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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

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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