c 27 c 2 optics. 



object is distinctly visible to the eye E, placed any 

 where before the lens D. Consequently, it will 

 appear as much larger through the lens than to the 

 naked eye, as C D is less than A B. 



If an object, A B (Fig. 10.), be placed in one 

 focus C, of a lens D E, and the eye in the other 

 focus F, the eye will see just so much of the object 

 as is equal to the diameter of the lens ; for the rays 

 A D and B E, which go from the object to the 

 extremities of the lens D and E, and are united at 

 the focus F, must necessarily proceed from the 

 object to the lens parallel to the axis FC, and, 

 therefore, parallel to each other ; consequently, 

 the part of the object A B, seen by the rays D F, 

 E F, will be equal to the diameter D E of the said 

 lens. 



If only the part d e of the lens be open, then 

 only so much of the object a b, as is equal to it, 

 will be perceived by the eye. Now, since A B is 

 equal to D E, or a b to d e, therefore the angle 

 D F E, or d F e, is the optic angle under which the 

 part of the object A B or a b appears to the eye at 

 F ; and since G F is but one half F C, therefore 

 the angle D F E, or dF e, is double that under 

 which the part A B or a b, would appear to the 

 naked eye at the distance F C ; that is, the eye 

 sees the object situated as above, twice as large 

 with the lens as it would do without it. 



If you would see a portion of an object larger 

 than the lens, your eye must be placed nearer the 

 lens than its focus. Let the lens be D E (Fig. 11), 

 and its two foci F and C ; in the focus C, let there be 

 an object, A B, larger than the lens ; suppose the 

 rays AD, BE, proceed from the extremities of 

 the object to those of the lens ; it is evident from 

 the figure, they will be convergent, and, therefore, 



