This MCQ module is based on: The Human Eye and the Colourful World – NCERT Exercises
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The Human Eye and the Colourful World – NCERT Exercises
🎓 Class 10
Science
CBSE
Theory
Ch 10 — The Human Eye and the Colourful World
⏱ ~16 min
🌐 Language: [gtranslate]
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Chapter 10 — Summary
Key Ideas at a Glance
- The human eye is a natural optical instrument: cornea + lens + retina. Most refraction takes place at the cornea.
- The image on the retina is real, inverted and diminished; the brain interprets it as erect.
- The iris controls the size of the pupil and hence the light entering; rods and cones on the retina convert light into electrical signals.
- The eye focuses objects at different distances by changing the focal length of its lens — this is accommodation, driven by the ciliary muscles.
- A normal eye has far point at infinity and near point at 25 cm.
- Myopia — distant objects blur; corrected by a concave lens of focal length equal to the person's far-point distance (with a minus sign).
- Hypermetropia — nearby objects blur; corrected by a convex lens.
- Presbyopia — age-related loss of accommodation; treated with bifocal lenses.
- Cataract — cloudiness of the lens; treated by surgery (IOL).
- A prism disperses white light into VIBGYOR because the refractive index of glass depends on colour.
- A rainbow forms by refraction + total internal reflection + refraction inside water droplets.
- Atmospheric refraction causes the twinkling of stars, advance sunrise and delayed sunset (day lengthens by ~4 min).
- Rayleigh scattering (∝ 1/λ⁴) makes the sky blue and the setting Sun red. Danger signals are red because red scatters the least.
- Persistence of vision (~1/16 s) enables continuous motion perception in films.
Keyword Grid
AccommodationAbility of the lens to change focal length.
Ciliary musclesMuscles that change the curvature of the eye lens.
RetinaLight-sensitive inner membrane of the eye.
Rods & ConesPhotoreceptors for dim-light and colour vision.
Near pointLeast distance of distinct vision (25 cm).
Far pointFarthest point of clear vision (infinity for normal eye).
MyopiaShort-sightedness; corrected with concave lens.
HypermetropiaLong-sightedness; corrected with convex lens.
PresbyopiaAge-related loss of accommodation; bifocal lens.
CataractCloudiness of the eye lens; surgical treatment.
DispersionSplitting of white light into VIBGYOR.
SpectrumThe coloured band produced by dispersion.
RainbowNatural spectrum formed in the sky after rain.
Atmospheric refractionBending of light by air layers of varying density.
Twinkling of starsFluctuation caused by atmospheric refraction.
ScatteringRe-emission of light in all directions by particles.
Tyndall effectScattering by colloidal particles.
Rayleigh scatteringIntensity ∝ 1/λ⁴; makes the sky blue.
Persistence of visionImage retention on retina for ~1/16 s.
Bifocal lensCombines concave (top) and convex (bottom) parts.
NCERT Exercises — Complete Solutions
Q1. The human eye can focus on objects at different distances by adjusting the focal length of the eye lens. This is due to
(a) presbyopia (b) accommodation (c) near-sightedness (d) far-sightedness
Answer: (b) Accommodation — the change of focal length of the eye lens by the ciliary muscles.
Q2. The human eye forms the image of an object at its
(a) cornea (b) iris (c) pupil (d) retina
Answer: (d) Retina — the light-sensitive screen at the back of the eye.
Q3. The least distance of distinct vision for a young adult with normal vision is about
(a) 25 m (b) 2.5 cm (c) 25 cm (d) 2.5 m
Answer: (c) 25 cm — the position of the near point for a normal eye.
Q4. The change in focal length of an eye lens is caused by the action of
(a) pupil (b) retina (c) ciliary muscles (d) iris
Answer: (c) Ciliary muscles — they contract or relax to change the curvature of the eye lens.
Q5. A person needs a lens of power −5.5 D for correcting his distant vision. For correcting his near vision he needs a lens of power +1.5 D. What is the focal length of the lens required for correcting (i) distant vision, (ii) near vision?
Solution:
(i) For distant vision: \(f_1=\dfrac{1}{P_1}=\dfrac{1}{-5.5\,\text{D}}=-0.1818\,\text{m}\approx -18.18\,\text{cm}\). A concave lens of focal length about 18.2 cm.
(ii) For near vision: \(f_2=\dfrac{1}{P_2}=\dfrac{1}{+1.5\,\text{D}}=+0.667\,\text{m}\approx +66.7\,\text{cm}\). A convex lens of focal length about 66.7 cm.
(i) For distant vision: \(f_1=\dfrac{1}{P_1}=\dfrac{1}{-5.5\,\text{D}}=-0.1818\,\text{m}\approx -18.18\,\text{cm}\). A concave lens of focal length about 18.2 cm.
(ii) For near vision: \(f_2=\dfrac{1}{P_2}=\dfrac{1}{+1.5\,\text{D}}=+0.667\,\text{m}\approx +66.7\,\text{cm}\). A convex lens of focal length about 66.7 cm.
Q6. The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?
Solution: For a parallel ray from infinity the corrective lens must form a virtual image at 80 cm.
\(u=-\infty\), \(v=-80\,\text{cm}=-0.8\,\text{m}\).
\(\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{-0.8}-0=-1.25\,\text{m}^{-1}\).
So \(f=-0.8\,\text{m}\), \(P=-1.25\,\text{D}\). The lens is a concave lens of power −1.25 D.
\(u=-\infty\), \(v=-80\,\text{cm}=-0.8\,\text{m}\).
\(\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{-0.8}-0=-1.25\,\text{m}^{-1}\).
So \(f=-0.8\,\text{m}\), \(P=-1.25\,\text{D}\). The lens is a concave lens of power −1.25 D.
Q7. Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.
Solution: See Fig 10.5 of Part 2 for the ray diagram. The convex corrective lens must form a virtual image at the person's near point (1 m) when the object is at 25 cm.
\(u=-25\,\text{cm}=-0.25\,\text{m}\); \(v=-100\,\text{cm}=-1\,\text{m}\).
\(\dfrac{1}{f}=\dfrac{1}{-1}-\dfrac{1}{-0.25}=-1+4=+3\,\text{m}^{-1}\).
\(f=+0.333\,\text{m}\), \(P=+3\,\text{D}\) (convex lens).
\(u=-25\,\text{cm}=-0.25\,\text{m}\); \(v=-100\,\text{cm}=-1\,\text{m}\).
\(\dfrac{1}{f}=\dfrac{1}{-1}-\dfrac{1}{-0.25}=-1+4=+3\,\text{m}^{-1}\).
\(f=+0.333\,\text{m}\), \(P=+3\,\text{D}\) (convex lens).
Q8. Why is a normal eye not able to see clearly objects placed closer than 25 cm?
Answer: 25 cm is the near point of a normal eye. For objects closer than this, the ciliary muscles cannot contract further to make the lens any more convex. The focal length cannot be reduced enough, so the image would form behind the retina and appear blurred. The eye also feels strain because muscles are already at maximum contraction.
Q9. What happens to the image distance in the eye when we increase the distance of an object from the eye?
Answer: The image distance (lens-to-retina distance) remains essentially unchanged — it is fixed at about 2.5 cm. Instead, the ciliary muscles relax so that the focal length of the lens increases, keeping the image on the retina.
Q10. Why do stars twinkle?
Answer: Stars are extremely far and behave as point sources. Light from a star passes through many air layers of continuously varying refractive index. Atmospheric currents keep disturbing these layers so the apparent position and brightness of the star fluctuate — producing the twinkling effect.
Q11. Explain why the planets do not twinkle.
Answer: Planets are much nearer and appear as extended disks rather than points. Each point on the disk sends light independently; when one point dims due to atmospheric refraction, another brightens. The net brightness received by the eye averages out, so planets appear steady.
Q12. Why does the Sun appear reddish early in the morning?
Answer: At sunrise the Sun is near the horizon, so its light traverses a long column of the atmosphere before reaching us. Short-wavelength colours (blue, violet) are strongly scattered out by air molecules (Rayleigh scattering, intensity ∝ 1/λ⁴). The transmitted light, rich in long wavelengths, makes the Sun look reddish.
Q13. Why does the sky appear dark instead of blue to an astronaut?
Answer: At the altitudes where astronauts orbit (or on the Moon), the atmosphere is essentially absent. With no air molecules to scatter sunlight, no blue light is redirected to the observer, so the sky appears pitch-black. Only objects directly illuminated by the Sun shine brightly against this black background.
Frequently Asked Questions — NCERT Exercises & Intext Questions
How do I solve NCERT Class 10 Science Chapter 10 (The Human Eye and the Colourful World) exercise questions for the CBSE board exam?
Solve NCERT Chapter 10 — The Human Eye and the Colourful World — exercise questions by first reading the question carefully, writing down the given data, recalling the relevant concepts like human eye, defects, dispersion, and applying them step by step. This Part 4 covers every intext and end-of-chapter exercise from the NCERT textbook. Write balanced equations, label diagrams clearly and show each step — CBSE Class 10 board examiners award step marks even if the final answer has a small slip. Practising these solutions strengthens conceptual clarity and builds speed for the board exam.
Are the NCERT intext questions from The Human Eye and the Colourful World important for the Class 10 board exam?
Yes, NCERT intext questions for Chapter 10 The Human Eye and the Colourful World are highly important for the CBSE Class 10 Science board exam. Many board questions are directly lifted or only slightly modified from these intext questions, and they test the foundational concepts — human eye, defects, dispersion — that chapter-end questions build on. Attempt every intext question first, then move on to the exercises. This practice ensures complete NCERT coverage, which is the CBSE exam's primary source.
What types of questions from The Human Eye and the Colourful World are asked in the CBSE Class 10 Science board exam?
The CBSE Class 10 board paper asks a mix of question types from The Human Eye and the Colourful World: 1-mark MCQ and assertion-reason, 2-mark short answers, 3-mark explanations, 5-mark long answers with diagrams or derivations, and 4-mark competency-based / case-study questions. These test understanding of human eye, defects, dispersion, scattering. Practising every NCERT exercise and intext question prepares you to answer all of these formats with confidence.
How many marks does Chapter 10 — The Human Eye and the Colourful World — carry in the Class 10 Science CBSE paper?
Chapter 10 — The Human Eye and the Colourful World — is part of the Class 10 Science syllabus and typically contributes 5–9 marks in the CBSE board paper, depending on the annual weightage. Questions are drawn from definitions, reasoning, numerical/descriptive problems and diagrams on topics like human eye, defects, dispersion. Solving the NCERT exercises in this part is essential because CBSE directly references NCERT for question design.
Where can I find step-by-step NCERT solutions for Chapter 10 The Human Eye and the Colourful World Class 10 Science?
You can find complete, step-by-step NCERT solutions for Chapter 10 The Human Eye and the Colourful World Class 10 Science on MyAiSchool. Every intext and end-of-chapter exercise question is solved with full working, labelled diagrams and CBSE-aligned mark distribution. Solutions highlight key points about human eye, defects, dispersion that examiners look for. This makes revision quick and exam-focused for Class 10 CBSE board students.
What is the best way to revise The Human Eye and the Colourful World before the Class 10 Science board exam?
The best way to revise The Human Eye and the Colourful World for the CBSE Class 10 Science board exam is a three-pass approach. First pass: skim the chapter and note down key terms like human eye, defects, dispersion in a one-page mind map. Second pass: solve every NCERT intext and exercise question without looking at the solution, then self-check. Third pass: attempt previous CBSE board questions and competency-based questions under timed conditions. This structured revision secures full marks for this chapter.
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