n n=1 (air) n 1 sin 2 r =

Transcription

1 Physis 55 Fall 7 Homework Assignment #11 Solutions Textbook problems: Ch. 7: 7.3, 7.4, 7.6, Two plane semi-infinite slabs of the same uniform, isotropi, nonpermeable, lossless dieletri with index of refration n are parallel and separated by an air gap (n 1) of width d. A plane eletromagneti wave of frequeny ω is indient on the gap from one of the slabs with angle of indiene i. For linear polarization both parallel to and perpendiular to the plane of inidene, a) alulate the ratio of power transmitted into the seond slab to the inident power and the ratio of refleted to inident power; We introdue (omplex) eletri field vetors of the form E i e i k x and E r e i k x on the inident side, E+ e i k x and E e i k x in the air gap, and E t e i k ( x d) on the transmitted side. (We have removed an unimportant phase from the transmitted side by shifting x by the vetor d pointing from the inident to the transmitted side of the air gap). n n1 (air) n k k i k k r k d If i is the inident angle, then the angle r from the normal in the air gap is given by Snell s law, n sin i sin r, and the transmitted angle is also i (beause it is the same dieletri). We see that os r 1 sin r 1 n sin i and that os r is purely imaginary in the event that i is greater than the ritial angle for total internal refletion. To obtain and in terms of, we may math the parallel omponents of E as well as the parallel omponents of H. We onsider two ases. For E perpendiular to the plane of inidene, the mathing beomes first interfae seond interfae E : + E + + E, E + e iφ + E e iφ H : n( ) os i (E + E ) os r, (E + e iφ E e iφ ) os r n os i

2 where we have introdued the phase φ k d k d os r ωd os r The mathing onditions at the first interfae may be written as E + 1 (1 + α) + 1 (1 α) E 1 (1 α) + 1 (1 + α) (1) where we have defined α n os i os r n os i 1 n sin i Similarly, the mathing onditions at the seond interfae yield E + 1 e iφ (1 + α) E 1 eiφ (1 α) () Equating (1) and () allows us to solve for the ratios 4α (1 + α) e iφ (1 α) e iφ α α os φ i(1 + α ) sin φ (1 α )(e iφ e iφ ) (1 + α) e iφ (1 α) e iφ i(1 α ) sin φ α os φ i(1 + α ) sin φ (3) where α n os i 1 n sin i, ωd os r φ ωd 1 n sin i So long as i is below the ritial angle, both α and φ are real. In this ase, the transmission and refletion oeffiients are T 4α 4α os φ + (1 + α ) sin φ 4α 4α + (1 α ) sin φ (4) R (1 α ) sin φ 4α os φ + (1 + α ) sin φ (1 α ) sin φ 4α + (1 α ) sin φ Note that T + R 1, as expeted. However, this exhibits a lassi interferene behavior, where T osillates between (α/(1 + α )) and 1 as the number of wavelengths in the gap vary. For E parallel to the plane of inidene, we find instead the mathing onditions first interfae seond interfae E : ( ) os i (E + E ) os r, (E + e iφ E e iφ ) os r os i H : n( + ) (E + + E ), E + e iφ E e iφ n

3 These equations have the same struture as the perpendiular ase, but with the index of refration entering somewhat differently. We find the mathing onditions n 1 E + 1 (1 + β) + 1 (1 β) and where this time n 1 E 1 (1 β) + 1 (1 + β) β n 1 E + 1 e iφ (1 + β) n 1 E 1 eiφ (1 β) os i n os r os i n 1 n sin i These expressions are similar to (1) and () above, exept with the replaement E ± n 1 E ± and α β. Hene the transmission and refletion oeffiients are given by expressions idential to (4), exept with the replaement α β. b) for i greater than the ritial angle for total internal refletion, sketh the ratio of transmitted power to inident power as a funtion of d measured in units of wavelength in the gap. To be onrete, onsider the ase for E perpendiular to the plane of inidene. Sine i is greater than the ritial angle, both α and φ will be purely imaginary. Whatever values they are, define α iγ, φ iξ Then the ratios / and / in (3) beome iγ iγ osh ξ + (1 γ ) sinh ξ (1 + γ ) sinh ξ iγ osh ξ + (1 γ ) sinh ξ so that where T R 4γ 4γ + (1 + γ ) sinh ξ (1 + γ ) sinh ξ 4γ + (1 + γ ) sinh ξ n os i γ n sin i 1, ξ ωd n sin i 1 In this ase, there is no osillatory behavior in the transmitted power, but only exponential suppression as the air gap is widened. It is easy to see that T 1

4 when d (orresponding to ξ ) and that T falls exponentially to when d (whih is the same as ξ ). For n 1.5 (approximately the index of refration of glass), the ritial angle for total internal refletion is i 4. A sketh of T as a funtion of d looks like T 1 n i 45 i 6 i 75. d/( π / k ) A plane-polarized eletromagneti wave of frequeny ω in free spae is inident normally on the flat surfae of a nonpermeable medium of ondutivity σ and dieletri onstant ɛ. a) Calulate the amplitude and phase of the refleted wave relative to the inident wave for arbitrary σ and ɛ. A medium of dieletri onstant ɛ and ondutivity σ may be desribed by an effetive dieletri onstant ε ɛ + i σ ω Sine the medium is nonpermeable, we have µ µ. As a result, for normal inidene, the ratio of the refleted to the inident eletri field is where n 1 n 1 + n ε ɛ + i σ ɛ ɛ ɛ ω The amplitude A and phase ϕ of the refleted wave is defined by Ae iϕ 1 ɛ/ɛ + iσ/ɛ ω 1 + ɛ/ɛ + iσ/ɛ ω This expression impliitly defines A and ϕ. To be more expliit, we deompose the omplex index of refration into a magnitude and phase n ηe iα/

5 where ( ) ɛ η + The amplitude of the refleted wave is then (1 n)(1 n A ) (1 + n)(1 + n ) ɛ ( ) σ, tan α σ ɛ ω ɛω 1 + n Rn 1 + n + Rn 1 + η η os(α/) 1 + η + η os(α/) (5) (6) while the phase is ϕ arg 1 n 1 + n arg (1 n)(1 + n ) (1 + n)(1 + n ) In η sin(α/) tan 1 tan 1 1 n 1 η (7) Note that some are must be taken when extrating the phase. In partiular, for ϕ tan 1 (y/x), we must ensure that the angle ϕ lies in the proper quadrant defined by the point (x, y). This is why we hoose to keep the minus sign in the numerator inside the artan. Sine the numerator is always negative, ϕ must lie in either the 3rd or the 4th quadrant. For η > 1, whih is the ase for all onventional dieletris, ϕ lies in the 3rd quadrant. To highlight this, we may write ϕ π + tan 1 η sin(α/) η 1 b) Disuss the limiting ases of a very poor and a very good ondutor, and show that for a good ondutor the refletion oeffiient (ratio of refleted to inident intensity) is approximately where δ is the skin depth. R 1 ω δ We begin with the ase of a very poor ondutor, σ ɛω. In this ase, η and α in (5) simplify to η ɛ ɛ, α σ ɛω where we have kept only linear terms in σ. Substituting this into (6) and (8) gives A 1 η 1 + η 1 n 1 + n η σ ϕ π + η 1 ɛω π + n σ n 1 ɛω where n ɛ/ɛ. Here we have assumed that n > 1 and that σ/ɛω n 1. For a very good ondutor, we take the opposite limit, σ ɛω. In this ase η σ ɛ ω 1, α π (8)

6 Inserting this into (6) and (8) gives η ( η + 1 A η + η ) 1/ /η /η η ϕ π + tan 1 η 1 π + tan 1 η π η 1 ɛ ω σ The amplitude A may be rewritten in terms of the skin depth δ /µ σω A 1 ωδ µ ɛ 1 ω δ This gives the refletion oeffiient R A 1 ω δ (9) 7.6 A plane wave of frequeny ω is inident normally from vauum on a semi-infinite slab of material with a omplex index of refration n(ω) [n (ω) ɛ(ω)/ɛ ]. a) Show that the ratio of refleted power to inident power is R 1 n(ω) 1 + n(ω) while the ratio of power transmitted into the medium to the inident power is T 4Rn(ω) 1 + n(ω) While this problem involves a omplex dieletri onstant ɛ(ω), we note that the mathing onditions for inident and refleted waves at an interfae hold for arbitrary (inluding omplex) values of µ and ɛ. For normal inidene, the expressions are simply 1 n(ω) 1 + n(ω), 1 + n(ω) where we have furthermore assumed that the material is non-permeable so that µ µ. For harmoni waves, the power is obtained from the real part of the Poynting vetor S 1 E H 1 ɛ µ E ˆn (1)

7 The refletion oeffiient is then straightforward R R(ˆn S r ) R(ˆn S i ) 1 n(ω) 1 + n(ω) For the transmission oeffiient, we also have to aount for the different material T R(ˆn S t ) R(ˆn S i ) R ɛ(ω) ɛ 4R[n(ω)] 1 + n(ω) b) Evaluate R[iω( E D B H )/] as a funtion of (x, y, z). Show that this rate of hange of energy per unit volume aounts for the relative transmitted power T. We write the eletri and magneti fields inside the material as E e ikˆn x, B µ ɛ n(ω)ˆn e ikˆn x where the omplex wavenumber k is given by k(ω) ωn(ω) In this ase, the power per unit volume expression beomes [ ] [ )] iω R ( E D B H iω ) R (ɛ(ω) E 1µ B [ ] iω R (ɛ(ω) ɛ n(ω) ) E t e I[k(ω)]ˆn x [ ] iɛ ω R (n(ω) n(ω) ) E t e I[k(ω)]ˆn x ɛ ωi[n(ω) ] E t e I[k(ω)]ˆn x ɛ ωr[n(ω)]i[n(ω)] E t e I[k(ω)]ˆn x ɛ R[n(ω)]I[k(ω)] E µ t e I[k(ω)]ˆn x (11) The power per area transmitted into the material may then be alulated as [ ] iω P t /A R ( E D B H ) dz ɛ R[n(ω)]I[k(ω)] E µ t 1 ɛ R[n(ω)] E µ t e I[k(ω)]z dz

8 On the other hand, the inident power per area may be obtained from (1) P i /A 1 ɛ E µ i (1) This gives P t R[n(ω)] P i R[n(ω)] n(ω) whih agrees with the above alulation of the transmission oeffiient. Note that the omplex Poynting vetor inside the material is Hene S 1 ɛ µ n(ω) e I[k(ω)]ˆn xˆn R( S ɛ ) R[n(ω)]I[k(ω)] E µ t e I[k(ω)]ˆn x (13) Comparing this with (11) demonstrates that the real part of the omplex Poynting s theorem holds S + iω ( E D B H ) + 1 J E so long as we take J (ie no free urrents). ) For a ondutor, with n 1 + i(σ/ωɛ ), σ real, write out the results of parts a and b in the limit ɛ ω σ. Express your answer in terms of δ as muh as possible. Calulate 1 R( J E) and ompare with the result of part b. Do both enter the omplex form of Poynting s theorem? For a ondutor with σ ωɛ, we make the approximation n 1 + i σ σ (1 + i) (1 + i) ωɛ ωɛ ωδ where we have introdued the skin depth δ /µ σω. In this ase, the refletion oeffiient is approximately R 1 n 1 + n 1 n n 1 1 (1 i)ωδ/ 1 + (1 i)ωδ/ 1 (1 i)ωδ ( 1 (1 i)ωδ 1 ωδ ) + i ωδ 1 ωδ Not surprisingly, this is the same result as (9). The transmission oeffiient is approximately T 4R n 1 + n 4/ωδ 1 + (1 + i)/ωδ 4/ωδ (1 + i)/ωδ ωδ (14)

9 Note that R + T 1 as expeted. For the power per unit volume of part b, we have from (11) [ ] iω R ( E D B H ) ɛ ωr(n)i(n) E t e I(k)ˆn ˆx ( ) ɛ ω Et e ˆn x/δ ωδ ɛ ωδ E t e ˆn x/δ Integrating this along z gives a power per area transmitted into the ondutor P t /A ɛ ωδ E t e z/δ dz ɛ ωδ E t Comparing this with the inident power per area (1) gives P t P i ωδ ωδ n 4 ωδ 1 n 4 ωδ whih agrees with the transmission oeffiient (14). 1 (1 + i)/ωδ ωδ For the divergene of the Poynting vetor, note that (13) beomes R( S) ɛ On the other hand, using J σ E, we see that ωδ e ˆn x/δ R[ 1 ( J E)] R[ 1 σ E ] 1 σ e I[k(ω)]ˆn x ɛ ωδ e ˆn x/δ This expression gives the same value as the power per unit volume term R[iω( E D B H )/]. Hene if we inlude both the power per unit volume term and the work term R[ 1 ( J E)] in the omplex Poynting s theorem S + iω ( E D B H ) + 1 J E we would double ount the ontribution of the urrent, and this theorem would appear to be violated. To get the orret result, we reall that we have a hoie of where the urrent J should be ounted. In partiular, by writing n 1 + i(σ/ωɛ ), we have played the trik of hiding the urrent J in the eletri displaement D D eff D + i ( J ω ɛ + iσ ) E ɛ effe ω

10 In this ase, the Ampère-Maxwell law beomes simply H iω D eff In partiular, one we assume D eff ɛ eff E, we have set the expliit urrent to zero in this equation (although it is hidden in D eff ). In this ase, the orret omplex Poynting s theorem reads S + iω ( E D eff B H ) where the work term is hidden in the power per unit volume term. Sine expression (11) was atually alulated with D eff, this is the form of the Poynting s theorem that we have diretly shown for the ondutor. On the other hand, if we treat the urrent J as an expliit quantity, then it enters through the work term R[ 1 ( J E)], but does not enter the power per unit volume term. In this ase, we return to the full Poynting s theorem S + iω ( E D B H ) + 1 J E where D is the honest eletri displaement, without the addition of an effetive urrent ontribution. (For n 1 + i(σ/ωɛ ), we have D ɛ E). Using this expression for D modifies the alulation of (11). In partiular R [ iω ( E D B H ) ] R [ iω )] (ɛ E 1µ B sine ɛ and µ are both real. As a result, when treating J expliitly, power onservation is a balane between S and 1 J E only. 7.8 A monohromati plane wave of frequeny ω is indident normally on a stak of layers of various thiknesses t j and lossless indies of refration n j. Inside the stak, the wave has both forward and bakward moving omponents. The hange in the wave through any interfae and also from one side of a layer to the other an be desribed by means of transfer matries. If the eletri field is written as E E + e ikx + E e ikx in eah layer, the transfer matrix equation E T s expliitly ( ) ( ) ( ) E + t11 t 1 E+ t 1 t E E a) Show that the transfer matrix for propagation inside, but aross, a layer of index of refration n j and thikness t j is ( ) e ik j t j T layer (n j, t j ) e ik jt j I os(k j t j ) + iσ 3 sin(k j t j )

11 where k j n j ω/, I is the unit matrix, and σ k are the Pauli spin matries of quantum mehanis. Show that the inverse matrix is T. Normal inidene makes this problem straightforward. For a right moving plane wave of the form e ikjz passing through a layer of thikness t j, one piks up a phase e ik jt j, while for a left moving wave, one piks up a phase e ik jt j. More preisely E + E + (z t j ) E + (z )e ik jt j E + e ik jt j E E (z t j ) E (z )e ik jt j This diretly leads to the transfer matrix ( ) e ik j t j T layer (n j, t j ) e ik jt j where the inverse is obviously the omplex onjugate. E e ik jt j b) Show that the transfer matrix to ross an interfae from n 1 (x < x ) to n (x > x ) is T interfae (, 1) 1 where n n 1 /n. ( ) n + 1 (n 1) (n + 1) (n 1) I σ (n 1) n For the mathing aross layers, we take the E perpendiular to plane of inidene onventions. This gives simply E : whih may be solved to give E + + E E + + E H : n 1 (E + E ) n (E + E ) E + 1 E +(1 + n) + 1 E (1 n) E 1 E +(1 n) + 1 E (1 + n) where n n 1 /n. This yields the transfer matrix T interfae (, 1) 1 ( n + 1 ) (n 1) (n 1) n + 1 ) Show that for a omplete stak, the inident, refleted, and transmitted waves are related by rans det(t ) n, efl t 1 n t t

12 where t ij are the elements of T, the produt of the forward-going transfer matries, inluding from the material filling spae on the inident side into the first layer and from the last layer into the medium filling the spae on the transmitted side. It ought to be lear that the omplete effet of going through several layers is to take a produt of transfer matries. For example E T E, where T T (4, 3)T (n 3, t 3 )T (3, )T (n, t )T (, 1) The transmitted and refleted eletri fields are obtained by solving This gives expliitly ( Et ) T ( Ei ) ( ) ( ) t11 t 1 Ei t 1 t t 11 + t 1, t 1 + t whih may be solved to obtain t 1, t 11t t 1 t 1 det(t ) t t t