The Human Eye and the Colourful World · Lesson 3 of 8
Defects of Vision and Their Correction
“When the eye misses the retina, the right lens helps light find its way home.”
• Distinguish myopia, hypermetropia and presbyopia by their symptoms and image positions. • Explain the physical causes of common refractive defects. • Select concave, convex or bifocal lenses for appropriate vision needs. • Interpret ray diagrams showing defective and corrected vision. • Calculate focal length and lens power for common correction problems. • Describe the purpose and basic process of eye donation.
Clear vision requires the eye’s optical system to make rays meet exactly on the retina. If they meet before the retina, would meet behind it, or cannot be adjusted properly across viewing distances, the image appears blurred. These are refractive defects: the incoming light is not focused at the required retinal position under particular viewing conditions.
The three common defects considered here are myopia, hypermetropia and presbyopia. Myopia mainly affects distant vision, hypermetropia mainly affects nearby vision, and presbyopia develops when ageing reduces accommodation. Suitable spherical lenses alter the rays before they enter the eye, allowing the eye to form the image on the retina.
| Defect | Main difficulty | Uncorrected image tendency | Typical correction |
|---|---|---|---|
| Myopia | Distant objects | In front of retina | Concave lens |
| Hypermetropia | Nearby objects | Behind retina | Convex lens |
| Presbyopia | Nearby objects as accommodation weakens | Near point recedes | Convex reading portion or bifocal lens as needed |
Myopia
A refractive defect in which nearby objects can be seen clearly but distant objects cannot be seen distinctly. It is also called near-sightedness.
A myopic eye has a far point nearer than infinity. Depending on the severity, a person may see clearly only up to a few metres. Parallel rays from a distant object are converged too strongly and form an image in front of the retina. By the time the rays reach the retina, they have begun to diverge again, so each object point produces a blurred patch rather than a sharp point.
Myopia may arise because the eye lens has excessive curvature and therefore too much converging power, or because the eyeball is elongated so that the retina lies farther behind the position where the rays meet. Both causes produce the same optical result for distant rays: focus occurs before the retina.
A concave lens of suitable power corrects myopia. It diverges the nearly parallel rays before they enter the eye. The corrected rays appear to come from the person’s far point, which the myopic eye can focus. The combined action of spectacle lens and eye lens then brings the image onto the retina. A convex lens would increase convergence and worsen the defect.
Hypermetropia
A refractive defect in which distant objects can be seen clearly but nearby objects cannot be seen distinctly. It is also called far-sightedness.
In hypermetropia, the near point lies farther away than the normal near point of about 25 cm. A nearby book must be held unusually far away for comfortable reading. Rays from a nearby object are strongly divergent, but the eye does not converge them enough. They would meet behind the retina, so the retinal image is blurred.
The defect may occur because the focal length of the eye lens is too long, meaning its converging power is insufficient, or because the eyeball is too small and the retina lies too close to the lens. A convex lens supplies additional converging power. It makes the rays entering the eye less divergent, or suitably convergent, so the eye can bring them to the retina.
Lens Power And Correction Strategy
Lens power measures how strongly a lens converges or diverges light. A short focal length means strong bending and therefore a large magnitude of power. Always convert focal length to metres before using the relationship. The sign is as important as the number: positive power identifies a converging convex lens, while negative power identifies a diverging concave lens.
• Identify whether distant or nearby vision is affected. • Decide where the correcting lens must make the rays appear to originate. • Assign signs using the lens sign convention: a real object on the incident-light side has negative u; a virtual image on that same side has negative v. • Use 1/f = 1/v − 1/u when image and object distances are involved. • Convert f to metres before using P = 1/f. • Interpret the sign of P to confirm the lens type.
Problem
A lens has power −5.5 D for distant vision and another lens has power +1.5 D for near vision. Find both focal lengths and identify the lenses.
- 1.For distant vision, P = −5.5 D. Use f = 1/P = 1/(−5.5) m = −0.1818 m.
- 2.Convert to centimetres: −0.1818 m × 100 = −18.18 cm, approximately −18.2 cm. The negative sign identifies a concave lens.
- 3.For near vision, P = +1.5 D. Then f = 1/(+1.5) m = +0.6667 m.
- 4.Convert to centimetres: +0.6667 m × 100 = +66.67 cm, approximately +66.7 cm. The positive sign identifies a convex lens.
- 5.Check: the larger magnitude 5.5 D corresponds to the shorter focal-length magnitude, as expected.
Problem
The far point of a myopic person is 80 cm in front of the eye. Find the nature and power of a thin correcting lens used for distant objects.
- 1.A distant object is effectively at infinity, so rays reaching the correcting lens are parallel.
- 2.The lens must form a virtual image at the person’s far point, 80 cm in front of the lens. Therefore f = v = −80 cm = −0.80 m.
- 3.Use P = 1/f: P = 1/(−0.80) D = −1.25 D.
- 4.The negative power confirms that the required lens is concave.
- 5.Reasonableness check: a concave lens diverges parallel rays so they appear to come from the finite far point that the eye can focus.
Problem
A hypermetropic person has a near point of 1.0 m. Find the power of the lens that allows an object at the normal near point of 25 cm to be seen clearly.
- 1.The real object is 25 cm in front of the correcting lens, so u = −25 cm = −0.25 m.
- 2.The lens must form a virtual image at the person’s near point, 1.0 m in front of the lens, so v = −1.0 m.
- 3.Use 1/f = 1/v − 1/u = 1/(−1.0) − 1/(−0.25).
- 4.Thus 1/f = −1 + 4 = 3 m⁻¹, so f = +1/3 m = +0.333 m.
- 5.Power P = 1/f = +3.0 D. The positive sign confirms a convex lens.
- 6.Check: the lens must add convergence for a nearby object, so a positive answer is physically sensible.
Problem
A person uses −2.0 D for distant vision and +2.5 D for near vision. Find the focal length of each portion of a bifocal lens and state where each is placed.
- 1.Distant portion: f = 1/(−2.0) m = −0.50 m = −50 cm. It is concave and is placed in the upper portion for looking ahead.
- 2.Near portion: f = 1/(+2.5) m = +0.40 m = +40 cm. It is convex and is placed in the lower portion for reading.
- 3.The signs show that the two portions perform opposite optical actions: divergence for distant myopic correction and extra convergence for nearby viewing.
Do not use centimetres directly in P = 1/f; power in dioptres requires metres. Do not discard the sign of focal length. Do not place a real corrected image at the defective eye’s far or near point: the spectacle lens generally forms a virtual image there for the eye to focus.
Presbyopia
An age-related defect in which the power of accommodation decreases and the near point gradually moves farther away, making nearby objects difficult to see clearly.
Presbyopia develops as the ciliary muscles gradually weaken and the eye lens becomes less flexible. The eye can no longer increase lens curvature as effectively for nearby objects, so the near point recedes. Presbyopia is not identical to hypermetropia: both can cause difficulty with nearby vision, but presbyopia specifically results from age-related loss of accommodation.
Some people require correction for both distant and nearby vision. A common bifocal lens combines a concave upper portion for distant vision with a convex lower portion for near vision. Looking forward naturally uses the upper portion; lowering the gaze for reading uses the lower portion. Depending on the person’s condition, refractive defects may also be corrected with contact lenses or surgical interventions.
Think It Over
The transparent cornea can remain useful after a person’s death. Donated corneal tissue may be used in corneal transplantation to restore vision for people with corneal blindness. A single pair of donated eyes can provide corneal tissue for up to four people. Eye donation therefore allows the gift of sight to continue and also supports valuable medical research and education when tissue is unsuitable for transplantation.
• Donors may belong to different age groups and sexes. The use of spectacles or previous cataract surgery does not automatically prevent donation. • People with diabetes, hypertension or asthma may also be considered when communicable disease exclusions do not apply. • The nearest eye bank should be informed immediately because removal is required within four to six hours after death. • A trained eye-bank team can carry out the brief removal procedure at a home or hospital without disfigurement. • People infected with or dying from conditions such as AIDS, hepatitis B or C, rabies, acute leukaemia, tetanus, cholera, meningitis or encephalitis cannot donate eyes. • Eye banks evaluate and distribute donated tissue under strict standards, and donor and recipient identities remain confidential.
Suitability is evaluated by trained professionals. The central scientific idea is that corneal transparency, not the person’s spectacle prescription, determines whether donated tissue can help someone with corneal blindness.
Quiz
Where is the image of a distant object formed in an uncorrected myopic eye?
Which lens corrects hypermetropia?
A correcting lens has power −2.0 D. What does the sign show?
Why does the near point recede in presbyopia?
A myopic person can see only up to 2 m. What should a distant-object correcting lens do?
Practice Problems
- A learner sees a book clearly but the distant board is blurred. Identify the defect, describe the image position and state the correction. Answer: The defect is myopia. Distant rays focus in front of the retina, and a suitable concave lens diverges them so the final image forms on the retina.
- Find the focal length of a +4.0 D lens. Solution: f = 1/P = 1/4.0 m = +0.25 m = +25 cm. The positive sign shows it is convex.
- The far point of a myopic eye is 50 cm. Find the correcting power for distant vision. Solution: A distant object must have its virtual image at the far point, so f = −50 cm = −0.50 m. P = 1/f = 1/(−0.50) = −2.0 D. A concave lens is required.
- A hypermetropic eye has a near point of 50 cm. Find the lens power needed to read at 25 cm. Solution: u = −0.25 m and v = −0.50 m. Then 1/f = 1/v − 1/u = −2 − (−4) = +2 m⁻¹. Therefore P = +2.0 D, so a convex lens is required.
- Explain why a bifocal lens may contain a concave upper portion and a convex lower portion. Answer: The concave upper part corrects distant myopic vision when looking ahead. The convex lower part supplies extra convergence for nearby reading when the gaze is lowered. Each portion serves a different viewing distance.
Key Takeaways
• Myopia causes distant rays to focus in front of the retina and is corrected with a concave lens. • Hypermetropia causes nearby rays to tend to focus behind the retina and is corrected with a convex lens. • Presbyopia results from weakening ciliary muscles and reduced lens flexibility as accommodation decreases. • Bifocal lenses can combine a concave upper part for distant vision and a convex lower part for near vision. • Lens power is P = 1/f when focal length is measured in metres. • Negative power identifies a concave lens, while positive power identifies a convex lens. • Ray diagrams should be interpreted by locating where rays meet relative to the retina before and after correction. • Eye banks evaluate donated corneal tissue for transplantation, research or medical education.