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    [转载]2010年IODC的照明设计问题 [复制链接]

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    离线mang2004
     
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    只看楼主 倒序阅读 楼主  发表于: 2011-05-14
    去年国际光学设计的照明问题。 Y Xn)?  
    转载自: http://iodc.info/illumination-design-problem/39-problems/58-2010-illumination-design-problem c9f~^}jNb  
    有兴趣的同好可以看看。光行坛有人参加吗? O`|'2x{[O  
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    Problem description: :ykQ[d`:|  
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    Transfer maximum monochromatic flux from a 1-mm-square Lambertian source in air to an equal-etendue nonimmersed target. The target surface is rectangular with a 16:9 aspect ratio. The surface area of the target must be at least 4 mm&sup2;. The target is defined such that only rays incident on the target surface at angles of θmax or less, relative to the surface normal, are considered to be within the phase space of the target, where the value of θmax is determined by the equal-étendue requirement. ^*}L9Ot~  
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    Problem design degrees of freedom: Ibpk\a?A{  
    Can be any combination of idealized refractive and reflective components. H#wn3O  
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    Assumptions and constraints n%Nf\z  
    The only media allowed are air (index of 1) or dielectrics with refractive index in the range 1.33 - 1.59. D97oS!*  
    j: ]/AReOL  
    The coupling efficiency will be computed in the geometrical optics approximation using 100,000 pseudo-randomly generated rays. Optical losses produced by the following material characteristics will be included in the efficiency computation: yuKfhg7  
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    Mirror reflectivity = 95% at all angles. Isoqs(Oi  
    TIR reflectivity = 100% at all angles. F/@#yQv?  
    Bulk absorption loss for all dielectric materials = 0.5% per cm. Mgf80r=  
    Fresnel losses at air-dielectric interfaces = 2% at all angles. i"|'p/9@q  
    Fresnel losses between two different dielectric interfaces = 0.2% at all angles. uX`Jc:1q3  
    No Fresnel losses within a gradient index material. pm<<!`w"  
    Minimum size of a component and edge thickness = 0.1 mm. "L{;=-e  
    Minimum air space between components (including source) = 0.1 mm Ep?a1&b  
    No volume or surface scattering. `\O[9.B  
    Light that finds its way back into the source is fully absorbed. 8]2S'm xE  
     
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