Looking for accurate SEBA Class 10 Science Chapter 11 (The Human Eye and the Colourful World) textual solutions for the 2026–27 academic year? Access step-by-step textbook exercise answers, ray diagrams for eye defect corrections (Myopia, Hypermetropia, Presbyopia), atmospheric refraction, and prism dispersion explanations designed to help you score top marks in your HSLC board exams.
Blue question
Q1. What is meant by the power of accommodation of the eye?
Answer: The power of accommodation is the ability of the eye lens to adjust its focal length using ciliary muscles. When viewing nearby objects, the ciliary muscles contract to increase lens curvature (making it thicker), and when viewing distant objects, they relax to decrease lens curvature (making it thinner), ensuring clear image formation on the retina.
Q2. A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What type of corrective lens should be used to restore proper vision?
Answer: The person requires a concave (diverging) lens. A concave lens diverges light rays coming from distant objects before they enter the eye, shifting the focal point back onto the retina to restore clear vision.
Q3. What are the far point and near point of a normal human eye?
Answer:
- Near Point: The minimum distance at which an object can be seen clearly without strain. For a normal human eye, it is 25 cm.
- Far Point: The maximum distance up to which the eye can see objects clearly. For a normal human eye, it is at infinity.
The Human Eye and Colourful World Class 10 SEBA Textual Exercise Solutions
Q4. A student sitting in the last row has difficulty reading the blackboard. What eye defect is the child suffering from, and how can it be corrected?
Answer: The child is suffering from myopia (short-sightedness), where nearby objects are clear but distant objects appear blurred because images form in front of the retina. This defect is corrected using spectacles with a concave lens of appropriate focal length.
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 varying distances accurately onto the retina.
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 acts as a light-sensitive screen at the back of the eye containing photoreceptor cells (rods and cones) where real and inverted images are focused.
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: The least distance of distinct vision (near point) is the minimum distance at which an object can be focused clearly on the retina without causing eye strain.
HSLC Science Chapter 11 Question Answer
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: Ciliary muscles alter the shape and thickness of the eye lens; their contraction increases lens curvature for near vision, while relaxation flattens the lens for distant vision.
5. 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?
Answer:
Formula:
P = 1/f
where P is power (in dioptres) and f is focal length (in metres).
(i) For distant vision:
P = -5.5 D
f = 1 / (-5.5) = -0.1818 m = -18.18 cm
So, focal length = –18.18 cm (concave lens)
(ii) For near vision:
P = +1.5 D
f = 1 / 1.5 = 0.6667 m = 66.67 cm
So, focal length = +66.67 cm (convex lens)
Class 10 Science Chapter 11 Question Answer SEBA 2026-27
6. 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?
Answer:
For a myopic person, u=∞u = \inftyu=∞, v=−80 cm=−0.8 mv = -80 \,cm = -0.8 \,mv=−80cm=−0.8m.
Using 1f=1v+1u=1−0.8+0=−1.25.\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{-0.8} + 0 = -1.25.f1=v1+u1=−0.81+0=−1.25.
So, f=−0.8 m.f = -0.8 \,m.f=−0.8m.For a myopic person:
u = ∞
v = -80 cm = -0.8 m
Using the lens formula:
1/f = 1/v + 1/u
1/f = 1/(-0.8) + 0
1/f = -1.25
Therefore:
f = -0.8 m
Power:
P = 1/f = -1.25 D
Lens required:
Concave lens of power –1.25 D
7. 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.
Answer:
Given:
Near point of defective eye:
v = -1 m (virtual image)
Object at normal near point:
u = -25 cm = -0.25 m
Lens formula:
1/f = 1/v – 1/u
Substitute values:
1/f = 1/(-1) – 1/(-0.25)
1/f = -1 + 4
1/f = 3
Therefore:
f = 1/3 = 0.333 m
Power, P=1f=+3DP = \dfrac{1}{f} = +3DP=f1=+3D.
Hence, a convex lens of power +3 dioptres is required.
Diagram description:
Draw:
- Eye lens and retina.
- Show parallel rays from near object diverging.
- After passing through convex lens, they converge, forming a virtual image at 25 cm (normal near point).
SEBA Class 10 Science Chapter 11 Notes
Q8. Why is a normal eye not able to see objects placed closer than 25 cm clearly?
Answer: Seeing objects closer than 25 cm requires the ciliary muscles to contract excessively to increase the curvature of the eye lens. Because ciliary muscles cannot contract beyond a maximum physical limit, the focal length of the lens cannot be decreased further, preventing the image from focusing sharply on the retina and causing strain or blurriness.
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 because the distance between the eye lens and the retina is fixed. When the object distance increases, the ciliary muscles relax to flatten the eye lens (increasing its focal length), ensuring that the image continues to form precisely on the retina.
Q10. Why do stars twinkle?
Answer: Stars twinkle due to atmospheric refraction of starlight. As starlight passes through varying layers of Earth’s atmosphere with continuously changing densities and temperatures, the light rays bend unpredictably. This fluctuating bending causes the apparent position and brightness of the point-source star to flicker continuously.
SEBA Class 10 Science Chapter 11 Solutions
Q11. Explain why planets do not twinkle.
Answer: Planets are much closer to Earth than stars and act as extended light sources (a collection of point sources). Although individual light rays from a planet undergo atmospheric refraction, the variations from all point sources average out, neutralizing the overall brightness fluctuation and keeping the planet’s light steady.
Q12. Why does the Sun appear reddish early in the morning during sunrise?
Answer: During sunrise, sunlight travels a longer distance through the thickest layers of Earth’s atmosphere to reach an observer. Shorter wavelengths of light (like blue and violet) are scattered away by atmospheric particles, while longer wavelengths (like red) undergo less scattering and penetrate through, making the Sun appear reddish.
Q13. Why does the sky appear dark or black instead of blue to an astronaut in space?
Answer: The sky appears blue on Earth because atmospheric gas molecules scatter short-wavelength blue light in all directions. In outer space, there is no atmosphere to scatter sunlight, so no light reaches the astronaut’s eyes from the surrounding space, making it appear pitch black.
🔬 Updated Solutions Notice: This page features complete, step-by-step SEBA Class 10 Science Chapter 11 (The Human Eye and the Colourful World) Textual Exercise Solutions updated for the 2026–27 academic session. All optical ray diagrams, eye power corrections, prism dispersion, and scattering concepts strictly follow the latest revised Assam Board (SEBA) textbook.
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