SPHERICAL ABERRATION. 59 



refraction a. 2 ', and the angle which the refracted ray makes with 

 the axis (</> 3 ). The triangle a 2 c n (Fig. 20) gives 



. . 



sm a. 7 = -r- sin 9 = - --- sin 2 , 



en r 



and similarly, in the triangle 3 a 2 n the sum of the two acute 

 angles equals < 2 , therefore 



03 = 02 W ~ <*%)' 



Finally, w r e obtain from the triangle a B c n the length a B c, and 

 hence, by the addition of r, the distance of the point 3 from the 

 last refracting surface, that is, / 3 , we get 



sin a/ 



f = r. -r + r. 

 sm 3 



By this equation the path of the marginal ray through the first 

 double-lens is determined. These formulae are also applicable 

 for the other double-lenses, for which the calculation must be 

 repeated exactly. The point 3 is to be taken as the object-point 

 for the anterior surface of the second lens ; its distance = / 3 + e v 

 where the latter quantity denotes the distance of the two lenses. 

 The angle </> 3 is the angle of incidence for the plane anterior 

 surface. 



By the last refraction the angle <f> receives a negative value, i.e., 

 the ray of light is turned again to the axis, and therefore exhibits 

 the opposite inclination to it. Consequently the sign + becomes 

 in the expression for the corresponding f 3 , as shown by the 

 construction, and we get 



sin a/ 



/, = r. - --- - r. 

 sm 3 



When the influence of the flint-lenses is excessive, the distance of 

 the marginal rays from the axis, after their passage through the 

 objective, is increased, and either their radii must be greater, or 

 their indices of refraction lower. 



If we now suppose that the extreme marginal rays make an 

 angle of 30 with the axis (e.g., with an angular aperture of 60), 

 and if we denote their focal length, calculated from the posterior 



