Отрывок: r r kA r ef r e − σ − σ   ′′ ρ = = − ⋅ σ − σ + σ +      + σ + σ   (19) With disregard for the latter summand, we get the following: 2 1 2 2 2 2 2 2 2 1( 0) ( 2 ) rkA r ef − σ     ′′ ρ = = − σ ⋅ σ − + σ       , (19а) (0) 0A′′ = , when 1 1.832r ≈ σ , (20) where 1.832 is the root of the equation 1 = (x2+2)e–x2/2. This corresp...
Полная запись метаданных
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dc.contributor.authorUstinov, A.V.-
dc.date.accessioned2017-10-25 12:05:17-
dc.date.available2017-10-25 12:05:17-
dc.date.issued2017-08-
dc.identifierDspace\SGAU\20171020\65751ru
dc.identifier.citationUstinov AV. Focal-plane field when lighting double-ring phase elements. Computer Optics 2017; 41(4): 515-520.ru
dc.identifier.urihttps://dx.doi.org/10.18287/2412-6179-2017-41-4-515-520-
dc.identifier.urihttp://repo.ssau.ru/handle/Zhurnal-Komputernaya-optika/Focalplane-field-when-lighting-doublering-phase-elements-65751-
dc.description.abstractThe focal-plane field amplitude is calculated when lighting double-ring phase elements by flat and Gaussian beams. Emerging conditions in the minimum or maximum centers, including flat-top maxima, are given. For the field amplitude, we obtain equations that define the radius of the first zero-intensity ring based on the deduced expressions. The root values are listed for several parameters of optical elements and incident beams due to the lack of analytical solutions. Numerical simulation results are given for flat incident beams; they are fully consistent with the theoretical calculations.ru
dc.description.sponsorshipThis work was financially supported by the Russian Foundation for Basic Research (grant No. 16-07-00825).ru
dc.language.isoenru
dc.publisherСамарский университетru
dc.relation.ispartofseries41;4-
dc.subjectphase optical elementsru
dc.subjectdouble-ring phase elementsru
dc.subjectfocal spot sizeru
dc.titleFocal-plane field when lighting double-ring phase elementsru
dc.typeArticleru
dc.textpartr r kA r ef r e − σ − σ   ′′ ρ = = − ⋅ σ − σ + σ +      + σ + σ   (19) With disregard for the latter summand, we get the following: 2 1 2 2 2 2 2 2 2 1( 0) ( 2 ) rkA r ef − σ     ′′ ρ = = − σ ⋅ σ − + σ       , (19а) (0) 0A′′ = , when 1 1.832r ≈ σ , (20) where 1.832 is the root of the equation 1 = (x2+2)e–x2/2. This corresp...-
dc.classindex.scsti29.31.15-
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