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References

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  1. Drude, Lehrbuch der Optik (Verlag von S. Hirzel, Leipzig, 1906), II, 4, p. 322 sqq.
  2. M. Czerny, Z. Physik 65, 600 (1930).
    [CrossRef]
  3. R. Bowling Barnes, M. Czerny, Phys. Rev. 38, 338 (1931).
    [CrossRef]
  4. H. Y. Fan, M. Becker, Symposium Volume of the Reading Conference on Semi-Conducting Materials, H. K. Henisch, ed. (Butterworth, London, 1951).

1931

R. Bowling Barnes, M. Czerny, Phys. Rev. 38, 338 (1931).
[CrossRef]

1930

M. Czerny, Z. Physik 65, 600 (1930).
[CrossRef]

Becker, M.

H. Y. Fan, M. Becker, Symposium Volume of the Reading Conference on Semi-Conducting Materials, H. K. Henisch, ed. (Butterworth, London, 1951).

Bowling Barnes, R.

R. Bowling Barnes, M. Czerny, Phys. Rev. 38, 338 (1931).
[CrossRef]

Czerny, M.

R. Bowling Barnes, M. Czerny, Phys. Rev. 38, 338 (1931).
[CrossRef]

M. Czerny, Z. Physik 65, 600 (1930).
[CrossRef]

Drude,

Drude, Lehrbuch der Optik (Verlag von S. Hirzel, Leipzig, 1906), II, 4, p. 322 sqq.

Fan, H. Y.

H. Y. Fan, M. Becker, Symposium Volume of the Reading Conference on Semi-Conducting Materials, H. K. Henisch, ed. (Butterworth, London, 1951).

Phys. Rev.

R. Bowling Barnes, M. Czerny, Phys. Rev. 38, 338 (1931).
[CrossRef]

Z. Physik

M. Czerny, Z. Physik 65, 600 (1930).
[CrossRef]

Other

Drude, Lehrbuch der Optik (Verlag von S. Hirzel, Leipzig, 1906), II, 4, p. 322 sqq.

H. Y. Fan, M. Becker, Symposium Volume of the Reading Conference on Semi-Conducting Materials, H. K. Henisch, ed. (Butterworth, London, 1951).

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Figures (1)

Fig. 1
Fig. 1

The abac chart. Here ρ = r and τ = eKd. In this figure T runs from 0 to 100%, R runs from 0 to 100%, τ = eKd runs from 0.01 to 1.0 and ρ = r runs from 0.01 to 0.71.

Equations (9)

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r = ( n 1 ) 2 + k 2 ( n + 1 ) 2 + k 2 ,
K = 4 π k λ .
T = ( 1 r ) 2 + 4 r sin 2 ψ ( e 2 π k d / λ r e 2 π k d / λ ) 2 + 4 r sin 2 [ ( 2 π n d / λ ) + ψ ] ,
R = r ( e 2 π k d / λ e 2 π k d / λ ) 2 + 4 sin 2 ( 2 π n d / λ ) ( e 2 π k d / λ r e 2 π k d / λ ) 2 + 4 r sin 2 [ ( 2 π n d / λ ) + ψ ] ,
tan ψ = 2 k n 2 + k 2 1 ,
T = ( 1 r ) 2 + 4 r sin 2 ψ e K d r 2 e K d ,
T = ( 1 r ) 2 e K d 1 r 2 e 2 K d ,
R = r ( 1 + T e K d ) .
( e K d ) 3 + [ ( 1 R T ) 2 1 + 1 T 2 ] T ( e K d ) 2 + [ ( 1 R T ) 2 2 ] e K d 1 T = 0.

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