Your eye is a two-lens system: the curved cornea does most of the bending, and the flexible lens behind it fine-tunes the focus. Muscles squeeze the lens to focus on near things and relax it for far things. A healthy young eye can focus anywhere from its near point (about 25 cm) out to its far point (infinity). Glasses help when that range is shifted. Optometrists describe lenses by their power, and power is the thin lens equation in disguise.
Key idea
A lens's power is $$P = \frac{1}{f}\quad(\text{diopters, D, with } f \text{ in meters}).$$ A nearsighted eye has a far point that is too close: it needs a diverging lens ($P < 0$) that makes a virtual image of distant objects at the far point. A farsighted eye has a near point that is too far: it needs a converging lens ($P > 0$) that makes a virtual image, at the near point, of an object 25 cm away.
Why it works
Glasses do not change your eye; they change what the eye is asked to look at. A corrective lens takes the object you want to see and forms an image of it somewhere your eye can focus. For a nearsighted eye with far point FP, a distant object ($d_o = \infty$) must have its image at $d_i = -\text{FP}$ (virtual, on the same side as the object): $\frac{1}{f} = \frac{1}{\infty} + \frac{1}{-\text{FP}}$, so $f = -\text{FP}$. For a farsighted eye with near point NP, an object at 25 cm must have its image at $-\text{NP}$: $$P = \frac{1}{0.25\ \text{m}} - \frac{1}{\text{NP}}.$$ Powers are handy because thin lenses touching each other simply add their powers.
Worked examples
Worked example: Nearsighted, with contact lenses
A nearsighted person's far point is 2.0 m. What contact lens power lets them see distant objects clearly?
- The lens must image infinity at the far point: $f = -2.0$ m.
- $P = \frac{1}{-2.0\ \text{m}} = -0.50$ D.
Answer: $P = -0.50$ D (diverging)
Worked example: Farsighted, with contact lenses
A farsighted person's near point is 1.00 m. What contact lens power lets them read at 25 cm?
- The lens must image an object at 0.25 m at the near point, $d_i = -1.00$ m.
- $P = \frac{1}{0.25} + \frac{1}{-1.00} = 4.00 - 1.00 = 3.00$ D.
Answer: $P = +3.00$ D (converging)
Worked example: Glasses sit in front of the eye
The person with a 2.0 m far point would rather wear glasses 2.0 cm in front of the eye. What power do the glasses need?
- Distances must be measured from the glasses: the far point is $2.00 - 0.02 = 1.98$ m from them.
- $f = -1.98$ m, so $P = -0.505$ D. For a weak lens the difference is tiny; for strong prescriptions it matters.
Answer: $P \approx -0.51$ D
Common mistake
Thinking a nearsighted person needs a converging lens “to help them see far away”. Their eye already bends light too much: distant objects focus in front of the retina. The fix is to spread the light a little first, with a diverging lens. Check with the image: the lens must move a distant object's image closer, to the far point, and only a diverging lens makes a virtual image of a far object close by.