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NCERT Class 10 Science The Human Eye and the Colourful World: Exercise Solutions

Step-by-step working for every question of End-of-chapter exercises in NCERT's Class 10 Science book, chapter “The Human Eye and the Colourful World”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. Exercises Q1

   The eye focuses at different distances by changing the eye lens focal length. This is (a) presbyopia (b) accommodation (c) near-sightedness (d) far-sightedness.

   1. The ability of the human eye to adjust its focal length to focus on objects at varying distances is known as accommodation.
   2. This process is achieved by the ciliary muscles changing the curvature of the eye lens.
   3. Presbyopia is the age-related loss of this ability, while near-sightedness and far-sightedness are defects of vision.
   4. Therefore, the correct term for the described phenomenon is accommodation.

   Answer: (b) accommodation

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2. Exercises Q2

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

   1. The eye works like a camera where light enters and forms an image.
   2. The cornea and lens bend the light rays to focus them.
   3. The iris controls the size of the pupil to regulate light entry.
   4. The focused light rays finally fall on the light-sensitive layer at the back of the eye.
   5. This light-sensitive layer is called the retina, which converts light into nerve signals.
   6. Therefore, the image is formed on the retina.

   Answer: (d) retina

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3. Exercises Q3

   The least distance of distinct vision of a normal young adult is about (a) 25 m (b) 2.5 cm (c) 25 cm (d) 2.5 m.

   1. The least distance of distinct vision is the closest distance at which the eye can see an object clearly without strain.
   2. For a normal young adult, this distance is approximately 25 cm.
   3. This distance is also known as the near point of the eye.
   4. Options like 25 m or 2.5 m are too far for the near point, while 2.5 cm is too close for a normal eye to focus on.

   Answer: (c) 25 cm

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4. Exercises Q4

   The focal length of the eye lens is changed by the (a) pupil (b) retina (c) ciliary muscles (d) iris.

   1. The eye lens is a flexible structure that changes its shape to focus light on the retina.
   2. The ciliary muscles are ring-shaped muscles attached to the lens by suspensory ligaments.
   3. When these muscles contract or relax, they change the curvature of the lens.
   4. This change in curvature alters the focal length of the eye lens, a process called accommodation.
   5. The pupil and iris control the amount of light entering the eye, not the focal length.
   6. The retina is the light-sensitive screen where the image is formed.

   Answer: (c) ciliary muscles

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5. Exercises Q5

   A person needs −5.5 D for distant vision and +1.5 D for near vision. Find the focal lengths of the two lenses.

   1. The power of the lens for distant vision is given as P₁ = −5.5 D.
      why?

      Power of a lens P = 1/f with f in metres; the unit is the dioptre (D). A convex lens has positive power and a concave lens negative.

   2. Using the formula P = 1/f, the focal length f₁ = 1/P₁.
      why?

      Power of a lens P = 1/f with f in metres; the unit is the dioptre (D). A convex lens has positive power and a concave lens negative.

   3. Calculating f₁ = 1/(−5.5) gives f₁ = −0.18 m.
   4. The power of the lens for near vision is given as P₂ = +1.5 D.
      why?

      Power of a lens P = 1/f with f in metres; the unit is the dioptre (D). A convex lens has positive power and a concave lens negative.

   5. Using the formula P = 1/f, the focal length f₂ = 1/P₂.
      why?

      Power of a lens P = 1/f with f in metres; the unit is the dioptre (D). A convex lens has positive power and a concave lens negative.

   6. Calculating f₂ = 1/(1.5) gives f₂ = 0.67 m.

   Answer: (i) -0.18 m; (ii) 0.67 m

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6. Exercises Q6

   A myopic person's far point is 80 cm. What lens (nature and power) corrects it?

   1. The far point of the myopic person is 80 cm, which is the image distance v = -80 cm.
   2. The corrective lens must form the image of a distant object at the person's far point.
   3. Using the lens formula 1/f = 1/v - 1/u with u = -∞, we get 1/f = 1/(-80) - 0.
      why?

      Lens formula 1/v − 1/u = 1/f, with distances measured from the optical centre using the New Cartesian sign convention.

   4. This gives the focal length f = -80 cm or -0.8 m.
   5. The power of the lens is P = 1/f in metres, so P = 1/(-0.8).
      why?

      Power of a lens P = 1/f with f in metres; the unit is the dioptre (D). A convex lens has positive power and a concave lens negative.

   6. Calculating the power gives P = -1.25 D.
   7. The negative sign indicates that the lens is concave.

   Answer: concave, P = -1.25 D

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7. Exercises Q7

   A hypermetropic eye has its near point at 1 m. What power of lens corrects it if the normal near point is 25 cm? Describe the correction.

   1. A hypermetropic eye sees distant objects but not nearby ones: its near point (1 m) is farther than the normal 25 cm, because the image of a near object forms behind the retina.
   2. Correction: a convex (converging) lens is placed in front of the eye. It takes an object at the normal near point, 25 cm, and forms its virtual image at the eye's near point, 100 cm, which the eye can see clearly.
   3. With the sign convention: u = −25 cm and v = −100 cm.
   4. Lens formula: 1/f = 1/v − 1/u = 1/(−100) − 1/(−25) = −1/100 + 4/100 = 3/100, so f = 100/3 ≈ 33.3 cm = 0.333 m.
      why?

      Lens formula 1/v − 1/u = 1/f, with distances measured from the optical centre using the New Cartesian sign convention.

   5. Power P = 1/f (f in metres) = 100/33.3 = +3 D.
      why?

      Power of a lens P = 1/f with f in metres; the unit is the dioptre (D). A convex lens has positive power and a concave lens negative.

   6. Diagram: rays from the object at 25 cm pass through the convex lens and enter the eye as if they came from a point 1 m away, so the eye lens focuses them on the retina.

   Answer: A convex lens of power P = +3 D (focal length about 33.3 cm) corrects it.

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8. Exercises Q8

   Why can a normal eye not see objects closer than 25 cm clearly?

   1. The ciliary muscles of the eye control the shape of the lens to focus light from objects at different distances.
   2. To see a nearby object clearly, the lens must become more convex (thicker) to increase its refractive power.
   3. This change in shape is achieved by the contraction of the ciliary muscles.
   4. However, the ciliary muscles have a limit to how much they can contract.
   5. When the muscles are fully contracted, the lens reaches its maximum convexity and cannot bend more.
   6. This maximum focusing power corresponds to a specific minimum distance from the eye.
   7. If an object is placed closer than this limit, the lens cannot focus the image sharply on the retina.
   8. As a result, the image appears blurred to the normal eye.
   9. This minimum distance at which the eye can see an object most distinctly is called the near point.
   10. For a normal human eye, the near point is approximately 25 cm.

   Answer: A normal eye cannot see objects closer than 25 cm clearly because the ciliary muscles cannot contract further to make the lens more convex, so the image cannot be focused on the retina.

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9. Exercises Q9

   What happens to the image distance in the eye when an object moves farther away?

   1. When an object moves farther away from the eye, the light rays entering the eye become more parallel.
   2. The ciliary muscles relax, which causes the suspensory ligaments to pull the lens.
   3. This action makes the eye lens thinner and increases its focal length.
   4. The increased focal length allows the lens to focus the parallel rays exactly on the retina.
   5. Since the distance between the lens and the retina is fixed in the human eye, the image distance remains constant.
   6. The eye adjusts only the focal length of the lens to keep the image distance unchanged.

   Answer: The image distance remains constant.

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10. Exercises Q10

    Why do stars twinkle?

    1. Stars appear to twinkle because their light passes through the Earth's atmosphere.
    2. The atmosphere consists of layers of air with different temperatures and densities.
    3. As starlight enters these layers, it undergoes refraction at each boundary.
    4. The air layers are constantly moving and changing due to convection currents.
    5. This causes the path of the light to bend in varying directions continuously.
    6. The apparent position of the star therefore shifts slightly up and down.
    7. This rapid change in position and brightness is perceived as twinkling.
    8. The effect is more noticeable for stars because they are point sources of light.
    9. Planets do not twinkle as much because they appear as extended sources of light.

    Answer: Stars twinkle due to the continuous refraction of their light by the varying layers of the Earth's atmosphere, causing their apparent position and brightness to fluctuate.

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11. Exercises Q11

    Why do planets not twinkle?

    1. Planets are much closer to Earth than stars, so they appear as small discs rather than point sources of light.
    2. As light from a planet passes through the turbulent layers of the atmosphere, different parts of the disc are refracted by different air currents.
    3. The bright and dark regions of the disc get mixed up due to this refraction, which averages out the fluctuations in brightness.
    4. Because the total amount of light reaching our eyes remains almost constant, we do not see the rapid changes in brightness that cause twinkling.

    Answer: Planets do not twinkle because they are close enough to appear as small discs, so the atmospheric refraction averages out the light fluctuations instead of causing rapid brightness changes.

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12. Exercises Q12

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

    1. The blue colour of the sky is caused by the scattering of sunlight by the molecules of air in the atmosphere.
    2. When sunlight enters the atmosphere, the shorter wavelength blue light is scattered in all directions by the air molecules.
    3. This scattered blue light reaches our eyes, making the sky appear blue to observers on the Earth's surface.
    4. An astronaut in space is located above the Earth's atmosphere, where there are no air molecules to scatter sunlight.
    5. Without the scattering of light by air molecules, the blue light does not reach the astronaut's eyes from the surrounding space.
    6. Since space is a vacuum with no particles to scatter light, the background appears dark to the astronaut.

    Answer: The sky looks dark to an astronaut because there is no atmosphere in space to scatter sunlight.

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