1 sin 2 r = 1 n 2 sin 2 i

Transcription

1 Physis 505 Fall 005 Homework Assignment #11 Solutions Textbook problems: Ch. 7: 7.3, 7.5, 7.8, 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 k x and e i k x on the inident side, E + e i k 0 x and E e i k 0 x in the air gap, and 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. 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 where we have introdued the phase φ k 0 d k 0 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)

2 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 R 4α 4α os φ + (1 + α ) sin φ 4α 4α + (1 α ) sin φ (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 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) n 1 E 1 (1 β) + 1 (1 + β) (4)

3 and where this time 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, and 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 T R 4γ 4γ + (1 + γ ) sinh ξ (1 + γ ) sinh ξ 4γ + (1 + γ ) sinh ξ 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 when d 0 (orresponding to ξ 0) and that T falls exponentially to 0 when d (whih is the same as ξ ). 7.5 A plane polarized eletromagneti wave E e i k x iωt is inident normally on a flat uniform sheet of an exellent ondutor (σ ωɛ 0 ) having thikness D. Assuming that in spae and in the onduting sheet µ/µ 0 ɛ/ɛ 0 1, disuss the refletion and transmission of the inident wave.

4 a) Show that the amplitudes of the refleted and transmitted waves, orret to the first order in (ɛ 0 ω/σ) 1/, are: where (1 e λ ) (1 e λ ) + γ(1 + e λ ) γe λ (1 e λ ) + γ(1 + e λ ) ɛ0 ω ωδ γ (1 i) (1 i) σ λ (1 i)d/δ and δ /ωµσ is the penetration depth. So long as we treat the ondutor as a medium with omplex dieletri onstant ɛ/ɛ i σ ωɛ 0 we may proeed as if everything were a dieletri. Sine there are two boundaries, this problem is very muh like the above Problem 7.3, exept the expressions are even simpler beause of the normal inidene. As above, we introdue eletri field vetors of the form e i k x and e i k x on the inident side, E + e i k 1 x and E e i k 1 x in the ondutor, and e i k ( x D) on the transmitted side. We use mathing for E perpendiular to the plane of inidene (whih orresponds to a sign onvention of having all eletri fields pointing in the same diretion). In this ase, the mathing beomes first interfae seond interfae E : + E + + E, E + e iφ + E e iφ H : ( ) n(e + E ), n(e + e iφ E e iφ ) where n is the omplex index of refration ɛ n 1 + i σ (5) ɛ 0 ωɛ 0 and φ is the phase hange for going through the dieletri φ k 1 D ωn D ωd Solving for and in terms of, we obtain 1 + i σ ωɛ 0 (6) 4/n (1 + 1/n) e iφ (1 1/n) e iφ 4/ne iφ (1 + 1/n )(1 e iφ ) + /n(1 + e iφ ) (1 1/n )(e iφ e iφ ) (1 + 1/n) e iφ (1 1/n) e iφ (1 1/n )(1 e iφ ) (1 + 1/n )(1 e iφ ) + /n(1 + e iφ )

5 whih is essentially equivalent to (3), up to redefining α 1/n. (In fat, this problem an easily be generalized for inidene at an arbitrary angle i by taking 1/n os i/n os r.) We now take the limit where this is an exellent ondutor, σ/ωɛ 0 1. In this ase, the index of refration (5) and phase hange (6) may be approximated by n i σ σ (1 + i) ωɛ 0 ɛ 0 ω γ φ ωd n (1 + i)ωd σ ɛ 0 ω µ0 σω (1 + i)d iλ For γ 1 (equivalent to n 1) we drop terms of O(1/n ) ompared to 1 to arrive at γe λ (1 e λ ) + γ(1 + e λ ) (1 e λ ) (1 e λ ) + γ(1 + e λ ) where we have kept the O(γ) term in the denominator whih beomes important in the limit λ 0. b) Verify that for zero thikness and infinite thikness you obtain the proper limiting results. The zero thikness limit orresponds to λ 0. In this ase, it is easy to see from (7) that λ 0 : 1, 0 In the infinite thikness limit, we find instead (7) λ : 0, γ Note that the refletion oeffiient does not go to unity, as some of the power is dissipated in the ondutor. A perfet ondutor (σ ) has γ 0, so all the power is refleted in the perfet ondutor limit. ) Show that, exept for sheets of very small thikness, the transmission oeffiient is 8(Rγ) e D/δ T 1 e D/δ os(d/δ) + e 4D/δ Sketh log T as a funtion of (D/δ), assuming Rγ 10. Define very small thikness. To ompute the transmission oeffiient from (7), we keep in mind that both γ and λ are omplex. As long as we are not in the very small thikness limit, the

6 O(γ) term in the denominator an be ignored. In this ase so that T γe λ (1 e λ ) 4 γ e Rλ 1 R(e λ ) + e 4Rλ Taking γ (Rγ) as well as e λ e id/δ e D/δ then gives T 8(Rγ) e D/δ 1 e D/δ os(d/δ) + e 4D/δ The very small thikness limit orresponds to when the O(γ) term beomes important. This ours when Expanding for small λ yields 1 e λ γ(1 + e λ ) λ γ Hene small thiknesses orrespond to D δ ωδ D < ωδ 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 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 0 T layer (n j, t j ) 0 e ik jt j I os(k j t j ) + iσ 3 sin(k j t j ) E

7 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. Again, 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 0)e ik jt j E + e ik jt j E E (z j ) E (z 0)e ik jt j This diretly leads to the transfer matrix ( ) e ik j t j 0 T layer (n j, t j ) 0 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 0 ) to n (x > x 0 ) is T interfae (, 1) 1 ( ) n + 1 (n 1) (n + 1) (n 1) I σ (n 1) n where n n 1 /n. For the mathing aross layers, we again 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 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

8 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 0 ) T ( Ei ) ( ) ( ) t11 t 1 Ei t 1 t t 11 + t 1, 0 t 1 + t whih may be solved to obtain t 1, t 11t t 1 t 1 det(t ) t t t 7.16 Plane waves propagate in a homogeneous, nonpermeable, but anisotropi dieletri. The dieletri is haraterized by a tensor ɛ ij, but if oordinate axes are hosen as the priniple axes, the omponents of displaement along these axes are related to the eletri-field omponents by D i ɛ i (i 1,, 3), where ɛ i are the eigenvalues of the matrix ɛ ij. a) Show that plane waves with frequeny ω and wave vetor k must satisfy k ( k E) + µ0 ω D 0 This is in fat the general Maxwell wave equation, and does not depend on the details of the dieletri tensor. This may be derived from the url equations, using i k and / t iω. In a soure-free region, the Ampère-Maxwell and Faraday laws give i k H iω D, i k E iω B 0 Taking i k ross Faraday s law, and using B µ 0 H gives i k (i k E) iµ 0 ω(i k H) 0 It is then straightforward to substitute in Ampère s law in the seond term to arrive at k ( k E) + µ0 ω D 0

9 b) Show that for a given wave vetor k kˆn there are two distint modes of propagation with different phase veloities v ω/k that satisfy the Fresnel equation 3 i1 n i v v i 0 where v i 1/ µ 0 ɛ i is alled a prinipal veloity, and n i is the omponent of ˆn along the ith prinipal axis. Letting k kˆn, and using the BAC CAB rule, we find ˆn(ˆn E) E + µ 0 v D 0 By working with the priniple axes, this equation may be entirely written in terms of E. Introduing the real symmetri matries A ij n i n j δ ij, W ij δ ij µ 0 ɛ j δ ij /v j we arrive at a generalized eigenvalue problem A E v W E or (A + v W) E 0 (8) The veloities of propagation are then the eigenvalues of this problem, and may be obtained by solving the seular equation 0 det(a + v W) v 6 n 1 + n + n 3 v 1 v v 3 + v ( n 1 [ v v1 v v 3 v 1 + n v v 4 ( n + n 3 + n 3 v 3 ) v v 3 + n 1 + n 3 v 1 v 3 + n 1 + n ) v1 v n 1(v v)(v v3) + n (v v1)(v v3) ] + n 3(v v1)(v v) Other than a trivial solution, v 0 (whih does not orrespond to a propagating mode), we find two veloities, v a and v b, orresponding to the two roots of the quadrati equation for v in the square brakets. In fat, taking the equation in brakets and dividing out by the produt Π i (v vi ) immediately gives the Fresnel equation n i v vi 0 i ) Show that D a D b 0, where D a, D b are the displaements assoiated with the two modes of propagation.

10 Here we may use standard linear algebra tehniques related to the orthogonality of eigenvetors. Considering first the generalized eigenvalue problem (8), we take distint eigenvalues v a and v b. Then the orresponding eigenvetors satisfy the equations (A + v aw) E a 0, (A + v b W) E b 0 Left-multiplying the first equation by E b and the seond by E a gives E b A E a + v a E b W E a 0, Ea A E b + v b E a W E b 0 Sine A and W are symmetri (real Hermitian), we may transpose the first equation and subtrat it from the seond. The result is (v b v a) E a W E b 0 whih implies E a W E b 0, sine v a v b (in the ase that v a v b, we may instead Gram-Shmidt orthogonalize to make the eigenvetors orthogonal). Finally, sine W is µ 0 times the dieletri matrix Σ diag(ɛ 1, ɛ, ɛ 3 ), and sine D Σ E, we may equivalently rewrite this orthogonality (with respet to the measure or metri W) as E a D b 0 or Eb D a 0 However, we an in fat learn more than this. Sine the matrix A ˆn ˆn I is not arbitrary, it satisfies the (almost) projetion ondition A A. As a result D a D b E a Σ Eb 1 µ 0 E a W Eb But sine A E b v b W E b, we obtain 1 µ 0 v av b E a A Eb 1 µ 0 v av b E a A E b D a D b 1 E µ a WE b 1 E 0 v a µ 0 va a D b 0 (Note, however, that in general E a E b 0.)