[求助]请大家帮帮忙!色度学里的相对光功率谱怎么求?
偶要做一个光源的色度测试,手头有光谱仪等等,可以测到光功率谱,但是在数据处理计算其色度坐标时需要用到待测光源的 相对光功率谱 ,这个相对光功率谱怎么求?是相对于所用标准光源的光功率谱?但是标准A 标准C 标准D65 这三种光源的相对光功率谱又是相对什么光源或者什么照明体做出来的呢?恳求大家不吝指教啊!!!
<p>光谱仪前放置一个积分球,光谱能量分布为Px(i)的被测器件和Pa(i)的标准A光源分别在球内点亮,光谱仪输出读数分别为Ix(i),Ia(i). </p><p></p><p> Px(i)=Pa(i)*Ix(i)/Ia(i);</p><p>其中Pa(i)已知,符合普朗克公式,</p><p><span style="FONT-SIZE: 10.5pt; FONT-FAMILY: 宋体; mso-bidi-font-size: 12.0pt; mso-bidi-font-family: Times New Roman; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA; mso-ascii-font-family: Times New Roman; mso-hansi-font-family: Times New Roman;">以数值最大时的波长为峰值波长</span><span style="FONT-SIZE: 10.5pt; FONT-FAMILY: 宋体; mso-bidi-font-size: 12.0pt; mso-bidi-font-family: Times New Roman; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA; mso-ascii-font-family: Times New Roman; mso-hansi-font-family: Times New Roman;">,以峰值波长的读数为</span><span lang="EN-US" style="FONT-SIZE: 10.5pt; FONT-FAMILY: Times New Roman; mso-bidi-font-size: 12.0pt; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA; mso-fareast-font-family: 宋体;">100</span><span style="FONT-SIZE: 10.5pt; FONT-FAMILY: 宋体; mso-bidi-font-size: 12.0pt; mso-bidi-font-family: Times New Roman; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA; mso-ascii-font-family: Times New Roman; mso-hansi-font-family: Times New Roman;">%,就得到被测器件</span><span style="FONT-SIZE: 10.5pt; FONT-FAMILY: 宋体; mso-bidi-font-size: 12.0pt; mso-bidi-font-family: Times New Roman; mso-font-kerning: 1.0pt; mso-ansi-language: EN-US; mso-fareast-language: ZH-CN; mso-bidi-language: AR-SA; mso-ascii-font-family: Times New Roman; mso-hansi-font-family: Times New Roman;">的相对光谱功率分布。</span></p> 赞同,楼上说的对!也就是与峰值波长点的功率相比 <p>mark1234567890000000000000</p>
谢谢指点!
我也颇受用!
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