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

The eye lens is made of a fibrous, jelly-like material whose curvature the ciliary muscles can modify. When the muscles relax, the lens thins, its focal length increases, and distant objects come to focus on the retina. When the muscles contract, the lens thickens, its focal length decreases, and nearby objects come to focus on the retina.

This ability of the eye lens to adjust its focal length is called the power of accommodation (NCERT, p. 162).

In myopia the far point is closer than infinity; the image of a distant object forms in front of the retina. To push that image back onto the retina, you need a lens that diverges incoming light before it enters the eye — a concave lens of suitable power beings that image back onto the retina.

The diagrams below shows the far point of a myopic eye and the correction using a concave lens.

The far point of the eye is the farthest distance up to which the eye can see objects clearly. For a normal eye it is at infinity. The near point (or least distance of distinct vision) is the minimum distance at which objects can be seen most distinctly without strain.

For a young adult with normal vision, it is about 25 cm (NCERT, p. 162).

The blackboard is a distant object for a student sitting in the last row. Difficulty seeing distant objects while nearby objects remain clear is the signature of myopia (near-sightedness). In this defect the image of a distant object forms in front of the retina.

It can be corrected by using a concave lens of suitable power, which diverges the incoming light so the image falls back on the retina (NCERT, p. 163). The diagram below shows the myopic eye forming the image in front of the retina.

Correct answer: (b) accommodation. The eye adjusts its focal length so objects at different distances focus on the retina — this ability is accommodation (NCERT, p. 162).

• (a) presbyopia is the loss of accommodation with age — the opposite of the ability described.
• (c) near-sightedness (myopia) is a refractive defect, not a focussing ability.
• (d) far-sightedness (hypermetropia) is also a defect, not an adjustment.

Correct answer: (d) retina. The retina is the light-sensitive screen on which the eye lens forms an inverted real image (NCERT, p. 161/162). The light-sensitive cells then send signals to the brain.

• (a) cornea is where most refraction occurs, but it does not itself form the image.
• (b) iris is the muscular diaphragm that controls pupil size — it regulates light entry, not image formation.
• (c) pupil is simply the hole that controls the amount of light entering.

Correct answer: (c) 25 cm. The least distance of distinct vision (near point) for a young adult with normal vision is about 25 cm (NCERT, p. 162). The other three options are unit/order-of-magnitude traps: 25 m and 2.5 m are too far, 2.5 cm is too close.

Correct answer: (c) ciliary muscles. The ciliary muscles change the curvature of the eye lens, which changes its focal length. When they contract the lens thickens and the focal length decreases; when they relax the lens thins and the focal length increases (NCERT, p. 162).

The link between power and focal length is \( P = \frac{1}{f} \), so \( f = \frac{1}{P} \). With \( f \) in metres, \( P \) comes out in dioptres. The sign of power tells you the lens type: negative power means a concave (diverging) lens, positive power means a convex (converging) lens.

Part (i): Distant vision, \( P = -5.5\ \text{D} \).

\[ f = \frac{1}{P} = \frac{1}{-5.5} = -0.1818\ \text{m} \approx -0.18\ \text{m} \]

The negative sign shows the lens is concave — it corrects myopia (defective distant vision).

Part (ii): Near vision, \( P = +1.5\ \text{D} \).

\[ f = \frac{1}{P} = \frac{1}{+1.5} = +0.6667\ \text{m} \approx +0.67\ \text{m} \]

The positive sign shows the lens is convex — it corrects hypermetropia/presbyopia (defective near vision).

Final answers: (i) \( f \approx -0.18\ \text{m} \) (concave lens); (ii) \( f \approx +0.67\ \text{m} \) (convex lens).

The corrective concave lens must make a distant object appear to sit at the person’s far point, so the eye can then focus it. The object is at infinity, so \( u = \infty \); the virtual image must form at the far point, 80 cm in front of the lens, so \( v = -80\ \text{cm} = -0.80\ \text{m} \) (negative, same side as the object).

Using the lens formula \( \frac{1}{f} = \frac{1}{v} – \frac{1}{u} \):

\[ \frac{1}{f} = \frac{1}{-0.80} – \frac{1}{\infty} = -1.25 – 0 \]

\[ f = -0.80\ \text{m} \quad \Rightarrow \quad P = \frac{1}{f} = \frac{1}{-0.80} = -1.25\ \text{D} \]

Final answer: A concave lens of power \( -1.25\ \text{D} \).

The convex lens must take an object at the normal near point (25 cm) and form a virtual image at this person’s near point (1 m = 100 cm), so the defective eye can then focus it. So \( u = -25\ \text{cm} \) and \( v = -100\ \text{cm} \) (both negative — virtual image on the same side as the object). Using the lens formula:

\[ \frac{1}{f} = \frac{1}{v} – \frac{1}{u} = \frac{1}{-100} – \frac{1}{-25} = -\frac{1}{100} + \frac{1}{25} = \frac{-1 + 4}{100} = \frac{3}{100} \]

\[ f = \frac{100}{3}\ \text{cm} = 33.3\ \text{cm} = 0.333\ \text{m} \quad \Rightarrow \quad P = \frac{1}{f} = \frac{1}{0.333} = +3.0\ \text{D} \]

Final answer: A convex lens of power \( +3.0\ \text{D} \).

How to draw the diagram (three steps):

1. Draw the uncorrected, defective eye (as in Fig. 10.3(b)): rays from a nearby object converge to a point behind the retina — the blurred case.
2. Place a convex lens in front of the eye and show the corrected rays (as in Fig. 10.3(c)): the lens adds converging power, so the rays now meet on the retina.
3. Label the object distance (25 cm, the normal near point) and mark \( N \) (the hypermetropic near point of 1 m) and \( N’ \) (the normal near point).

Accommodation has a physical limit. The ciliary muscles cannot contract enough to thicken the lens beyond a certain curvature, so there is a minimum focal length the eye lens can achieve (NCERT, p. 162).

For an object closer than 25 cm, the required focal length is shorter than this minimum, so the image forms behind the retina and appears blurred, and the eye feels strain.

The image distance — from the eye lens to the retina — is fixed by the anatomy of the eyeball and does not change. Instead, the eye lens adjusts its focal length through accommodation: as the object moves farther away, the ciliary muscles relax, the lens thins, and the focal length increases, so the image still forms on the retina (NCERT, p. 161-162).

Starlight passes through layers of the earth’s atmosphere with different densities and temperatures, so the refractive index changes continuously. The path of the light ray keeps shifting slightly, and the apparent position of the star fluctuates — the amount of light entering the eye flickers, so the star appears brighter at one moment and fainter the next. This is twinkling.

Because stars are very distant, they act as point-sized sources, so this variation is noticeable (NCERT, p. 168). As the figure above shows, the apparent position is slightly higher than the actual position near the horizon.

Planets are much closer to the earth than stars, so they appear as extended sources (small discs) rather than point sources. If we think of a planet as a collection of many point-sized sources, each point twinkles, but the variations from the many points average out to zero, so the total brightness appears steady (NCERT, p. 168).

The blue colour of the sky is due to scattering of shorter (blue) wavelengths by air molecules and fine particles in the atmosphere (NCERT, p. 169). At the very high altitude where astronauts orbit, there is almost no atmosphere — no particles, so no scattering of sunlight occurs. The sky therefore appears dark to them, even though sunlight is present.