90 Mr. R. Potter on a particular Modification 



7 , sin <pd<p ju,* — 1 —d. cos <p „ a 2 — 1 



or, dr = — e — ^~^- 2 t ~~- = - e - — ^~^± ^- 2 - 



COS^ ^ COS^<j> jX 8 



and / dr =z l — ed. 2. „ ; 



•y */ cos <p jk* 



or, r = C — e 2. 



COS <p ]X 2 



which is the equation of the curve we seek ; and from the 

 circumstances in which we consider the experiments made, 

 we may take r and <p as the coordinates referred to the pole £ 

 or e in fig. 5 ; and p, which was originally the angle of inci- 

 dence, being counted from the line a £z/, or a parallel to it, 

 this line making with the axes of the pencils after refraction 

 an angle equal to the angle of minimum deviation. 



For the constant, let r become R when <p = 0, we have 



then a^CvWfcl C =R + 2 ,^i; 



hence r = R + 2 e £—^- -2e ^— ^ . — — , 

 \f? p* cos <p 



which we may write thus, 



r = R + x — — — , or r = Q 



cos <p cos 4, 



and we see that our equation holds good for any values of R, 

 as the motion of the rays of light requires. 



To calculate the points of interference on the transcenden- 

 tal curves given by the light supposed to set out simultaneously 

 from the images of the luminous point, we must return as 

 before to rectangular coordinates, and follow an analogous 

 process. 



Then t and £ being the secondary images of the luminous 

 point and the poles of the curves, we will take the lines a %y 

 and a s x for the axes of the rectangular coordinates ; and it 

 will be required to find the distance a £, which we will put 

 = a, and the distance a s, which we will call /3. 



Now writing the equation of the upper curve 



r'=Q *Z % 



cos <p 



and that of the lower one 3 £ 



cos % 

 counting y' from the point ?, and x' from the point e, we shall 

 have • j/' = y — a, .*•' = x — 0, 



