurities.

USE - Ferritic heat-resistant steel used as ferrite heat transfer component (claimed) which is used for thermal power generation boilers, turbines, and steam pipes.

ADVANTAGE - The ferritic heat-resistant steel has excellent heat transfer characteristics and steam oxidation resistance.

DETAILED DESCRIPTION - A ferritic heat-resistant steel comprises a base material, and an oxide layer (A) provided on the surface of the base material. The base material comprises a chemical composition comprising 0.01-0.3 %mass carbon, 0.01-2.0 %mass silicon, 0.01-2.0 %mass manganese, 0.10 %mass phosphorus, 0.03 %mass sulfur, 7.0-14.0 %mass chromium, 0.005-0.15 %mass nitrogen, 0.001-0.3 %mass aluminum, and further comprises 0.5-7.0 %mass elements selected from 0-5.0 %mass molybdenum, 0-5.0 %mass tantalum, 0-5.0 %mass tungsten, and 0-5.0 %mass rhenium. The oxide layer (A) comprises a chemical composition comprises 0-5.0 %mass copper, 0-5.0 %mass nickel, 0-5.0 %mass cobalt, 0-1.0 %mass titanium, 0-1.0 %mass vanadium, 0-1.0 %mass niobium, 0-1.0 %mass hafnium, 0-0.1 %mass calcium, 0-0.1 %mass magnesium, 0-0.1 %mass zirconium, 0-0.1 %mass boron, 20-45 %mass chromium and manganese, 0.5-10 %mass elements selected from molybdenum, tantalum, tungsten and rhenium, and 0-0.1 %mass rare earth elements, and remaining iron and impurities. An INDEPENDENT CLAIM is included for ferrite heat transfer component, which comprises the base material, an oxide layer (B) arranged on the base material, an oxide layer (C) arranged between the oxide layer (B) and the substrate. The oxide layer (B) comprises 80 vol.% iron (II,III) oxide and iron (III) oxide. The oxide layer (C) comprises 5-30 mass% chromium and manganese, 1-15 %mass elements selected from molybdenum, tantalum, tungsten and rhenium.

DESCRIPTION OF DRAWING(S) - The drawing shows a schematic view of ferritic heat-resistant steel.

Ferritic heat-resistant steel (1)

Base material (2)

Oxide layer (A)

权利要求

1 . A fuel filter element, comprising a filter element main body and a protective cover, a fuel return column extending from an upper end surface of the filter element main body, the protective cover being formed with a U-shaped groove adapted to the fuel return column, the protective cover being adaptively installed on the filter element main body, the protective cover being formed with an annular hollow groove, and the U-shaped groove being configured to accommodate the fuel return column to form a protective assembly.

2 . The fuel filter element according to claim 1 , wherein a top end of the fuel return column is adaptively provided with a plug.

3 . The fuel filter element according to claim 2 , wherein the plug is provided with a raised edge, and the raised edge is configured to guide the plug to smoothly enter a slide way during replacement and installation of the filter element.

4 . The fuel filter element according to claim 1 , wherein a height of the fuel return column is 20-60 mm.

5 . The fuel filter element according to claim 1 , wherein a movable gap is formed between the fuel return column and an edge of the U-shaped groove.

6 . The fuel filter element according to claim 1 , wherein the filter element main body comprises filter paper, a water filtering channel, a water-repellent filtering layer, and a central tube.

7 . The fuel filter element according to claim 1 , wherein the protective cover is in clamped connection with the filter element main body.

8 . A fuel filter element, comprising a filter element main body and a protective cover, a fuel return column extending on an upper end surface of the filter element main body, the protective cover being formed with an adaptive hole adapted to the fuel return column, the protective cover being adaptively installed on the filter element main body, the protective cover being formed with an annular hollow groove, and the adaptive hole being configured to accommodate the fuel return column to form a protective assembly.

9 . The fuel filter element according to claim 8 , wherein the adaptive hole is circular, U-shaped or special-shaped.

10 . The fuel filter element according to claim 9 , wherein the filter element main body comprises filter paper and a water filtering channel, the filter paper communicates with the water filtering channel, the filter element main body further comprises a central tube and a water-repellent filtering layer, and the water-repellent filtering layer is arranged around the central tube; and the fuel return column is adaptively provided with a plug, the plug is provided with a raised edge, and the raised edge is configured to guide the plug to smoothly enter a slide way during replacement and installation of the filter element.

说明书

[0001]单圆锥面双折射率光纤

[0002]技术领域

[0003]本发明涉及一种单锥面双折射率光纤，其包括有一定长度要求的单面圆锥形光纤和与该单面圆锥形光纤连接在一起的普通光纤。

[0004]背景技术

[0005]早期生产的光纤其数值孔径过小，也就是集光能力不强。虽然普通的圆锥形光纤集光能力很强，但在远距离传输光信号时无法解决多模色散的问题，故而无法进行远距离传输光信号。

[0006]早期生产的液晶显示器由于液晶本身的特性，可视范围较小。

[0007]发明内容

[0008]本发明要解决的技术问题是提供一种既具有很强的集光能力，又能进行远距离传输光信号的光纤，并且这种光纤可增大液晶显示器的可视范围。

[0009]为解决上述技术问题，本发明采用了这样一种将圆锥形光纤和普通光纤结合起来的光纤，其包括一段圆锥形光纤和一个与该段圆锥形光纤连接起来的普通光纤，圆锥形光纤的纤芯的折射率不同于普通光纤的的纤芯的折射率。

[0010]通过控制圆锥形光纤的长度，可以避免该段圆锥形光纤与普通光纤连接处的端面对光纤集光能力的影响，从而使两种光纤的优点结合起来并且没有不利的影响。

[0011]通过控制圆锥形光纤的纤芯和包层的折射率以及普通光纤的纤芯和包层的折射率，可以避免由于该段圆锥形光纤与普通光纤连接处的端面两侧折射率不同引起的反射对于光纤集光能力的影响。

[0012]通过将本发明用于液晶显示器上，可增大液晶显示器的可视范围。

[0013]附图说明

[0014]下面结合附图和具体实施方式对本发明作进一步详细的说明：

[0015]图1为本发明光纤的纵向切面图。

[0016]图2、图3、图4、图5、图6、图7、图8为说明本发明光纤所须满足条件的光路图。

[0017]具体实施方式

[0018]图1所示单锥面双折射率光纤由圆锥形光纤1与普通光纤2组成。

[0019]图1所示角θmax为可在光纤中传播的光的最大入射角，光线3为以角θmax入射的光线，角θ0为光线3在圆锥形光纤1的光芯与包层界面的入射角。n1与n2分别为圆锥形光纤1的光芯与包层的折射率，n1′与n2′分别为普通光纤2的光芯与包层的折射率，D为单锥面双折射率光纤光芯的直径，L为圆锥形光纤1的圆柱形结构的长度。

[0020]为避免圆锥形光纤1与普通光纤2连接处的端面对光纤集光能力的影响，要求L≥Dtgθ0。

[0021]为避免由于该段圆锥形光纤1与普通光纤2连接处的端面两侧折射率不同引起的反射对于光纤集光能力的影响，要求(n12-n22)≤(n1′2-n2′2)，n1′＞n1，(n2′/n1′)≤(n2/n1)。

[0022]可传播的光线与水平方向的最大偏转角度为(θmax+α)且有

[0023]sin(θmax+α)＝cosα-(n2/n0)sin2α+2sinα{(n2/n0)sinα[1-(n2/n0)sinα]}1/2

[0024]对n1、n2、n0、α有如下限制条件

[0025]n0≥2n2时，sinα＝(n0n2/n12)-{[1-(n2/n1)2][1-(n0/n1)2]}2

[0026]在具体生产中，可由光纤的具体使用环境决定n0，再由光纤的常用材料决定n2′，n1′，n1。再确定实际所需的集光能力，既(θmax+α)的大小，由现有设备决定α的可能大小，从而得出θmax的可能大小，因为圆锥形光纤1的包层可为液体，从而n2可由下式

[0027]sinθmax下＝1-((2n2)/n0)sinα

[0028]决定，n2的改变可采用改变液体浓度的方法，当然也可采用其它方法。

[0029]在具体生产中，也可由已有的集光角度要求以及参数n1、n2来选择参数n2′、n1′。

[0030]确定本发明涉及的单锥面双折射率光纤的参数也可由实际情况来确定，不必严格按照上述过程来确定。

[0031]圆锥形光纤1与普通光纤2的连接可用熔接法，也可用其它方法。

[0032]圆锥形光纤1的锥面可用研磨法，也可用其它方法。

[0033]由光路可逆原理，将本发明涉及的单锥面双折射率光纤的平面端接于液晶显示器上，则液晶显示器的可视角度会增大。

[0034]在具体生产中，可将本发明涉及的单锥面双折射率光纤平行排列，制成平板形状，再密接于液晶显示器，或者与液晶显示器分离而成为可拆卸的装置。

[0035]对本发明单圆锥面双折射率光纤的参数限制条件可说明如下：

[0036]1.如图2所示，当光线由法线下方入射时：

[0037]п/2-α＝θ1+θ0

[0038]θ1＝п/2-(α+θ0)

[0039]n1sinθ1＝n1sin[п/2-(α+θ0)]＝n1cos(α+θ0)

[0040]n0sinθ＝n1sinθ1＝n1cos(α+θ0)＝n1(cosαcosθ0-sinαsinθ0)

[0041]若要光在光纤中传播：

[0042]θ0≥θ0临(θ0临为介质n1和n2之间的临界角)

[0043]则要

[0044]n0sinθ≤n1cos(α+θ0临

[0045]＝n1(cosαcosθ0临-sinαsinθ0临)

[0046]又sinθ0临＝n2/n1

[0047]则

[0048]n0sinθmax下＝(n12-n22)1/2cosα-n2sinα........................①

[0049]2.如图3所示，当光线由法线下方入射时：

[0050]п/2+α＝θ1+θ0

[0051]θ1＝п/2+α-θ0

[0052]n1sinθ1＝n1sin(п/2+α-θ0)

[0053]＝n1cos(θ0-α)

[0054]＝n1(cosθ0cosα+sinθ0sinα)

[0055]若要光在光纤中传播：

[0056]θ0≥θ0临

[0057]n0sinθ＝n1sinθ1≤n1cos(θ0临-α)

[0058]则

[0059]n0sinθmax上＝(n12-n22)1/2cosα+n2sinα........................②

[0060]由①、②，且令θmax上＝п/2可得：

[0061]n0＝(n12-n22)1/2cosα+n2sinα....................................③

[0062]sinθmax下＝1-(2n2/n0)sina.......................................④

[0063]sina＝(n0n2)/n12±{[1-(n2/n1)2][1-n0/n1]}1/2(详细过程见附录1).........⑤

[0064]既：

[0065]sinθmax下＝1-n22/n12(±(2n2/n0)±{[1-(n2/n1)2][1-n0/n1]}1/2)

[0066]仅对光纤做此改变，可增强光纤的集光能力。

[0067]如若制成图4形状，则还可减小多模色散。

[0068]为使两个端面之间无相互影响，既n1′与n2′的取值不影响集光能力。具体讨论见下。

[0069](1)当入射光的反射点在n1与n2的交界处，既如图4所示时，

[0070]θ0+＝п/2

[0071]β+η＝п/2

[0072]n1sin＝n1′sinβ

[0073]则可得

[0074]n12(1-sin2θ0)＝n1′2(1-sin2η)

[0075]为保证光的传播

[0076]sinη≥sinη0(η0为光线由n1′向n2′入射时的全反射临界角)

[0077]＝n2′/n1′

[0078]1-sin2η≤1-(n2′/n1′)2

[0079]则有

[0080]n12(1-sin2θ0)＝n1′2(1-sin2η)≤n1′2-n2′2

[0081]sin2θ0≥1-(n1′2-n2′2)/n12

[0082]又有

[0083]sinθ0≥n2/n1，既sin2θ0≥n22/n12

[0084]当

[0085](n22/n12)≥1-(n1′2-n2′2)/n12

[0086]既(n12-n22)≤(n1′2-n2′2)........................⑥

[0087]当上式满足时，图4所示的光可在光纤中传播。

[0088](2)当入射光的反射点在n1与n1′交界处时，

[0089]I.若既如图5所示时

[0090]n1′与n1之间的端面有可能影响光纤的集光能力

[0091]此时若有n1′＞n1............................⑦

[0092]则有

[0093]1.不发生全反射

[0094]2.β＜

[0095]由

[0096]可得η′＜η

[0097]光纤所能收集的最大角度光线如图5中点划线所示，明显

[0098]η′＞θ0

[0099]即η＞η′＞θ0

[0100]若n1与n2′构成的光纤的临界角η0临小于θ0临

[0101]即η0临≤θ0临

[0102]则可得(n2′/n1′)≤(n2/n1)..............................⑧

[0103]则有η＞η′＞θ0≥θ0临≥η0临

[0104]则光线必可在光纤中传播，

[0105]为保证n1′与n2′的取值不影响光纤的集光能力，则L1的长度必满足

[0106]L1≥(Dtgθ0)/2(D为光纤的直径)

[0107]II.当光线如图6所示时，n1′与n1之间的端面有可能会影响光纤的集光能力，

[0108]此时n1′＞n1，则

[0109]1.不发生全反射

[0110]2.β＜

[0111]可得η′＜η

[0112]光纤所能收集的最大角度光线如图6点划线所示，明显：

[0113]η′＞θ0

[0114]即η＞η′＞θ0

[0115]若n1与n2构成的光纤的临界角η0临小于θ0临

[0116]即η0临≤θ0临

[0117]则可得(n2′/n1′)≤(n2/n1)

[0118]则有η＞η′＞θ0≥θ0临≥η0临

[0119]光线必可在光纤中传播，

[0120]此时，

[0121]L≥Dtgθ0

[0122]则综合(1)、(2)两种情况，只要

[0123]n1′＞n1，(n2′/n1′)≤(n2/n1)，L≥Dtgθ0，且(n12-n22)≤(n1′2-n2′2)，

[0124]则n1′与n1之间的端面不影响光的传输。

[0125]附录1

[0126]n0＝(n12-n22)1/2cosα+n2sinα

[0127]n0-n2sinα＝(n12-n22)1/2cosα

[0128]n02-2n0n2sinα+n22sin2α＝(n12-n22)cos2α

[0129]＝(n12-n22)(1-sin2α)

[0130]n12sin2α-2n0n2sinα+n02+n22-n12＝0

[0131]Δ＝b2-4ac＝(-2n0n2)2-4n12(n02+n22-n12)

[0132]＝4n02n22-(4n12n02+4n12n22-4n14)

[0133]＝4[n02(n22-n12)+n12(n12-n22)]

[0134]＝4(n12-n22)(n12-n02)

[0135]注意：n1＞n2，n1＞n0

[0136]sinα＝(-b±Δ1/2)/(2a)

[0137]＝{2n0n2±2[(n12-n22)(n12-n02)]1/2}/(2n12)

[0138]＝(n0n2/n12)±{[1-(n2/n1)2][1-(n0/n1)2]}1/2…………………⑤

[0139]由于α≥0，当取负号时，要有

[0140]{n0n2-[(n12-n22)(n12-n02)]1/2}/n12≥0

[0141]则有

[0142]n02≥(n12-n22)...........................................................⑥

[0143]附录2

[0144]对n1、n2、n0、α的限制条件

[0145]1.sinθmax下＝1-((2n2)/n0)sinα..........................................④

[0146]0≤sinθmax下≤1

[0147]2.sinα＝((n0n2)/n12)±{[1-(n2/n1)2][1-(n0/n1)2]}1/2.....................⑤

[0148]0≤sinα≤1

[0149]3.(n12-n22)≤(n1′2-n2′2)...............................................⑥

[0150]4.n1′＞n1...............................................................⑦

[0151]5.(n2′/n1′)≤(n2/n1)...................................................⑧

[0152]则对于1.

[0153]0≤sinθmax下≤1

[0154]0≤1-((2n2)/n0)sinα≤1

[0155]-1≤-((2n2)/n0)sinα≤0

[0156]0≤sinα≤(n0/2n2)

[0157]则对于sinα

[0158](1)当(n0/2n2)＜1时

[0159]0≤sinα≤(n0/2n2)

[0160](2)当1≤(n0/2n2)时

[0161]0≤sinα≤1

[0162]对于(1)

[0163]①先取正号，既

[0164]sinα＝(n0n2/n12)+{[1-(n2/n1)2][1-(n0/n1)2]}1/2

[0165]则sinα≥0成立

[0166]由sinα≤(n0/2n2)

[0167][1-(n2/n1)2][1-(n0/n1)2]≤[(n0/2n2)-(n0n2/n12)]2

[0168]可得

[0169]4n24+n02n12-4n12n22≥0.............................................⑨

[0170]②再取负号，既

[0171]sinα＝(n0n2/n12)-{[1-(n2/n1)2][1-(n0/n1)2]}1/2

[0172]要求sinα≥0

[0173]则可得

[0174]0≥n12-n02-n22......................................................⑩

[0175]又要求sinα≤(n0/2n2)

[0176]既要求

[0177](n0n2/n12)-(n0/2n2)≤{[1-(n2/n1)2][1-(n0/n1)2]}2

[0178]当

[0179][(n0n2/n12)-(n0/2n2)]≤0，既2n22≤n12时

[0180]由于n1＞n2，n1＞n0，则等式成立。

[0181]当

[0182][(n0n2/n12)-(n0/2n2)]＞0，既2n22＞n12时

[0183]有

[0184]n02n12≤4n12n22-4n24

[0185]综合得

[0186]sinα＝(n0n2/n12)+{[1-(n2/n1)2][1-(n0/n1)2]}2时

[0187]要求n0＜2n2

[0188]4n24+n02n12-4n12n22≤0

[0189]sinα＝(n0n2/n12)-{[1-(n2/n1)2][1-(n0/n1)2]}2时

[0190]要求

[0191]或

[0192]对(2)

[0193]①sinα＝(n0n2/n12)+{[1-(n2/n1)2][1-(n0/n1)2]}2时

[0194]sinα≥0恒成立

[0195]由(恒成立)

[0196]②sinα＝(n0n2/n12)-{[1-(n2/n1)2][1-(n0/n1)2]}1/2时

[0197]由

[0198]由但当n0＝n2时，不满足1≤n0/(2n2)

[0199]则sinα≠(n0n2/n12)-{[1-(n2/n1)2][1-(n0/n1)2]}1/2

[0200]由①、②得

[0201]sinα＝(n0n2/n12)+{[1-(n2/n1)2][1-(n0/n1)2]}1/2时，要求n0≥2n2

[0202]sinα≠(n0n2/n12)-{[1-(n2/n1)2][1-(n0/n1)2]}1/2

[0203]则由影响因素1、2所确定的折射率之间的关系为

[0204]n0≥2n2时，sinα＝(n0n2/n12)-{[1-(n2/n1)2][1-(n0/n1)2]}2

[0205]附录3对集光能力极大值的讨论

[0206]集光角度＝θmax+α

[0207]sin(θmax+α)

[0208]＝sinθmaxcosα+cosθmaxsinα

[0209]＝[1-(2n2/n0)sinα]cosα+2sinα{(n2/n0)sinα[1-(n2/n0)sinα]}1/2

[0210]＝cosα-(n2/n0)sin2α+2sinα{(n2/n0)sinα[1-(n2/n0)sinα]}1/2

[0211]令0＝[sin(θmax+α)]/(α)

[0212]要求sinα≠0

[0213]则有

[0214](n2/n0)+4(n2/n0)3sin2α-[5(n2/n0)2+4(n2/n0)4sinα]+4(n2/n0)3＝0

[0215]Δ＝16(n2/n0)16-24(n2/n0)6+9(n2/n0)4

[0216]由Δ≥0

[0217]sinα＝{[5(n2/n0)2+4(n2/n0)4]±Δ1/2}/{2[(n2/n0)+4(n2/n0)3]}

[0218]一般n0＝1

[0219]则sinα＝[5n2+4n23±n2(16n212-24n22+9)]/(2+8n22)

[0220]取n2＝1.1

[0221]n1＝1.26

[0222]为判定此值为极大值还是极小值，

[0223]取sinα＝0.38，n2＝1.1，n1＝1.27

[0224]取sinα＝0.4，n2＝1.1，n1＝1.25

[0225]则所求得(θmax+α)的极值为极小值。

[0226]附录4锥面对光线的反射

[0227]当锥面把光线反射到n1与n2的界面上时，

[0228]既如图7所示，明显∠3大于∠4，则此时锥面对光线的反射可不用讨论。

[0229]当锥面把光线反射到n1与n1′的界面上时，

[0230]既如图8所示，可把锥面的反射光线平移至如虚线所示，则此种情况与入射光线为虚线的情况一样。则不用重复讨论。

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专

Organic solvent type coating compsn. used as primer for plastics - contg. epoxide adduct of graft polymer of olefin! polymer and maleic acid or anhydride.

126200803982A

发明人

MATOBA T, IWASE O, NEZU T, SUGIURA S H, SUGIURA S

受让人

KANSAI PAINT CO LTD (KAPA-C)

申请人

LEE JIN SEOK

申请号

1020197009089

申请日

2020-06-02

公开(公告)号

126200803982A

公开(公告)日

2003-09-12

IPC分类号

C07D501/28C07D499/64C12D009/14C07D499/46C07B029/02C07D501/60C12D013/00C12D009/04A61K031/395C07D501/20C12D001/02C12D009/06C12D009/00A61K031/43A61K031/545C07D499/44C12P035/04C12P037/04

CPC分类号

优先权号

0350231

优先权日

2003-06-17

摘要

USE - Ferritic heat-resistant steel used as ferrite heat transfer component (claimed) which is used for thermal power generation boilers, turbines, and steam pipes.

ADVANTAGE - The ferritic heat-resistant steel has excellent heat transfer characteristics and steam oxidation resistance.

DESCRIPTION OF DRAWING(S) - The drawing shows a schematic view of ferritic heat-resistant steel.

Ferritic heat-resistant steel (1)

Base material (2)

Oxide layer (A)

权利要求

2 . The fuel filter element according to claim 1 , wherein a top end of the fuel return column is adaptively provided with a plug.

4 . The fuel filter element according to claim 1 , wherein a height of the fuel return column is 20-60 mm.

5 . The fuel filter element according to claim 1 , wherein a movable gap is formed between the fuel return column and an edge of the U-shaped groove.

6 . The fuel filter element according to claim 1 , wherein the filter element main body comprises filter paper, a water filtering channel, a water-repellent filtering layer, and a central tube.

7 . The fuel filter element according to claim 1 , wherein the protective cover is in clamped connection with the filter element main body.

9 . The fuel filter element according to claim 8 , wherein the adaptive hole is circular, U-shaped or special-shaped.

说明书

[0001]单圆锥面双折射率光纤

[0002]技术领域

[0003]本发明涉及一种单锥面双折射率光纤，其包括有一定长度要求的单面圆锥形光纤和与该单面圆锥形光纤连接在一起的普通光纤。

[0004]背景技术

[0006]早期生产的液晶显示器由于液晶本身的特性，可视范围较小。

[0007]发明内容

[0008]本发明要解决的技术问题是提供一种既具有很强的集光能力，又能进行远距离传输光信号的光纤，并且这种光纤可增大液晶显示器的可视范围。

[0012]通过将本发明用于液晶显示器上，可增大液晶显示器的可视范围。

[0013]附图说明

[0014]下面结合附图和具体实施方式对本发明作进一步详细的说明：

[0015]图1为本发明光纤的纵向切面图。

[0016]图2、图3、图4、图5、图6、图7、图8为说明本发明光纤所须满足条件的光路图。

[0017]具体实施方式

[0018]图1所示单锥面双折射率光纤由圆锥形光纤1与普通光纤2组成。

[0020]为避免圆锥形光纤1与普通光纤2连接处的端面对光纤集光能力的影响，要求L≥Dtgθ0。

[0022]可传播的光线与水平方向的最大偏转角度为(θmax+α)且有

[0023]sin(θmax+α)＝cosα-(n2/n0)sin2α+2sinα{(n2/n0)sinα[1-(n2/n0)sinα]}1/2

[0024]对n1、n2、n0、α有如下限制条件

[0025]n0≥2n2时，sinα＝(n0n2/n12)-{[1-(n2/n1)2][1-(n0/n1)2]}2

[0027]sinθmax下＝1-((2n2)/n0)sinα

[0028]决定，n2的改变可采用改变液体浓度的方法，当然也可采用其它方法。

[0029]在具体生产中，也可由已有的集光角度要求以及参数n1、n2来选择参数n2′、n1′。

[0030]确定本发明涉及的单锥面双折射率光纤的参数也可由实际情况来确定，不必严格按照上述过程来确定。

[0031]圆锥形光纤1与普通光纤2的连接可用熔接法，也可用其它方法。

[0032]圆锥形光纤1的锥面可用研磨法，也可用其它方法。

[0033]由光路可逆原理，将本发明涉及的单锥面双折射率光纤的平面端接于液晶显示器上，则液晶显示器的可视角度会增大。

[0035]对本发明单圆锥面双折射率光纤的参数限制条件可说明如下：

[0036]1.如图2所示，当光线由法线下方入射时：

[0037]п/2-α＝θ1+θ0

[0038]θ1＝п/2-(α+θ0)

[0039]n1sinθ1＝n1sin[п/2-(α+θ0)]＝n1cos(α+θ0)

[0040]n0sinθ＝n1sinθ1＝n1cos(α+θ0)＝n1(cosαcosθ0-sinαsinθ0)

[0041]若要光在光纤中传播：

[0042]θ0≥θ0临(θ0临为介质n1和n2之间的临界角)

[0043]则要

[0044]n0sinθ≤n1cos(α+θ0临

[0045]＝n1(cosαcosθ0临-sinαsinθ0临)

[0046]又sinθ0临＝n2/n1

[0047]则

[0048]n0sinθmax下＝(n12-n22)1/2cosα-n2sinα........................①

[0049]2.如图3所示，当光线由法线下方入射时：

[0050]п/2+α＝θ1+θ0

[0051]θ1＝п/2+α-θ0

[0052]n1sinθ1＝n1sin(п/2+α-θ0)

[0053]＝n1cos(θ0-α)

[0054]＝n1(cosθ0cosα+sinθ0sinα)

[0055]若要光在光纤中传播：

[0056]θ0≥θ0临

[0057]n0sinθ＝n1sinθ1≤n1cos(θ0临-α)

[0058]则

[0059]n0sinθmax上＝(n12-n22)1/2cosα+n2sinα........................②

[0060]由①、②，且令θmax上＝п/2可得：

[0061]n0＝(n12-n22)1/2cosα+n2sinα....................................③

[0062]sinθmax下＝1-(2n2/n0)sina.......................................④

[0063]sina＝(n0n2)/n12±{[1-(n2/n1)2][1-n0/n1]}1/2(详细过程见附录1).........⑤

[0064]既：

[0065]sinθmax下＝1-n22/n12(±(2n2/n0)±{[1-(n2/n1)2][1-n0/n1]}1/2)

[0066]仅对光纤做此改变，可增强光纤的集光能力。

[0067]如若制成图4形状，则还可减小多模色散。

[0068]为使两个端面之间无相互影响，既n1′与n2′的取值不影响集光能力。具体讨论见下。

[0069](1)当入射光的反射点在n1与n2的交界处，既如图4所示时，

[0070]θ0+＝п/2

[0071]β+η＝п/2

[0072]n1sin＝n1′sinβ

[0073]则可得

[0074]n12(1-sin2θ0)＝n1′2(1-sin2η)

[0075]为保证光的传播

[0076]sinη≥sinη0(η0为光线由n1′向n2′入射时的全反射临界角)

[0077]＝n2′/n1′

[0078]1-sin2η≤1-(n2′/n1′)2

[0079]则有

[0080]n12(1-sin2θ0)＝n1′2(1-sin2η)≤n1′2-n2′2

[0081]sin2θ0≥1-(n1′2-n2′2)/n12

[0082]又有

[0083]sinθ0≥n2/n1，既sin2θ0≥n22/n12

[0084]当

[0085](n22/n12)≥1-(n1′2-n2′2)/n12

[0086]既(n12-n22)≤(n1′2-n2′2)........................⑥

[0087]当上式满足时，图4所示的光可在光纤中传播。

[0088](2)当入射光的反射点在n1与n1′交界处时，

[0089]I.若既如图5所示时

[0090]n1′与n1之间的端面有可能影响光纤的集光能力

[0091]此时若有n1′＞n1............................⑦

[0092]则有

[0093]1.不发生全反射

[0094]2.β＜

[0095]由

[0096]可得η′＜η

[0097]光纤所能收集的最大角度光线如图5中点划线所示，明显

[0098]η′＞θ0

[0099]即η＞η′＞θ0

[0100]若n1与n2′构成的光纤的临界角η0临小于θ0临

[0101]即η0临≤θ0临

[0102]则可得(n2′/n1′)≤(n2/n1)..............................⑧

[0103]则有η＞η′＞θ0≥θ0临≥η0临

[0104]则光线必可在光纤中传播，

[0105]为保证n1′与n2′的取值不影响光纤的集光能力，则L1的长度必满足

[0106]L1≥(Dtgθ0)/2(D为光纤的直径)

[0107]II.当光线如图6所示时，n1′与n1之间的端面有可能会影响光纤的集光能力，

[0108]此时n1′＞n1，则

[0109]1.不发生全反射

[0110]2.β＜

[0111]可得η′＜η

[0112]光纤所能收集的最大角度光线如图6点划线所示，明显：

[0113]η′＞θ0

[0114]即η＞η′＞θ0

[0115]若n1与n2构成的光纤的临界角η0临小于θ0临

[0116]即η0临≤θ0临

[0117]则可得(n2′/n1′)≤(n2/n1)

[0118]则有η＞η′＞θ0≥θ0临≥η0临

[0119]光线必可在光纤中传播，

[0120]此时，

[0121]L≥Dtgθ0

[0122]则综合(1)、(2)两种情况，只要

[0123]n1′＞n1，(n2′/n1′)≤(n2/n1)，L≥Dtgθ0，且(n12-n22)≤(n1′2-n2′2)，

[0124]则n1′与n1之间的端面不影响光的传输。

[0125]附录1

[0126]n0＝(n12-n22)1/2cosα+n2sinα

[0127]n0-n2sinα＝(n12-n22)1/2cosα

[0128]n02-2n0n2sinα+n22sin2α＝(n12-n22)cos2α

[0129]＝(n12-n22)(1-sin2α)

[0130]n12sin2α-2n0n2sinα+n02+n22-n12＝0

[0131]Δ＝b2-4ac＝(-2n0n2)2-4n12(n02+n22-n12)

[0132]＝4n02n22-(4n12n02+4n12n22-4n14)

[0133]＝4[n02(n22-n12)+n12(n12-n22)]

[0134]＝4(n12-n22)(n12-n02)

[0135]注意：n1＞n2，n1＞n0

[0136]sinα＝(-b±Δ1/2)/(2a)

[0137]＝{2n0n2±2[(n12-n22)(n12-n02)]1/2}/(2n12)

[0138]＝(n0n2/n12)±{[1-(n2/n1)2][1-(n0/n1)2]}1/2…………………⑤

[0139]由于α≥0，当取负号时，要有

[0140]{n0n2-[(n12-n22)(n12-n02)]1/2}/n12≥0

[0141]则有

[0142]n02≥(n12-n22)...........................................................⑥

[0143]附录2

[0144]对n1、n2、n0、α的限制条件

[0145]1.sinθmax下＝1-((2n2)/n0)sinα..........................................④

[0146]0≤sinθmax下≤1

[0147]2.sinα＝((n0n2)/n12)±{[1-(n2/n1)2][1-(n0/n1)2]}1/2.....................⑤

[0148]0≤sinα≤1

[0149]3.(n12-n22)≤(n1′2-n2′2)...............................................⑥

[0150]4.n1′＞n1...............................................................⑦

[0151]5.(n2′/n1′)≤(n2/n1)...................................................⑧

[0152]则对于1.

[0153]0≤sinθmax下≤1

[0154]0≤1-((2n2)/n0)sinα≤1

[0155]-1≤-((2n2)/n0)sinα≤0

[0156]0≤sinα≤(n0/2n2)

[0157]则对于sinα

[0158](1)当(n0/2n2)＜1时

[0159]0≤sinα≤(n0/2n2)

[0160](2)当1≤(n0/2n2)时

[0161]0≤sinα≤1

[0162]对于(1)

[0163]①先取正号，既

[0164]sinα＝(n0n2/n12)+{[1-(n2/n1)2][1-(n0/n1)2]}1/2

[0165]则sinα≥0成立

[0166]由sinα≤(n0/2n2)

[0167][1-(n2/n1)2][1-(n0/n1)2]≤[(n0/2n2)-(n0n2/n12)]2

[0168]可得

[0169]4n24+n02n12-4n12n22≥0.............................................⑨

[0170]②再取负号，既

[0171]sinα＝(n0n2/n12)-{[1-(n2/n1)2][1-(n0/n1)2]}1/2

[0172]要求sinα≥0

[0173]则可得

[0174]0≥n12-n02-n22......................................................⑩

[0175]又要求sinα≤(n0/2n2)

[0176]既要求

[0177](n0n2/n12)-(n0/2n2)≤{[1-(n2/n1)2][1-(n0/n1)2]}2

[0178]当

[0179][(n0n2/n12)-(n0/2n2)]≤0，既2n22≤n12时

[0180]由于n1＞n2，n1＞n0，则等式成立。

[0181]当

[0182][(n0n2/n12)-(n0/2n2)]＞0，既2n22＞n12时

[0183]有

[0184]n02n12≤4n12n22-4n24

[0185]综合得

[0186]sinα＝(n0n2/n12)+{[1-(n2/n1)2][1-(n0/n1)2]}2时

[0187]要求n0＜2n2

[0188]4n24+n02n12-4n12n22≤0

[0189]sinα＝(n0n2/n12)-{[1-(n2/n1)2][1-(n0/n1)2]}2时

[0190]要求

[0191]或

[0192]对(2)

[0193]①sinα＝(n0n2/n12)+{[1-(n2/n1)2][1-(n0/n1)2]}2时

[0194]sinα≥0恒成立

[0195]由(恒成立)

[0196]②sinα＝(n0n2/n12)-{[1-(n2/n1)2][1-(n0/n1)2]}1/2时

[0197]由

[0198]由但当n0＝n2时，不满足1≤n0/(2n2)

[0199]则sinα≠(n0n2/n12)-{[1-(n2/n1)2][1-(n0/n1)2]}1/2

[0200]由①、②得

[0201]sinα＝(n0n2/n12)+{[1-(n2/n1)2][1-(n0/n1)2]}1/2时，要求n0≥2n2

[0202]sinα≠(n0n2/n12)-{[1-(n2/n1)2][1-(n0/n1)2]}1/2

[0203]则由影响因素1、2所确定的折射率之间的关系为

[0204]n0≥2n2时，sinα＝(n0n2/n12)-{[1-(n2/n1)2][1-(n0/n1)2]}2

[0205]附录3对集光能力极大值的讨论

[0206]集光角度＝θmax+α

[0207]sin(θmax+α)

[0208]＝sinθmaxcosα+cosθmaxsinα

[0209]＝[1-(2n2/n0)sinα]cosα+2sinα{(n2/n0)sinα[1-(n2/n0)sinα]}1/2

[0210]＝cosα-(n2/n0)sin2α+2sinα{(n2/n0)sinα[1-(n2/n0)sinα]}1/2

[0211]令0＝[sin(θmax+α)]/(α)

[0212]要求sinα≠0

[0213]则有

[0214](n2/n0)+4(n2/n0)3sin2α-[5(n2/n0)2+4(n2/n0)4sinα]+4(n2/n0)3＝0

[0215]Δ＝16(n2/n0)16-24(n2/n0)6+9(n2/n0)4

[0216]由Δ≥0

[0217]sinα＝{[5(n2/n0)2+4(n2/n0)4]±Δ1/2}/{2[(n2/n0)+4(n2/n0)3]}

[0218]一般n0＝1

[0219]则sinα＝[5n2+4n23±n2(16n212-24n22+9)]/(2+8n22)

[0220]取n2＝1.1

[0221]n1＝1.26

[0222]为判定此值为极大值还是极小值，

[0223]取sinα＝0.38，n2＝1.1，n1＝1.27

[0224]取sinα＝0.4，n2＝1.1，n1＝1.25

[0225]则所求得(θmax+α)的极值为极小值。

[0226]附录4锥面对光线的反射

[0227]当锥面把光线反射到n1与n2的界面上时，

[0228]既如图7所示，明显∠3大于∠4，则此时锥面对光线的反射可不用讨论。

[0229]当锥面把光线反射到n1与n1′的界面上时，

[0230]既如图8所示，可把锥面的反射光线平移至如虚线所示，则此种情况与入射光线为虚线的情况一样。则不用重复讨论。

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暂无同族专利