502

the Mie extinction efficiency factor (Van de Hulst, 1957) . Moreover, the haze

composition is integrated into the complex index of refraction m(A).

where e

Since most hazes are composed of small particles (0

conversion will be arranged to emphasize this point, i.e.

To

mize the

3(A) =ir /“2 10 3 “ N(10°) In 10 Q ext ’ m(A)}da (2)

Here we have transformed the variable r by a = log^o r * Finite limits are given

Z(f

i

Incorpoi

to the integral to remove a source of singularity from the quadrature matrix.

Limits are chosen so that N(10 a ) = 0, aa 2 .

E (f

i

A factor 10 a 10 a , appears in the Yamamoto and Tanaka (1969) form of the

where y

integral (2) which tends to bring the factor In 10 Q Bxt (2TrlO a /A, m(A)} into the

family of "well-behaved" functions the equation (2), in this form, becomes

Mir

8(A) = tt / a2 f(a) K(a,A)da,

a i

f(a) = 10 4a N(10 a ),

— a 1 0 a

K(O.X) = 10 “in 10 Q ,m(A)} (3).

A

This is an example of the Fredholm integral equation of the first kind.

where A*

The method of solving this integral equation given by Phillips and Twomey

utilizes a notion of constraining the error in measurement of input data to extract

Solving

a member of a family of solutions. We introduce matrix notation for the equation

(3) with the errors in 8 included:

f =

g + e = Af (4)

Data Aqc

where g = (8(Aj)}, e is the error vector, and f is the solution. A is the quadra

ture matrix whose elements are given by

a.. =w. K(a., A.)

Ji i v i y

where ok are quadrature weights for integration with respect to the ou .

Clearly the system (4) does not generally provide a unique solution, there

fore, by the criterion established by Phillips, we select f as a solution to satisfy

The

continer

1973) ar.

the phot

depths v\

Haze opt

braicall

to Rayle

gible, s

r** d 2 f^ 2 j fT d 2 f 2 j

¿0 57^ dr = y" Jo dr 7 dr

f eF

where r* = 10 2 and F is a family of solutions of the system (4).

An additional constraint is placed on the error vector, namely that its mag

nitude by constant or

E e 2 = e 2 (6)

i

3

Tat

observat

input ve

( 6 )