282 



Mr. Ainger on the 



that its existence is easily deducible from the descriptions 

 usually given of the circumstances under which the rainbow 

 is produced. 



I will not occupy your time with an enumeration of the 

 various mistakes which seem to exist on this subject, but refer 

 you at once to the Traite de Physique of M. Biot, whose 

 description and analysis, as far as I am able to appreciate 

 them, are the best I have seen. I think, however, with great 

 deference, that even these are, to a certain extent, imperfect 

 and incomplete. M. Biot says 



* The phenomenon of the rainbow is produced by the coloured spectra 

 which issue from different drops of water after two refractions, separated 

 by one or two intermediate reflexions. But how,' he proceeds, ' does the 

 superposition of these partial spectra compose the colours of the bow 

 and determine its magnitude ? This is what we have to examine. 



* To do this simply, let us first consider a single incident ray of simple 

 colour for example, red ; then, supposing that it emerges from the drop 

 after a certain number of reflexions and refractions, let us calculate the 

 angle it forms with its primitive direction. 



' Let S I, Fig. 1 . be such a ray entering at I and escaping at I' after a 



N. 



Fig. 1. 



second refraction, without intermediate reflexion. From the centre of 

 the globe draw C I N, C I' N', which will be perpendicular to the sur- 

 face. Then SIN will be the angle of incidence, which we will call i ; 

 and C 1 1' will be the angle of reflexion, which we will call r. Further, 

 in consequence of the symmetry of the figure, the interior incidence at I' 

 will also be r, and the emergence will be i. Prolong the incident and 

 emergent rays till they meet at T, forming the angle I T I', which will be 

 the deviation produced by refraction; we will call this A, Now it will 

 be easy to find its value in functions of the angles i and r ; for in the 

 quadrilateral C I T I', all the angles are known, except A. In short, the 

 angles at I and I' are both equal to i ; further, the triangle C 1 1' being 

 isosceles, the angle I C I' is equal to 180 -2 r ; then, since the sum of 

 the angles of a quadrilateral are equal to four right angles, we have 

 A + 2 i + 180 - 2 r = 2, 180, or A = 180 +27'- 2 i. 



