674. J.Opt.Soc.Am.A/Vol.4,No.4/April 1987 Y.-F.Liand J.W.Y.Lit n3 tan(hodo) (for even modes) (27a) 4 -cot(hodo) (for odd modes) n2 (27b) d2 If we set n d no nor n2>B>ni ng (28) 2do and n1 p:=l82-h2n,22 (6=1,2,3), (29) n2 then we have n3 P1=P1, Fig.2.Geometry of a symmetrical seven-layer waveguide D2=jp2, (30) The above verifications show that,for the cases of three P3=p3 and four layers,our general formulas give results that agree with the published results. Equation(26)becomes Symmetrical Seven-Layer Waveguide ho A1+ P3 tanh(pid)+ tanh(P2d2) Ruschin and Marom14obtained the eigenvalue equations for a symmetrical seven-layer waveguide by using matrix treat- ment.In this section we shall get the eigenvalue equations P3+tanh(pidi) P2 P3 tanh(pdi) tanh(p2d2).(31) for the same structure by using our formulas and then com- P2 pare our results with theirs. The symmetrical seven-layer waveguide is shown in Fig.2. By using our parameters and performing lengthy algebraic We have dropped all the pluses and minuses from the sub- manipulations,it can be shown that our eigenvalue equation scripts because the structure is symmetrical.Layers 0 and 2 (31)will lead to the same solution for both the odd and even are the guiding layers with high refractive indices.n is the modes given by Eqs.(21a)and(21b)in the paper by Ruschin index of refraction of the clad,and n3 is that of the surround- and Marom.14 Here we wish to point out that in that paper ing medium there is a typographical error,namely,the minus between the For a symmetrical waveguide,the eigenvalue Eq.(8)can two main terms in Eq.(21a)should be a plus.23 be written as We can check the eigenvalue equations quickly by reducing them to those for the symmetrical five-or three-layer wave- 2hodo=200+q (g=0,1,.), (22) guides. that is, Symmetrical Five-Layer Waveguide tan(hodo)=tan o (23) We put d2=0,so Eq.(31)yields ho for the even modes and A p 1+P3 tanh() P3 tanh(p d). (32) -cot(hodo)=tanφpo (24) Symmetrical Three-Layer Waveguide for the odd modes. The half-phase shift o is given by We put di=d2=0,so Eq.(31)yields tan o=(pi/ho)tanh vi, (25a) A=p3/ho (33) tanh(p:d)+(pi+/p)tanhit which is a well-known result. tanh(pptanh(p:d)tanh (i=1,2,.,l-2), (25b) FORMULAS IN TERMS OF THE NORMALIZED VARIABLES tanh(pi-1d1)+(pilpi-1) tanh1+(p/p-tanh(p-d-1) (25c) To obtain the properties of a multilayer slab waveguide,we have to evaluate the eigenvalue Eq.(8)for TE modes and Eq. By using Eqs.(23)-(25),we can get the two eigenvalue equa- (13)for TM modes to get the propagation constant B.To tions for the symmetrical seven-layer waveguide: make the results of such a numerical evaluation more broadly applicable,we introduce a normalized thickness v and other tanh(pd) normalized parameters: u2=h2d2n02-n-nm2), (34a) +++ae]时 u2=h2d02(n,2-N. (34b) tanh(p2d2),(26) w:2=h2d2W2-n,3)(i=-m,,-1,0,+1,.,+0, where (34c)674. J. Opt. Soc. Am. A/Vol. 4, No. 4/April 1987 n3 T n, d2 fli~~~~~~~~~~~~~~~~~~ ni di no 2do ni n2 n3 Fig. 2. Geometry of a symmetrical seven-layer waveguide. The above verifications show that, for the cases of three and four layers, our general formulas give results that agree with the published results. Symmetrical Seven-Layer Waveguide Ruschin and Marom14 obtained the eigenvalue equations for a symmetrical seven-layer waveguide by using matrix treat￾ment. In this section we shall get the eigenvalue equations for the same structure by using our formulas and then com￾pare our results with theirs. The symmetrical seven-layer waveguide is shown in Fig. 2. We have dropped all the pluses and minuses from the sub￾scripts because the structure is symmetrical. Layers 0 and 2 are the guiding layers with high refractive indices. nj is the index of refraction of the clad, and n3 is that of the surround￾ing medium. For a symmetrical waveguide, the eigenvalue Eq. (8) can be written as 2hod0 2='20 +q-zr (q =0,1I....), that is, tan(hodo) = tan 00 for the even modes and -cot(hodo) = tan 'o for the odd modes. The half-phase shift 0' is given by (22) (23) (24) A = an(hodo) -cot(hodo) (for even modes) (for odd modes) (27a) (27b) If we set no n2 > 1 > n,, n3 and Pi= I13-k 2 ni2 1 2 (i = 1, 2, 3), then we have Pi = Pi, P2 = JP2, (28) (29) (30) P3 = P3- Equation (26) becomes °A 11+ 3 tanh(pd,) +JP - tanh(pld,) tanh(p 2A2)r Pi Pi L P J j = + tanh(pldj) - [r = tanh(pidj) tanh(P 2d2 ). (31) Pi ~ ~ L'P2J By using our parameters and performing lengthy algebraic manipulations, it can be shown that our eigenvalue equation (31) will lead to the same solution for both the odd and even modes given by Eqs. (21a) and (21b) in the paper by Ruschin and Marom.14 Here we wish to point out that in that paper there is a typographical error, namely, the minus between the two main terms in Eq. (21a) should be a plus.23 We can check the eigenvalue equations quickly by reducing them to those for the symmetrical five- or three-layer wave￾guides. Symmetrical Five-Layer Waveguide We put d2 = 0, so Eq. (31) yields ° A [1 + 3 tanh(pldj) = 3 + tanh(pldl). (32) Symmetrical Three-Layer Waveguide We put d1 = d2 = 0, so Eq. (31) yields tan 'P = (p,/ho)tanh 4s,, tanh(pid;) + (Dp:_/p:)tanh tib:- (25a) tanh - = r ti *1 - I " 1-1 i1 + (pi+,/pj)tanh(pjdj)tanh 4i+i (i = 1, 2,...,1 - 2), (25b) tanh(pj_1dj_1 ) + (pl/IP-,) 1 + (pj/p1_I)tanh(p1.. dj..) By using Eqs. (23)-(25), we can get the two eigenvalue equa￾tions for the symmetrical seven-layer waveguide: °A 1+ P3 tanh(pdj) + [-+ p2tanh(pldl) utanh(p2A2)| Pi 1 Pi L2 PI J = Pp + tanh(pldl) + +-tanh(pdl)J p2 tanh(pAd2 ), (26) where A = P3/hoX (33) which is a well-known result. FORMULAS IN TERMS OF THE NORMALIZED VARIABLES )To obtain the properties of a multilayer slab waveguide, we have to evaluate the eigenvalue Eq. (8) for TE modes and Eq. (13) for TM modes to get the propagation constant 13. To make the results of such a numerical evaluation more broadly applicable, we introduce a normalized thickness v and other normalized parameters: V2= k 2 d0 2 (n 0 2 -n 2), Ui2 = k 2 d0 2 (n 2 - N2 ). wi2 = k 2 d0 2 (N2 -n 2) (34a) (34b) (i =-m,. .. .,-1, O. +1, ..., +1), (34c) Y.-F. Li and J. W. Y. Lit