Fr=%Eosine Real(ve-sin0)|+r FEo F:=vE(cos0+lE-sin))|1+rs FEO (6b) The above results are valid for all values of e whether real the incident beams cross-sectional area must be A=cose to produce a unit area footprint at the interface, the time rate of change of the incident beams momentum upon reflection from the surface gives rise to F=hEine cose (1-Ps 2)E. and F:=2Eocos20(1+ks 2E2 These can be readily shown to agree with Eq (6), which has been obtained by direct integration of the force density through the thickness of the medium Next we allow e to be real, and set the refractive index n=VE. Since E1=|I+rs leO sine=nsine, and VE-sin-0=n cose, we may write Eq (6)as follows F=% esine′cose'E1P, (7a) F:=E(l-esin'e+ Ecos)E,R Note that the cross-sectional area A of the transmitted beam is not unity, but cose, which means that the above expressions for Fx and F: must be divided by cose if force per unit area is desired. The direction of the force F=Fxx+ F: z is not the same as the propagation direction (i. e, at angle A to the surface normal); the reason for this will become clear shortly From section 4 we know that. inside the dielectric. the force propagation direction must be F=/4E(E+ D)E, I(sinex cose z). Multiplying this force by the beam's cross-sectional area A=cose, then subtracting it from the previously calculated force in Eq (7)yields the following residual force: AF=E(E-1)sine(cos0x-sinA'z)E,F cOS 6 △F cose 6 Left edge Right edge and cose, respectively. A segment from the beams left edge(area propc a force AF on the dielectric; this force is not compensated by an equal and opposite force the right-hand edge of the beam, as is the case elsewhere at the opposite edges of the beam essive AF shown here corresponds to an s-polarized beam.( For p-light AF retains the same magnitude but reverses direction, so the edge force becomes expa #5025-S1500US Received 10 August 2004; revised 13 October 2004; accepted 20 October 2004 (C)2004OSA November 2004/Vol 12. No 22/OPTICS EXPRESS 5384Fx = ½εosinθ Real(√ε – sin2 θ ) |1 + rs | 2 Eo 2 (6a) Fz = ¼εo(cos2 θ + |ε – sin2 θ| ) |1 + rs| 2 Eo 2 (6b) The above results are valid for all values of ε, whether real or complex. Considering that the incident beam’s cross-sectional area must be A = cosθ to produce a unit area footprint at the interface, the time rate of change of the incident beam’s momentum upon reflection from the surface gives rise to Fx = ½εosinθ cosθ (1 − |rs | 2 )Eo 2 and Fz = ½εocos2 θ (1 + |rs | 2 )Eo 2 . These can be readily shown to agree with Eq. (6), which has been obtained by direct integration of the force density through the thickness of the medium. Next we allow ε to be real, and set the refractive index n = √ε. Since |Et | = |1 + rs | 2 Eo 2 , sinθ = n sinθ′, and √ε – sin2 θ = n cosθ′, we may write Eq. (6) as follows: Fx = ½εoε sinθ′ cosθ′ |Et |2 , (7a) Fz = ¼εo(1 − ε sin2 θ′ + ε cos2 θ′) |Et |2 . (7b) Note that the cross-sectional area A of the transmitted beam is not unity, but cosθ′, which means that the above expressions for Fx and Fz must be divided by cosθ′ if force per unit area is desired. The direction of the force F = Fx x + Fz z is not the same as the propagation direction (i.e., at angle θ′ to the surface normal); the reason for this will become clear shortly. From Section 4 we know that, inside the dielectric, the force per unit area along the propagation direction must be F = ¼εo(ε + 1)|Et |2 (sinθ′x + cosθ′z). Multiplying this force by the beam’s cross-sectional area A = cosθ′, then subtracting it from the previously calculated force in Eq. (7) yields the following residual force: ∆F = ¼εo(ε − 1)sinθ′(cosθ′x − sinθ′z) |Et | 2 . (8) Fig. 4. Oblique incidence on a semi-infinite dielectric at angle θ. The beam’s footprint at the surface has unit area, while the incident and transmitted beams’ cross-sectional areas are cosθ and cosθ′, respectively. A segment from the beam’s left edge (area proportional to sinθ′) exerts a force ∆F on the dielectric; this force is not compensated by an equal and opposite force on the right-hand edge of the beam, as is the case elsewhere at the opposite edges of the beam. The compressive ∆F shown here corresponds to an s-polarized beam. (For p-light ∆F retains the same magnitude but reverses direction, so the edge force becomes expansive.) θ θ′ cosθ′ sinθ′ cosθ θ′ ∆F n Left edge Right edge (C) 2004 OSA 1 November 2004 / Vol. 12, No. 22 / OPTICS EXPRESS 5384 #5025- $15.00 US Received 10 August 2004; revised 13 October 2004; accepted 20 October 2004