Exercise — Human Eye and Colourful World

1. 11 markNCERT Cl-10 Science, Exercise 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.

   Hint. Three of these four options are defects; only one is the normal ability being described.

   Step 1 — Classify the options. Presbyopia, near-sightedness and far-sightedness are all defects of vision, not the normal focusing ability.

   Step 2 — Identify the correct term. Since the eye is doing exactly this — adjusting its lens's focal length for different distances — this ability is called accommodation.

   ✦ Answer: (b) accommodation.

   Where students slip. Picking presbyopia — that's specifically the age-related loss of accommodation, not accommodation itself.

2. 21 markNCERT Cl-10 Science, Exercise Q2

   The human eye forms the image of an object at its (a) cornea. (b) iris. (c) pupil. (d) retina.

   Hint. This structure is described as a light-sensitive membrane at the back of the eye.

   Step 1 — Recall each structure's role. The cornea and lens mainly refract light; the iris and pupil control how much light enters.

   Step 2 — Identify where the image actually forms. The light-sensitive retina, at the back of the eye, is where the image is formed, since that's where the light-sensitive cells that generate signals to the brain are located.

   ✦ Answer: (d) retina.

   Where students slip. Picking the cornea — the cornea does most of the refracting, but the image itself forms on the retina, further back in the eye.

3. 31 markNCERT Cl-10 Science, 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.

   Hint. This is the comfortable reading distance you'd hold a book at.

   Step 1 — Recall the value. The near point (least distance of distinct vision) for a normal young adult is about 25 cm.

   ✦ Answer: (c) 25 cm.

   Where students slip. Picking 25 m — that's a thousand times too large; 25 cm is roughly a comfortable book-reading distance, not metres away.

4. 41 markNCERT Cl-10 Science, 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.

   Hint. The pupil and iris control light intake, not the lens shape.

   Step 1 — Rule out light-control structures. The pupil and iris regulate how much light enters, but don't change the lens's shape.

   Step 2 — Identify what does change the lens shape. Since the ciliary muscles contract or relax, they change the curvature — and hence the focal length — of the eye lens.

   ✦ Answer: (c) ciliary muscles.

   Where students slip. Picking the iris — the iris controls pupil size (and hence light intake), not the eye lens's curvature.

5. 53 marksNCERT Cl-10 Science, 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?

   Hint. Focal length is just the reciprocal of power — apply it twice, once per lens.

   Step 1 — (i) Distant vision. f = 1/P = 1/(−5.5) ≈ −0.182 m ≈ −18.2 cm.

   Step 2 — (ii) Near vision. f = 1/P = 1/(1.5) ≈ +0.667 m ≈ +66.7 cm, since a positive power always gives a positive (converging) focal length.

   ✦ Answer: (i) ≈ −18.2 cm (concave lens, for myopia) (ii) ≈ +66.7 cm (convex lens, for hypermetropia).

   Where students slip. Mixing up which sign goes with which correction — the negative power/focal length corrects distant vision (myopia), and the positive one corrects near vision (hypermetropia), matching the concave/convex pattern throughout this chapter.

6. 63 marksNCERT Cl-10 Science, 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?

   Hint. The lens must make a distant object's image appear to originate exactly at this person's own far point.

   Step 1 — Set up the values for this myopic eye. The lens should form a virtual image of an object at infinity exactly at the person's far point: u = −∞, v = −0.8 m.

   Step 2 — Apply the lens formula. 1/f = 1/v − 1/u = 1/(−0.8) − 0 = −1.25, so f = −0.8 m.

   Step 3 — Find the power. P = 1/f = 1/(−0.8) = −1.25 D.

   ✦ Answer: A concave lens of power −1.25 D (focal length −0.8 m) is required.

   Where students slip. Forgetting to convert the far point from cm to metres before taking the reciprocal for power — power in dioptres specifically requires focal length in metres.

7. 73 marksNCERT Cl-10 Science, 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.

   Hint. The lens should let the person read a book held at the normal near point, even though their eye alone can't focus that close.

   Step 1 — Set up the values. The lens must take an object placed at the normal near point (u = −0.25 m) and form a virtual image at the person's actual near point (v = −1 m), where their eye can focus it.

   Step 2 — Apply the lens formula. 1/f = 1/v − 1/u = 1/(−1) − 1/(−0.25) = −1 + 4 = 3, so f = 1/3 m.

   Step 3 — Find the power. P = 1/f = 3 D.

   Step 4 — Diagram (description). A convex lens placed in front of the hypermetropic eye takes diverging rays from an object at 25 cm and refracts them so they appear to diverge from 1 m instead, exactly where the defective eye can bring them to a focus on the retina.

   ✦ Answer: A convex lens of power +3 D is required.

   Where students slip. Setting v = +1 m instead of −1 m — the corrective lens still forms a virtual image (on the same side as the object), just at a farther distance than the object itself; the image is never real here.

8. 82 marksNCERT Cl-10 Science, Exercise Q8

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

   Hint. The ciliary muscles can only contract so much before the lens simply can't curve any further.

   Step 1 — Recall the limit on accommodation. The eye lens's focal length cannot be decreased below a certain minimum limit, no matter how much the ciliary muscles contract.

   Step 2 — Apply this to very close objects. Focusing an object closer than 25 cm would require an even shorter focal length than the lens can achieve, so the image can no longer be brought sharply onto the retina.

   Step 3 — State the result. This is why viewing something closer than 25 cm causes the image to blur or strain the eye, and why 25 cm is defined as the near point.

   ✦ Answer: Because the eye lens's focal length can only decrease so far — beyond that limit (objects closer than about 25 cm) it can no longer focus the image sharply on the retina, causing blur or strain.

   Where students slip. Treating 25 cm as an arbitrary round number — it specifically marks the physical limit of how much the lens can curve, not just a convenient reference distance.

9. 92 marksNCERT Cl-10 Science, Exercise Q9

   What happens to the image distance in the eye when we increase the distance of an object from the eye?

   Hint. The retina doesn't move — so what has to change instead to keep the image landing there?

   Step 1 — Recall where the image must always form. The image must always land on the retina, which is at a fixed distance from the eye lens.

   Step 2 — State what stays constant. Since the retina doesn't move, the image distance in the eye stays the same regardless of how far the object is.

   Step 3 — State what changes instead. It is the focal length of the eye lens (via the ciliary muscles adjusting its curvature) that changes with object distance, not the image distance.

   ✦ Answer: The image distance doesn't change — it stays fixed at the retina; instead, the eye lens's focal length adjusts (accommodation) to keep the image focused there as object distance changes.

   Where students slip. Assuming image distance increases along with object distance, as in a simple lens problem — the eye's retina is fixed in place, so it's the focal length that adapts, not the image distance.

10. 102 marksNCERT Cl-10 Science, Exercise Q10

    Why do stars twinkle?

    Hint. Starlight has to pass through a lot of constantly shifting air before it reaches your eye.

    Step 1 — Recall what happens to starlight in the atmosphere. Starlight continuously refracts through atmospheric layers of gradually (and unpredictably) changing density.

    Step 2 — Explain the effect on apparent position. This makes the star's apparent position shift slightly and constantly.

    Step 3 — Connect this to a point source. Since stars are extremely distant, they act as point sources, so this shifting causes the amount of starlight reaching our eye to fluctuate continuously, making the star appear alternately brighter and fainter.

    ✦ Answer: Continuous, unpredictable atmospheric refraction shifts the apparent position of a star (a point source) moment to moment, making the light reaching our eyes fluctuate — this fluctuation is the twinkling.

    Where students slip. Attributing twinkling to the star's own light output varying — the star's actual brightness is steady; it's Earth's constantly shifting atmosphere that causes the apparent flickering.

11. 112 marksNCERT Cl-10 Science, Exercise Q11

    Explain why the planets do not twinkle.

    Hint. A planet isn't really a single point of light the way a star is.

    Step 1 — Compare planets and stars. Planets are much closer to Earth than stars, so they appear as extended sources (like a collection of many point sources close together), not as a single point.

    Step 2 — Explain what happens to the fluctuations. Each of these many points still varies slightly due to atmospheric refraction, but with so many points making up the disc, their individual variations average out to nearly zero overall.

    Step 3 — State the result. Since the net brightness stays effectively constant, planets don't show the twinkling effect that a single point-source star does.

    ✦ Answer: Planets are extended sources (not single points) because they're much closer to Earth, so the many small variations across their disc average out, leaving their overall brightness steady — hence no twinkling.

    Where students slip. Saying planets 'don't twinkle because they're not stars' without explaining the actual mechanism — the real reason is the point-source vs. extended-source distinction, which is what causes fluctuations to average out for planets but not for stars.

12. 122 marksNCERT Cl-10 Science, Exercise Q12

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

    Hint. The blue sky depends entirely on something being present around Earth that isn't present in space.

    Step 1 — Recall why the sky is blue at all. The sky's blue colour comes from sunlight being scattered by the fine particles and molecules in Earth's atmosphere.

    Step 2 — Apply this to an astronaut in space. An astronaut is above/outside the atmosphere, where there is no medium to scatter sunlight.

    Step 3 — State the consequence. With no scattering, there is no scattered light reaching the astronaut's eyes from all directions, so the sky appears dark, even though the Sun itself is shining brightly.

    ✦ Answer: With no atmosphere to scatter sunlight in space, there's no scattered blue light reaching the astronaut's eyes from all directions, so the sky appears dark rather than blue.

    Where students slip. Assuming the Sun not being overhead is the reason for a dark sky — the Sun is still visible and shining; the sky's blueness (or lack of it) depends specifically on whether there's an atmosphere present to scatter its light, not on the Sun's position.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 10 Science textbook, Reprint 2026-27 (jesc110.pdf) — one in-text question set (4 questions, not 10 as some older manifests claim) plus one end-of-chapter Exercise (12 questions, not 17). Unchanged by rationalisation; the chapter describes the retina generically (no rods/cones, blind spot or fovea named) and never uses the term 'Rayleigh scattering.'. Questions are referenced from the NCERT textbook for identification.