Inside the slab, the density of bound charges is zero, because V. d=EEVE=0. The bound currents have a density of ,=dp/dt=EE- l)dE/dt. However, the force exerted by the magnetic field of the light on these currents is zero because of the 90 phase between H and dE/at. Thus the Lorentz force inside the volume of the slab is zero. The only force is due to the bound charges induced on the two surfaces of the plate, the density of which is obtained from the discontinuity in E En across the interface, namely O= EEo(1-1/n)sine, (17 We digress momentarily to look at the induced charges on the two facets of the slab, and erify their consistency with Maxwell's equations. The polarization density is P=E(E-lE, where E inside the medium has magnitude Eo/n. Consider a(tilted) cylindrical volume, aligned with the internal E-field and stretched between the two facets as shown in If the base area of this cylinder is denoted by a, its volume will be ad, where d is the thickness of the slab. The electric dipole moment of this cylindrical volume is thus adP, which must be lal to the surface charge ao on either base multiplied by the length of the cylinder, d/sine=nd/sine. The charge density on each surface is thus o=f(E-1)E sine/n EEo(1-1/n)sine, consistent with our earlier finding in Eq (17) In the horizontal direction, Ex=EocosBB is continuous across the boundary, and the Lorentz force on the charges is The factor v accounts for time-averaging(also spatial-averaging over each surface, since the field and the charges vary sinusoidally both with time and with the coordinate x). If the incident beams cross-sectional area is denoted by A, its footprint on the slab will have an ea A/coseB. The force density Fx, when integrated over the footprint and multiplied by the distance d between the two surfaces, yields the following torque Ton the slab T=hE Ad(1-1/n)sine (19) Note that the net force on the slab is zero, because the top and bottom surfaces cancel each other out. However, the Fr component yields a torque that tends to rotate the slab around the y-axis. Now, the electromagnetic field's momenta before and after the slab are the same, both in magnitude(p=vAEEo)and in direction, resulting in no net imparted force However, upon transmission through the slab, the incoming p is displaced parallel to itself by A=d(1-1/m)sineB, as shown in Fig. 6(a). The change pA in the angular momentum of the beam is thus seen to be identical with the torque T exerted by Fx, as given by Eq (19).As a numerical example, consider the a glass slab having A=1.0 mm, d= 10 um, and n=1.5, illuminated at BB=56.3 with a 1.0 W/mm beam of light. Using Eq (19), the torque on the slab is found to be t= 154fN. m Next, we consider the perpendicular force F: on the top surface of the slab. Since E:is discontinuous across the boundary, we must average E: just above and just below the interface. Thus F:=hoe again to account for time- and space-averaging. Integrating over the footprint of the beam (area=A/coseB, we obtain F:= EEo(n-n)A cosBB. The forces F: on the top and bottom facets of the slab, having equal magnitudes and opposite signs, cancel out Note that Fs, being laterally displaced by d tan0'B=d/n between the top and bottom facets, must also exert a torque on the plate. This torque, however, is exactly cancelled out by an equal and opposite torque originating from the force of the beam exerted at its right and left edges. As derived in Section 6, the force density F edge)= E(n2-DEEt is normal to edge and expansive in the case of p-light. This force is of eowa magnitude and opposilenthe #5025-S1500US Received 10 August 2004; revised 13 October 2004; accepted 20 October 2004 (C)2004OSA November 2004/Vol 12. No 22/OPTICS EXPRESS 5388Inside the slab, the density of bound charges is zero, because ∇ · D = εoε ∇ · E = 0. The bound currents have a density of Jb = ∂P/∂t = εo(ε – 1)∂E/∂t. However, the force exerted by the magnetic field of the light on these currents is zero because of the 90° phase between H and ∂E/∂t. Thus the Lorentz force inside the volume of the slab is zero. The only force is due to the bound charges induced on the two surfaces of the plate, the density of which is obtained from the discontinuity in εoE⊥ across the interface, namely, σ = εoEo(1 − 1/n2 )sinθB. (17) We digress momentarily to look at the induced charges on the two facets of the slab, and verify their consistency with Maxwell’s equations. The polarization density is P =εo(ε – 1)E, where E inside the medium has magnitude Eo/n. Consider a (tilted) cylindrical volume, aligned with the internal E-field and stretched between the two facets, as shown in Fig. 6(b). If the base area of this cylinder is denoted by a, its volume will be ad, where d is the thickness of the slab. The electric dipole moment of this cylindrical volume is thus adP, which must be equal to the surface charge aσ on either base multiplied by the length of the cylinder, d/sinθ′B = nd/sinθB. The charge density on each surface is thus σ = εo(ε – 1)EosinθB/n2 = εoEo(1 – 1/n2 )sinθB, consistent with our earlier finding in Eq. (17). In the horizontal direction, Ex = EocosθB is continuous across the boundary, and the Lorentz force on the charges is Fx = ½σEx = ½εoEo 2 (1 − 1/n2 ) sinθB cosθB. (18) The factor ½ accounts for time-averaging (also spatial-averaging over each surface, since the field and the charges vary sinusoidally both with time and with the coordinate x). If the incident beam’s cross-sectional area is denoted by A, its footprint on the slab will have an area A/cosθB. The force density Fx, when integrated over the footprint and multiplied by the distance d between the two surfaces, yields the following torque T on the slab: T = ½εoEo 2 Ad(1 − 1/n2 )sinθB. (19) Note that the net force on the slab is zero, because the top and bottom surfaces cancel each other out. However, the Fx component yields a torque that tends to rotate the slab around the y-axis. Now, the electromagnetic field’s momenta before and after the slab are the same, both in magnitude ( p = ½AεoEo 2 ) and in direction, resulting in no net imparted force. However, upon transmission through the slab, the incoming p is displaced parallel to itself by ∆ = d(1 − 1/n2 )sinθB, as shown in Fig. 6(a). The change p∆ in the angular momentum of the beam is thus seen to be identical with the torque T exerted by Fx, as given by Eq. (19). As a numerical example, consider the case of a glass slab having A = 1.0 mm2 , d = 10 µm, and n = 1.5, illuminated at θB = 56.3° with a 1.0 W/mm2 beam of light. Using Eq. (19), the torque on the slab is found to be T = 15.4 f N.m. Next, we consider the perpendicular force Fz on the top surface of the slab. Since Ez is discontinuous across the boundary, we must average Ez just above and just below the interface. Thus Fz = ½σEz = ¼εoEo 2 (1 − 1/n2 )(1 + 1/n2 ) sin2 θB, with the factor ½ introduced again to account for time- and space-averaging. Integrating over the footprint of the beam (area = A/cosθB), we obtain Fz = ¼εoEo 2 (n2 − n–2)A cosθB. The forces Fz on the top and bottom facets of the slab, having equal magnitudes and opposite signs, cancel out. Note that Fz, being laterally displaced by d tanθ′B = d/n between the top and bottom facets, must also exert a torque on the plate. This torque, however, is exactly cancelled out by an equal and opposite torque originating from the force of the beam exerted at its right and left edges. As derived in Section 6, the force density F (edge) = ¼εo(n2 − 1)|Et | 2 is normal to the edge and expansive in the case of p-light. This force is of equal magnitude and opposite sign (C) 2004 OSA 1 November 2004 / Vol. 12, No. 22 / OPTICS EXPRESS 5388 #5025- $15.00 US Received 10 August 2004; revised 13 October 2004; accepted 20 October 2004