VISION 979 



Reduced Eye 



Radius of curvature of the single refracting surface - 5-1 mm. 



Index of refraction of the single refracting medium - - i*35* ,, 

 An tero -posterior diameter of reduced eye (distance of 



principal focus from the single refracting surface) - 20-0 ,, 

 Distance of the single refracting surface behind the 



anterior surface of the cornea - 2*2 ,, 



Distance of the nodal point of the reduced eye from 



jits anterior surface - - 5-0 ,, 



Distance of the nodal point from the principal focus 



(retina) - - 15-0 ,, 



Knowing the position of the centre of curvature of the single 

 ideal refracting surface i.e., the nodal point of the reduced eye 

 all that is necessary in order to determine the position of the 

 image of an object on the retina is to draw straight lines from its 

 circumference through the nodal point. Each of these lines cuts 

 the refracting surface at right angles, and therefore passes through 

 without any devi- 

 ation. The retinal 

 image is accord- 

 ingly inverted and 

 its size is propor- 

 tional to the solid 

 angle contained 

 between the lines 

 drawn from the 



boundary of the Fig. 401. Figure to show how the Visual Angle and Size 

 object to the nodal * Retinal Image varies with the Distance of an Object 



of Given Size. For the distant position of AB the visual 

 , ' angle is a, for the near position (dotted lines) B. 



angle contained by 



the prolongations of the same lines towards the retina. This angle 

 is called the visual angle, and evidently varies directly as the size 

 of the object, and inversely as its distance. Thus the visual angle 

 under which the moon is seen is much larger than that under which 

 we view any of the fixed stars, because the comparative nearness of 

 the earth's satellite more than makes up for its relatively small size. 



The dimensions of the retinal image of an object are easily calculated 

 when the size of the object and its distance are known. For let AB 

 in Fig. 401 represent one diameter of an object, A'B' the image of this 

 diameter, and let AB', BA', be straight lines passing through the nodal 

 point. Then AB and A'B' may be considered as parallel lines, and 

 the triangles of which they form the bases, and the nodal point the 

 common apex, as similar triangles. Accordingly, if D is the distance 

 of the nodal point from A, and d its distance from B', we have 



AB A'B' XT 



-=r-= -| . Now, d may approximately be taken as 15 mm. Suppose, 



then, that the size of the moon's image on the retina is required. Here 

 0=238,000 miles, and AB (the diameter of the moon)= 2,160 miles. 



* Or a little more than that of the aqueous humour. 



