[求助]zemax物理光学能量问题 [复制链接]

请问一下，我在zemax物理光学传播（physical opticals propagation 里看到了以total power的量，这个量代表的是光束的能量吗 gAViwy9{  
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Under POP propagation you will see Beam Definition, under this tab you can define either the "Total Power" of the beam or the "Peak Irradiance". So the total power returns under the window (as you captured) will be reporting the total power with respect to the energy defined under this "Beam Definition". f8
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hopebox:Under POP propagation you will see Beam Definition, under this tab you can define either the "Total Power" of the beam or the "Peak Irradiance". So the total power returns under the window (as you captured) will be reporting the total power .. (2016-07-20 16:07)  +Y  

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选择total power 和 peak irradiance会有什么不同呢，还有我想改变镜子大小来看光的通过率，设定选择surface type里的 aperture type 里的通光孔径的大小，再看total power 的大小，这样是不是可以说明不同半径下光的通过率啊。

诸葛随风:选择total power 和 peak irradiance会有什么不同呢，还有我想改变镜子大小来看光的通过率，设定选择surface type里的 aperture type 里的通光孔径的大小，再看total power 的大小，这样是不是可以说明不同半径下光的通过率啊。 (2016-07-21 09:49)  BrQXSN$i  

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Whether choosing Total power or peak irradance, the energy/power of beam as a function of radius will just be calculated relatively to how you define it. So it shouldn't matter, and just depends on how the source beam is defined though (since the beam profile is there). &Jr~)o   
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I think the method you proposed shall return the physical quantity you want to study. Hope this helps.

hopebox:Whether choosing Total power or peak irradance, the energy/power of beam as a function of radius will just be calculated relatively to how you define it. So it shouldn't matter, and just depends on how the source beam is defined though (since the beam pro .. (2016-07-21 15:47)  iUv
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我想了解不同镜子尺寸光的损失，不知道能不能通过看total power损失来研究呢

为什么不直接看透过率呢

alien21:为什么不直接看透过率呢 (2016-07-22 15:10)  oe<9CK:?>  

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透过率？怎么弄啊？

诸葛随风:透过率？怎么弄啊？ (2016-07-24 09:30)  s bf\;_!  

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Yes, Alien21 is right, you could check Polarization > Transmission as a way to check it, too. 9$-V/7@)  
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If using POP, make sure you tick the box of "Use Polarization". So the total power shall account all sorts of energy loss.

谢谢楼主

学习一下 tx=~bm"*?