Hi everyone,
I added and weighted the intensities of the PSFs of three different wavelengths in the system, but the resulting PSF is different from directly using a multi-wavelength PSF.
Did I miss something?
Thanks,
-Shun
Solved
combining the PSF of three different wavelengths
Best answer by Jeff.Wilde
From Fourier optics we know that the 2D intensity PSF for a quasi-monochromatic incoherent imaging system is proportional to the magnitude squared of the 2D Fourier transform of the exit pupil wavefront. For purposes of computing an image irradiance (via a 2D convolution of the ideal image irradiance with the PSF), the proper constant of proportionality should be included. This constant is A/(z*lambda)^2, where A is the pupil area, z is the distance from the exit pupil to the image plane, and lambda is the wavelength. This result applies to a quasi-monochromatic case (note: a perfectly monochromatic imaging system is always spatially coherent, while incoherent imaging requires a finite spectral bandwidth, but it can be quite narrow) -- the important point here is to note that the proportionality factor scales as (1/lambda)^2.
So, in your model with three system wavelengths, if you capture the three quasi-monochromatic (normalized) 2D PSF intensity functions found using the FFT PSF tool, then you can find the corresponding polychromatic PSF by calculating a spectrally weighted average of these three normalized PSF’s. In your case try this weighting:
PSF1 = (Normalized PSF for lambda_1) * (lambda_p / lambda_1)^2
PSF2 = (Normalized PSF for lambda_2) * (lambda_p / lambda_2)^2
PSF3 = (Normalized PSF for lambda_3) * (lambda_p / lambda_3)^2
Here lambda_p is the primary wavelength (which can be chosen to be any one of the three system wavelengths, but in your model it’s lambda_1 = 0.532 um). Note, this assumes that all three system wavelengths have equal weighting in the wavelength data editor. Also, use a non-zero Image Delta value (say 0.5 um) so that all three PSF’s have the same spatial sampling.
Then, calculate the polychromatic 2D PSF as:
PSF_poly = (PSF1 + PSF2 + PSF3)/3
If you compare this to the polychromatic PSF computed directly by OpticStudio, you should find a close match.
Regards,
Jeff
+3How much different? Was the sampling the same? Did you use FFT or Huygens? If FFT, try it with Huygens, making sure the sampling is the same for all. Subtract your 3-PSF sum from the polychromatic PSF to quantify any differences.
Mike
+3From Fourier optics we know that the 2D intensity PSF for a quasi-monochromatic incoherent imaging system is proportional to the magnitude squared of the 2D Fourier transform of the exit pupil wavefront. For purposes of computing an image irradiance (via a 2D convolution of the ideal image irradiance with the PSF), the proper constant of proportionality should be included. This constant is A/(z*lambda)^2, where A is the pupil area, z is the distance from the exit pupil to the image plane, and lambda is the wavelength. This result applies to a quasi-monochromatic case (note: a perfectly monochromatic imaging system is always spatially coherent, while incoherent imaging requires a finite spectral bandwidth, but it can be quite narrow) -- the important point here is to note that the proportionality factor scales as (1/lambda)^2.
So, in your model with three system wavelengths, if you capture the three quasi-monochromatic (normalized) 2D PSF intensity functions found using the FFT PSF tool, then you can find the corresponding polychromatic PSF by calculating a spectrally weighted average of these three normalized PSF’s. In your case try this weighting:
PSF1 = (Normalized PSF for lambda_1) * (lambda_p / lambda_1)^2
PSF2 = (Normalized PSF for lambda_2) * (lambda_p / lambda_2)^2
PSF3 = (Normalized PSF for lambda_3) * (lambda_p / lambda_3)^2
Here lambda_p is the primary wavelength (which can be chosen to be any one of the three system wavelengths, but in your model it’s lambda_1 = 0.532 um). Note, this assumes that all three system wavelengths have equal weighting in the wavelength data editor. Also, use a non-zero Image Delta value (say 0.5 um) so that all three PSF’s have the same spatial sampling.
Then, calculate the polychromatic 2D PSF as:
PSF_poly = (PSF1 + PSF2 + PSF3)/3
If you compare this to the polychromatic PSF computed directly by OpticStudio, you should find a close match.
Regards,
Jeff
Thanks for your resonse
I use huygens psf , i choose the most central point . All the samples are the same as the photos .
The weights for all three wavelengths are 1. The value obtained for 3-PSF is 0.6256, while the value for the polychromatic PSF is 0.4856.They differ by 0.14.
-shun
Jeff,
Based on your response , i choose the most central point .The weights for all three wavelengths are 1.
PSF1 = 0.37605*(0.532/0.457)^2 = 0.5096
PSF2 = 0.95525*(0.532/0.532)^2 = 0.95525
PSF3 = 0.54177*(0.532/0.630)^2 = 0.38632
(PSF1+PSF2+PSF3)/3 = 0.61706
But the polychromatic PSF is 0.51653. They differ by 0.10053.
Is this value considered close?
thanks for your help
-shun
+3When I take the model you posted, and use the Huygens PSF
I find for the central points:
PSF1 = 0.953*(0.532/0.532)^2 = 0.953
PSF2 = 0.714*(0.532/00.630)^2 = 0.509
PSF3 = 0.120*(0.532/0.457)^2 = 0.163
(PSF1+PSF2+PSF3)/3 = 0.541
and the built-in polychromatic PSF is 0.529, which is reasonably close. The two polychromatic functions match quite well out on the wings of the PSF.
-Jeff
Thanks Jeff
I found where I was wrong.
Thanks for your help
Have a nice day
-shun
Hi Jeff
I'm sorry to bother you, but I have encountered a problem.
I can get similar results when FIELD is 0, but when FIELD is not 0, the results can differ significantly.
When off-axis, is there anything else that needs to be taken into consideration?
-shun
+3Hi Shun,
I opened your new file, made some changes, and then for off-axis Field 2 compared the built-in polychromatic PSF to the manually generated version (using the three individual monochromatic PSF’s). It looks like everything is okay to me. Here are cross sections taken from the 2D PSF datasets.
Perhaps you are making a mistake somewhere. My results show that the match is fairly good, but not perfect. I’m not completely sure why it’s not perfect, but it’s hard to know because the details for the Zemax code are not available. My version of your latest model is attached.
Regards,
Jeff
Thanks Jeff
I am only taking values from the central part and there is a significant difference in the central values.
The rest of the values are the same as yours.
but thanks for your help.
-shun
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