330 PROFESSOR L. BECKER ON THE DISTRIBUTION OP 



The left side can be developed into a power series of g 



(3) (1 -<?)- = i P^, 

 therefore F and f must also be power series of g : 



(4) , . 



(5) f(p)=f(g) = C'<fZn n g* > where 



Let 



d n = f 2 cos" e do. 



Jo 



,^ , 1.3...2a-l7r , 2.4... 2a 



rf2a = 2^r 



These substituted in (2) give 



(7) 

 or 



The second integral in (2) gives the polarised light, while the left side equals the 

 total light. Their ratio was designated by q (p) ; hence 



Provided q (p) be observed as a function of p, i.e. of g, R B can be calculated by (II), 

 and Q by (1), i.e. F (p) and f(p) become known functions. Their values are, if 

 C" = C' (360/500) 5 , 



FO) = NG>)ETG)+8G>)]- 



(III) 



f(p) = K(p)p(p) 



Substitute T, S, and P and find N (p) and c,/c 2 . The problem can therefore be 

 solved if q (p) were known. I am unable to say whether the measurements of the 

 polarised light made at the last eclipse suffice to determine this function. 



With reference to (I), 2P n /3e? ll+3 very nearly equal the coefficients of a binomial 

 series, and it is not difficult to prove that 



The exponent y = 4, of formula (D), is derived from the observations with an error 

 of 0*3 (assuming x = 140 to be correct), hence the errors of the exponents in the 



