Abstract

The Elsasser integral expression representing absorption by a periodic line pattern is evaluated analytically and compared with a numerical integration of the exact absorption expression. A series is found which represents the integral for all values of the parameters.

© 1964 Optical Society of America

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References

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  1. W. M. Elsasser, Phys. Rev. 54, 126 (1938).
    [Crossref]
  2. Lewis D. Kaplan, J. Meteorol. 10, 100 (1953).
    [Crossref]
  3. Gilbert N. Plass, J. Opt. Soc. Am. 48, 690 (1958).
    [Crossref]

1958 (1)

1953 (1)

Lewis D. Kaplan, J. Meteorol. 10, 100 (1953).
[Crossref]

1938 (1)

W. M. Elsasser, Phys. Rev. 54, 126 (1938).
[Crossref]

Elsasser, W. M.

W. M. Elsasser, Phys. Rev. 54, 126 (1938).
[Crossref]

Kaplan, Lewis D.

Lewis D. Kaplan, J. Meteorol. 10, 100 (1953).
[Crossref]

Plass, Gilbert N.

J. Meteorol. (1)

Lewis D. Kaplan, J. Meteorol. 10, 100 (1953).
[Crossref]

J. Opt. Soc. Am. (1)

Phys. Rev. (1)

W. M. Elsasser, Phys. Rev. 54, 126 (1938).
[Crossref]

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Figures (1)

Fig. 1
Fig. 1

Idealized variation of k(v) with the frequency ν.

Tables (2)

Tables Icon

Table I The first few an’s of the series solution of Elsasser’s integral.

Tables Icon

Table II Comparison of numerical integration of exact expression with series solution of Elsasser’s integral.

Equations (21)

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k ( ν ) = ( S / π ) α ( ν - ν i ) 2 + α 2 ,
k ( ν ) = ( S / d ) sinh β cosh β - cos σ ,
β = 2 π α / d             σ = 2 π ν / d .
A = 1 - exp [ - k ( ν ) u ]
A ( β , y ) = sinh β 0 y exp ( - t cosh β ) J 0 ( i t ) d t .
A ( β , y ) = sinh β 0 y exp ( - t cosh β ) J 0 ( i t ) d t .
a = cosh β , J 0 ( i t ) = n = 0 b n t n ,
I = 0 y exp ( - a t ) J 0 ( i t ) d t .
I = 0 y exp ( - a t ) n = 0 b n t n d t .
I = n = 0 b n 0 y t n exp ( - a t ) d t .
a n = 0 y t n exp ( - a t ) d t .
I = n = 0 b n a n .
a n = ( n / a ) a n - 1 - y n / a exp ( - a y ) .
a n = n ( n - 1 ) a n - 2 / a 2 - exp ( - a y ) ( y n / a + n y n - 1 / a 2 ) .
b n = 1 / [ ( n / 2 ) ! ] 2 2 n n even , b n = 0 n odd ,
J 0 ( i x ) = m = 0 x 2 m / [ ( m ) ! ] 2 2 2 m .
I = n = 0 a 2 n / ( n ! ) 2 2 2 n ,
a 0 = 0 y exp ( - a t ) d t = ( 1 / a ) [ 1 - exp ( - a y ) ] .
A ( β , y ) = sinh β n = 0 a 2 n / ( n ! ) 2 2 2 n .
a n = [ n ! / a n + 1 ] - exp ( - a y ) m = 0 n ( n ! y m ) / ( m ! a n + 1 - m ) .
a n = [ n ! / a n + 1 ] [ 1 - exp ( - a y ) m = 0 n ( a y ) m / m ! ] .