480 Transactions of the Society. 



We now pass to the case of a self-luminous ground interrupted 

 by a dark bar. As in (22), we have for the illumination at any 

 point u within the geometrical image 



T , v f + w sin 2 u 7 

 I(u) = — 5— du 



J _» u" 



_ r«-«sin*u du _ C a + U ^ du . . (90) 

 J o u 2 J o u 2 



and for any point on the positive side beyond the geometrical 

 image 



- 



u 2 



du 



+ c"-^u dK _ r* •****, . . (9i) 



J o u J o u 



2 a denoting as before the width of the geometrical image of the 

 bar, while u is reckoned from the centre of symmetry. If a = 0, 



I ( U ) = f + "" Si ^ du = 7T. . . (92) 



The integrals in (90), (91) may be reduced to dependence upon 

 the sine-integral. It may be proved * that 



'■ x <=i" »> sin 2 x 



J ,!B sin 2 %i 7 C 2x sin u , 

 _ du = • du — 

 o u 2 Jo u 



= si (2aj) - ^^. . . . (93) 



Thus, inside the geometrical image, 



I («) = tt - si(2 a - 2«) + Sin2(a -^ 



a — u 



-si(2a + 2?0 + sin2(a + ^ ; . (94) 



a + ^* 



and beyond it, 



T / \ I • /O O \ Sm2 ( M — tt ) 



I (%) = 7T + Si (2 u — 2 a) i '- 



u — a 



- si (2 u + 2 a) + sm2 fo + g ) . . (95) 



u + a 



* E.g. by •writing rw for m in the integral to be examined and differentiating 

 with respect to r. Or (93) may be verified by differentiating with respect to x. 



