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NATIONAL BUREAU OF STANDARDS REPORT

Fire Research Information Services

National Bureau of Standards /

Bldg. 225, Rm. A4S f

Washington, D.C. 20234

^

1673

THE SCATTERING OF LIGHT BY A CONICAL SURFACE

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wRflStjsoi

by

C. H. Page

U. S. DEPARTMENT OF COMMERCE

NATIONAL BUREAU OF STANDARDS
U. S. DEPARTMENT OF COMMERCE

Charles Sawyer, Secretary

NATIONAL BUREAU OF STANDARDS

A. V. Astin, Acting Director

THE NATIONAL BUREAU OF STANDARDS

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NBS PROJECT NBS REPORT

1002-43-1030 May 15, 1952 1673

THE SCATTERING OF LIGHT BY A CONICAL SURFACE

by

C. H. Page

Approved for public release by the

Tfcis report is ia.u Director of the National Institute of

' tion

: r°PriBtin s- reproduction

in any form, either c , permission in writing is obtained

from the Office of t Standards and Technology (NIST) , Washington 25 , D. C. on October 9, 2015.

THE SCATTERING OF LIGHT BY A CONICAL SURFACE

by

Chester H. Page

ABSTRACT

The multiple scattering of light from the Boundary surface

of a conical region is studied, under the assumption that

the scattering follows Lambert *s Law„ For uniform primary

illumination, the increased- Illumination at the tip is

found analytically, and the illumination at other points

computed “by numerical methods

„

Let us consider a conical surface illuminated on its inner face By light

or other radiant energy „ Let the surface Be such that the fraction e of

the incident radiation is absorbed, and the reminder scattered according to

Lambert 5 s Law; i„e„, the scattered energy per unit solid angle is proportional

to cos @, with 0 the angle between the direction of scattering and the normal

to the surface

.

Now radiation of intensity S per unit area, arriving at the angle

from the normal, gives the surface illumination I = R cos „ The total

scattered radiation is (l - e )R cos ©,, and the scattered intensity in the

direction ©^ is

(l - € )R COS 0>2 cos ©p

per unit solid angle

or

(l - € )R COS COS ©0

(l - € )l COS Op

per unit area at

2

distance r„

Suppose that the surface exalts radiation at the rate Se per unit area,

with P a function of temperature only „ Assume this radiation to also follow

the cosine distribution law„ Then the total power appearing to he emitted

from a unit area is

( 1 ) E = Pe + (l - e)l.

The power incident per unit area at a point (l) is the summation of the

radiated and scattered power from all parts of the surface 'Visible" from the

point in questions

HD = j

E(2) K(i,2) ; dtf

2

where ddf is the element of area at point (2), and

COS 0 . COS ©p

K(l,2) = (see. Figure l)

Jtr12

In addition, we must allow for direct illumination from an external source,

making the complete equation

1(1) - D(l) + f E ( 2 ) K(l,2) dd

2

. We are interested in two special cases

s

I. No external, source of illumination and the conical surface at a uniform

temperature

„

Then 1(1) = j

E(2) K(l,2) d 6,

or E(l) = Pe + (l - €

) j

1(2) K(l,2) d<?2

by using equation (l).

Define the "apparent" blackness of the surface as £ = g- , giving

€(1) « € + (1 - e )J

%(2) K(l,2) dd

2

and define the gain of apparent blackness by virtue of the scattering ae G s e/e, giving

G(l) = 1 + (1 - €) f 0(2) K(l,2) d(52 (3) e

3 o

II. Let the emission be zero and the primary illumination uniform

1(1) * D + (1 - e)

J

1(2) K(l,2) ddg

"by substituting (l) into (2).

The gain in illumination by virtue of scattering is G

8 = l/D and is

determined by the equation

G

s (l) *5=1 + (1 - e)'J K(l,2)d<5^

or G

8 (1 ) = 1 + ( 1 — €

) j

G»(2) K(l,2) dd

2

Note that this is identical with equation '(3 ), hence G

8 and G are identical.

The specific problem at hand is to solve this integral equation for a conical surface

.

Let the cone have the semi -apex angle a. Locate a point on the cone by

its distance s from the apex, and the azimuth angle 0 from a convenient

generator. The cartesian coordinates of a point on the cone are

The element of area is dtf.*®sin a d0 ds. Without loss of generality

let have the azimuth

0^ - 0^ so that

x. - s sin a cos 0

y = s sin a sin 0

z = a cos a.

The normal at P has the direction cosines

^ - -cos Q COS 0 a = -cos Q sin 0

, n - sin a.

1+

.X_ — X= Zp — Zj coa ©, = - ———- cos a + ~ sin a T T* 12 12

and

COS 0p =

Sp sin a cos a (l ~ cos 0O

'

r

12

s,, sin a cos a (l - cos 0O )

12

K(l, 2) = cos cos ©p sin a cos “ a cos 0

2

J s 1

s 2

jcr12 nr

12

and

2 2 2 ^ } 2 ,2 ^ r . ~ + Sp ’ - 2a . s . cos a + sin a cos 0

g

The “basic equation (3 J becomes

T ,2k

,, ^ x , 1 = € ,3 2 i j G(t) st

2

(l - cos 0)

2

d0 dt

(4) O(s) = 1 + —— sinJ a cos a f - -V- - - y-1———

V - - - r -g “

o

£5' + t - 2stf@os ffi^c©s 0 sin a}]

o I o

where T is the slant length of the cone

The denominator (r^ p

) can be written

Rs - t)

2 + 2st sin2

a (l

Changing the variable from 0 to 0/2, using

the angle integration into the forms cos 0=2 sin 0/2, puts

h sin ©

(a

2 + b sin"" ©)

2

n/2

c/2

sin © d©

i c+ b sin2

©)

2

A second change of variable to x = sin o vxexa

(5) x dx

/ 2 . 2x2 ,r t (a + bx ) y 1 = X

which can be evaluated in the complex plane by the following procedure

„

Consider

oo z dz along the contour of Figure 2.

5°

-oo

/ 2 .2

(a + hz “

2

‘ z

For -l^i x 4 1, the integrand is even, hut on the remainder of the real axis

the integrand is odd because of the branch points . The contribution of the

semicircles around +1 is zero, making the infinite integral equal to the

desired (5)» Deforming the contour into an infinite semicircle plus a circle

around the pole at z ~ ia/ ’^/b allows straightforward integration, and yields

it/b“ + 2jt 1 Res at ia/l/fc,

yields

Evaluating the residue and substituting for a, b.

(6) G(s) = 1 + s-t (s-tr + 6st sin a

|7s-t f+ Ust sirlccj

dt

.

The expression in braces is a function of the ratio t/s only, and is not

affected 'by Interchanging s and t„

From physical intuition, we would expect that the illumination (hence G)- as we approach the apex would be independent of the length of the cone; i.e.,

the cone appears infinite . This is indeed the case . The cone of infinite

length has the solution G - constant = G

q

2

( 7 ) G

o cos asin a

00

where x ~ t/s.

This integral can be evaluated by brute force, yielding

G = 1 + — G 2(1 - sin3 a)

o 2 o

'

or

(8) G = 1

y- = — —v

- 1 - (l - s)(l - singes?) € + (l - € ) sin" a

and

£ = €G = ———

£

r— 4l.

€ + (l - e )' sin"’ a
.

.

'

.

f
6 .

For the finite cone, conveniently scaled to 'unit slant length

2 f l

(9) 0(a) - 1 + ~~ fg-f I 0(t) K(s,t) dt.

OO o ~ Note that f K(s,t)dt = 2 (l - sin'll) independent of s. cos a

Now G >0, K j>Q, so

1 00

G E dt lG.max 1 Kdt < G f K dt — max

°

cos a

or

) <» 1 + sin a,

max

and

G rl 1 + max —

slrr* a) G

max

or G <

max —

€ +|

*3 € ) sinJ a

G

i ) G(q). sG(s)<c!.e&(Q}'^“, X,

Therefore the absorption is nowhere greater than it would he for a perfectly

black surface (c - l)>9 although the intensity of Illumination (l = GD ) may

become high as we approach the tip. (As € G^ ^ 1/sin- a).

The straightforward approximation of the integral equation (6 or 9) by

a set of simultaneous equations would be achieved by representing the

integral by a formula such as Simpson's rules

( 10 ) j' G(t) K(s,t)dt h

3

;o) 0) + lG(h) K(s/h) + 2G(2h) K(s,2h)

+ + to(l - h) K(s, 1 - h) + G(l) K(s,l)| r expressing the integral in terms of the unknown values of G ( nb ) . Substitution in (9) yields an equation holding for all sj if we let s = 0, h,

2h 0 * °1 successively, we obtain a set of N = 1 + (l/h) simultaneous equations

for the N unknowns G(nh)j n - 0

° 0 • N - 1. Unfortunately, this procedur’e

will not lead to correct answers. The Simpson rule integration assumes the

integrand to be fitted piecewise by parabolas „ The second equation in the set.

7

for s = h, assumes the product G(t) K(h, t) to he approxime.ted by a parabola from t = 0 to t = 2h. The occurrence of js - tj in the

kernel produces a cusp at s = t (Figure 3)# hence a parabolic fit for

the points t = 0, h, 2h cannot yield a reasonable approximation

.

This phenomenon is independent of the mesh size h chosen, since the

kernel is a function of t/s„

The difficulty can be avoided by using the trapezoidal rule for

numerical integration, but the loss of accuracy seems too high a price

to pay.

Using Simpson's rule, it appears that we need more divisions in t

than in s; hence assuming values of G(t) at (say) 7 points yields

(say) k equations for G(s). This leaves us with more unknowns than

equations, allowing subsidiary conditions. The obvious choice is to

determine G(t) at the extra 3 points by interpolation. That is, we

set up b equations for G(s) at k points, express the integral as a

7 point Simpson integration expressing the values of G(t) at the

additional intermediate points as linear combinations of the values of

G(t) at the additional intermediate points as linear combinations of

the values of G(t) at the first k points. This procedure can be

expressed as the matric equation

(11) G = l+ XKwLG

where G is the vector whose components are the values of G(s) at

the points s = 0, l/3, 2/3, lj K is the non-square matrix of values

of K(s,t) at s = n/3 (n = 0

0 • ) and t = m/6 (m = 0 °««6); w is a diagonal matrix of the Simpson weights; L is an interpolation

matrix of seven rows and four columns; and

\ =

Since the values of G(s)

the values at s = m/6

1 - € ?

2 cos a

sin a

at s = n/3 determine a cubic from which

can be found, we have

L =

1 0 0 0

5/16 15/16 -5/16 1/16

0 1 0 0

1/16 9/16 9/16 -l/l6

0 0 1 0

1/16 - 5/16 15/16 5/16

0 0 0 1

For a cone with semi-apex angle 15°, equation (.10 ) is test solved

by the minimized iteration technique of Lanczos and yields the

results of Figures 1 and 5 for several values of €

.

1/ C. Lanczos, NBS Jour, of Res. 45, p.255 (1950)

8o

For a cone with semi-apex angle 45°? the contribution of multiple

scattering falls off rapidly with the order of the scatter^, so the

simple Liouville-Neumann iteration converges fairly rapidly . This solv

ing procedure yields results shown in Figures 6 and 7°

APPENDIX

It is of some interest to compare the cone with a wedge of infinite

transverse length. The equation corresponding to (4) is readily found

(12) G(s) = 1 + ~~ sin

2

2a

• T

f oo G(t ) st dy dt

•CD {sd

+ - 2st cos 2a + y }

27~2

where a is the semi-apex angle of the wedge. Integration with respect

to y is straightforward and yields

( 13 ) G(s) = 1 + sin2

2a G(t) st dt

{s

'2 + t^ - 2st cos 2d }^ ,/2

The solution at the tip is again found by making T - oc* „ and is

(14)

Comparison with (8) shows that G for the wedge is always less than

G

q

for the corresponding cone.

FIG.3

o>

€

- Vfe £ ” %.

6 =

!/2

FIG* 5

<D>

e--'k

e*'A

e =‘/2

s

FIG.?

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