7.3 Eyeglasses

First a little nomenclature. Optometrists (and opthamologists) use the dioptric power measured in diopters. A diopter is $1/f$ where f is measured in meters. The focal length of lenses in contact is $$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$$ or using dioptric power $$D=D_1+D_2$$ A "normal" eye will focus an object at infinity onto the retina with the lens relaxed. As objects come closer the lens of the eye is made to bulgeto keep the object in focus. A nearsighted person (such as certain Phys 201 prof's) has an eye that focuses the object at infinity in front ofthe retina. This person has a far point beyond which things are focused incorrectly. By using a lens that casts the object at infinity onto the farpoint this problem can be corrected. Say someone's far point is at 2m. In this case we want $$\frac{1}{f}=\frac{1}{s_o}+\frac{1}{s_i}$$ $$\frac{1}{f}=\frac{1}{\infty }+\frac{1}{-2}$$ a corrective lens with $f=-2m$ or $D=-\frac{1}{2}$ Diopters. Notice that the desired focal length is the far point (with the correct choiceof sign)Of course in practice we need to take into account the distance between the glasses and the eye, but the above is valid for a contactlens. In the case of eyeglasses you need to subtract off the distance between the lense and the eye. For example if the distance between the eye andthe glasses is 2cm then the above becomes $$\frac{1}{f}=\frac{1}{\infty }+\frac{1}{-1.98}$$

A far sighted person focuses objects at infinity behind the retina. Their eye can accommodate the object at infinity but they have trouble with nearbypoints. A normal eye should be able to focus objects as close as 25cm. A far sighted person can focus to a near point that is greater than 25cm and so weneed to take the 25cm point and make it look like it is at the near point. Say some one has a near point of 125cm then $$\frac{1}{f}=\frac{1}{-1.25}+\frac{1}{.25}$$ which is $+3.2$ Diopters. Of course again we should correct for the distance between the glasses and theeye, which if that is 2cm makes the equation: $$\frac{1}{f}=\frac{1}{-1.23}+\frac{1}{.23}$$ Finally their are old guys like me who need bifocals. As one ages, the eye lens becomes more rigid and it is harder and harder for it to deform and focuson close in objects. Then if you are also nearsighted, then you have to resort to bifocals.