This MCQ module is based on: Defects of Vision, Their Correction and Cataract
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Defects of Vision, Their Correction and Cataract
🎓 Class 10
Science
CBSE
Theory
Ch 10 — The Human Eye and the Colourful World
⏱ ~19 min
🌐 Language: [gtranslate]
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Introduction — When Vision Fails
In Part 1 we saw that accommodation keeps images sharp on the retina. But ciliary muscles, the eye lens or even the eyeball can malfunction — producing blurred images. Such conditions are called defects of vision (or refractive errors). In this part we study the four common defects — myopia, hypermetropia, presbyopia and cataract — their causes, ray diagrams, and corrective lenses, and we solve worked numericals on power.
10.5 Myopia — Short-sightedness (Near-sight)
Myopia, also called short-sightedness, is a defect in which distant objects appear blurred while nearby objects are clear. The far point shifts from infinity to a nearer distance.
Causes of Myopia:
- Excessive curvature of the eye lens (lens too converging).
- Elongation of the eyeball (retina too far from the lens).
Correction: A concave lens of suitable focal length is placed before the eye. It diverges the incoming parallel rays so that they appear to come from the person's far point. The eye then focuses them accurately on the retina.
Focal length of the correcting lens
For an object at infinity, the concave lens should form a virtual image at the person's far point F. Using the sign convention: \(u = -\infty\), \(v = -F\) (far point distance). The lens formula \(\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\) gives
\[ \frac{1}{f}=\frac{1}{-F}-0 \;\Rightarrow\; f=-F \]So the focal length of the concave lens equals the negative of the far-point distance.
Worked Example 10.1: A myopic person can see clearly only up to 1.5 m. Find the power of the lens required to correct his vision.
Solution: Far point \(F = 1.5\,\text{m}\). Focal length of concave lens \(f = -1.5\,\text{m}\).
Power \(P = \dfrac{1}{f(\text{m})} = \dfrac{1}{-1.5} = -0.67\,\text{D}\). The person needs a lens of power −0.67 D.
Solution: Far point \(F = 1.5\,\text{m}\). Focal length of concave lens \(f = -1.5\,\text{m}\).
Power \(P = \dfrac{1}{f(\text{m})} = \dfrac{1}{-1.5} = -0.67\,\text{D}\). The person needs a lens of power −0.67 D.
10.6 Hypermetropia — Long-sightedness (Far-sight)
Hypermetropia, or long-sightedness, is the opposite of myopia. A person can see distant objects clearly but cannot see nearby objects without strain. The near point shifts to more than 25 cm.
Causes of Hypermetropia:
- Focal length of the eye lens is too long (lens too flat).
- Eyeball becomes too small (retina is too close to the lens).
Correction: A convex lens of suitable focal length is used. It provides extra convergence so that rays from an object at 25 cm appear to come from the person's actual near point and can be focused on the retina.
Worked Example 10.2: The near point of a hypermetropic eye is 1 m. What is the power of the lens required to read a book at 25 cm?
Solution: The lens must form a virtual image at the near point when the object is at 25 cm.
\(u = -25\,\text{cm} = -0.25\,\text{m}\); \(v = -100\,\text{cm} = -1\,\text{m}\).
\(\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{-1}-\dfrac{1}{-0.25}=-1+4=+3\,\text{m}^{-1}\).
So \(f = +0.33\,\text{m}\) and \(P = +3\,\text{D}\) (a convex lens).
Solution: The lens must form a virtual image at the near point when the object is at 25 cm.
\(u = -25\,\text{cm} = -0.25\,\text{m}\); \(v = -100\,\text{cm} = -1\,\text{m}\).
\(\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{-1}-\dfrac{1}{-0.25}=-1+4=+3\,\text{m}^{-1}\).
So \(f = +0.33\,\text{m}\) and \(P = +3\,\text{D}\) (a convex lens).
10.7 Presbyopia — The Ageing Eye
Presbyopia is an age-related defect in which the power of accommodation decreases. With age, ciliary muscles weaken and the eye lens loses its flexibility, so the near point gradually recedes. Some older people suffer both myopia and hypermetropia together — they need a bifocal lens. A bifocal lens has a concave (upper) portion for distant vision and a convex (lower) portion for reading.
10.8 Astigmatism (Brief)
If the cornea is not perfectly spherical, a person cannot focus vertical and horizontal lines at the same time. This defect is called astigmatism. It is corrected using a cylindrical lens.
10.9 Cataract
With age, the crystalline eye lens of some people gradually becomes cloudy and milky. This condition is called cataract. Vision becomes foggy and eventually is lost. Spectacles do not help because the defect is in the transparency of the lens itself, not its focal length. The cloudy lens is surgically removed and replaced by an artificial intraocular lens (IOL) — vision is usually restored fully.
Summary — the four common defects:
| Defect | What blurs | Cause | Correction |
|---|---|---|---|
| Myopia | Distant objects | Lens too curved / eyeball too long | Concave lens (−P) |
| Hypermetropia | Near objects | Lens too flat / eyeball too short | Convex lens (+P) |
| Presbyopia | Near objects (age) | Loss of accommodation | Bifocal lens |
| Cataract | All objects (foggy) | Lens becomes opaque | Surgery (IOL) |
Worked Example 10.3 (CBSE model): A person cannot see distinctly objects kept beyond 2 m. What is the nature, focal length and power of the corrective lens?
Solution: The defect is myopia; far point = 2 m.
Corrective lens: concave, \(f=-2\,\text{m}\).
\(P=\dfrac{1}{-2}=-0.5\,\text{D}\).
Solution: The defect is myopia; far point = 2 m.
Corrective lens: concave, \(f=-2\,\text{m}\).
\(P=\dfrac{1}{-2}=-0.5\,\text{D}\).
Worked Example 10.4: A person needs a lens of power +1.5 D for correcting his distant vision. For correcting his near vision he needs a lens of power +2.0 D. What is the focal length of the lens required for correcting (i) distant vision, (ii) near vision?
Solution:
(i) \(f_1=\dfrac{1}{+1.5}=+0.667\,\text{m}=+66.7\,\text{cm}\).
(ii) \(f_2=\dfrac{1}{+2.0}=+0.5\,\text{m}=+50\,\text{cm}\).
He uses a bifocal lens (+1.5 D for top, +2.0 D for bottom) — here both parts are convex because he is hypermetropic + presbyopic.
Solution:
(i) \(f_1=\dfrac{1}{+1.5}=+0.667\,\text{m}=+66.7\,\text{cm}\).
(ii) \(f_2=\dfrac{1}{+2.0}=+0.5\,\text{m}=+50\,\text{cm}\).
He uses a bifocal lens (+1.5 D for top, +2.0 D for bottom) — here both parts are convex because he is hypermetropic + presbyopic.
Activity 10.2 — Testing for Myopia in ClassL3 Apply
Aim: Estimate the far point of a classmate's eye.
- Hang a large printed chart (e.g. a newspaper headline) on the wall.
- Ask a classmate to stand just close enough to read the largest letters clearly.
- Move the chart away gradually and note the greatest distance at which she can still read it clearly.
- Repeat the experiment wearing her spectacles (if she has any).
Predict: Will the distance increase or decrease after she puts on her spectacles?
If she is myopic, her uncorrected far point will be finite (less than infinity). With concave spectacles, parallel rays from distant objects are diverged so they appear to come from her far point; her eye then focuses them sharply on the retina, and she can read the chart from much farther away — confirming correction.
Competency-Based Questions
Scenario: Mrs Sharma, aged 55, visits an optometrist. Her eye test shows that she cannot see objects clearly beyond 2 m and also cannot read a book placed at 25 cm. The doctor prescribes bifocal lenses.
Q1. (MCQ) The upper half of Mrs Sharma's bifocal lens will be L2 Understand
(b) Concave — the upper part corrects her distant-vision problem (myopia).
Q2. Calculate the power of the lens needed to correct Mrs Sharma's distant vision. L3 Apply
Far point = 2 m. Focal length \(f=-2\,\text{m}\). Power \(P=1/(-2)=-0.5\,\text{D}\).
Q3. Mrs Sharma's near point is at 50 cm. What power of lens does she need for comfortable reading at 25 cm? L3 Apply
u=−0.25 m, v=−0.50 m. \(1/f=1/v-1/u=1/(-0.5)-1/(-0.25)=-2+4=+2\,\text{m}^{-1}\). Power = +2 D (convex lens).
Q4. (True/False + justify) "A cataract can be corrected by wearing high-power convex spectacles." L4 Analyse
False. Cataract is an opacity (cloudiness) of the lens itself — light cannot pass through clearly regardless of focal length. The only treatment is surgical replacement of the lens with an IOL.
Q5. (Short answer) Distinguish between myopia and hypermetropia in a tabular form (any two points). L2 Understand
(i) Myopia — cannot see distant objects; hypermetropia — cannot see near objects. (ii) In myopia image forms in front of retina; in hypermetropia image would form beyond retina. (iii) Correction: myopia uses concave (negative) lens, hypermetropia uses convex (positive) lens.
Assertion–Reason Questions
Options: (A) Both A & R true, R correctly explains A. (B) Both A & R true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
Assertion (A): A myopic person uses a concave lens of appropriate focal length.
Reason (R): A concave lens forms a virtual image at the person's far point for a distant object.
(A) — Both true; R correctly explains A.
Assertion (A): Old-age presbyopia can often be corrected using bifocal spectacles.
Reason (R): The crystalline lens of an elderly person loses flexibility and its power of accommodation decreases.
(A) — Both true; R correctly explains A.
Assertion (A): Cataract is corrected by wearing concave spectacles.
Reason (R): In cataract, the eye lens develops excessive curvature.
(D) — A is false and R is false. Cataract is due to loss of transparency of the lens (not curvature), and it is treated by surgery, not by spectacles.
Frequently Asked Questions — Defects of Vision & Correction
What is defects of vision & correction in Class 10 Science (CBSE board)?
Defects of Vision & Correction is a key topic in NCERT Class 10 Science Chapter 10 — The Human Eye and the Colourful World. It explains common defects of vision — myopia, hypermetropia, presbyopia — their causes and correction using lenses. Core ideas covered include myopia, hypermetropia, presbyopia, cataract. Mastering this subtopic is essential for scoring well in the CBSE Class 10 Science board exam because board papers repeatedly test these concepts through MCQs, short answers and long-answer questions. This part gives a complete, exam-ready explanation with activities, diagrams and competency-based practice aligned to NCERT.
Why is myopia important in NCERT Class 10 Science?
Myopia is important in NCERT Class 10 Science because it forms the foundation for understanding defects of vision & correction in Chapter 10 — The Human Eye and the Colourful World. Without a clear idea of myopia, students cannot answer higher-order CBSE board questions involving hypermetropia, presbyopia, cataract. Board papers regularly include 2-mark and 3-mark questions on this concept, and competency-based questions often link myopia to real-life situations. Building clarity here pays off directly in board marks.
How is defects of vision & correction tested in the Class 10 Science CBSE board exam?
The CBSE Class 10 Science board exam tests defects of vision & correction through a mix of 1-mark MCQs, 2-mark short answers, 3-mark explanations with examples, 5-mark descriptive questions (often with diagrams or balanced equations) and 4-mark competency-based questions. Expect direct questions on myopia, hypermetropia, presbyopia and application-based questions drawn from NCERT activities. Students who follow NCERT thoroughly and practice this chapter's questions consistently score in the 90%+ range.
What are the key terms to remember for defects of vision & correction in Class 10 Science?
The key terms to remember for defects of vision & correction in NCERT Class 10 Science Chapter 10 are: myopia, hypermetropia, presbyopia, cataract, concave lens, convex lens. Each of these concepts carries exam weightage and regularly appears in the CBSE board paper. Write clear one-line definitions of every term in your revision notes and revisit them before the exam. Linking these terms visually through a flowchart or concept map makes recall easier during the Class 10 Science board exam.
Is Defects of Vision & Correction included in the Class 10 Science syllabus for 2025–26 CBSE board exam?
Yes, Defects of Vision & Correction is a part of the NCERT Class 10 Science syllabus (2025–26) prescribed by CBSE. It falls under Chapter 10 — The Human Eye and the Colourful World — and is examined in the annual board paper. The current syllabus retains the full treatment of myopia, hypermetropia, presbyopia as per the NCERT textbook. Because CBSE bases every board question on NCERT, studying this part thoroughly ensures complete syllabus coverage and guarantees marks from this chapter.
How should I prepare defects of vision & correction for the CBSE Class 10 Science board exam?
Prepare defects of vision & correction for the CBSE Class 10 Science board exam in three steps. First, read this NCERT part carefully, highlighting definitions and diagrams of myopia, hypermetropia, presbyopia. Second, solve every in-text question and end-of-chapter exercise — CBSE questions often come directly from NCERT. Third, practice competency-based and assertion-reason questions to sharpen reasoning. Write answers in the exam-style format (point-wise with diagrams) and time yourself. This method delivers confidence and full marks in the board exam.
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