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