150 Mr Swan's Formula for constructing 



But if the ray C A H totally reflected 



sin C K A :i^i, or sin ( 45°- ^ ~ sin"' ^-A 1 ^~- 



Therefore the equation of condition to insure the total 



reflexion of all the rays is 



•vl/ . —1 1 . — 1 / 



4X)° -^ :i -p^ sin — + sin ^, 

 4 iCi fi a 



from which it is easy to find -^^ having previously assumed d 



the radius of the mirror,/ the radius of the flame, and ^ the 



refractive index of the glass. It is obviously necessary to 



assume the value of /a for the least refrangible rays of the 



spectrum, in order to insure the total reflexion of all the 



incident light. 



The following examples will illustrate the manner of using 

 the formulse which have now been investigated : — 



1. Suppose we assume the radius of the mirror to be 20 

 inches, that the flame is 1*4 inches in diameter, and that the 

 index of refraction of the glass is 1-57, we have first to deter- 

 mine '^ from the formula, 



45» ^ 4: > gin~^ i + sin""^-^. 

 4 fi fji.d 



In this, putting /= -7, c? = 20, and ^=1-57, we obtain 



4=16° 38' nearly. Now, since "^=16° 21' 49", it follows 



there must be at least five zones and a central conoid. 



Having thus obtained v)/, we have next to find r, a, and 6, • 

 from the formulae at (1). 



4- 



e/ sin <^ 



2 sin / 45-^Uin'* 



(«-i) 



8 vr^'f «^^ 



a= — G?cos ^ + r sin ( 45— ^j 



6 = — r sin 45 

 From these it will be found that 



r = 58-512 a = 18-523 6 = - 41-375. 

 To compare this result with that given by the formulae 

 (2), it must be borne in mind that 2 m in the second Set of 

 formulae is the same as x" in the first. Now, we have al- 

 ready found, p. 146, that 



