Abstract

The theory of optical modes guided by a homogeneous thin film is developed in optical terminology. The properties of the modes are determined from the zeros of characteristic equations which describe the phenomena. When there are no losses, these equations are solved exactly. For the lossy case, simple exact solutions can not be found. A perturbation procedure is introduced which yields explicit first-order expressions for the optical constants N and K of any mode which can be supported by a lossy system. Under stated circumstances, the attenuation of the mode will be much less than that suggested by the loss tangents of the dielectrics comprising the system.

© 1964 Optical Society of America

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  1. N. S. Kapany, J. J. Burke, and C. C. Shaw, J. Opt. Soc. Am. 53, 929 (1963).
    [Crossref]
  2. If the substrate is chosen as a perfect conductor, then we already have developed the theory since it would be equivalent to a plane of symmetry for TM modes, and a plane of antisymmetry for TE modes.

1963 (1)

J. Opt. Soc. Am. (1)

Other (1)

If the substrate is chosen as a perfect conductor, then we already have developed the theory since it would be equivalent to a plane of symmetry for TM modes, and a plane of antisymmetry for TE modes.

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Figures (5)

Fig. 1
Fig. 1

The geometry of the symmetrical two-media problem. The x axis is centered in the film, and the mode travels in a direction perpendicular to the polarization vector (Hz or Ez) which is parallel to the layer.

Fig. 2
Fig. 2

The geometry of the boundary value problems, in a reduced domain, which are equivalent by reasons of symmetry to that posed by Fig. 1.

Fig. 3
Fig. 3

The variation of N as a function of the total optical thickness 2ℓ/λ for the geometry of Fig. 1, where n1=1.52 and n2=1.70. There is no significant variation in the N if κ1 and/or κ2 are nonzero.

Fig. 4
Fig. 4

The effect of dielectric film losses on the first three modes’ attenuation if n1=1.52, n2=1.70, and κ1=0.0. The vertical scale is so normalized that it is independent of any particular choice of κ2.

Fig. 5
Fig. 5

The geometry of the three-media problem. The particular mode sketched is a quasieven one since it is symmetric about the x axis in the k2 region.

Tables (1)

Tables Icon

Table I Results of exact calculations of N and K for the TM0 mode if n1=1.52, n2=1.70, κ1=0.0, and a range of 2l0. No discernable difference was observed in the value of N as κ2 varied from 0.0 to 10−2.

Equations (55)

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curl E = i k 0 H ,
curl H = - i k 0 m j 2 E ,
k 0 = 2 π / λ 0 = ω / c ;
m j 2 = n j 2 ( 1 + i κ j ) 2 = j + i 4 π σ j / ω ,
TM :             E x = - 1 i k 0 m j 2 V y ,             E y = + 1 i k 0 m j 2 V x , E z = H x = H y = 0 ,
TE :             H x = + 1 i k 0 U y ,             H y = - 1 i k 0 U x , H z = E x = E y = 0 ,
( 2 + k j 2 ) U , V = 0 ,
k j = k 0 m j = k 0 n j ( 1 + i κ j ) .
U and U / y shall be continuous across the interfaces y = ± l for T E modes ,
V and ( 1 / k j 2 ) ( V / y ) shall be continuous across the interfaces y = ± l for T M modes .
U and V = 0             at y = 0             for odd modes .
U y and V y = 0             at y = 0             for even modes .
exp ( i α x - β y ) ; β = ( α 2 - k 1 2 ) 1 2
exp ( i α x ) cos [ ( k 2 2 - α 2 ) 1 2 y ] ,             exp ( i α x ) sin [ ( k 2 2 - α 2 ) 1 2 y ]
V = A exp ( i α x ) cos [ ( k 2 2 - α 2 ) 1 2 y ]             y l
V = exp ( i α x - β y )             y l
A = exp ( - β l ) / cos [ ( k 2 2 - α 2 ) 1 2 l ] ,
( k 2 2 - α 2 ) 1 2 k 2 2 tan [ ( k 2 2 - α 2 ) 1 2 l ] - ( α 2 - k 1 2 ) 1 2 k 1 2 = 0 ,             even T M modes .
( k 2 2 - α 2 ) 1 2 k 2 2 cot [ ( k 2 2 - α 2 ) 1 2 l ] + ( α 2 - k 1 2 ) 1 2 k 1 2 = 0 ,             odd T M modes ,
( k 2 2 - α 2 ) 1 2 tan [ ( k 2 2 - α 2 ) 1 2 l ] - ( α 2 = k 1 2 ) 1 2 = 0 ,             even T E modes ,
( k 2 2 - α 2 ) 1 2 cot [ ( k 2 2 - α 2 ) 1 2 l ] + ( α 2 - k 1 2 ) 1 2 = 0 ,             odd T E modes .
α = k 0 N ( 1 + i K ) ,
l / λ 0 = a tan - 1 [ ( n 2 / n 1 ) 2 b ] ,             T M even ,
l / λ 0 = - a cot - 1 [ ( n 2 / n 1 ) 2 b ] ,             T M odd ,
l / λ 0 = a tan - 1 b ,             T E even ,
l / λ 0 = - a cot - 1 b ,             T E odd ,
a = 1 / [ 2 π ( n 2 2 - N 2 ) 1 2 ] ,
b = ( N 2 - n 1 2 ) 1 2 / ( n 2 2 - N 2 ) 1 2 .
n 1 < N < n 2 .
2 l j λ 0 j 2 ( n 2 2 - n 1 2 ) 1 2 ,             j = 0 , 1 , 2 , 3 , 4 ,
F ( α , [ k 1 , k 2 , l ] ) = 0.
F 0 F ( α 0 , [ k 1 0 , k 2 0 , l 0 ] ) = 0.
k 1 = k 1 0 + Δ k 1 .
α = α 0 + Δ α
s 2 = k 1 2 - α 2 , l 2 = k 1 2 + α 2 .
( F 0 / s ) Δ s + ( F 0 / t ) Δ t = 0.
F 0 s = F 0 α α s + F 0 k 1 k 1 s ,
F 0 t = F 0 α α t + F 0 k 1 k 1 t ,
Δ s = s α Δ α + s k 1 Δ k 1 ,
Δ t = t α Δ α + t k 1 Δ k 1 .
k 1 s α = - α s k 1 ,
k 1 t α = α t k 1 ,
Δ α k 2 fixed = - ( F 0 k 1 / F 0 α ) Δ k 1 .
Δ α k 1 fixed = - ( F 0 k 2 / F 0 α ) Δ k 2 ,
Δ k 1 = i k 0 n 1 κ 1 ,
Δ k 2 = i k 0 n 2 κ 2 ,
Δ α = i k 0 N K ,
N K n 1 κ 1 | κ 2 fixed = - ( F 0 k 1 / F 0 α ) ,
N K n 2 κ 2 | κ 1 fixed = - ( F 0 k 2 / F 0 α ) .
N K / n 2 κ 2 0             if N n 1 ,
TE { F 0 k 1 = n 1 R 1 , F 0 k 2 = n 2 ( n 2 2 - N 2 ) [ R 1 + d ( n 2 2 - n 1 2 ) ] , F 0 α = - N R 1 R 2 2 [ ( n 2 2 - n 1 2 ) ( 1 + R 1 d ) ] ,
R 1 = ( N 2 - n 1 2 ) 1 2 ,             R 2 = ( n 2 2 - N 2 ) 1 2 ,             d = 2 π l / λ 0 ,
TM { F 0 k 1 = 2 N 2 - n 1 2 n 1 3 R 1 F 0 k 2 = d n 1 2 n 2 R 2 2 [ N 2 n 1 2 ( n 2 4 - n 1 4 ) + n 2 2 ( n 1 2 - n 2 2 ) ] - R 1 n 1 2 n 2 R 2 2 ( R 2 - N 2 ) , F 0 α = - d N 3 n 1 4 n 2 2 R 2 2 ( n 2 4 - n 1 4 ) - N ( 1 - R 1 d ) n 1 2 R 2 2 R 1 ( n 2 2 - n 1 2 ) .
D i j / λ 0 = l i 12 / λ 0 + l j 23 / λ 0
D i j λ 0 = 1 4 [ i ( n 2 2 - n 1 2 ) 1 2 + j ( n 2 2 - n 3 2 ) 1 2 ] i , j = 1 , 3 , 5 , , quasiodd . i , j = 0 , 2 , 4 , , quasieven ,