130 THE MICROSCOPE AND VISUAL ANGLE [Ch. V 



is situated, in order that its two extremities shall appear separate. 

 And so with any two details; they must be far enough apart to make 

 the visual angle one minute. 



To determine the actual length in millimeters required to subtend 

 one minute of arc in any case, it is only necessary to remember that 

 the entire circumference is 6.2832 times its radius (2 irr), and that 

 this circumference is divided into 360 or 21,600 minutes. 



If, now, the radius of the circle, or the distance of the eye from 

 the object, is 1 meter, the circumference of the circle will be 6.2832 

 meters or 6283.2 millimeters. As there are 21,600 minutes in the 

 entire circumference, the actual length of one minute with a circle 

 having a radius of one meter is 6283.2 mm. divided by 21,600 

 equals 0.29088 mm. That is, the eye at one meter distance requires 

 two points or two lines to be separated a distance of 0.29088 mm. 

 in order that they may be seen as two and not appear to be fused 

 together. 



It is assumed by workers with the microscope that the distance of 

 most distinct vision for adults when looking at objects for details of 

 structure is 254 mm. or 10 inches. This is the standard distance 

 selected for the determination of magnifying power in microscopy 

 also. 



The question now is, how large a retinal image will be formed by 

 an object giving a visual angle of 1 minute at the standard distance of 

 254 mm. 



First must be found the actual size of the object to give a visual 

 angle of 1 minute at 254 mm. distance. It is known from the above 

 calculation that for one meter or 1000 mm. the object must have a 

 size of 0.29088 mm. Now for 254 mm. the length must be T Vu 4 u of 

 this number or 0.073S8352 mm., that is, a little more than one-fourth 

 the size at 1 meter. 



Now to determine the size of the retinal image at 254 mm. image 

 distance, the distance from the center or nodal point of the eye must 

 be known as well as the image distance and the size of the object. The 

 distance of the retinal image from the nodal point is assumed to be 

 15 mm. (Howell, p. 306); then the size of the retinal image will be: 

 0.073S8352: x: : 254: 15=0.00436 mm. or 4.36/a, and this size would 



