ScienceCBSE Class 10 · The Human Eye and the Colourful World

Defects of Vision and Their Correction

Updated 14 October 2026 · Fact-checked

The three common refractive defects are myopia (cannot see far objects; image forms in front of the retina; corrected by a concave lens), hypermetropia (cannot see near objects; image forms behind the retina; corrected by a convex lens) and presbyopia (age-related loss of accommodation; often needs bifocal lenses).

Understand Defects of Vision and Their Correction

Why defects arise. A normal eye focuses objects from 25 cm to infinity on the retina. If the eye gradually loses its power of accommodation, or its shape or lens curvature is not right, images do not fall on the retina and vision becomes blurred. Such refractive defects are corrected with suitable spherical lenses (or, these days, contact lenses or surgery).

1. Myopia (near-sightedness).

The person sees nearby objects clearly but distant objects are blurred; the far point is nearer than infinity, perhaps a few metres.
The image of a distant object forms in front of the retina.
Causes: (i) excessive curvature of the eye lens, or (ii) elongation of the eyeball.
Correction: a concave (diverging) lens of suitable power, which spreads the rays slightly so the image moves back onto the retina. The lens is chosen so that it forms a virtual image of a very distant object at the person's far point. If the far point is at distance x, the lens has f = −x.

2. Hypermetropia (far-sightedness).

The person sees distant objects clearly but nearby objects are blurred; the near point is farther than 25 cm, so reading material must be held far away.
Light from a nearby object is focused behind the retina.
Causes: an eye lens whose focal length is too long, or an eyeball that is too small (too short from front to back).
Correction: a convex (converging) lens of suitable power, which supplies the extra focusing power. The lens is chosen so that an object at 25 cm forms a virtual image at the person's near point. Use the lens formula with u = −25 cm and v = −(near point).

3. Presbyopia.

With ageing, the power of accommodation decreases and the near point gradually moves away, so nearby objects are hard to see without glasses.
Cause: gradual weakening of the ciliary muscles and reduced flexibility of the eye lens.
Some people have both myopia and hypermetropia and need bifocal lenses: the upper part is concave (for distant vision) and the lower part is convex (for near vision).

Calculating the corrective lens.

Myopia, far point at distance x: $f = - x$ , $P = \frac{1}{f}$ (negative).
Hypermetropia, near point at distance d: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ with $u = - 25$ cm and $v = - d$ ; P positive.

Remember to convert to metres for power.

Defects of Vision and Their Correction: formulas and key facts

Myopia correction: $f = - (far point distance); P = \frac{1}{f} < 0$; Concave lens: image of a distant object at the far point.
Hypermetropia correction: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}, u = - 25 cm, v = - (near point)$; Convex lens; P > 0.
Power: $P = \frac{1}{f ( m )}$; In dioptres.
Bifocal: $upper: concave (far); lower: convex (near)$; For people with both defects.

How to solve Defects of Vision and Their Correction questions

For defect questions, identify the defect from the symptom, then design the lens.

1Symptom → defect: cannot see far (myopia); cannot see near (hypermetropia); near point recedes with age (presbyopia).
2Give the image position (in front of or behind the retina) and the two causes.
3Name the corrective lens: concave for myopia, convex for hypermetropia, bifocal for both.
4For myopia power: f equals minus the far point distance.
5For hypermetropia power: use u = −25 cm and v = −near point in the lens formula; solve for f.
6Convert f to metres and find P = 1/f with the correct sign.
7Draw the ray diagrams (defective eye and corrected eye) if asked.

Defects of Vision and Their Correction in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Myopia and its correction

Suppose a student sitting in the last row cannot read the blackboard clearly but reads her notebook easily. An eye test shows that she sees clearly only up to 50 cm.

Name the defect, state where the image of the blackboard forms in her eye, give two possible causes, and find the type and power of the corrective lens.

Try it first, then show the solutionSolution

Clear near vision but blurred distant vision: myopia (near-sightedness).
The image of the distant blackboard forms in front of the retina.
Possible causes: excessive curvature of the eye lens, or elongation of the eyeball.
Correcting lens must form a virtual image of a distant object at her far point: f = −50 cm = −0.50 m.
Power P = 1/f = 1/(−0.50) = −2 D: a concave lens of power −2 D. Check: u = −∞, v = −50 cm gives 1/f = 1/v = −1/50.

Answer: Myopia; image forms in front of the retina; caused by excessive lens curvature or an elongated eyeball; corrected by a concave lens of −2 D (f = −50 cm).

Memory hook: Can't see the board? Myopia: minus lens.

Tricks to remember Defects of Vision and Their Correction

Short memory aids. Say each one aloud once, then use it on a question the same day.

Myopia = Minus
MYopia: Minus (concave) lens; HYPERmetropia: Plus (convex) lens.
Use it when: Choosing the lens.
Short-sighted, short far point
Myopic people's far point is short; hypermetropic people's near point is far.
Use it when: Identifying the defect from data.
Bifocal: Up far, Down near
Look up through the concave top for distance; look down through the convex bottom to read.
Use it when: Bifocal structure.

Defects of Vision and Their Correction: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Blackboard problem · 3 marks · Short answerBoard level

From the back bench, Ravi finds the writing on the blackboard blurred, though his textbook is perfectly clear. Name the likely defect and say how it is corrected. Draw ray diagrams of the defective and corrected eye.

Try it first, then show the solutionSolution

Defect: myopia (near-sightedness); distant objects are blurred (1 mark).
Correction: a concave lens of suitable power, which diverges the rays so the image moves back onto the retina (1 mark).
Diagrams: (i) parallel rays from a distant object meeting in front of the retina; (ii) the same rays passing through a concave lens before the eye and meeting on the retina (1 mark).

Answer: Myopia; corrected by a concave lens.

Example 2: Hypermetropia lens power · 3 marks · Short answerHarder board level

The near point of a hypermetropic eye is 75 cm. Find the power of the lens needed so that the person can read at 25 cm. State the type of lens.

Try it first, then show the solutionSolution

Object at u = −25 cm must give a virtual image at the near point: v = −75 cm (1 mark).
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = - \frac{1}{75} + \frac{1}{25} = \frac{- 1 + 3}{75} = \frac{2}{75}$ , so f = +37.5 cm (1 mark).
P = $\frac{1}{0.375}$ ≈ +2.67 D: a convex lens. Check: $- \frac{1}{75} + \frac{1}{25} = \frac{2}{75} = \frac{1}{37.5}$ (1 mark).

Answer: Convex lens, f = +37.5 cm, P ≈ +2.67 D.

Example 3: Bifocal prescription · 3 marks · Short answerHarder board level

A person needs a lens of power −2.5 D for distant vision and +2.0 D for near vision. (a) Find the focal length of each lens. (b) What kind of spectacles are suitable and how are the lenses arranged? (c) Name the condition that commonly leads to this need.

Try it first, then show the solutionSolution

(a) Distant: f = 1/(−2.5) = −0.40 m = −40 cm (concave); near: f = 1/2.0 = +0.50 m = +50 cm (convex) (1½ marks).
(b) Bifocal lenses: upper part concave for distant vision, lower part convex for near vision (1 mark).
(c) A person may suffer from both myopia and hypermetropia; with age, presbyopia adds difficulty with near vision (½ mark).

Answer: (a) −40 cm and +50 cm. (b) Bifocals: concave top, convex bottom. (c) Both defects together (often with presbyopia).

Example 4: Causes and image position · 5 marks · Long answerHarder board level

Compare myopia and hypermetropia under the headings: what the person cannot see, where the image forms, two causes, and correcting lens.

Try it first, then show the solutionSolution

Cannot see: myopia — distant objects; hypermetropia — nearby objects (1 mark).
Image: myopia — in front of the retina; hypermetropia — behind the retina (1 mark).
Causes of myopia: excessive curvature of the eye lens; elongation of the eyeball (1 mark).
Causes of hypermetropia: focal length of the eye lens too long; eyeball too small (1 mark).
Correction: myopia — concave lens; hypermetropia — convex lens (1 mark).

Answer: Myopia: far blurred, image in front, curved lens or long eyeball, concave lens. Hypermetropia: near blurred, image behind, long focal length or small eyeball, convex lens.

Example 5: Design check · 3 marks · Short answerHOTS / competency

A myopic person's far point is 2.0 m. (a) Find the power of the lens needed. (b) The person is wrongly given a lens of −1.0 D. Where does this lens form the image of a very distant object? Explain why distant objects can still be seen, but why the prescription is wrong.

Try it first, then show the solutionSolution

(a) The lens must image a distant object at the far point: f = −2.0 m, P = 1/(−2.0) = −0.5 D (1 mark).
(b) With f = −1.0 m and u = −∞, the lens formula gives v = f = −1.0 m: the image is 1.0 m in front of the eye (1 mark).
This image lies within the clear range (closer than the 2.0 m far point), so the eye can focus on it by accommodating. But the lens is twice as strong as needed: with −0.5 D the image would sit exactly at the far point and the ciliary muscles could stay relaxed; with −1.0 D they must work even for distant viewing, causing strain (1 mark).

Answer: (a) −0.5 D. (b) Image at 1.0 m in front of the eye; it can be seen, but the lens over-corrects and strains the eye; −0.5 D is correct.

Where students lose marks in Defects of Vision and Their Correction

The slips examiners see most often, and what to write instead.

Mistake: Swapping the lenses
Students link 'near-sighted' with 'convex'.
Wrong working
Myopia is corrected with a convex lens.
Right working
Myopia: concave lens. Hypermetropia: convex lens.
Do this: Myopia = Minus.
Usually costs: 1-2 marks.
Mistake: Wrong u and v in hypermetropia calculations
Students put the near point as the object.
Wrong working
u = −100 cm, v = −25 cm.
Right working
u = −25 cm (book), v = −100 cm (image at the near point).
Do this: The lens must make the book (25 cm) appear at the near point.
Usually costs: The answer mark and the sign of power.
Mistake: Forgetting metres in power
f left in cm.
Wrong working
f = −50 cm, P = −1/50 = −0.02 D.
Right working
f = −0.50 m, P = −2 D.
Do this: Convert to metres before 1/f.
Usually costs: 1 mark.
Mistake: Incomplete causes
Only one cause is given.
Wrong working
Myopia is caused by a long eyeball.
Right working
Excessive curvature of the eye lens, or elongation of the eyeball.
Do this: Give both NCERT causes.
Usually costs: ½ to 1 mark.

Exam tips

Draw three diagrams for each defect if asked: normal near/far point, defective eye, corrected eye.
Mark the near point N and far point F clearly in ray diagrams.
State the image position (in front of or behind the retina) for each defect.
For bifocals, say which part is concave and which convex and why.
Show the lens formula substitution with signs; examiners give step marks.

Practice questions on Defects of Vision and Their Correction

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
An eye test shows that a 30-year-old can see distant hills clearly, but his near point is 1 m away. Which lens should his reading glasses have?
(a) concave (b) convex (c) bifocal (d) a plane glass sheet
Hint
Near vision is affected; the lens must add focusing power.
Answer
(b) convex. His near point is farther than 25 cm, so he is hypermetropic; a converging lens supplies the extra focusing power for near objects.
Q2 · 1 mark · Assertion-reasonBoard level
Assertion (A): Hypermetropia can be corrected using a convex lens.
Reason (R): In hypermetropia, light from a nearby object is focused behind the retina.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
A converging lens adds focusing power.
Answer
(a).
Q3 · 2 marks · Very short answerHarder board level
The far point of a myopic person is 1.25 m. Find the nature and power of the corrective lens.
Hint
f = −far point.
Answer
Concave lens; f = −1.25 m; P = 1/(−1.25) = −0.8 D.
Q4 · 4 marks · Case-basedHOTS / competency
Read the passage and answer.
Suppose a 60-year-old grandfather holds the newspaper at arm's length to read it, though he sees distant hills clearly. His optician measures his near point at 100 cm.
(a) Name the defect most likely to be developing with his age. (1 mark)
(b) Give the cause of this defect. (1 mark)
(c) Calculate the power of the reading lens that would let him read comfortably at 25 cm. (2 marks)
Hint
u = −25 cm, v = −100 cm.
Answer
(a) Presbyopia (age-related; near point receding). (b) Gradual weakening of the ciliary muscles and reduced flexibility of the eye lens. (c) 1/f = 1/v − 1/u = −1/100 + 1/25 = 3/100, f = 33.3 cm = 0.333 m, P = +3 D (convex).
Q5 · 3 marks · Short answerHOTS / competency
Why does a myopic eye's far point move closer if the eyeball is elongated? Explain using where the image forms.
Hint
The lens focuses parallel rays at its focal point.
Answer
Parallel rays from a distant object are focused at the lens's focal point. In an elongated eyeball the retina lies beyond this point, so the image forms in front of the retina and is blurred. Only rays from a nearer object, which focus farther behind the lens, reach a focus on the retina; so the farthest point seen clearly is closer than infinity.
Q6 · 3 marks · Short answerHarder board level
List three ways in which refractive defects can be corrected today, and explain why a myopic person reading a book may not need glasses.
Hint
Myopic people see near objects clearly.
Answer
Spectacles with suitable spherical lenses, contact lenses, and surgical interventions. A myopic person sees nearby objects clearly (only the far point is too close), so reading at a normal distance is usually comfortable without glasses.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Lens for myopia?Show answer
Concave lens.
2.Where is the image in hypermetropia?Show answer
Behind the retina.
3.Two causes of myopia?Show answer
Excessive curvature of the eye lens; elongation of the eyeball.
4.Two causes of hypermetropia?Show answer
Focal length of the eye lens too long; eyeball too small.
5.Cause of presbyopia?Show answer
Weakening of ciliary muscles and reduced flexibility of the eye lens with age.
6.Upper part of a bifocal lens?Show answer
Concave, for distant vision.

Defects of Vision and Their Correction: frequently asked questions

What is the difference between myopia and hypermetropia?

In myopia distant objects are blurred because the image forms in front of the retina; it is corrected by a concave lens. In hypermetropia nearby objects are blurred because light focuses behind the retina; it is corrected by a convex lens.

What is presbyopia?

Loss of the eye's power of accommodation with age, due to weakening ciliary muscles and a less flexible eye lens, so nearby objects are hard to see.

How do I calculate the power of the lens for myopia?

The focal length equals minus the far point distance. Convert to metres and use P = 1/f; the power is negative.

What are bifocal lenses?

Lenses with a concave upper part for distant vision and a convex lower part for near vision, for people who have both defects.

Can vision defects be corrected without spectacles?

Yes. NCERT notes that contact lenses and surgical interventions are now possible.