Subscript $t$ expresses the transverse plane, which is perpendicular to the $z$ axis. According to these, Maxwell’s equation can be expressed as^{17}^{,}^{18}Display Formula
$D1(\u2207t)\xb7Hz+D3(\u2207t)\xb7Ez=0D2(\u2207t)\xb7Ez\u2212D4(\u2207t)\xb7Hz=0,$(3)
where $D1(\u2207t)$, $D2(\u2207t)$, $D3(\u2207t)$, and $D4(\u2207t)$ are quadratic functions of $\u2207t$, and $\u2207t$ is a differentiating operator. By using plane wave expansion method, we can solve Eq. (3) to get internal electromagnetic field quantities $Ez$ and $Hz$, and get $E\u2192t$ and $H\u2192t$. Using the continuity conditions on the boundary, we can get scattered field $E\u2192s$ (parallel component $E\u2192||s$ and perpendicular component $E\u2192\u22a5s$) Display Formula$(E||sE\u22a5s)=1sin\u2009\zeta 2\pi k\rho \u2009sin\u2009\zeta ej\pi 4e\u2212jk(\rho \u2009sin\u2009\zeta +z\u2009cos\u2009\zeta )\u2062(J1J4J3J2)(E||iE\u22a5i),$(4)
Display Formula${J1=\u2211\u2212\u221e\u221ebnIe\u2212in\Theta =b0I+2\u2211n=1\u221ebnI\u2009cos(n\Theta )J2=\u2211\u2212\u221e\u221eanIIe\u2212in\Theta =a0II+2\u2211n=1\u221eanII\u2009cos(n\Theta )J3=\u2211\u2212\u221e\u221eanIe\u2212in\Theta =\u22122i\u2211n=1\u221eanI\u2009sin(n\Theta )J4=\u2211\u2212\u221e\u221ebnIIe\u2212in\Theta =\u22122i\u2211n=1\u221ebnII\u2009sin(n\Theta ),$(5)
Display Formula${anI=\u22122\pi \tau Fn(2)Jn(\xi +)/{\delta \xb7Hn(2)(\tau )}bnI={2\pi \tau \eta Fn(1)Jn(\xi \u2212)/\delta \u2212sin\u2009\zeta Jn(\tau )}/Hn(2)(\tau )anII={\u2212sin\u2009\zeta Jn(\tau )\u22122\pi \tau \eta Fn(4)Jn(\xi +)/\delta}/Hn(2)(\tau )bnII=2\pi \tau Fn(3)Jn(\xi \u2212)/{\delta \xb7Hn(2)(\tau )}\delta =Fn(1)\xb7Fn(4)\u2212Fn(2)\xb7Fn(3),$(6)
Display Formula${Fn(1)=\u2212j\eta sin\u2009\zeta Hn(2)\u2032(\tau )Jn(\xi +)+gn(1)Jn\u2032(\xi +)Fn(2)=cos\u2009\zeta (n02\u2212n\u22a52)ak\u2009sin2\u2009\zeta (n\u22a52\u2212n02\u2009\u2009cos2\u2009\zeta )nHn(2)(\tau )Jn(\xi \u2212)Fn(3)=cos\u2009\zeta (n02\u2212n\u22a52)ak\u2009sin2\u2009\zeta (n\u22a52\u2212n02\u2009cos2\u2009\zeta )nHn(2)(\tau )Jn(\xi +)Fn(4)=j1\eta \u2009sin\u2009\zeta Hn(2)\u2032(\tau )Jn(\xi \u2212)\u2212gn(1)Jn\u2032(\xi \u2212),$(7)
Display Formula${gn(1)=j\eta n02\chi +k(n\u22a52\u2212n02\u2009cos2\u2009\zeta )\xb7Hn(2)(\tau )gn(2)=jn\u22a52\chi \u2212\eta k(n\u22a52\u2212n02\u2009cos2\u2009\zeta )\xb7Hn(2)(\tau ),$(8)
Display Formula$\eta =1n0\mu 0\epsilon 0{\tau =ak\u2009sin\u2009\zeta \xi \xb1=a\xb7\chi \xb1{\chi +2=k2(n\u22a52n02\u2212cos2\u2009\zeta )\chi \u22122=k2(n||2n02\u2212n||2n\u22a52\u2009cos2\u2009\zeta )\u2062(n||\u2265n\u22a5){\chi +2=k2(n||2n02\u2212n||2n\u22a52\u2009cos2\u2009\zeta )\chi \u22122=k2(n\u22a52n02\u2212cos2\u2009\zeta )(n||<n\u22a5),$(9)
where $\rho $ and $z$ represent the radial and axial distances in the cylindrical coordinates, $a$ is the radius of the cylinder, $n0$ is the refractive index and $\eta $ is the wave impedance of the ambient medium, $k=1/\lambda $ is the wave number, $\zeta $ is the angle between the direction of incident light and the direction of cylinder, and $\Theta $ is the azimuth angle of the scattered light on the scattering cone. $Jn$ is the Bessel function, $Hn(1)=Jn+iYn$ and $Hn(2)=Jn\u2212iYn$, respectively, expressing the first and second kind of Hankel function.