04.06.2013 Views

P7 – Scattering of Surface Plasmon Polaritons by Gold ... - repetit.dk

P7 – Scattering of Surface Plasmon Polaritons by Gold ... - repetit.dk

P7 – Scattering of Surface Plasmon Polaritons by Gold ... - repetit.dk

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

kSPP<br />

S in =<br />

<br />

2 −2akSPPz 2<br />

a − 1 e cos (kSPPx + ωt) ˆx<br />

2.4. VECTORIAL MODEL<br />

µ0ω<br />

<br />

akSPP 2<br />

+ 1 − a<br />

µ0ω<br />

e −2akSPPz <br />

sin (kSPPx + ωt) cos (kSPPx + ωt) ˆy<br />

= kSPP 2 −2akSPPz<br />

a − 1 e ˆx<br />

2µ0ω<br />

since the time average <strong>of</strong> the y component is zero. Below the surface the poynting vector is<br />

<br />

4<br />

< −a kSPP<br />

S in =<br />

µ0ω<br />

4 a kSPP<br />

+<br />

µ0ωa<br />

= −a4 kSPP<br />

2µ0ω<br />

<br />

1<br />

− 1<br />

a2 <br />

1<br />

− 1<br />

a2 <br />

1<br />

− 1<br />

a2 e 2kSPPz/a cos 2 <br />

(kSPPx + ωt) ˆx<br />

e 2kSPPz/a <br />

sin (kSPPx + ωt) cos (kSPPx + ωt) ˆy<br />

e 2kSPPz/a ˆx<br />

(2.94)<br />

(2.95)<br />

By integrating these expressions according to Eqn. 2.86 the following expression for the incident<br />

power is obtained<br />

Pin = 1<br />

2k0<br />

ε0<br />

µ0<br />

1 − a2 4<br />

1 − a<br />

2a<br />

<br />

(2.96)<br />

By doing the same calculation but with the field above the surface only, the same result without<br />

the −a 4 term is obtained. This shows that the −a 4 term stems from the flux <strong>of</strong> energy below<br />

the surface, and that the energy flux can be concentrated above the surface <strong>by</strong> making −εm<br />

significantly larger than εr. The numerator <strong>of</strong> Eqn. 2.86 can be calculated <strong>by</strong> a procedure<br />

similar to the one just shown. The scattered field is first determined, then the magnetic field,<br />

and the poynting vector can then be calculated and averaged in time. The calculation is not<br />

made here. The result is [Evlyukhin and Bozhevolnyi, 2005]<br />

〈SSPP〉 = 1<br />

k0<br />

ε0<br />

µ0<br />

|A| 2<br />

2 α0<br />

e<br />

1 + 2ξβ<br />

−2akSPP(z+2zp) (1 + ηp cos φ) 2 1 − a2 πρ<br />

By carrying out the integration and using Eqn. 2.86 the scattering cross section becomes<br />

(2.97)<br />

σSPP (φ) dφ = 2 |A| 2<br />

2 α0 (1 + ηp cos φ)<br />

1 + 2ξβ<br />

2<br />

e<br />

πkSPP<br />

−4akSPPzp (2.98)<br />

From this expression it is seen that the cross section decays exponentially as the scatterersurface<br />

distance zp increases. This makes good sense because the SPP field decreases exponentially<br />

away from the surface into the dieletric, so that the scattering will also decrease exponentially.<br />

The angular dependence <strong>of</strong> the scattering cross section is determined <strong>by</strong> (1 + ηp cos φ)<br />

37

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!