CHROMATIC ANISOPLANATISM

If the beacon beam that senses the turbulence has a wavelength different from that of the laser beam that is sent out, then the two beams will follow different paths through the atmosphere because of the dispersive properties of the atmosphere. The analysis given here parallels that given by Belsher and Fried.[#!1!#]

... The change of refractive index with wavelength has been given by Allen[#!16!#] as

Δn0 = $\displaystyle \left(\vphantom{ {\lambda _1^2-\lambda _2^2} }\right.$λ12-λ22$\displaystyle \left.\vphantom{ {\lambda _1^2-\lambda _2^2} }\right)$$\displaystyle \left[\vphantom{ {{{29\,498.1} \over {\left( {146\lambda _2^2-1} ... ...{\left( {41\lambda _2^2-1} \right)\left( {41\lambda _1^2-1} \right)}}} }\right.$$\displaystyle {{{29\,498.1} \over {\left( {146\lambda _2^2-1} \right)\left( {14... ...4} \over {\left( {41\lambda _2^2-1} \right)\left( {41\lambda _1^2-1} \right)}}}$$\displaystyle \left.\vphantom{ {{{29\,498.1} \over {\left( {146\lambda _2^2-1} ... ...{\left( {41\lambda _2^2-1} \right)\left( {41\lambda _1^2-1} \right)}}} }\right]$10-6.     (40)

The atmospheric density versus altitude is given by Cole.[#!17!#] The ratio of the ... . Thus the beam displacement along the path is
dc(z) = - $\displaystyle {{{ \rm\bbox{\xi}} \,\sin \left( \xi \right)\,\Delta \kern 1ptn_0} \over {\xi \,\cos ^2\left( \xi \right)}}$ $\displaystyle \left[\vphantom{ {\int\limits_{\,\,\, 0}^{\,\,\,\,\,\, z} {\rm d}... ...imits_{\,\,\, 0}^{\,\,\,\,\,\, L} {\rm d}z'\alpha \left( {z'} \right)} }\right.$$\displaystyle {\int\limits_{\,\,\, 0}^{\,\,\,\,\,\, z} {\rm d}z' \alpha \left( ... ... L}\int\limits_{\,\,\, 0}^{\,\,\,\,\,\, L} {\rm d}z'\alpha \left( {z'} \right)}$$\displaystyle \left.\vphantom{ {\int\limits_{\,\,\, 0}^{\,\,\,\,\,\, z} {\rm d}... ...imits_{\,\,\, 0}^{\,\,\,\,\,\, L} {\rm d}z'\alpha \left( {z'} \right)} }\right]$.     (41)

Define the integral of the air density as
I$\displaystyle \left(\vphantom{ z }\right.$z$\displaystyle \left.\vphantom{ z }\right)$ = $\displaystyle \int\limits_{{\,\,\, 0}}^{{\,\,\,\,\,\, z}}$dz'α$\displaystyle \left(\vphantom{ {z'} }\right.$z'$\displaystyle \left.\vphantom{ {z'} }\right)$.     (42)

Evaluating the integral and ...

The moments of this displacement are

dm = $\displaystyle \left[\vphantom{ {{{\sin \left( \xi \right)\Delta \kern 1ptn_0} \over {\cos ^2\left( \xi \right)}}} }\right.$$\displaystyle {{{\sin \left( \xi \right)\Delta \kern 1ptn_0} \over {\cos ^2\left( \xi \right)}}}$$\displaystyle \left.\vphantom{ {{{\sin \left( \xi \right)\Delta \kern 1ptn_0} \over {\cos ^2\left( \xi \right)}}} }\right]^{m}_{}$Tm,     (43)

where
Tm = 2.91 k02sec$\displaystyle \left(\vphantom{ \xi }\right.$ξ$\displaystyle \left.\vphantom{ \xi }\right)$$\displaystyle \int\limits_{{\,\,\, 0}}^{{\,\,\,\,\,\, H}}$dh Cn2(h) $\displaystyle \left[\vphantom{ {I(h)-{{h\sec \left( \xi \right)} \over L}I(L)} }\right.$$\displaystyle {I(h)-{{h\sec \left( \xi \right)} \over L}I(L)}$$\displaystyle \left.\vphantom{ {I(h)-{{h\sec \left( \xi \right)} \over L}I(L)} }\right]^{m}_{}$.     (44)

H is the altitude of the target. The last term in brackets goes to zero as the range becomes infinite. ...For the infinite range, this reduces to
Tm = 2.91 k02sec$\displaystyle \left(\vphantom{ \xi }\right.$ξ$\displaystyle \left.\vphantom{ \xi }\right)$$\displaystyle \int\limits_{{\,\,\, 0}}^{{\,\,\,\,\,\, H}}$dh Cn2(h) Im(h).     (45)

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