Page:Handbook of Ophthalmology (3rd edition).djvu/24

18 lenses the object may still be distinctly seen. If, for instance, concave 12 is the strongest concave glass with which binocular vision at a distance of 12 inches is still possible, it can easily be calculated upon what distance the accommodation must be adjusted. We employ again the formula $$\frac{1}{a} + \frac{1}{\alpha} = \pm\frac{1}{f}$$ in which $$a$$ expresses the distance of the luminous point, $$\alpha$$ the distance of the image, and $$f$$ the focal distance of the concave or convex lens. The value of $$f$$ is negative when the focal distance is a virtual one, as in the case of concave glasses. Upon our supposition that $$a=12$$ and $$f=12$$, the formula $$\frac{1}{a} + \frac{1}{\alpha} = -\frac{1}{f}$$ becomes $$\frac{1}{12} + \frac{1}{\alpha} = -\frac{1}{12}$$. The virtual image of $$a$$ lies at $$\alpha$$, 6 inches from the lens,—that is, after their refraction in the concave lens the rays diverge as if they had proceeded from the point $$\alpha$$, 6 inches in front of the lens. The accommodation must be adjusted for this distance in order to see distinctly the image at $$\alpha$$, while the visual axes still intersect at the point $$a$$. That is to say, if in our experiment the visual axes converge toward a point 12 inches distant, and $$-\frac{1}{12}$$ is the strongest concave lens with which the fixation-object can still be distinctly seen, it follows that with the same convergence of the visual axes the accommodation can still be adjusted on a much nearer point. We find for this case the relative near point to lie 6 inches in front of the concave lens; or supposing the distance between the lens and the optical centre of the eye to be $$\frac{1}{2}$$ inch, the relative near point lies $$6\frac{1}{2}$$ inches distant from the latter.

We can determine in the same manner and with the same degree of convergence the strongest convex glass with which the point $$a$$ can still be distinctly seen. In this case the accommodation must, of course, be relaxed and adjusted for rays of less divergence, or no distinct retinal image can be formed. If we find that while maintaining the convergence of the visual axes upon a distance of 12 inches, convex 16 is the strongest convex lens with which the fixed point can still be distinctly seen, we can, by the help of the same formula, calculate the distance of the point upon which, during the experiment, the accommodation must be adjusted. The formula $$\frac{1}{a} + \frac{1}{\alpha} = \frac{1}{f}$$ becomes $$\frac{1}{12} + \frac{1}{\alpha} = \frac{1}{16}$$ or $$\frac{1}{\alpha} = -\frac{1}{48}$$; that is, the virtual image-point of the fixed point $$a$$ lies 48 inches from the convex lens, or the rays of light proceeding from $$a$$ diverge after their refraction by the convex lens as if they had