Abstract

The solution of the multilayer filter is obtained as a boundary value problem. A recursion formula is developed, which, applied once for each layer gives the reflected beam. A graphical means for determining the unknown quantity in the recursion formula has been shown to apply to dielectric layers.

The solution has been used to explain an irregular fringe shift in the reflected pattern from an interferometer, one of whose plates has a uniform wedge of dielectric deposited on it. It is shown that this irregular shifting of the fringes with thickness of the dielectric must be considered if interferometric determinations of index of refraction are to be made.

This method of determining the reflected beam has also been applied in finding the transmission characteristics of the frustrated total reflection filter. It is found that the width of the transmission band which, theoretically, could be arbitrarily small without loss of peak transmission, is limited by the uniformity in the thickness of the spacer layer.

© 1949 Optical Society of America

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References

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  1. A. Herpin, Comptes Rendus 225, 182–183 (1947).
  2. W. Weinstein, J. Opt. Soc. Am. 37, 576–581 (1947).
    [Crossref] [PubMed]
  3. L. N. Hadley and D. M. Dennison, J. Opt. Soc. Am. 37, 451–465 (1948).
    [Crossref]
  4. R. L. Mooney, J. Opt. Soc. Am. 35, 574–583 (1945).
    [Crossref]
  5. S. A. Schelkunoff, Electromagnetic Waves (D. Van Nostrand and Company, Inc., New York, 1943), p. 199.
  6. P. Leurgans and A. F. Turner, J. Opt. Soc. Am. 37, 983A (1947).

1948 (1)

1947 (3)

A. Herpin, Comptes Rendus 225, 182–183 (1947).

W. Weinstein, J. Opt. Soc. Am. 37, 576–581 (1947).
[Crossref] [PubMed]

P. Leurgans and A. F. Turner, J. Opt. Soc. Am. 37, 983A (1947).

1945 (1)

Dennison, D. M.

Hadley, L. N.

Herpin, A.

A. Herpin, Comptes Rendus 225, 182–183 (1947).

Leurgans, P.

P. Leurgans and A. F. Turner, J. Opt. Soc. Am. 37, 983A (1947).

Mooney, R. L.

Schelkunoff, S. A.

S. A. Schelkunoff, Electromagnetic Waves (D. Van Nostrand and Company, Inc., New York, 1943), p. 199.

Turner, A. F.

P. Leurgans and A. F. Turner, J. Opt. Soc. Am. 37, 983A (1947).

Weinstein, W.

Comptes Rendus (1)

A. Herpin, Comptes Rendus 225, 182–183 (1947).

J. Opt. Soc. Am. (4)

Other (1)

S. A. Schelkunoff, Electromagnetic Waves (D. Van Nostrand and Company, Inc., New York, 1943), p. 199.

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Figures (7)

F. 1
F. 1

The notation used in determining the reflected wave E1, having an incident wave E0. The unit vector, n, gives the z axis. The propagation vector, n2s, is associated with the electric vector E2s.

F. 2
F. 2

Construction for determining t ¯ s / t s, from the known values of t ¯ s + 1 and δs. Drop a perpendicular from t ¯ s + 1 to the axis and then up to the circle, as shown. Lay off the angle 2δs around the circle, drop a perpendicular to the axis, and locate the point on the circle that makes the radius vector perpendicular to the line from the axis point. The radius vector is t ¯ s / t s, and if multiplied by ts will give t ¯ s.

F. 3
F. 3

The coating of the interferometer plate to produce the anomalous fringe shift. First, an opaque layer of aluminum; then, a wedge of high index dielectric; and, finally, over half the plate another opaque aluminum layer.

F. 4
F. 4

The reflected fringe pattern. The upper, uniform shift indicated the regularity of the wedge, while the lower irregular shift shows the anomalous effect.

F. 5
F. 5

The amplitude and phase shift of the reflected beam from an aluminum surface covered with ZnS, as a function of the thickness of the ZnS. The dotted line indicates the phase shift due to a double passage through the ZnS layer.

F. 6
F. 6

The problem of determining the transmission of an FTR filter is reduced to finding the reflected amplitude and phase of light incident in the spacer layer.

F. 7
F. 7

Values of δ0 for s and p peaks for an FTR filter using a high index glass prism (1.72), cryolite (1.34) frustrating layer, and ZnS (2.38) spacer layer. N is the order number.

Tables (1)

Tables Icon

Table I Dispersion of FTR filter (normal incidence in air, 60° incidence on FTR surface).

Equations (40)

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R = ( 1 t ¯ 1 ) / ( 1 + t ¯ 1 ) .
E ( s ) = E 2 s exp ( i k s n 2 s r ) + E 2 s + 1 exp ( i k s n 2 s + 1 r ) H ( s ) = ( k s / ω ) [ ( n 2 s × E 2 s exp ( i k s n 2 s r ) + n 2 s + 1 × E 2 s + 1 × exp ( i k s n 2 s + 1 r ) ]
n × E ( s 1 ) = n × E ( s ) n × H ( s 1 ) = n × H ( s ) at z = z s
n E 0 = n E 1 = = n E 2 p + 2 = 0
α s 1 , s E 2 ( s 1 ) + α s 1 , s E 2 s 1 α s , s E 2 s α s , s E 2 s + 1 = 0 α s 1 , s E 2 ( s 1 ) α s 1 , s E 2 s 1 t s α s , s E 2 s + t s α s , s E 2 s + 1 = 0
α s , s = exp ( i k s n 2 s r s ) α s , s + 1 = exp ( i k s n 2 s r s + 1 ) α s , s = exp ( i k s n 2 s + 1 r s ) α s , s + 1 = exp ( i k s n 2 s + 1 r s + 1 )
t s = k s cos θ s k s 1 cos θ s 1 ,
k s sin θ s = k s 1 sin θ s 1 s = 1 , 2 , , p + 1 .
δ s = k s ( z s + 1 z s ) cos θ s = ( 2 π μ s d s cos θ s ) / λ ,
b s = exp i k s { n 2 s × n } { n × ( r s + 1 r s ) } ,
R = ( 1 t 1 ) D 2 + t 1 D 3 exp i δ 1 ( 1 t 1 ) D 2 + t 1 D 3 exp i δ 1 = 1 t ¯ 1 1 + t ¯ 1
t ¯ 1 = t 1 ( 1 exp i δ 1 D 2 / D 3 ) .
D 2 s 1 / D 2 s = 1 t ¯ s = 1 t s ( 1 exp i δ s i sin δ s + [ exp i δ s / ( D 2 s + 1 / D 2 s + 2 ) ] )
t ¯ s = t s t ¯ s + 1 i tan δ s 1 i t ¯ s + 1 tan δ s .
t ¯ p = t p t p + 1 i tan δ p 1 i t p + 1 tan δ p
n H 0 = n H 1 = = n H 2 p + 2 = 0 .
H ( s ) = H 2 s exp ( i k s n 2 s r ) + H 2 s + 1 exp ( i k s n 2 s + 1 r ) E ( s ) = ( ω / k s ) [ n 2 s × H 2 s exp ( i k s n 2 s r ) + n 2 s + 1 × H 2 s + 1 exp ( i k s n 2 s + 1 r ) ] .
t s = k s 1 cos θ s k s cos θ s 1 .
t ¯ s = t s t ¯ s + 1
t ¯ s = t s / t ¯ s + 1 .
k s = 2 π ( μ s + i η s ) / λ
T = [ 1 + 4 | R | 2 sin 2 ( δ 0 + Δ ) ( 1 | R | 2 ) 2 ] 1 ,
t ¯ 1 = t 1 ( t 2 i tan δ 1 ) / ( 1 i t 2 tan δ 1 ) .
cos θ 1 = i ( sin 2 θ 1 1 ) 1 2 .
t 2 = i t 2 t 1 = i t 1 δ 1 = i γ .
t ¯ 1 = i t 1 ( i t 2 + tanh γ ) / ( 1 i t 2 tanh γ ) .
tanh γ = 1 .
R = ( 1 i t 1 ) / ( 1 + i t 1 )
tan ( Δ / 2 ) = t 1 .
tan Δ = csch β ,
β = log t 1 .
δ 0 + Δ = n π , n an integer .
d δ 0 + d Δ = 0 .
d δ 0 / δ 0 = ( d λ / λ ) tan θ 0 d θ 0 + d d 0 / d 0
d Δ = sech β d β = sech β ( tan θ 1 d θ 1 + tan θ 0 d θ 0 )
tan θ s d θ s = tan 2 θ s cot θ 2 d θ 2 s = 0 , 1
tan 2 θ 1 = sin 2 θ 1 / ( sin 2 θ 1 1 ) .
d δ 0 + d Δ = 0 = ( d λ / λ ) + ( d d 0 / d 0 ) [ ( 1 + sech β δ 0 ) tan 2 θ 0 tan 2 θ 1 ( sech β ) / δ 0 ] cot θ 2 d θ 2 .
d λ / d θ 2 = λ [ ( 1 + sech β δ 0 ) tan 2 θ 0 tan 2 θ 1 ( sech β ) / δ 0 ] cot θ 2 .
d λ / d d 0 = λ / d 0 .