## Abstract

The angular distribution of radiation intensity emitted from a plane circular source is altered by an attached cylindrical shield. Under the assumption of cylindrical symmetry, explicit expressions are obtained for the angular intensity distribution of the radiation emerging from the mouth of the shield. The source is characterized by an intensity which is a function of the point of emission and the direction of emission. The shield is characterized by its length-to-diameter ratio and by the reflectivity of its inner surface. The reflectivity is not necessarily uniform. The general result is an infinite series of integrals over subregions of the source area. For certain cases these integrals are evaluated in terms of elementary mathematical functions.

© 1956 Optical Society of America

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### Figures (4)

Fig. 1

The reflection a typical ray.

Fig. 2

The source regions Rk(θ) of radiation reflected k times for α tan θ=5.85.

Fig. 3

The function F(θ) for r=0.0 and α=1,2,4. Note: F(0)=1.

Fig. 4

The function F(θ) for r=0.6 and α=1,2,4. Note: F(0)=1.

### Equations (51)

$Q 0 ( ω ) = ∫ ∫ R I 0 ( P ; ω ) d A .$
$G 0 ( θ ) = ∫ 0 2 π Q 0 ( θ , ψ ) sin θ d ψ .$
$G 0 ( θ ) = 2 π sin θ Q 0 ( θ , ψ ) .$
$z j ( P ; ω ) ≦ 2 α for j ≦ k ( P ; ω ) , z j ( P ; ω ) > 2 α for j > k ( P ; ω ) .$
$G ( θ ) = ∫ 0 2 π Q ( θ , ψ ) sin θ d ψ .$
$G ( θ ) = 2 π sin θ Q ( θ , ψ ) .$
$∏ j = 1 k ( P ; ω ) r ( z j , i ) I 0 d A d ω$
$G ( θ ) = sin θ ∫ 0 2 π ∫ R ∫ ∏ j = 1 k ( P ; ω ) r ( z j , i ) I 0 ( P ; ω ) d A d ψ .$
$G ( θ ) = 2 π sin θ ∫ R ∫ ∏ j = 1 k ( P ; ω ) r ( z j , i ) I 0 ( P ; ω ) d A ,$
$Q ( θ , ψ ) = ∫ R ∫ ∏ j = 1 k ( P ; ω ) r ( z j , i ) I 0 ( P ; ω ) d A ,$
$Q ( θ , ψ ) = ∫ - 1 1 ∫ - ( 1 - y 2 ) 1 2 ( 1 - y 2 ) 1 2 ∏ j = 1 k ( P ; θ , 0 ) r ( z j , i ) I 0 ( P ; θ , 0 ) d x d y ,$
$y 1 = y , x 1 = ( 1 - y 2 ) 1 2 , z 1 = ( x 1 - x ) cot θ = [ ( 1 - y 2 ) 1 2 - x ] cot θ . }$
$cos i ( P ; θ , 0 ) = ( 1 - y 2 ) 1 2 sin θ ,$
$z j + 1 - z j = 2 ( 1 - y 2 ) 1 2 cot θ .$
$z j = ( z j - z j - 1 ) + ⋯ + ( z 2 - z 1 ) + z 1 ,$
$z j ( P ; θ , 0 ) = [ ( 2 j - 1 ) ( 1 - y 2 ) 1 2 - x ] cot θ ,$
$2 α < z ≦ 2 α + 2 ( 1 - y 2 ) 1 2 cot θ .$
$2 α < { [ 2 ( k + 1 ) - 1 ] ( 1 - y 2 ) 1 2 - x } cot θ ≦ 2 α + 2 ( 1 - y 2 ) 1 2 cot θ ,$
$2 α tan θ + x - ( 1 - y 2 ) 1 2 2 ( 1 - y 2 ) 1 2 < k ( P ; θ , 0 ) ≦ 2 α tan θ + x + ( 1 - y 2 ) 1 2 2 ( 1 - y 2 ) 1 2 ,$
$k ( P ; θ , 0 ) = greatest integral value of f ( x , y , θ ) = giv [ f ( x , y , θ ) ] ,$
$f ( x , y , θ ) = 2 α tan θ + x + ( 1 - y 2 ) 1 2 2 ( 1 - y 2 ) 1 2 .$
$k ( x , y , 0 ; θ , 0 ) = f ( x , y , θ ) when x = ζ ( y , θ ) ,$
$k ( x , y , 0 ; θ , 0 ) = giv [ f ( - ( 1 - y 2 ) 1 2 , y , θ ) ] for - ( 1 - y 2 ) 1 2 < x < ζ ( y , θ ) , k ( x , y , 0 ; θ , 0 ) = giv [ f ( ( 1 - y 2 ) 1 2 , y , θ ] + 1 for ζ ( y , θ ) ≦ x < ( 1 - y 2 ) 1 2 .$
$k ( x , y , 0 ; θ , 0 ) = giv [ α ( 1 - y 2 ) - 1 2 tan θ ] for - ( 1 - y 2 ) 1 2 < x < ζ ( y , θ ) ,$
$k ( x , y , 0 ; θ , 0 ) = giv [ α ( 1 - y 2 ) - 1 2 tan θ ] + 1 for ζ ( y , θ ) ≦ x < ( 1 - y 2 ) 1 2 .$
$2 α tan θ + ζ ( y , θ ) + ( 1 - y 2 ) 1 2 2 ( 1 - y 2 ) 1 2 = giv [ α ( 1 - y 2 ) - 1 2 tan θ ] + 1.$
$ζ ( y , θ ) = { 2 giv [ α ( 1 - y 2 ) - 1 2 tan θ ] + 1 } ( 1 - y 2 ) 1 2 - 2 α tan θ .$
$ζ ( y , θ ) = { 1 - 2 fp [ α ( 1 - y 2 ) - 1 2 tan θ ] } ( 1 - y 2 ) 1 2 ,$
$k ( x , 0 , 0 ; θ , 0 ) = giv [ α tan θ ] for - 1 < x < ζ ( 0 , θ ) , k ( x , 0 , 0 ; θ , 0 ) = giv [ α tan θ ] + 1 for ζ ( 0 , θ ) ≦ x < 1 ,$
$ζ ( 0 , θ ) = 2 giv [ α tan θ ] + 1 - 2 α tan θ = 1 - 2 fp [ α tan θ ] .$
$k 0 ( θ ) = giv [ α tan θ ] .$
$y k ( θ ) = 0 for k = k 0 ( θ ) , y k ( θ ) = + [ 1 - ( α tan θ k ) 2 ] 1 2 for k > k 0 ( θ ) .$
$x k ( θ ) = 1 for k = k 0 ( θ ) , x k ( θ ) = α tan θ k for k > k 0 ( θ ) .$
$x k 2 ( θ ) + y k 2 ( θ ) = 1 for k ≧ k 0 ( θ ) ,$
$k ( x , y , 0 ; θ , 0 ) = [ k for - ( 1 - y 2 ) 1 2 < x < ζ ( y , θ ) k + 1 for ζ ( y , θ ) ≦ x < ( 1 - y 2 ) 1 2 ]$
$ζ ( y , θ ) = ( 2 k + 1 ) ( 1 - y 2 ) 1 2 - 2 α tan θ for y k ( θ ) ≦ ∣ y ∣ < y k + 1 ( θ ) , k ≧ k 0 ( θ ) .$
$ζ ( y , θ ) = 2 k 0 ( θ ) + 1 - 2 α tan θ .$
$ζ ( y , θ ) = ( 2 k + 1 ) x k ( θ ) - 2 α tan θ = x k ( θ ) .$
$ζ ( y , θ ) → ( 2 k + 1 ) x k + 1 ( θ ) - 2 α tan θ = - x k + 1 ( θ ) .$
$( x + 2 α tan θ ) 2 ( 2 k + 1 ) 2 + y 2 = 1.$
$- ( 1 - y 2 ) 1 2 < x < ( 2 k + 1 ) ( 1 - y 2 ) 1 2 , y k ( θ ) ≦ ∣ y ∣ < y k + 1 ( θ ) }$
$( 2 k - 1 ) ( 1 - y 2 ) 1 2 - 2 α tan θ ≦ x < ( 1 - y 2 ) 1 2 , y k - 1 ( θ ) ≦ ∣ y ∣ < y k ( θ ) , }$
$Q ( θ , ψ ) = ∑ k = k 0 ( θ ) ∞ ∫ ∫ R k ( θ ) ∏ j = 1 k r ( z j , i ) I 0 ( x , y , 0 ; θ , 0 ) d x d y ,$
$F ( ω ) = Q ( ω ) / Q 0 ( ω )$
$Q ( θ , ψ ) = ∑ k = k 0 ( θ ) ∞ r k ∫ ∫ R k ( θ ) I 0 ( x , y , 0 ; θ , 0 ) d x d y ,$
$Q ( θ , ψ ) = ∫ ∫ R 0 ( θ ) I 0 ( x , y , 0 ; θ , 0 ) d x d y .$
$Q ( θ , ψ ) = I 0 ( θ ) ∑ k = k 0 ( θ ) ∞ r k A k ( θ ) ,$
$A k ( θ ) = ∫ ∫ R k ( θ ) d x d y .$
$A k ( θ ) = 2 [ C k + 1 ( θ ) - 2 C k ( θ ) + C k - 1 ( θ ) ] ,$
$C k ( θ ) = k [ cos - 1 x k ( θ ) - x k ( θ ) y k ( θ ) ] for k > k 0 ( θ ) ,$
$F ( ω ) = 1 π ∑ k = k 0 ( θ ) ∞ r k A k ( θ ) .$