{p-[(vE-1)/(e+1)l/{1-p[(E-1)/(e+1)]} (25c) E/I(e+12 exp(-i2Ivedn)-(Ve-1 exp(i2IcvEC (25d) The force per unit area of the slab can be calculated in a way similar to that described in the preceding section, namely, by calculating the Lorentz force of the H-field on the dipolar current distribution, then integrating through the thickness of the slab. The final result F2=%n2+x2+1)[1-exp(- +%sa(n2+x2+1)[1-exp(+4兀K)2 +h2E(n+x-1)[cos(4Tnd/o+QEI-PE2)-Cos(QE 1-OE2)] E1E2l H=E。/z rEZo E1 eEz veE,/Z E1=。 H=tE。/z0 thickness d and dielectric constant a is illuminated at normal incidence by a plane wave. The magnetic fields Lorentz force standing wave with he slab se to a downward force that is precisely equal to the difference between the and outgoing momenta of the light beam. In the special case when a is real, K goes to zero and the above formula reduces to EE。2 F (27) (n2-1)sin(2兀nd^) This can be shown to be consistent with the balance of the incident, reflected and transmitted F=%EE(1+|r2-|t2)=ckE2 where we have used [rF+|tP=l for a non-absorbing slab. According to Eq(27), the net force is zero when d=2n. This is consistent with the fact that a half-wave-thick plate does #5025-S1500US Received 10 August 2004; revised 13 October 2004; accepted 20 October 2004 (C)2004OSA November 2004/Vol 12. No 22/OPTICS EXPRESS 5391r = {ρ – [(√ε – 1) / (√ε + 1)]}/{1 − ρ [(√ε – 1) / (√ε + 1)]} (25c) t = 4√ε / [(√ε + 1)2 exp (−i2π√ε d/λo) − (√ε − 1)2 exp (+i2π√ε d/λo)] (25d) The force per unit area of the slab can be calculated in a way similar to that described in the preceding section, namely, by calculating the Lorentz force of the H-field on the dipolar current distribution, then integrating through the thickness of the slab. The final result is Fz = ¼εo(n2 + κ2 + 1) [1 − exp (− 4πκ d/λo)] |E1| 2 + ¼εo(n2 + κ2 + 1) [1 − exp (+4πκ d/λo)] |E2| 2 + ½εo(n2 + κ2 − 1) [cos(4πnd/λo + φE1 − φE 2) − cos(φE 1 − φE 2)] |E1E2| (26) Fig. 8. A plate of thickness d and dielectric constant ε is illuminated at normal incidence by a linearly-polarized plane wave. The magnetic field’s Lorentz force on the standing wave within the slab gives rise to a downward force that is precisely equal to the difference between the rates of incoming and outgoing momenta of the light beam. In the special case when ε is real, κ goes to zero and the above formula reduces to εoEo 2 Fz = · (27) 2n 2 1 + (n2 – 1)sin(2πnd/λo) This can be shown to be consistent with the balance of the incident, reflected, and transmitted momenta, namely, Fz = ½ εoEo 2 (1 + | r |2 − | t |2 ) = εo |r |2 Eo 2 , (28) where we have used |r | 2 + | t | 2 = 1 for a non-absorbing slab. According to Eq. (27), the net force is zero when d = λo/2n. This is consistent with the fact that a half-wave-thick plate does Ho=Eo/Zo Eo Et = tEo Ht = tEo/Zo X Z -√εE2 /Zo E2 √εE1/Zo E1 d ε -rEo/Zo rEo (C) 2004 OSA 1 November 2004 / Vol. 12, No. 22 / OPTICS EXPRESS 5391 #5025- $15.00 US Received 10 August 2004; revised 13 October 2004; accepted 20 October 2004