Defects of vision and their correction

A normal eye brings both a nearby page held at 25 cm and a distant hill on the horizon into sharp focus on the retina, because the ciliary muscles can adjust the eye lens's curvature within a working range called the power of accommodation. When that range shifts too close, shrinks, or fails with age, the image no longer lands exactly on the retina, and a defect of vision results. CBSE Class 10 asks you to know three named defects — myopia, hypermetropia, and presbyopia — what causes each, where the mismatched image actually forms, which single spherical lens fixes it, and how to run the lens-formula numericals that turn near-point or far-point data into the focal length and power of the correcting lens. This article stays inside that scope: no new physics beyond the thin-lens formula and the sign convention you already use for mirrors and lenses, just applied more carefully to a biological instrument.

The core insight

Myopia, hypermetropia, and presbyopia are all mismatches between where the eye's own accommodation range can focus and where a clear image is needed, and each is corrected by choosing a spherical lens whose focal length — found from the lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ — relocates the effective object exactly onto the eye's accessible focusing range.

The core idea

A normal eye has a near point at about 25 cm (the closest distance of distinct vision) and a far point at infinity. Anything outside that range cannot be brought to a sharp focus on the retina no matter how the ciliary muscles strain, and that is exactly what happens in each defect below.

Myopia (near-sightedness). The eyeball is too long, or the eye lens too strongly curved, so parallel rays from a distant object converge and cross in front of the retina before they even reach it — by the time the light hits the retina it has already spread back out into a blur. The far point has moved in from infinity to some finite distance. Correction: a concave (diverging) lens, placed before the eye, spreads the incoming parallel rays out just enough that they now appear to originate from the myopic eye's own far point, so the eye's normal focusing machinery can complete the job.

Hypermetropia (far-sightedness). The eyeball is too short, or the eye lens too weakly curved, so rays from a nearby object are still converging when they reach the retina — they would meet behind it if the retina were not in the way. The near point has receded beyond the normal 25 cm. Correction: a convex (converging) lens, which pre-converges the rays from a close object (say, one held at the normal 25 cm) just enough that the eye perceives them as coming from its own, more distant, near point.

Presbyopia. With age the ciliary muscles weaken and the eye lens loses elasticity, so the power of accommodation itself shrinks — the near point recedes, as in hypermetropia, and sometimes the far point is affected too if myopia is already present. Many older people therefore need bifocal lenses, with a convex portion for close work and a concave (or plain) portion for distance, in the same pair of spectacles. This is qualitative in CBSE — no numericals are set on presbyopia. In every numerical, the same tool does the work: the thin-lens formula, applied to the corrective lens alone (not to the eye lens itself), with the standard Cartesian sign convention.

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}, \qquad P = \frac{1}{f}\ (f\text{ in metres},\ P\text{ in dioptre, D})

where: $u$ is the object distance from the corrective lens, $v$ is the image distance it must produce, $f$ is the lens's focal length, and $P$ is its power. Distances are measured from the optical centre; distances in the direction of the incident light are positive, distances measured against it are negative — so a real object in front of the lens always gives $u$ a negative value, and the virtual images used throughout this topic also come out negative.

Figure

Cross sections comparing where the image forms in a normal eye, a myopic eye and a hypermetropic eye, and how a concave or convex correcting lens moves the image back onto the retina

Derivation AFocal length and power of a corrective lens from near or far point data

The goal is to turn near-point or far-point data for a defective eye into the focal length and power of the single corrective lens that fixes it, using nothing beyond the thin-lens formula and a careful choice of object and image distance for the two defect types, since that choice — not the algebra — is where most marks are lost.

Step 1. Identify the defect and state, in words, what the corrective lens must achieve. For myopia, an object at infinity must appear, after the lens, to come from the eye's own far point, so that the myopic eye can then focus it normally. For hypermetropia, an object at the normal near point of 25 cm must appear, after the lens, to come from the eye's own (farther) near point.

Step 2. Fix the object distance $u$ for the corrective lens. For myopia, $u = -\infty$ (a distant object). For hypermetropia, $u = -25\ \text{cm}$ , converted to metres, since the normal reading distance is what the lens must accept as input.

Step 3. Fix the image distance $v$ . In both cases the lens must form a virtual image — on the same side as the object, so $v$ is negative — located exactly at the eye's own defective point: $v = -(\text{far point})$ for myopia, $v = -(\text{near point})$ for hypermetropia.

Step 4. Substitute into $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ . For myopia the $1/u$ term vanishes since $u \to -\infty$ , leaving $\frac{1}{f} = \frac{1}{v}$ , so $f$ simply equals the (negative) far point — a concave lens. For hypermetropia both terms survive and typically partially cancel, leaving a smaller, positive $f$ — a convex lens.

Step 5. Report $f$ with its sign and unit (metres), then compute power $P = \frac{1}{f}$ in dioptre. A negative $f$ (and $P$ ) confirms a concave lens for myopia; a positive $f$ (and $P$ ) confirms a convex lens for hypermetropia — a sign check that catches most arithmetic slips.

Worked example

Question: A person is myopic and cannot see objects clearly beyond 2 m. Find the power and nature of the corrective lens needed so that the person can see distant objects clearly. (Extends the same numerical to also check the power needed if the far point were instead 5 m.)

Step 1. The far point is 2 m, so the corrective lens must form a virtual image of an object at infinity exactly at that far point.

Step 2. Take $u = -\infty$ and $v = -2\ \text{m}$ (virtual image, same side as the object).

Step 3. Apply the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{-2} - 0 = -\frac{1}{2}\ \text{m}^{-1}$ , so $f = -2\ \text{m}$ .

Step 4. Power $P = \frac{1}{f} = -0.5\ \text{D}$ . A concave (diverging) lens of power $-0.5\ \text{D}$ (focal length $-2\ \text{m}$ ) is required.

Step 5 (stretch check). If the far point had instead been 5 m, the same method gives $f = -5\ \text{m}$ and $P = -0.2\ \text{D}$ — a weaker concave lens, confirming that a farther (less severe) myopic far point needs a smaller-magnitude negative power.

Common traps

Mixing up which defect needs which lens: myopia ("sees near, not far") needs a concave lens; hypermetropia ("sees far, not near") needs a convex lens.
Forgetting the sign of $u$ and $v$ in the lens formula, which flips the sign of $f$ and hence the wrong lens type.
Thinking myopia means the far point is very close — it only means the far point is less than infinity, which can still be several metres away.
Treating presbyopia as "myopia plus hypermetropia" rather than a distinct, age-related loss of accommodation that may need bifocals.

Ask the tutor: want to try a hypermetropia numerical with different near-point data, or see how bifocal power is split between the two portions?

Why is the image distance taken as negative in these numericals?

Because the corrective lens is forming a virtual image, and the sign convention says distances measured against the direction of incident light are negative. In both myopia and hypermetropia the lens must make light appear to come from a point on the same side as the real object — that is exactly what a virtual image is — so $v$ is always negative in this topic, alongside $u$ , which is negative because the object is always real and in front of the lens.

Why does a concave lens correct myopia instead of a convex one?

A myopic eye already converges rays too strongly, bringing them to a focus before they reach the retina. Adding a convex lens would converge them even more, making things worse. A concave lens diverges the incoming parallel rays slightly before they enter the eye, so the eye's own (excess) converging power now lands the focus exactly on the retina instead of in front of it.

Can one pair of spectacles fix both myopia and presbyopia at once?

Yes, if a person who was already myopic also develops presbyopia with age, opticians prescribe bifocal lenses: the upper portion, typically concave, corrects the original myopia for distance viewing, while the lower portion, convex, compensates for the weakened accommodation and restores near vision for reading. The two corrections address different problems (a permanent eyeball/lens-shape issue versus an age-related loss of flexibility) but can share the same spectacle frame.

Defects of vision and their correction

The core idea

Worked example

Common traps

Before this

Where this leads