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

NCERT Class 10 Science — The Human Eye and the Colourful World, Chapter 10 Exercises. All 12 questions solved.

Also here. Solutions to the in-text questions for this chapter and the activities for this chapter are on their own pages.

The twelve questions at the end of Chapter 10 fall into four groups. Questions 1 to 4 are multiple-choice questions on the parts of the eye and on accommodation. Questions 5 to 7 work out the spectacle lens a person needs to correct a defect of vision. Questions 8 and 9 ask how the eye manages to focus on objects at different distances, and Questions 10 to 12 are about what the atmosphere does to light on its way to us — bending it and scattering it. The numerical questions run on three results from Chapter 9:

• Power of a lens — $P = \dfrac{1}{f}$, with $f$ in metres, measured in dioptres ($\text{D}$). A convex (converging) lens has positive power and a concave (diverging) lens negative power.
• Lens formula — $\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f}$, with the New Cartesian sign convention: distances are measured from the optical centre, positive in the direction the light travels. An object in front of the lens has a negative $u$, and so does a virtual image formed on the same side as the object.
• What a correcting lens does — it forms a virtual image of the object at a place the defective eye can see clearly: at the far point for a myopic eye, at the eye’s own near point for a hypermetropic eye.

Key insight. Spectacles do not do the eye’s focusing for it; they move the object, as far as the eye is concerned. A correcting lens takes an object the eye cannot focus on and forms a virtual image of it at a distance the eye can manage. Once you know where the object really is and where its image must appear, both $u$ and $v$ are known, and the lens formula gives the focal length and power in one line (Questions 6 and 7). The last three questions follow the same habit of asking what happens to light on its way to the eye: the atmosphere bends starlight (Questions 10 and 11) and scatters sunlight (Question 12).

Question 1

The human eye can focus on objects at different distances by adjusting the focal length of the eye lens. This is due to

Solution. The eye lens is made of a flexible, jelly-like material, and the ciliary muscles can change its curvature, and so its focal length. This ability of the eye lens to adjust its focal length is called accommodation. The other three options are all defects of vision, not abilities: presbyopia is the loss of accommodation with age, near-sightedness is another name for myopia, and far-sightedness is another name for hypermetropia.

Question 2

The human eye forms the image of an object at its

Solution. Each part of the eye has its own job. Light is refracted mostly at the cornea as it enters; the iris controls the size of the pupil; the pupil is the opening that lets light in. The eye lens then forms a real, inverted image on the retina, the light-sensitive screen at the back of the eye, whose cells send signals to the brain through the optic nerve.

Question 3

The least distance of distinct vision for a young adult with normal vision is about

Solution. The least distance of distinct vision, or near point, is the closest an object can be and still be seen clearly without strain. For a young adult with normal vision it is about $25\ \text{cm}$ — roughly where you naturally hold a book. The other options are easy to rule out by experience: nobody can read a page held $2.5\ \text{cm}$ from the eye, and someone who could not see clearly nearer than $2.5\ \text{m}$ or $25\ \text{m}$ could not read at all.

Question 4

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

Solution. The ciliary muscles hold the eye lens and change its shape. When they relax, the lens becomes thinner and its focal length increases, for seeing distant objects. When they contract, the lens becomes thicker and its focal length decreases, for seeing nearby objects. The iris and pupil only control how much light enters, and the retina only receives the image.

Question 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?

Solution. Power and focal length are linked by $P = \dfrac{1}{f}$, so $f = \dfrac{1}{P}$. With $P$ in dioptres, this gives $f$ in metres. The sign of the power carries over to the focal length, and it tells us the kind of lens.

Question 5 (i)

Solution. For distant vision, $P = -5.5\ \text{D}$:

$$f = \frac{1}{P} = \frac{1}{-5.5}\ \text{m} = -0.18\ \text{m} = -18.2\ \text{cm}$$

The negative focal length means a concave (diverging) lens, which is what a person who cannot see distant objects clearly — a myopic person — needs.

Question 5 (ii)

Solution. For near vision, $P = +1.5\ \text{D}$:

$$f = \frac{1}{P} = \frac{1}{1.5}\ \text{m} = +0.67\ \text{m} = +66.7\ \text{cm}$$

The positive focal length means a convex (converging) lens, for a person who cannot see nearby objects clearly. Someone who needs both lenses is often given bi-focal lenses: the concave part on top for distant vision, the convex part below for reading.

Question 6

The far point of a myopic person is $80\ \text{cm}$ in front of the eye. What is the nature and power of the lens required to correct the problem?

Solution. A myopic eye can see clearly only up to its far point, here $80\ \text{cm}$ away. Rays from anything farther are brought to a focus in front of the retina, as in panel (a), so distant objects look blurred. The correcting lens must take the parallel rays from a very distant object and make them spread out as though they came from the far point, $80\ \text{cm}$ away, which the eye can see clearly — panel (b). In other words, the lens must form a virtual image of a very distant object at the far point.

So the object is very far away, $u = -\infty$, and the image is on the same side as the object, $80\ \text{cm}$ in front of the lens, $v = -80\ \text{cm}$. (The spectacle lens is taken to be close to the eye.) Put these into the lens formula; since $u$ is enormous, $\dfrac{1}{u}$ is zero:

$$\frac{1}{f} = \frac{1}{v} – \frac{1}{u} = \frac{1}{-80} – 0 \quad\Rightarrow\quad f = -80\ \text{cm} = -0.80\ \text{m}$$

This says the focus of the lens must sit exactly at the far point, which is why the rays in panel (b) appear to come from $O = F$. The power is

$$P = \frac{1}{f} = \frac{1}{-0.80\ \text{m}} = -1.25\ \text{D}$$

The negative sign means a concave (diverging) lens, as expected for myopia.

Question 7

Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is $1\ \text{m}$. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is $25\ \text{cm}$.

Solution. The diagram. A hypermetropic eye can see distant objects clearly, but its near point $N$ has moved out to $1\ \text{m}$. When a book is held at the normal near point $N’$, $25\ \text{cm}$ away, the eye lens cannot bend the rays from it strongly enough, and they would meet at $I$, behind the retina — panel (a). The page looks blurred. A convex lens in front of the eye bends the rays inwards a little before they reach the eye, so that they leave the lens as though they had come from $N$, $1\ \text{m}$ away — panel (b). The eye can focus on $N$, so it now forms a sharp image on the retina.

The power. In the language of the lens formula, the lens must form a virtual image at $N$ of an object at $N’$. Taking the lens to be close to the eye, the object is $25\ \text{cm}$ in front of it and the image is $1\ \text{m} = 100\ \text{cm}$ in front of it, on the same side as the object, so both distances are negative:

$$u = -25\ \text{cm}, \qquad v = -100\ \text{cm}$$

$$\frac{1}{f} = \frac{1}{v} – \frac{1}{u} = \frac{1}{-100} – \frac{1}{-25} = -\frac{1}{100} + \frac{4}{100} = \frac{3}{100}\ \text{cm}^{-1}$$

$$f = \frac{100}{3}\ \text{cm} = 33.3\ \text{cm} = 0.333\ \text{m}, \qquad P = \frac{1}{f} = \frac{1}{0.333\ \text{m}} = +3.0\ \text{D}$$

The positive power means a convex (converging) lens. There is a quick check that this makes sense. The object at $25\ \text{cm}$ is closer to the lens than its focus at $33.3\ \text{cm}$, and Chapter 9 says a convex lens then forms a virtual, erect, enlarged image on the same side as the object — exactly the kind of image we needed. Its magnification is $m = \dfrac{v}{u} = \dfrac{-100}{-25} = 4$, so the print appears four times larger but also four times farther away, and so it looks just as it would to a normal eye at $25\ \text{cm}$.

Question 8

Why is a normal eye not able to see clearly the objects placed closer than $25\ \text{cm}$?

Solution. To see a nearby object, the eye has to shorten the focal length of its lens: the ciliary muscles contract, and the lens becomes thicker and more curved. The closer the object, the shorter the focal length needed to bring its image onto the retina. But the eye lens can only be made so thick — its focal length cannot be decreased below a certain minimum. For a normal eye that minimum is reached for an object about $25\ \text{cm}$ away, which is why $25\ \text{cm}$ is the near point. For an object any closer, the lens is already at its shortest focal length, the rays cannot be bent enough, and they would meet behind the retina. The image on the retina is blurred, and the effort to focus strains the eye.

Question 9

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

Solution. In a laboratory experiment with a lens, moving the object moves the image, and we move the screen to find it. The eye cannot do that: its screen, the retina, is fixed, so the distance from the eye lens to the retina — the image distance — never changes. Instead the eye changes its focal length.

The lens formula shows which way. Write the object’s distance from the eye as $d$, so that $u = -d$, and let $v$ be the fixed distance from the lens to the retina:

$$\frac{1}{f} = \frac{1}{v} – \frac{1}{u} = \frac{1}{v} + \frac{1}{d}$$

As the object moves away, $d$ grows and $\dfrac{1}{d}$ gets smaller, so $\dfrac{1}{f}$ gets smaller and $f$ gets larger. For a very distant object $\dfrac{1}{d}$ is practically zero and $f = v$. So as the object moves away, the ciliary muscles relax, the eye lens becomes thinner and its focal length increases, keeping the image on the retina.

Question 10

Why do stars twinkle?

Solution. Stars twinkle because of atmospheric refraction. Starlight travels through empty space in a straight line, but once it enters the earth’s atmosphere it passes through layers of air whose refractive index increases gradually as it gets closer to the ground, where the air is denser. At each layer it bends a little towards the normal, so it reaches us along a curved path. Our eye traces the light straight back along the direction in which it arrives, so we see the star at an apparent position slightly higher than its real one, as in the figure.

The air is not still: its temperature and density keep changing. So the path of the starlight keeps changing slightly, and with it the apparent position of the star. A star is so far away that it is practically a point source of light, so all of its light reaches us along essentially one narrow path. When that path wavers, the amount of starlight entering the eye changes — sometimes more, sometimes less — and the star appears brighter at one moment and fainter at the next. That flickering is twinkling.

Question 11

Explain why the planets do not twinkle.

Solution. Light from a planet passes through the same restless atmosphere as starlight, and each ray from it wavers in just the same way. The difference is size. The planets are much closer to the earth than the stars, so they are seen as small discs — extended sources — rather than points. A planet can be thought of as a collection of a very large number of point sources, side by side. At any moment the light from some of these points is dimmed while the light from others is brightened, and these changes are independent of one another. Added together over the whole disc, they average out to zero, so the total amount of light entering the eye stays steady and the planet does not twinkle.

Question 12

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

Solution. First recall why the sky looks blue from the ground. The molecules of air and other very fine particles in the atmosphere are smaller than the wavelength of visible light, and they scatter light of shorter wavelength, at the blue end of the spectrum, much more strongly than red light. When sunlight passes through the atmosphere, this scattered blue light reaches our eyes from every part of the sky, so the whole sky looks blue.

An astronaut in space is above the atmosphere, where there is practically no air to scatter sunlight. The Sun shines just as brightly, but its light goes straight on, and no scattered light comes to the eye from the rest of the sky. With no scattering, the sky looks dark — just as it begins to look darker to passengers in aircraft flying at very high altitudes, where the air is thin.

Common mistakes

• Question 1: choosing presbyopia. It is about focusing, so it sounds right, but presbyopia is the loss of the power of accommodation in old age. The ability itself is accommodation.
• Question 4: choosing the iris. The iris does change as the eye adjusts to light, which makes it tempting, but it only controls the size of the pupil, and so how much light gets in. The focal length is changed by the ciliary muscles.
• Question 5: forgetting that $f = 1/P$ gives metres. Power is in dioptres because the focal length is in metres, so $1/(-5.5)$ is $-0.18\ \text{m}$, not $-0.18\ \text{cm}$. Dropping the minus sign is the other slip: the sign is what tells you the lens is concave.
• Question 6: putting $u = -80\ \text{cm}$. The far point is where the image must be formed, not where the object is. The object is a distant one, $u = -\infty$, and $v = -80\ \text{cm}$. Taking the far point as the object distance gives no sensible answer at all.
• Question 7: taking $v = +100\ \text{cm}$. The image the lens forms is virtual and on the same side as the object, so $v$ is negative. With $+100$ the working gives $\dfrac{1}{f} = \dfrac{1}{100} + \dfrac{1}{25}$ and a power of $+5\ \text{D}$, which is wrong.
• Question 7: swapping $u$ and $v$. The object is the book at $25\ \text{cm}$; the image is at the eye’s near point, $1\ \text{m}$. Swapping them gives $-3\ \text{D}$, a concave lens — the wrong kind of lens for hypermetropia, which is a good sign that something is back to front.
• Question 9: saying the image distance increases. That is what happens with a lens and a movable screen in the laboratory. In the eye the retina cannot move, so the image distance stays the same and the focal length changes instead.
• Question 12: saying the sky is dark because there is no sunlight in space. Sunlight is just as strong up there. What is missing is air to scatter it towards the eye from the rest of the sky.

Practise next

• The in-text questions of this chapter — one box of four, placed straight after the section on defects of vision, so all four ask the same thing from different angles: what the eye does by itself, where that runs out, and which lens takes over.
• Chapter 9, Light – Reflection and Refraction — the lens formula, the sign convention and the power of a lens used in Questions 5 to 7 all come from here; revise them if those questions felt shaky.
• Chapter 11, Electricity — the next chapter of the book, and a fresh start: current, potential difference and resistance.