Page:Optics.djvu/143

 159. When a very small object is to be examined, the first expedient that occurs, is to put it as near one's eye as is conveniently possible, because the angle it subtends at the center of the eye being thus increased, it so far appears larger. There is, however, a limit to this, though some persons, particularly those who are short-sighted, can avail themselves of it to a greater extent than others. The cause of the limitation is, as the student will remember, that when an object is placed too near the eye, the extreme rays of any pencil admitted by the pupil are not refracted to the focus on the retina, and the vision becomes indistinct.

This evil may be remedied, in the first place, by supplying the fault of the pupil, that is, by stopping the extreme rays, and admitting the others through a small aperture in a plate of some opaque substance. This accordingly is found to answer pretty well in some cases, but only where there is a good deal of light proceeding from the object.

In order to explain this more completely and correctly, we will suppose $PQR$, Fig. 170, to be a small object, $1⁄20$ of an inch long, for instance, which is to be examined by a person who cannot see distinctly any thing nearer to his eye than 6 inches. It is plain that the greatest angle the object can subtend at his eye, is that of which the tangent is $1⁄6×20$, or $1⁄120$.

Now let $A$ be a convex lens of half an inch principal focal length, placed at a quarter of an inch from the eye $O$, and let $AQ$ be $12⁄25$, and consequently, $Aq$ the distance of the image, 12 inches, its length $pr=Aq⁄AQ$. $PR$, that is, $12·25⁄12·1⁄20$, or $25⁄20$; which is an inch and a quarter.

The angle it subtends at the eye has for its tangent this length divided by the distance $Oq$, which is $121⁄4$, or $49⁄4$; it is, therefore, $25⁄20·4⁄49$, or $1⁄8$ nearly.