By Vladimir P. Lukin, Boris V. Fortes

ISBN-10: 0819443379

ISBN-13: 9780819443373

A result of huge software of adaptive optical structures, an figuring out of optical wave propagation in randomly inhomogeneous media has turn into crucial, and a number of other numerical types of person AOS elements and of effective correction algorithms were built. This monograph comprises specified descriptions of the mathematical experiments that have been designed and conducted in the course of greater than a decade's worthy of research.

**Contents **

- Preface to the English variation

- creation

- Mathematical Simulation of Laser Beam Propagation within the surroundings

- Modeling an Adaptive Optics method

- Adaptive Imaging

- Minimization and part Correction of Thermal Blooming of High-Power Beams

- A Reference Beacon as a Key section of an Adaptive Optics procedure

- end

- Index

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**Extra info for Adaptive Beaming and Imaging in the Turbulent Atmosphere**

**Example text**

24) Substituting Eqs. 24) into Eq. 21), we find the relation sought: Mathematical Simulation of Laser Beam Propagation in the Atmosphere n2 r 1 1 z 2 n1 r (1 z ) . 25) Let us use this relation in a problem of thermal blooming. 26) t CP where is the absorption coefficient, 0 is the medium density, CP is specific heat at constant pressure, V z is the wind velocity profile, and is heat conductivity. 27) ( z )nt , nt n / t . 0 C P Then consider the propagation of two high-power optical beams with the boundary conditions of Eqs.

Through an eikonal of the reference wave : E0 A0 exp ik d . 7) impose no restrictions on the reference wave, but they assume that the diffraction-limited phase of the reference wave is known. 8) 50 Chapter 2 E0 A0 exp i arg U 0 (for PC correction). 9) For the result of correction by Eq. , arg U 0d arg A0 . 11) where L is the length of the propagation path. In such a formulation of the boundary condition for the reference wave, the PC moves toward the exact WFR as optical distortions on the path decrease.

20) Dividing Eq. 18) by Eq. 20), we get U1 r (1 z ) n2 r n1 r (1 z ) U2 r 2ikU 2 z r U 2xx r U 2yy r . 21) The first fraction on the right-hand side of Eq. 21) follows from Eq. 11): U1 r (1 z ) 1 . 22) To calculate the second fraction, differentiate Eq. 11) with respect to z, x , and y2. 23) 1 z 2ikU1z r (1 z ) U1xx r (1 z ) U1yy r (1 z ) , 3 from which it is seen that the second fraction on the right-hand side of Eq.

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