Problem 1: To prove that under the assumptions at hand, the group velocity of an EM wave is less than c, I am going to show that

Transcription

1 PHY 387 K. Solutons for problem set #7. Problem 1: To prove that under the assumptons at hand, the group velocty of an EM wave s less than c, I am gong to show that (a) v group < v phase, and (b) v group v phases < c 2. (S.1) Together, these two ponts mmedately lead to v group < c. My startng pont s the refracton coeffcent n(ω) = ǫ(ω)µ(ω) and ts frequency dependence. Gven eq. (1) and µ 1, we have n 2 (ω) 1 + Ne2 f m e ǫ 0 ω 2, ω2 ωγ (S.2) whch at frequences not to close to any of the resonances ω.e., n the regme of normal dsperson becomes n 2 (ω) 1 + Ne2 f m e ǫ 0 ω 2. ω2 (S.3) As I have explaned n class, the phase velocty and the group velocty of the EM waves follow from ths refracton ndex accordng to v phase = ω k = c n, v group = dω dk = c n + ω(dn/dω). (S.4) In the normal dsperson regme, the refracton ndex has postve dervatve, dn 2 dω = Ne2 2f ω m e ǫ 0 (ω 2 > 0, (S.5) ω2 ) 2 hence c v group = n + ω dn dω > n = c v phase (S.6) and therefore v group < v phase. Ths proves pont (a). 1

2 To prove pont (b), consder the product c c ( = n n + ω dn ) v phase v group dω = n 2 + ω 2 dn2 dω. (S.7) For the refracton ndex (S.3), n 2 + ω 2 dn2 dω = 1 + Ne2 m e ǫ 0 ω 2 = 1 + Ne2 f ω 2 m e ǫ 0 (ω 2 ω2 ) 2 = 1 + (postve terms) > 1, f Ne2 f ω 2 + ω2 m e ǫ 0 (ω 2 ω2 ) (S.8) hence c v phase c v group > 1 = v group v phases < c 2. (S.9) Ths proves pont (b). Problem 2(a): For the electrc and magnetc felds as n eqs. (2), the space dervatve acts as (k+κ)n z whle the tme dervatve acts as ω. Consequently, the Gauss law equatons D = B = 0 = E = H = 0 (S.10) mply (k +κ)n z E = (k +κ)n z H = 0, (S.11) whch means that both of the ampltude vectors E and H must le n the (x,y) plane transverse to the wave s drecton. Next, the nducton law E = B t = µ 0 H t (S.12) 2

3 leads to (k +κ)n z E = +ωµ 0 H = H = (k +κ) µ 0 ω n z E. (S.13) Lkewse, the Maxwell Ampere equaton H = J + D t = σe + ǫǫ 0 E t (S.14) leads to (k +κ)n 2 H = σ E ωǫǫ 0 E = (σ ωǫǫ 0 ) E (S.15) and hence (k +κ) E = (ωǫǫ 0 + σ) n z H. (S.16) To make sure eqs. (S.13) and (S.16) are consstent wth each other, we need and therefore E = n z (n z E) = µ 0ω (k κ) n z H = + µ 0ω (ωǫǫ 0 + σ) (k +κ) 2 E (S.17) (k +κ) 2 = µ 0 ω (ωǫǫ 0 + σ). (S.18) For convenence, let s defne the complex refracton ndex accordng to n(ω) = ǫ + σ ǫ 0 ω. (S.19) Then n terms of ths complex ndex, eq. (S.18) becomes (k +κ) 2 = n2 (ω)ω 2 c 2 = k = ω c Ren(ω), κ = ω c Im(ω). (S.20) Also, the relaton (S.13) between the electrc and the magnetc ampltudes of the wave becomes H = n(ω) Z 0 n z E where Z 0 = µ0 ǫ Ω s the vacuum mpedance. (S.21) 3

4 Problem 2(b): Fresnel equatons for the reflecton coeffcent r = Ereflected E ncdent (S.22) work for any any refracton ndces, real or complex. In partcular, for an EM wave strkng the boundary head on from the vacuum sde, the Fresnel equaton gves us r = n 1 n+1. (S.23) For a real n of the second materal ths reflecton coeffcent would be real, for an purely magnary n we would have a unmodular r and hence 100% reflectvty R = r 2 = 1, whle for a general complex n. the reflectvty s R = r 2 = n Re(n) n Re(n). (S.24) In partcular, for the complex n from eq. (S.19), Re(n) n ( ) = cos phase(n) = 2 1 phase(n2 ) = 1 2 arccos ǫ n 2 = ǫ 2 n 2, (S.25) hence 2Re(n) = 2(ǫ+ n 2 ) (S.26) and therefore reflectvty R = n (ǫ+ n 2 ) n (ǫ+ n 2 ). (S.27) 4

5 Problem 2(c): For good conductors, the magnary part of n 2, Im(n 2 ) = σ ǫ 0 ω (S.28) s very large for any rado-wave or mcrowave frequences. For example, for the copper at ω = 2π 100 GHz we have Im(n 2 ) 10 7, and even at the vsble-lght frequences we would get Im(n 2 ) However, at the nfrared and hgher frequences we would need a frequency-dependent formula for the conductvty, so eq. (S.19) would no longer be vald. So for the purposes of ths problem, let s lmt ourselves to the rado-wave or mcrowave frequences, and for these frequences Im(n 2 ) 1 for any metal, semmetal, or even a good electrolyte lke the sea water. Thus, for good conductors the magnary part of n 2 s very large whle the real part ǫ s not too large. Consequently, n 2 Im(n 2 ) = σ ǫ 0 ω, or n terms of the skn depth n the conductor (S.29) δ = 2 µ 0 σω (S.30) and the wavelength n vacuum we have Consequently, and therefore reflectvty λ 0 = 2πc ω, n 2 λ2 2π 2 δ 2 1. Re(n) n 2 R 2(λ 0/2πδ) (λ 0 /2πδ) 2(λ 0 /2πδ) (λ 0 /2πδ) = = λ 0 2πδ, ( ) 2πδ λ 0 (S.31) (S.32) (S.33) ( ) 2 2πδ (S.34) λ 0 5

6 Ths reflectvty s very close to 100%, the leadng correcton beng 1 R 4πδ λ 1. (S.35) Problem 2(d): The USW frequency band reserved for the FM rado broadcasts all over the world s from 88 to 108 MHZ, so let work wth (ω/2π) = 100 MHz. At ths frequency, the vacuum wavelength s λ 0 = 3 meters, whle the skn depth s sea water of conductvty σ = 5 Ω /m s only δ = 2 µ 0 σω = 0.7 mm. (S.36) Consequently, 1 R 4πδ λ 0 = , (S.37) thus the sea water reflects 99.7% of the rado-wave s energy. Problem 3(a): To smplfy the notatons, let me ntroduce a few unt vectors, namely n = (+snα,0,+cosα) and n r = (+snα,0, cosα) (S.38) n the drectons of the ncdent and the reflected waves, and m = (+cosα,0, snα) and m r = (+cosα,0,+snα) (S.39) n the drectons to the ncdent/reflected waves wthn the plane of ncdence. In terms of these vectors, the ncdent wave (6) takes a more compact form E (x,t) = E 0 e exp(k 0 n x ωt)exp ( ( m x) 2 /2a 2). (S.40) Fourer transformng ths wave from the coordnate space to the k space n all 3 dmensons, 6

7 we get Ẽ(k,t) = d 3 xe k x E(x,t) = E 0 e e ωt (2π) 2 δ(k y )δ(k n k 0 ) A( m k) (S.41) where A( k) = dwe w k e w2 /2a 2 = 2πa exp( a 2 k 2 /2). (S.42) In terms of ths A( k), the reverse Fourer transform back to the coordnate space becomes E (x,t) = E 0 e exp(k 0 n x ωt) d k 2π A( k) exp( k m x). (S.43) Now consder the reflected wave. For each wave vector k = k 0 n + k m = (k,x,k,y,k,z ) (S.44) of the Fourer-transformed ncdent wave, the reflected wave has k r = (+k,x,+k I,y, k,z ) = k 0 n r + k m r. (S.45) In addton, the reflected wave has a dfferent polarzaton vector er nstead of e and ts ampltude has an extra phase exp(φ(k )). Apart from these dfferences, the reflected wave s completely smlar to the ncdent wave (S.43), thus E r (x,t) = E 0 e r exp(k 0 n r x ωt) d k 2π A( k) exp( k m x) exp(φ(k = k 0 n + k m )). (7) 7

8 Problem 3(b): The A( k) factor n eq. (7) s a very narrow Gaussan peak of wdth (1/a) k 0, so nsde the Fourer ntegral we may take the k = k 0 n + k m wave vector to be rather close to the nomnal k 0 n. Specfcally, we may treat t as havng a fxed magntude x = k 0 whle ts drecton vares n a small range α = α 0 + k k 0 = α 0 ± 1 ak 0. (S.46) Snce the phase φ of the reflecton coeffcent depends on the ncdence angle α, ts value becomes dependent on the k, that s why eq. (7) has the exp(φ) factor nsde the ntegral. But snce the angle α vares n a rather small range, we may approxmate φ(α) φ 0 + dφ dα α = φ 0 + dφ dα k k 0. (S.47) Let s gve the dervatve here a name, or rather let D def = φ(k = k 0 n + k m ) k = 1 k 0 dφ dα (S.48) to keeps the sgn as n eq. (9). Note: for the moment D s defned accordng to eq. (S.48), and we are yet to show that the reflected beam s shfted as on the dagram (4) by precsely D = D. To see how ths works, we need to perform the Fourer ntegral n eq. (7) for φ(k = k 0 n + k m ) φ 0 D k. (S.49) Thus, E r (x,t) = E 0 e r exp(k 0 n r x ωt) (ntegral) (S.50) where (ntegral) d k 2π A( k) exp( k m r x) exp ( φ( ) = φ 0 D k ) = e φ0 d k 2π A( k) exp( k( m r x D) ) = e φ0 exp ( ( m r x D) 2 /2a 2). (S.51) 8

9 Altogether, E r (x,t) = E 0 e r exp(φ 0 + k 0 n r x ωt) exp ( ( m r x D) 2 /2a 2), (S.52) whch s nothng by eq. (8) n slghtly dfferent notatons. And as promsed, the sdeways dsplacement D s precsely the D from eq. (S.48), n perfect agreement wth eq. (9). Quod erat demonstrandum. Problem 3(c): See the last page of my notes on refracton and reflecton. For the wave polarzed normally to the plane of ncdence sn 2 α (n 2 /n 1 ) 2 φ = 2arctan. (S.53) cosα For the wave polarzed parallel to the plane of ncdence φ = 2arctan sn 2 α (n 2 /n 1 ) 2. (n 2 /n 1 ) 2 cosα (S.54) Problem 3(d): Let s rewrte the phase-shft equatons (S.53) and (S.54) as φ = 2arctan ( g(α) ), φ = 2arctan ( (n 1 /n 2 ) 2 g(α) ), (S.55) where g(α) = sn2 α (n 2 /n 1 ) 2 cos 2 α. (S.56) Takng the dervatves, we fnd dg dα = 2snαcosα cos 2 α + 2snα(sn2 α (n 2 /n 1 ) 2 ) cos 3 α = 2(1 (n 2 /n 1 ) 2 ) snα cos 3 α. (S.57) 9

10 dφ dg = +2 = 1 1+( g) g = 1 g(1+g) cosα cos 2 α sn 2 α (n 2 /n 1 ) 2 1 (n 2 /n 1 ) 2, (S.58) dφ dg = +2 = (n 1 /n 2 ) (n 1 /n 2 ) 4 g 1 2 g cosα sn 2 α (n 2 /n 1 ) 2 (n 2 /n 1 ) 2 cos 2 α (n 2 /n 1 ) 4 cos 2 α + sn 2 α (n 2 /n 1 ) 2 = dφ dg 1 (1+(n 1 /n 2 ) 2 )sn 2 α + 1, (S.59) and consequently D = 1 k dφ dg dg dα = 2 k snα (10) sn 2 α (n 2 /n 1 ) 2 and D = 1 k dφ dg dg dα = D 1 (1+(n 1 /n 2 ) 2 )sn 2 α + 1. (11) Quod erat demonstrandum. 10