Class 10 Science Chapter 10: The Human Eye and the Colourful World — Important Questions & Sample Paper
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Reviewed by qpaper's CBSE curriculum team · Edited by Mohit · Updated June 2026
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Yes — this page has 44+ original Class 10 Science Chapter 10 (“The Human Eye and the Colourful World”) important questions with answers (Multiple Choice (MCQ), Assertion–Reason, Short Answer, Short Answer, Long Answer, Case Study). Practise them free, or generate a full CBSE board-pattern sample paper (80 marks) and export it to PDF or Word — in English & Hindi, for 2026-27.
CBSE Class 10 Science Chapter 10, 'The Human Eye and the Colourful World', bridges optics and real-life phenomena. It begins with eye anatomy—cornea, lens, retina, ciliary muscles—and explains accommodation: the eye’s ability to adjust for near and far objects. The near point (25 cm for a normal eye) and far point (infinity) are key benchmarks. Three vision defects are studied: myopia (nearsightedness), hypermetropia (farsightedness), and presbyopia. Numericals using the lens formula (P = 1/f) to find correcting lens power are a core skill. Ray diagrams showing how concave and convex lenses fix these defects frequently appear in exams.
The chapter then turns to the colourful world. White light’s dispersion through a prism produces a spectrum, which explains rainbow formation. Atmospheric refraction accounts for star twinkling, the apparent higher position of celestial bodies, and the advanced sunrise/delayed sunset. Light scattering—via the Tyndall effect—explains why the sky is blue, the sun is red at horizons, and clouds appear white. An intriguing concept is colour perception: objects reflect certain colours and absorb others, thus a red rose looks black in pure blue light.
Typical exam questions include power calculations, ray sketching, and conceptual reasoning on atmospheric effects and colour subtraction. Mastering these areas equips students for both straightforward and application-oriented questions.
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Science — The Human Eye and the Colourful World
SECTION A
- 1.1
A glass prism has a refracting angle of 60° and produces a minimum deviation of 30° for a certain colour of light. The angle of incidence for this colour, at minimum deviation, is:
(a) 30°(b) 45°(c) 60°(d) 90° - 2.1
A hollow glass prism filled with air is immersed in water. When a narrow beam of white light is passed through it, which of the following is most likely observed?
(a) Light is deviated but no dispersion occurs(b) Light is deviated and it disperses into a spectrum with violet deviating most(c) Light is deviated and it disperses into a spectrum with red deviating most(d) Light undergoes total internal reflection - 3.1
A person can see nearby objects clearly but cannot see distant objects clearly. This defect of vision is known as:
(a) Hypermetropia(b) Myopia(c) Presbyopia(d) Astigmatism
+ 41 more questions in the full paper
Generate full paperMarks distribution & blueprint
In a CBSE exam, this chapter typically contributes questions across the following types. The last column shows how many original questions of each type we have ready in our bank for this chapter:
| Question type | Marks each | In our bank |
|---|---|---|
| Multiple Choice (MCQ) | 1 mark | 13 |
| Assertion–Reason | 1 mark | 6 |
| Short Answer | 2 marks | 8 |
| Short Answer | 3 marks | 6 |
| Long Answer | 5 marks | 5 |
| Case Study | 4 marks | 6 |
44 original, exam-style questions in our bank for this chapter — with answers.
Important & sample questions (with answers)
Real, exam-style questions to practise and revise — each with its answer. Generate a full paper for unlimited more.
- Multiple Choice (MCQ)
Q1. A glass prism has a refracting angle of 60° and produces a minimum deviation of 30° for a certain colour of light. The angle of incidence for this colour, at minimum deviation, is:
1 mark(A) 30°(B) 45°(C) 60°(D) 90°▸ Answer▾ Answer
45°
- Multiple Choice (MCQ)
Q2. A hollow glass prism filled with air is immersed in water. When a narrow beam of white light is passed through it, which of the following is most likely observed?
1 mark(A) Light is deviated but no dispersion occurs(B) Light is deviated and it disperses into a spectrum with violet deviating most(C) Light is deviated and it disperses into a spectrum with red deviating most(D) Light undergoes total internal reflection▸ Answer▾ Answer
Light is deviated but no dispersion occurs
- Multiple Choice (MCQ)
Q3. A person can see nearby objects clearly but cannot see distant objects clearly. This defect of vision is known as:
1 mark(A) Hypermetropia(B) Myopia(C) Presbyopia(D) Astigmatism▸ Answer▾ Answer
Myopia
- Multiple Choice (MCQ)
Q4. A person can read a book placed at 25 cm, but cannot see clearly beyond 2 m. To see distant objects clearly, he should use a lens of power:
1 mark(A) −0.5 D(B) +0.5 D(C) −2 D(D) +2 D▸ Answer▾ Answer
−0.5 D
- Assertion–Reason
Q5. Assertion (A): A person with hypermetropia can see distant objects clearly. Reason (R): In hypermetropia, the image of a distant object is formed behind the retina.
1 mark(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.▸ Answer▾ Answer
Both A and R are true but R is not the correct explanation of A.
- Short Answer
Q6. Why does the sky appear dark to an astronaut in space while it appears blue to an observer on Earth? Explain.
2 marks▸ Answer▾ Answer
On Earth, the atmosphere scatters sunlight, with blue light being scattered the most, giving the sky a blue colour. In space, there is no atmosphere, hence no scattering of light, making the sky appear dark to an astronaut.
- Short Answer
Q7. The far point of a nearsighted person is 80 cm in front of the eye. What is the power of the corrective lens required? Show your calculation.
2 marks▸ Answer▾ Answer
The corrective lens should have a focal length such that an object at infinity appears to come from the far point (80 cm). So, f = -80 cm = -0.8 m. Power P = 1/f = 1/(-0.8) = -1.25 D. Hence, a concave lens of power -1.25 D is needed.
- Short Answer
Q8. A red rose appears bright red in white light, but when viewed in a room lit only by blue light, it looks almost black. Explain this observation using the concept of colour subtraction due to absorption and reflection of light.
3 marks▸ Answer▾ Answer
The colour of an object is determined by the wavelengths of light it reflects. A red rose appears red because it reflects mainly red light and absorbs other colours. In white light (containing all colours), it reflects red, so it looks red. When illuminated by pure blue light, there is no red light present in the incident light. The rose absorbs the blue light and reflects nothing (or very little), so it appears black (absence of reflected visible light).
- Short Answer
Q9. A man cannot clearly see objects beyond a distance of 2 m. Identify the vision defect he suffers from. Calculate the power of the lens required to correct this defect. Show your working.
3 marks▸ Answer▾ Answer
He suffers from myopia (nearsightedness). To correct it, a concave lens is used. For a distant object (at infinity), the lens must form its image at the person’s far point (2 m). Using lens formula, with u = –∞, v = –2 m (virtual image on same side as object), we get 1/f = 1/v – 1/u = –1/2 – 0 = –0.5 m⁻¹. So f = –2 m. Power P = 1/f (in m) = –0.5 D. Thus, a concave lens of power –0.5 D is needed.
- Long Answer
Q10. A person needs a lens of power +2 D for correcting his vision. Calculate the focal length of the lens required. Identify the defect and explain with a ray diagram how this lens corrects it. Mention one cause of this defect.
5 marks▸ Answer▾ Answer
Power P = +2 D = 1/f (in meters) => f = 1/P = 1/2 = 0.5 m = 50 cm (convex lens). The defect is hypermetropia (far-sightedness). One cause: the eyeball is too short, so the retina is closer to the lens than normal, causing light from near objects to focus behind the retina. Correction: a convex lens of suitable power is placed in front of the eye. The rays from a near object (at 25 cm) first pass through the convex lens which converges them, and then the eye lens further converges them to form a sharp image on the retina. Ray diagram: Without correction, rays from near point (farther) focus behind retina; with convex lens, rays from 25 cm are made to appear coming from the person's near point, so eye sees clear image.
- Long Answer
Q11. A person can see distant objects clearly but finds difficulty in reading a book. Name the defect of vision, list its possible causes, and explain with a ray diagram how it is corrected using a suitable lens.
5 marks▸ Answer▾ Answer
The defect is hypermetropia (long-sightedness). Causes: (1) The eyeball is too short from front to back, (2) The eye lens is too flat, resulting in low converging power. The near point is farther than 25 cm. Light from a near object focuses behind the retina. Correction: a convex lens (converging lens) of appropriate power is placed in front of the eye. The convex lens first converges the incoming light slightly before it enters the eye, allowing the eye lens to then focus the image exactly on the retina. Ray diagram: (without correction) parallel rays from near object diverge after eye lens, converging behind retina; (with correction) convex lens converges rays, then eye lens focuses on retina.
- Case Study
Q12. During a science fair, a student demonstrates the scattering of light using a colloidal solution. The student explains that the sky appears blue and the sun appears red at sunset. The wavelength of red light is about 700 nm and that of violet light is about 400 nm.
4 marks- (i) State the law of scattering that explains these observations.1 mark
- (ii) Calculate how many times more violet light is scattered than red light in the atmosphere.2 marks
- (iii) Why does the sky appear dark to an astronaut in space?1 mark
▸ Answer▾ Answer
Rayleigh scattering: I ∝ 1/λ⁴. Violet (400 nm) is scattered (700/400)⁴ ≈ 9.4 times more than red (700 nm). In space, absence of atmospheric particles means no scattering, making the sky dark.
Frequently asked questions
Which topics from this chapter carry the highest weightage in CBSE Class 10 board exams?
Vision defects (myopia, hypermetropia, presbyopia) and their correction using concave/convex lenses often appear as ray diagrams and numericals on lens power (P=1/f). Atmospheric refraction (twinkling, apparent star position) and scattering of light (blue sky, red sun) are high-scoring theory topics. Numerical problems on correcting lenses and conceptual questions on Tyndall effect and colour subtraction are also commonly tested.
How do I quickly recall which lens corrects myopia and which corrects hypermetropia?
Myopia (nearsightedness) is corrected with a concave lens, which diverges light before it enters the eye. Remember 'my-oh-needs-a-minus'—concave lenses have negative power. Hypermetropia (farsightedness) is corrected with a convex lens, which converges light; these lenses have positive power. Presbyopia is often corrected with bifocal lenses.
Why do stars twinkle while planets generally shine steady?
Stars are point sources of light; as their starlight passes through atmospheric layers of varying refractive index, it undergoes continuous refraction, causing the apparent position to shift and brightness to fluctuate—this is twinkling. Planets are closer and appear as extended sources; the variations in light from different points average out, so they appear steady.
More chapters
- Ch 1: Chemical Reactions and Equations
- Ch 2: Acids, Bases and Salts
- Ch 3: Metals and Non-metals
- Ch 4: Carbon and its Compounds
- Ch 5: Life Processes
- Ch 6: Control and Coordination
- Ch 7: How do Organisms Reproduce?
- Ch 8: Heredity
- Ch 9: Light – Reflection and Refraction
- Ch 10: The Human Eye and the Colourful World
- Ch 11: Electricity
- Ch 12: Magnetic Effects of Electric Current
- Ch 13: Our Environment