Optimization problem (imaging science)

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SUMMARY

The discussion focuses on optimizing a transmission imaging system by calculating the optimal source-to-object distance and the resulting Full Width at Half Maximum (FWHM) of the total Point Spread Function (PSF) at the object plane. The parameters include a focal spot PSF with a Gaussian FWHM of 0.5 mm and a detector PSF with a Gaussian FWHM of 0.1 mm, with a source-to-detector distance of 100 cm. The optimal source-to-object distance is determined to be 96.1 cm, resulting in a total PSF FWHM of 0.98 mm at this distance.

PREREQUISITES
  • Understanding of Gaussian functions and their properties
  • Knowledge of Point Spread Function (PSF) in imaging systems
  • Familiarity with calculus, specifically derivatives and optimization techniques
  • Basic concepts of optical imaging systems and distances in optics
NEXT STEPS
  • Study the derivation of the FWHM for combined Gaussian functions
  • Learn about optimization techniques in calculus, particularly for multivariable functions
  • Research the implications of PSF in imaging science and its impact on image quality
  • Explore practical applications of imaging systems in fields such as medical imaging or photography
USEFUL FOR

This discussion is beneficial for imaging scientists, optical engineers, and students studying imaging systems who seek to understand the optimization of PSF in practical applications.

richard7893
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Homework Statement



Assume a transmission imaging system with the following parameters: focal spot PSF
equal to a Gaussian with FWHM = 0.5 mm, a detector PSF equal to a Gaussian with
FWHM = 0.1 mm, source-to-detector distance = 100 cm. Compute:
1. the optimal source-to-object distance
2. the FWHM (in mm) of the total PSF at the object plane at this optimal source-to object distance.




Homework Equations



a= S1/(S1+S2); b=S2/(S1+S2); where S1 is the distance from focal spot to object, S2 is the distance from the object to the detector, & S1+S2 is the distance from the focal spot to detector (100cm).

a+b=1 (S1 + S2 have to add to 100cm, the detector-source distance)

FWHM_object plane = ( (an 'a' dependent factor *FWHM_focal spot)^2 + (an 'a' dependent factor *FWHM_detector)^2)^(1/2) ==> (pythagorean, in case the text looks confusing. Hard doing this without pencil & paper).


The Attempt at a Solution


I know I have to take the derivative wrt 'a' of the FWHM_obj plane equation, set it equal to 0 and solve for 'a'.

I know at this distance,'a', the PSF of the object is smallest because it's FWHM is at a minimum, making that location the optimum place to put the object.

Ultimately I'm trying to solve for S1 which I'll know once I find the value of 'a'.

The answer is supposed to come out to: 96.1cm, this is how far the object needs to be from the focal spot. And the FWHM of the object PSF is 0.98mm. I just don't know how to arrive at these numbers.

My teacher gave me the above equations, but I just don't know what the 'a' dependent factor needs to be.

And my prof. also said I didn't need to use any exponentials, even though the problem mentions gaussian functions in it.

Any help will be appreciated.
 
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