REFRACTION IN THE EYE 471 



y =/ 2 F z , the distance from the retina to the 



.. image behind it. 

 g, the distance from the anterior focus <f>i to the 



object. 



</>! lies 20 mm. from the nodal point K. 

 F% F v in the reduced eye, is 20 x 15 300 mm. 

 from K. 



Find the distance behind the retina of an 

 image whose >object is 320 mm. from K. 



The distance is 1 mm. 



2. If y is known, the diameter of the dispersion 

 circles .can be calculated. In the example just 

 given, the pencil of rays diverging from each 

 luminous point in the object was reunited in a 

 single point one millimetre behind the retina. 

 At the retina, the converging cone had a certain 

 section, i. e. the circle of dispersion. The base of 

 the ; cone is evidently the pupil, which in the 

 reduced 'eye is taken to be 19 mm. in front of the 

 retina and 4 mm. in diameter. 1 



The length of y divided by the length of the 

 whole cone (19 mm.), gives the proportion in 



1 The diameter of the cone is not precisely that of the pupil. 

 The rays in the 1 vitreous would appear to come from the image 

 of the pupil formed by the lens. Thus the diameter changes 

 from 4 to 4.23 mm. At the same time the position of the base 

 is changed from 3.6 mm. (the distance of the plane of the pupil 

 behind the cornea) to 3.7 behind the cornea. This brings the 

 base of the cone 19 mm. in front of the retina, which is the 

 position assumed for it in the reduced eye. 



