NCERT Solutions Class 10 Science Chapter 11 Human Eye and Colourful World

These NCERT Solutions Class 10 Science Chapter 11 Human Eye and Colourful World provide complete step-by-step answers and detailed scientific explanations to help students ace their CBSE board examinations. The human eye and the optical phenomena occurring in nature are among the most scoring and conceptual sections of the Class 10 Science curriculum. With these structured NCERT Science Class 10 solutions, students will build crystal-clear conceptual clarity on eye anatomy, visual defects, corrective lens calculations, prism dispersion, atmospheric refraction, and light scattering.

Whether you are revising for upcoming unit tests, completing homework assignments, or aiming for a 100% score in your board exams, this comprehensive guide offers verified NCERT Solutions step by step along with diagnostic ray diagram guidelines and numerical problem-solving templates.

Chapter Overview: Human Eye and the Colourful World

Chapter 11 of the CBSE Class 10 Science syllabus connects optical physics with biological mechanisms and natural atmospheric phenomena. In the board exam blueprint, the Light and Optical Phenomena unit carries substantial weightage (typically 10–12 marks combined with Chapter 10 Light: Reflection and Refraction). This chapter contains zero complex ray tracing derivations of multiple mirrors, making it a high-yield scoring opportunity for students who master ray diagrams and optical power calculations.

The chapter is broadly categorized into five foundational pillars:

• The Human Eye and Accommodation: Structural components (cornea, iris, pupil, crystalline lens, ciliary muscles, retina, optic nerve), power of accommodation, near point (25 cm), and far point (infinity).
• Defects of Vision and Their Correction: Myopia (near-sightedness), Hypermetropia (far-sightedness), Presbyopia (old-age hypermetropia), and Astigmatism, solved quantitatively using the lens formula.
• Refraction of Light Through a Triangular Prism: Deviation of light, angle of incidence, angle of emergence, and the relationship \(i + e = A + D\).
• Dispersion and Rainbow Formation: Splitting of white light into VIBGYOR, Newton's double prism recombination experiment, and the combined refraction-dispersion-internal reflection mechanism in raindrops.
• Atmospheric Refraction & Scattering of Light: Twinkling of stars, advance sunrise & delayed sunset, Tyndall effect, blue color of the clear sky, and reddish appearance of the Sun during sunrise and sunset.

In-Text Question Solutions (Page-by-Page)

In-Text Questions (Set 1)

Question 1: What is meant by the power of accommodation of the eye?

Answer:

The power of accommodation is the ability of the crystalline eye lens to adjust its focal length with the help of ciliary muscles, enabling the eye to focus clearly on both nearby and distant objects onto the retina.

• Viewing distant objects: Ciliary muscles relax, the eye lens becomes thin, its curvature decreases, and the focal length increases to focus parallel rays on the retina.
• Viewing nearby objects: Ciliary muscles contract, the eye lens becomes thicker, its curvature increases, and the focal length decreases to focus diverging rays on the retina.

Question 2: A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type of the corrective lens used to restore proper vision?

Step-by-Step Solution:

• Given:
  Far point of the myopic eye, \(d = 1.2\text{ m}\)
  Object distance for distant vision, \(u = -\infty\)
  Image distance to form at the person's far point, \(v = -1.2\text{ m}\)
• To Find: Nature and focal length (\(f\)) of the corrective lens
• Formula: Lens Formula: \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\)
• Substitution & Calculation: $$\frac{1}{f} = \frac{1}{-1.2} - \frac{1}{-\infty} = -\frac{1}{1.2} - 0 = -\frac{1}{1.2}\text{ m}^{-1}$$ $$f = -1.2\text{ m}$$ Power \(P = \frac{1}{f\text{ (in metres)}} = \frac{1}{-1.2} = -\frac{10}{12} = -0.83\text{ D}\)
• Answer: A concave lens of focal length \(-1.2\text{ m}\) (or power \(-0.83\text{ D}\)) is required to restore proper distant vision.

Question 3: What is the far point and near point of the human eye with normal vision?

Answer:

• Near Point (Least Distance of Distinct Vision): For a normal adult eye, the near point is 25 cm (\(0.25\text{ m}\)). Objects placed closer than 25 cm cause eye strain because ciliary muscles cannot contract beyond a certain threshold.
• Far Point: For a normal human eye, the far point is at infinity (\(\infty\)).

Question 4: A student has difficulty reading the blackboard while sitting in the last row. What could be the defect the child is suffering from? How can it be corrected?

Answer:

• Defect: The student is suffering from Myopia (short-sightedness or near-sightedness) because the student can see nearby objects distinctly but cannot focus clearly on distant objects like the blackboard.
• Causes: (i) Excessive curvature of the eye lens (thick lens), or (ii) Elongation of the eyeball.
• Correction: It is corrected using spectacles containing a concave lens of suitable focal length/power, which diverges incoming parallel rays so that they appear to originate from the student's actual far point.

NCERT Textbook Exercise Solutions

Multiple-Choice Questions (MCQs)

Exercise Q1: The human eye can focus 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

Explanation: Accommodation is the physiological process where ciliary muscles modify the curvature and focal length of the crystalline lens to focus light from objects at varied distances precisely onto the retina.

Exercise Q2: The human eye forms the image of an object at its:

(a) cornea    (b) iris    (c) pupil    (d) retina

Answer: (d) retina

Explanation: The retina is the light-sensitive inner coat of the eye containing photoreceptor cells (rods and cones) that convert light signals into electrical impulses sent via the optic nerve to the brain.

Exercise 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

Explanation: At 25 cm, the ciliary muscles are comfortably contracted without causing visual fatigue or blurring.

Exercise Q4: The change in focal length of an eye lens is caused by the action of the:

(a) pupil    (b) retina    (c) ciliary muscles    (d) iris

Answer: (c) ciliary muscles

Explanation: The ciliary muscles hold the crystalline lens in place and alter its thickness and curvature to adjust its focal length.

Descriptive & Numerical Questions

Exercise Q5: A person needs a lens of power -5.5 dioptres for correcting his distant vision. For correcting his near vision, he needs a lens of power +1.5 dioptre. What is the focal length of the lens required for correcting (i) distant vision, and (ii) near vision?

Step-by-Step Solution:

Part (i) For Distant Vision:

• Given: Power \(P_1 = -5.5\text{ D}\)
• Formula: \(f_1 = \frac{1}{P_1}\)
• Calculation: $$f_1 = \frac{1}{-5.5\text{ D}} = -\frac{10}{55}\text{ m} = -\frac{2}{11}\text{ m} \approx -0.1818\text{ m} = -18.18\text{ cm}$$
• Result: Focal length for distant vision is \(-0.182\text{ m}\) (or \(-18.18\text{ cm}\)), requiring a concave lens.

Part (ii) For Near Vision:

• Given: Power \(P_2 = +1.5\text{ D}\)
• Formula: \(f_2 = \frac{1}{P_2}\)
• Calculation: $$f_2 = \frac{1}{+1.5\text{ D}} = +\frac{10}{15}\text{ m} = +\frac{2}{3}\text{ m} \approx +0.667\text{ m} = +66.67\text{ cm}$$
• Result: Focal length for near vision is \(+0.667\text{ m}\) (or \(+66.67\text{ cm}\)), requiring a convex lens.

Exercise 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?

Step-by-Step Solution:

• Given:
  Far point distance, \(d = 80\text{ cm} = 0.8\text{ m}\)
  Object distance for distant vision, \(u = -\infty\)
  Image distance, \(v = -80\text{ cm} = -0.8\text{ m}\)
• To Find: Nature of lens and its power (\(P\))
• Formula: \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\) and \(P = \frac{1}{f\text{ (in m)}}\)
• Calculation: $$\frac{1}{f} = \frac{1}{-80} - \frac{1}{-\infty} = -\frac{1}{80} - 0 \implies f = -80\text{ cm} = -0.8\text{ m}$$ $$P = \frac{1}{-0.8\text{ m}} = -\frac{10}{8}\text{ D} = -1.25\text{ D}$$
• Answer: The corrective lens must be a concave (diverging) lens of power \(-1.25\text{ D}\) and focal length \(-80\text{ cm}\).

Exercise 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.

Step-by-Step Solution:

• Given:
  Near point of defective eye, \(v = -1\text{ m} = -100\text{ cm}\)
  Near point of normal eye (Object distance), \(u = -25\text{ cm} = -0.25\text{ m}\)
• To Find: Focal length (\(f\)) and Power (\(P\)) of the corrective convex lens
• Formula: \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\)
• Substitution & Working: $$\frac{1}{f} = \frac{1}{-100} - \left(\frac{1}{-25}\right) = -\frac{1}{100} + \frac{1}{25} = \frac{-1 + 4}{100} = \frac{3}{100}\text{ cm}^{-1}$$ $$f = +\frac{100}{3}\text{ cm} = +\frac{1}{3}\text{ m} \approx +33.33\text{ cm}$$ $$P = \frac{1}{f\text{ (in m)}} = \frac{1}{+(1/3)\text{ m}} = +3.0\text{ D}$$
• Ray Diagram Description:
  1. Hypermetropic eye: Light rays from a normal near point \(N\) (25 cm) converge behind the retina.
  2. Hypermetropic near point: Light rays starting from defective near point \(N'\) (1 m) focus on the retina.
  3. Correction with Convex Lens: A converging convex lens placed in front of the eye refracts rays from \(N\) (25 cm) so they appear to diverge from \(N'\) (1 m), focusing sharply onto the retina.
• Answer: A convex (converging) lens of power \(+3.0\text{ D}\) and focal length \(+33.33\text{ cm}\) is required.

Exercise Q8: Why is a normal eye not able to see clearly the objects placed closer than 25 cm?

Answer:

The focal length of the crystalline lens is adjusted by the ciliary muscles. When viewing nearby objects, ciliary muscles contract to increase lens curvature (making it thicker) and decrease its focal length. However, the ciliary muscles cannot contract beyond a biological maximum limit. When an object is placed closer than 25 cm, the eye lens cannot become convex enough to converge the sharply diverging rays onto the retina, forming a blurred image behind the retina and causing acute eye strain.

Exercise 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 inside the eye remains constant. In the human eye, the distance between the crystalline lens and the light-sensitive retina is fixed (approximately 2.3 cm to 2.5 cm). When the object distance increases, the ciliary muscles relax, making the eye lens thinner and increasing its focal length according to the lens formula, ensuring that the image is always formed exactly on the retina.

Exercise Q10: Why do stars twinkle?

Answer:

Stars twinkle due to atmospheric refraction of starlight:

1. Stars are situated extremely far away from the Earth, acting as effective point sources of light.
2. As starlight traverses Earth's atmosphere, it passes through layers of air whose physical conditions, temperature, and optical density (refractive index) change continuously.
3. Since the atmosphere bends starlight towards the normal (as optical density increases downwards), the path of rays fluctuates continuously.
4. This results in continuous fluctuations in the apparent position of the star and the amount of starlight entering the observer's pupil. Consequently, the star appears to flicker or twinkle.

Exercise Q11: Explain why the planets do not twinkle.

Answer:

Planets do not twinkle because of two physical reasons:

1. Extended Sources: Planets are much closer to Earth than distant stars. Therefore, they appear not as point sources, but as extended discs composed of a large collection of point sources of light.
2. Nullification of Variations: While individual light rays from different points of the planet undergo atmospheric refraction and fluctuation, the total variation in the amount of light entering our eye from all point-sized sources averages out to zero. Thus, the overall brightness remains steady.

Exercise Q12: Why does the Sun appear reddish early in the morning and at sunset?

Answer:

The reddish appearance of the Sun during sunrise and sunset is caused by the scattering of light (Rayleigh scattering):

• At sunrise and sunset, the Sun is situated near the horizon. The sunlight must travel through a much thicker layer of the atmosphere and a longer distance to reach our eyes compared to noon.
• According to Rayleigh's law of scattering, the intensity of scattered light is inversely proportional to the fourth power of its wavelength: $$I \propto \frac{1}{\lambda^4}$$
• Shorter wavelengths (blue, violet, indigo) are scattered away strongly in all directions by gas molecules and fine atmospheric particles.
• Longer wavelengths (red and orange) have larger wavelengths and undergo the least scattering, penetrating through the thick air layer to reach the observer. Thus, the Sun appears distinctly reddish.

Exercise Q13: Why does the sky appear dark instead of blue to an astronaut?

Answer:

The blue appearance of the daytime sky is caused by the scattering of sunlight by gas molecules and fine colloidal particles in Earth's atmosphere. In outer space (or at extremely high altitudes), there is no atmosphere and therefore no particles present to scatter sunlight. With no scattered light reaching the astronaut's eyes, space appears completely dark/black, and stars appear as steady, non-twinkling points of light against a black backdrop.

Important Formulas, Optical Laws, and Summary Table

To score full marks in numerical and theoretical questions, memorize these essential optical relations and sign conventions:

Concept / Phenomenon	Formula / Governing Relation	Key Variables & Sign Conventions
Power of a Lens	\(P = \frac{1}{f\text{ (in metres)}}\)	\(P\) in Dioptres (D), Concave lens: \(P < 0\), Convex lens: \(P > 0\)
Lens Formula	\(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\)	\(u\) is always negative; \(v\) is negative for virtual images in eye correction
Myopia Correction	\(f = -d\text{ (far point)}\)	Concave lens, \(u = -\infty\), \(v = -d\)
Hypermetropia Correction	\(\frac{1}{f} = \frac{1}{-d'} - \frac{1}{-25\text{ cm}}\)	Convex lens, \(u = -25\text{ cm}\), \(v = -d'\) (defective near point)
Prism Angle Relation	\(i + e = A + D\)	\(i =\) angle of incidence, \(e =\) emergence, \(A =\) prism angle, \(D =\) deviation
Rayleigh Scattering Law	\(I \propto \frac{1}{\lambda^4}\)	Shorter \(\lambda\) (blue) scatters \(\approx 16\times\) more than longer \(\lambda\) (red)

Common Mistakes and Tips for CBSE Board Exams

• Forgetting Sign Conventions in Defect Numericals: Always take object distance \(u = -\infty\) for myopia and \(u = -25\text{ cm}\) for hypermetropia. Both \(u\) and \(v\) are negative in eye correction numericals because the virtual image forms in front of the lens.
• Focal Length Unit in Power Formula: When computing \(P = 1/f\), ensure \(f\) is converted from centimetres to metres. Writing \(P = 1/(-80) = -0.0125\text{ D}\) instead of \(P = 1/(-0.8\text{ m}) = -1.25\text{ D}\) is the single most common student mistake.
• Total Internal Reflection vs Internal Reflection: In the explanation of rainbow formation, NCERT specifies that water droplets refract, disperse, and internally reflect light before refracting it out. Avoid overcomplicating with critical angles unless asked.
• Confusing Atmospheric Refraction with Scattering: Remember that twinkling of stars and advance sunrise/delayed sunset are caused by Atmospheric Refraction (variation in optical refractive index), whereas blue sky, red sunsets, and danger signal colors are caused by Scattering of Light.
• Labeling Ray Diagrams: In board exams, an unlabeled ray diagram loses 50% of the allocated marks. Always mark arrows indicating light direction, focal points, and normal lines.

CBSE Board Exam Relevance & Question Weightage

Chapter 11 carries around 4 to 6 marks in the CBSE Class 10 Science theoretical examination. Questions from this chapter typically appear in three formats:

• Section A (1 Mark): MCQs on near point, power of accommodation, and causes of blue sky / red sunrise.
• Section B & C (2 or 3 Marks): Numerical problems on calculating the power of corrective lenses for myopic and hypermetropic eyes, or reasoning questions explaining star twinkling and planet non-twinkling.
• Section D / E (4 or 5 Marks): Case-based or long-answer questions detailing dispersion through a glass prism, Newton's recombination experiment, or rainbow formation accompanied by complete labeled diagrams.

More NCERT Solutions and Practice

Consistent practice with real board exam style questions is essential to securing a top score. Explore chapter-wise CBSE question banks, verified sample question papers, marking scheme solutions, and customizable test generators by visiting Theorify QPTool to elevate your Class 10 board preparation today.