Abstract

A new analysis method has been developed for determining optical constants and thickness of a thin film given measured reflectance and transmission values. In contrast to other methods, the present inversion technique is essentially algebraic, not numerical. There is no requirement for an initial guess solution, and there are no missing solutions. The generic measurement configuration is a multilayer structure with no restriction placed on the number of films. However, the film with unknown optical parameters must be the last layer adjacent to the output medium. The analysis method is reviewed and examples presented for both single-film and multilayer structures.

© 1983 Optical Society of America

Full Article  |  PDF Article

References

  • View by:
  • |
  • |
  • |

  1. J. M. Bennett, M. J. Booty, Appl. Opt. 5, 41 (1966).
    [CrossRef] [PubMed]
  2. L. Ward et al., J. Phys. D 2, 301 (1969).
    [CrossRef]
  3. J. E. Nestell, R. W. Christy, Appl. Opt. 11, 643 (1972).
    [CrossRef] [PubMed]
  4. W. Hansen, J. Opt. Soc. Am. 63, 793 (1973).
    [CrossRef]

1973

1972

1969

L. Ward et al., J. Phys. D 2, 301 (1969).
[CrossRef]

1966

Cited By

OSA participates in CrossRef's Cited-By Linking service. Citing articles from OSA journals and other participating publishers are listed here.

Alert me when this article is cited.


Figures (6)

Fig. 1
Fig. 1

Optical constants analysis measurement configuration. Film with unknown optical parameters (film-X) is the outermost layer of a stratified medium. No restriction is placed on the number of films in the multilayer stack.

Fig. 2
Fig. 2

Addition of film modules indicating parameters for the reduced stack and film-X needed in the theory.

Fig. 3
Fig. 3

Example circle diagram in the complex amplitude reflectivity plane. All vectors rf from the origin to points on the circle are consistent with the reflectance data point R.

Fig. 4
Fig. 4

Example problems for testing the analysis. R and T data were calculated assuming n = 2.9, k = 0.1, and d = 0.3 μm for film-X.

Fig. 5
Fig. 5

Inversion results for structures A and B shown in Fig. 4. Each point (●,▲) corresponds to one θ value fixing rf, Eq. (3) in the text.

Fig. 6
Fig. 6

Thin-film configuration used for derivation of Eqs. (1) and (2).

Tables (1)

Tables Icon

Table I Example Problem Solutions

Equations (36)

Equations on this page are rendered with MathJax. Learn more.

R = | r 0 - ( r 0 r 0 - t 0 t 0 ) r f 1 - r 0 r f | 2 ,
T = n i - 1 | t 0 t f 1 - r 0 r f | 2 .
r f = C ^ + L exp ( j θ ) ,
C ^ = ( r 0 μ ^ * - R r 0 * ) / ( μ ^ 2 - R R 0 ) ,
L = T 0 R 1 / 2 / μ ^ 2 - R R 0 ,
μ ^ = r 0 r 0 - t 0 t 0 ,
R 0 = r 0 2 ,             T 0 = n i t 0 2 = n i - 1 t 0 2 .
T f = T T 0 - 1 1 - r 0 r f 2 .
μ ^ = r 0 / r 0 * ,             μ ^ 2 = 1.
C ^ = ( 1 - R ) r 0 * / ( 1 - R R 0 ) ,
L = ( 1 - R 0 ) R 1 / 2 / ( 1 - R R 0 ) ,
T f = T ( 1 - r f 2 ) / ( 1 - R ) .
f = ( n ^ - 1 ) / ( n ^ + 1 ) ,
= exp j ( 2 π n ^ d / λ ) ,
r f = - f ( 1 - 2 ) / ( 1 - 2 f 2 ) ,
t f = ( 1 - f 2 ) / ( 1 - 2 f 2 ) .
n ^ = [ ( 1 - r f ) 2 - t f 2 ( 1 + r f ) 2 - t f 2 ] 1 / 2 ,
= t f / ( 1 + f r f ) .
n = [ ( r f + r f * - 2 r f 2 ) / ( r f + r f * + 2 r f 2 ) ] 1 / 2 .
2 = ( 1 + f - 1 r f + f r f * + r f 2 ) / ( 1 + f r f + f r f * + f 2 r f 2 ) ,
n ^ 2 - 1 2 T f 2 = n ^ 2 ( 1 + r f ) 2 - ( 1 - r f ) 2 2 .
½ n ^ 2 = A ^ + B exp ( j ϕ ) .
A ^ = - 2 r f ( 1 + r f * ) 2 ( 1 + r f 4 - T f 2 ) + 1 / 2 ,
B = 2 T f r f 1 + r f 4 - T f 2 .
n 2 = ( A x + B cos ϕ ) + [ ( A x + B cos ϕ ) 2 + ( A y + B sin ϕ ) 2 ] 1 / 2 ,
k = ( A y + B sin ϕ ) / n
t f 2 = ( 1 + r f ) 2 + 4 r f / ( n ^ 2 - 1 ) ,
exp ( - 4 π k d / λ ) = T f / 1 + f r f 2 ,
exp ( j 4 π n d / λ ) = ( 1 + f r f ) * t f 2 / ( 1 + f r f ) T f .
F ( ϕ ) = ( n ) ( k d ) = ( k ) ( n d ) .
r = r 0 + t 0 b .
b = ( t 0 + r 0 b ) r f ,
b = t 0 r f / ( 1 - r 0 r f ) .
t = ( t 0 + r 0 b ) t f ,
r = r 0 - ( r 0 r 0 - t 0 t 0 ) r f 1 - r 0 r f ,
t = t 0 t f 1 - r 0 r f .

Metrics