Hyperopia: When the Near Point Is Pushed Out
1 · Predict
A hyperopic (far-sighted) eye can't accommodate enough to focus on nearby objects — its near point is farther than the normal 25 cm. Does a more severe hyperopia (a farther-out near point) require more or less corrective lens power?
- More — a farther-out near point needs a stronger converging lens to bring close objects into the eye's limited focusing range.
- A farther-out near point needs less corrective power.
- Corrective power doesn't depend on how far out the near point is.
2 · Set Up
- Open the hyperopia preset and press Reset. This eye's near point is pushed out to 100 cm — anything closer appears blurred.
- Enable the corrective-power readout.
- Set the eye's near point for each trial (exploring different hyperopia severities) and record the corrective lens power needed.
3 · Collect Data
| Near point (cm) | Corrective power P (D) |
|---|---|
| 60 | |
| 80 | |
| 100 |
Plot corrective power P (y-axis) against near point (x-axis) for your three trials.
4 · Analyze
- For one trial, compute P = 1/0.25 − 100/near point (cm), giving diopters. Compare to the table.
- Explain why P comes out positive — hyperopia is corrected with a CONVERGING lens, which images a normally-close object (at the standard 25 cm reading distance) out at the eye's actual (pushed-out) near point instead.
5 · Extend
- Reading glasses (a common over-the-counter form of hyperopia correction) come in standard positive-diopter strengths like +1.00, +1.50, +2.00 D. Explain, using your data, why someone whose near point is pushed out much farther than normal would need a higher-power pair.
- Age-related presbyopia (stiffening eye lens) and true hyperopia (an eyeball shape issue) both push the near point outward and are corrected the same way, with converging lenses — but they're different underlying causes. Why might both conditions still use the same 'add reading power' correction formula?
The Physics Behind This Experiment
Hyperopia Corrective Power
A converging corrective lens images an object at the standard reading distance (25 cm) out at the eye's actual near point, so its power is P = 1/0.25 m − 1/near point(m) — positive diopters, growing stronger as the near point moves farther out.