S(,1)>=E21Z乙 (The above equation will be exact, i.e., terms in Af and higher order will be absent, iff=NAf, e, if the center frequency happens to be an integer-multiple of Af) Each participating frequency thus contributes its own(time-averaged) Poynting vector <S:>= yEo Zo to the energy flux of the beam. The time-averaged momentum density of the beam (i.e, momentum per unit volume), P:=<S: >/c, is thus uniform throughout the free-space region, having equal contributions from the two frequency components of the beam Reflected Refractive index=以(/ Www Incident beam light beam consisting of two equal-amplitude plane-waves of slightly is linearly polarized, having its E-field along the x-axis and H-field along action of the beam is reflected at the surface. the remainder enters the ng it at the group velocity g=c/(n+nf). Heref=7(fi+f2)is the center frequency, and n=dn/df; both n and n are evaluated at the center frequency Suppose now that the beam reaches the interface with a dielectric medium of refractive index n(f), as shown in Fig. 1. Each frequency component gets reflected at the interface, with an(amplitude)reflection coefficient p=(1-n)(1+ n). The time rate of change of the optica momentum g on the vacuum side of the interface is then equal to the rate of arrival of the incident momentum, cp:=Eo/(cZ=Eo Eo plus the rate of departure of the reflected optical momentum,namely 1+(1-n1)(1+n2+k(1-m)(1+n2)2}eE2 In the limit -f2, the refractive indices n1, n2 in Eq (4a)are nearly identical, and the above expression simplifies to dg /dr=2(n+1)(n+1)jeE where n=7(n1 + n2). Momentum conservation requires the above dqs /dt to be balanced by the force exerted on the dielectric medium plus the time rate of change of any momentum taken by the transmitted beam into the dielectric 3. Superposition of two plane waves in a dielectric medium Inside the medium, each frequency component arrives with an amplitude transmission coefficient(for the E-field) given by T=2/(1+n). The transmitted E- and H-fields may thus be written as follows: E(=,D)=[2E(n1+1)sin{2xf[(m11)-l}-[2EM(m2+1)sin{2r/[(n2=lc)-l}(5a) 6629-$1500US Received 18 February 2005; revised 14 March 2005; accepted 15 March 2005 (C)2005OSA 21 March 2005/ Vol 13. No 6/ OPTICS EXPRESS 2247< Sz (z, t) > = Eo 2 /Zo (3) (The above equation will be exact, i.e., terms in ∆f and higher order will be absent, if f = N ∆f, i.e., if the center frequency happens to be an integer-multiple of ∆f.) Each participating frequency thus contributes its own (time-averaged) Poynting vector < Sz > = ½Eo 2 /Zo to the energy flux of the beam. The time-averaged momentum density of the beam (i.e., momentum per unit volume), pz = < Sz >/c 2 , is thus uniform throughout the free-space region, having equal contributions from the two frequency components of the beam. Fig. 1. A light beam consisting of two equal-amplitude plane-waves of slightly differing frequencies, f1 and f2, is normally incident on a semi-infinite dielectric of refractive index n( f ). The beam is linearly polarized, having its E-field along the x-axis and H-field along the y-axis. While a fraction of the beam is reflected at the surface, the remainder enters the dielectric, penetrating it at the group velocity Vg = c/(n + n′ f ). Here f = ½( f1 + f2) is the center frequency, and n′ = dn/d f ; both n and n′ are evaluated at the center frequency. Suppose now that the beam reaches the interface with a dielectric medium of refractive index n( f ), as shown in Fig. 1. Each frequency component gets reflected at the interface, with an (amplitude) reflection coefficient ρ = (1 – n)/(1 + n). The time rate of change of the optical momentum q on the vacuum side of the interface is then equal to the rate of arrival of the incident momentum, c pz = Eo 2 /(cZo) = εoEo 2 , plus the rate of departure of the reflected optical momentum, namely, d qz/dt = {1 + ½|(1 – n1)/(1 + n1)| 2 + ½|(1 – n2)/(1 + n2)| 2 }εo Eo 2 . (4a) In the limit f1 → f2, the refractive indices n1, n2 in Eq.(4a) are nearly identical, and the above expression simplifies to d qz /dt = 2[(n2 + 1)/(n + 1)2 ]εoEo 2 , (4b) where n = ½(n1 + n2). Momentum conservation requires the above d qz/dt to be balanced by the force exerted on the dielectric medium plus the time rate of change of any momentum taken by the transmitted beam into the dielectric. 3. Superposition of two plane waves in a dielectric medium Inside the medium, each frequency component arrives with an amplitude transmission coefficient (for the E-field) given by τ = 2/(1+ n). The transmitted E- and H-fields may thus be written as follows: Ex(z, t)= [2Eo/(n1 + 1)]sin{2π f1 [(n1 z /c) – t]}– [2Eo/(n2 + 1)]sin{2π f2 [(n2 z /c) – t]} (5a) Incident beam Transmitted X Reflected Refractive index = n( f ) Z (C) 2005 OSA 21 March 2005 / Vol. 13, No. 6 / OPTICS EXPRESS 2247 #6629 - $15.00 US Received 18 February 2005; revised 14 March 2005; accepted 15 March 2005