600 



Appendix B 



bent to pass through the front focal point f 2 . Line (b) is constructed 

 in a similar fashion. Finally, (c) is a line joining the end of the object 

 with the center of curvature r. This line is undeviated as it is normal 

 to the interface. Any two of the three lines (a), (b), and (c) are sufficient 

 to locate and to determine the size of the object. The three points — f 1} f 2 , 

 and r are sufficient with the optic center o to permit all constructions. 

 The point r is related to f x and f 2 by the equation 



r=f 2 -fi 



(10) 



In the eye, light passes through four approximately spherical surfaces, 

 separating media of different indices of refraction. Equation 9 could be 



Object 



n. 



Interface 



Optic Axis 



Figure 5. Ray diagram illustrating refraction by a curved 

 interface between two media of indices of refraction n x and n 2 . 

 The interface is a small spherical section with center of curva- 

 ture at r. Any two of the three rays, a, b, and c, are sufficient 

 to determine the location and size of the image. 



used successively at each of these surfaces, noting that the image formed 

 by one surface is the object of the next. It is quite possible to have 

 virtual objects under these conditions. 



Although the preceding is all that is necessary to discuss the geometrical 

 optics of the eye, a somewhat more generalized approach that lumps the 

 effects of several surfaces is neater. For example, by applying Equation 

 9 twice, one can solve the problem of a single lens. To illustrate this, 

 suppose a medium of index of refraction n 2 separates two others of 

 indices of refraction n x and n 3 , respectively. The surfaces of separation 

 are assumed to be small sections of spheres whose line of centers is 

 referred to as the optic axis. The radius of the surface separating the 

 media of indices of refraction n x and n 2 will be designated by r a , and the 

 radius of the other surface by r b . If the object is located at a distance 

 —p along the optic axis from the first surface, one may find the location 

 of the image q' by solving Equation 9 in the form 



4 



»i 



Ho - U-y 



(11) 



This image q' is the object for the second surface. If the distance between 



