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How to analyse irradiance versus total power in geometrical image analysis

  • 2 September 2021
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Hi,

In my spectrometer design, I want to understand how much percentage of incoming power will be detected at each wavelength and if that is above sensitivity of my detector. I have used the geometrical image analysis for that. However, I cannot make sense of the numbers here.

If the total power through the entrance pupil is 9e-15 over the 3mm width of the detector (which covers 250nm to 350nm wavelength) with 231pixels in each axis (sice the detector pixel size is 13um), I donot understand how come the irradiance power at each wavelength is higher than the total power! I assume image width is only in the width in x-axis or y-axis. let me know if that is also not a correct assumption.

 

 


 

I  set the total power to 1(equivalent to 100%), to have the outcome of the irradiance in fraction of Watts/mm^2, but I have the same issue. 

 

Secondly, I would like to know when my detetcor is not along global Y-axis, looking at the Cross Y is correct. Meaning is the Cross X and Cross Y in the image analysis is along the spectra on the detector, which will be Global_y * cos(angle between detetcor and global y-axis) ?

If looking at the geometrical image analysis is not the way to go, please let me know. I am looking at the geometrical analysis, since my system is limited by geometrical errors rather than diffraction. 

Thank you in advance

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Best answer by Angel Morales 8 September 2021, 00:19

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Hi Fatemeh,

When you define the input power, you do so in Watts. Irradiance is a measurement of Watts/mm^2, so the pixel size is taken into account in the GIA output. If I take your maximum irradiance in your first plot of 2.9e-13 W/mm^2 and multiply your pixel size to it (13 um x 13um = 1.69e-4 mm^2), we get 4.901e-17 W, which I think falls more along your expectations. The same is true when you input 1 W as your input power. Also note that each GIA output has a percent efficiency value which shows how much total power was collected on your detectors -- both are less than the input power you’ve defined.

As for the Cross X/Y display, this is done on the local coordinate surface of whatever surface you’re evaluating to, so I think your understanding is correct. If the Image surface is tilted, then GIA will use the X/Y axes of the tilted surface, not the global X/Y axes.

Thanks!

Great Angel. Now it is cleared up for me.

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