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TEXT-BOOK OF PHYSIOLOGY. 



them as such, is about 1.454 mm. Lines drawn from the extremities of 

 such an object or interval, to the nodal point, subtend an angle of 60 seconds. 1 

 Beyond this the two points are indistinguishable. In other words the 

 emmetropic eye possesses the power of distinguishing the correspondingly 

 small interval between the two images on the retina of the two objective 

 points. The size of the image or the interval between the two retinal points, 

 determined from the foregoing factors by the formulae on page 655 is 0.004 

 mm., which would correspond to a visual angle of 60 seconds. If the 

 retinal distance is less than this the two sensations fuse into one. The reason 

 assigned for this is, that the distance between the centers of two adjoining 

 cones in the macula is 0.004 mm - With a visual angle not less than 60 

 seconds, the two foci fall on separate cones. With a smaller visual angle 

 the two foci fall on, and excite but a single cone and hence there arises the 

 sensation of but a single point. The acuteness of vision, therefore, of the 

 emmetropic eye depends on its power of distinguishing the smallest retinal 

 image or the smallest interval between two cones on the retina, correspond- 

 ing to a visual angle of 60 seconds. 



In ophthalmic practice it is customary in testing the acuteness of vision 

 to employ test letters of specific sizes for specific distances. The letters are 

 so proportioned that when they are placed at the specified distances, the 

 extremities of the letters subtend an angle of 5 minutes. The letters have 

 been constructed on the following basis: Since to an angle of 60 seconds 

 there corresponds an object of 1.454 mm. at the distance of 5 meters as 

 shown before and as the object decreases in proportion to the distance 

 (for the same visual angle) it is evident that the object would have to be one- 

 fifth of 1.454 mm. or 0.2908 mm. in order to subtend an angle of 60 seconds 

 at one meter. From this the size for any other distance in meters is found 

 simply by multiplying 0.2908 mm. by the distance. The standard letters 

 are so constructed that each is inscribed within a square the sides of which 

 at a specific distance subtend an angle of 5 minutes and which is again sub- 

 divided into 25 small squares each side of which subtends an angle of i 

 minute. These partial little squares correspond to the details of the letter 

 while the whole letter of course, embraces an angle of 5 minutes both as to 

 height and to breadth. The letter that could be distinctly seen at a dis- 



1 The size of the visual angle, under which an object of this size and situated at a distance 

 of 5 meters is distinctly seen, can be determined from the following Fig. 316, in which A B represents 



the size of the object 1.454 mm. ; 

 N, the nodal point; CN, the 

 line which bisects the object, 

 represents the distance of the 

 object from the nodal point; 

 a, the visual angle subtended 

 and whose value it is desired 

 to know, and b one-half of the 

 angle a. By trigonometry the 

 size of the angle a can be de- 

 FIG. 316. FIGURE SHOWING THE METHOD OF OBTAINING term i n e d in the following way: 

 THE VISUAL ANGLE EXPRESSED IN DEGREES OR FRACTION one-half the size of the object 

 OF A DEGREE OF AN ARC A B > *? o^ded by its distance 



from the nodal point; the quo- 

 tient is the tangent of half the angle. Thus 0.727 -5- 5000 =0.0001454. By reference to tables of 

 natural tangents, it will be found that the angle or fraction of the circle corresponding to this 

 tangent is 30 seconds, and that therefore the whole angle is 60 seconds. 



