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1、 c I | :1002-2082(2011)01-0133-05j M h 邓键, 童卫红, 安晓强, 王茜( 2 / , + 610041)摘要: 研究了双视场红外变焦系统的光学被动消热差技术, 通过理论分析温度变化对变焦光学系统性能的影响, 推导出被动消热差公式。将热离焦和热色差作为两种附加的初级像差引入像差平衡关系式, 可通过常规光学系统消像差的4种途径: 分配光焦度、有效匹配光学材料、改变透镜形状和引入非球面实现全被动补偿。在此基础上建立了一个通用模型, 该系统有6片透镜、一个非球面, 在- 45+ 55温度范围内的2个视场均具有较好的像质、冷反射和较宽松的公差特性。在该模型中适当引
2、入衍射元件, 可减少镜片, 提高光学性能。关键词: 红外变焦光学系统; 光学补偿; 无热化; 光学设计 m s | : TN202 D S : AAthermalization of infrared zoom systemDENG Jian, TONG Wei-hong, AN Xiao-qiang, WANG Qian(Southernwest Institute of Technical Physics, Cheng Du 610041, China)Abstract: Passive athermalization technology of infrared zoom optical
3、 system is studied.Athermalization relationship was deduced and the condition expression for zoom systemathermalization was obtained. Defocusing and chromatic aberration caused by thermal changecan be taken as two kinds of primary aberrations, and can be eliminated with a passiveathermalization syst
4、em. There are four ways to implement athermalization in conventionalsystems, appropriatefocalpower arrangement, proper match of optical material, changeof lenssurfaceshape, anduseof asphericalelements. A dualfield ofview (FOV) infraredzoom opticalmodelwas built, which had 6lenses and one aspherical
5、surface. It achieved good image qualityin therange of - 45j+ 55in both FOV, with low Narcissus and less stringent tolerancerequirement. Diffraction surface can be easily introduced in this model, which reduces thenumber of lens used and improveoptical performance.Key words: infrared zoom optical sys
6、tem; optical compensate; athermalization; optical design v , M H | q | ? 3 v M ,F cn , t | Y ; d ? b !9 1 p P S = ; d T , 1 3 Z E :; $ T a T $ T 1b T $ T 1 Y V a . e # c d ; m M d Y , d ,O r b - 1 ; $ h / , Y V ; a c , M b Vl :2010-07-05; :2010-10-11T e : o(1976- ),3 , i , = ,pV , 1 V Y ; d ! 9 a _
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9、 d 9 ) 5 b v ; d ,h !lT e P K v o l s B o ,!lT d M :!lT2( f /# )2 (2) ; d M d ? M , n i F m B 1 3:m= n- 1, 1 P ; bm 1j ; dFig.1Dual FOV system m 1 U j d ,V s - % F aM F % F , ; ; 9 T , # M V l / 3:f (z)= f 1#1#2= f1f 2f 3- z2+ b1z+ b2b2= !12!23- f 22b1= !23- !12(3)T :f - % F ; f 2 M F ; f 3 % F ; f (z) M F M z ; #1 M F v bB t L
