Chapter 2 · The Human Eye and the Colourful World

2.1 The Human Eye and Defects of Vision

Your eye is a living camera with a lens that changes shape. Learn how it focuses near and far, why some eyes cannot, and how a spectacle lens of the right power puts the image back on the retina.

Look up from this screen at the farthest thing you can see — a tree, a building, the sky — and then back at these words. Both looked sharp, and you did nothing to make that happen. In that half-second a ring of tiny muscles inside each eye squeezed and then relaxed, changing the shape of a soft, clear lens so that the picture stayed in focus on the back of the eye. No camera lens made of glass can do that; a camera has to move its lens instead.

Not every eye manages perfectly. Some students can read a book but not the blackboard; many grandparents can see the road clearly but hold the newspaper at arm's length. In this section you will meet the parts of the eye, see how it focuses (its power of accommodation), and learn the three common defects of vision. Then you will use the lens formula and lens power from Section 1.4 to work out exactly which spectacle lens corrects each defect — the same calculation an optometrist does in a few seconds.

The Structure of the Human Eye

The eyeball is roughly a sphere, only about $2.3\ \text{cm}$ across, filled with clear jelly. Light from an object passes through a series of transparent layers before it reaches the light-sensitive screen at the back.

PartWhat it isWhat it does
Corneathe clear, curved front window of the eyelight enters here; it does most of the bending of the light
Iristhe coloured ring (brown, black, blue…) behind the corneaa muscle that changes the size of the pupil
Pupilthe dark hole in the middle of the iriscontrols how much light enters: small in bright light, wide in the dark
Eye lensa soft, transparent, jelly-like convex lensfine-tunes the focus by changing its shape
Ciliary musclesa ring of muscle holding the lenschange the curvature, and so the focal length, of the eye lens
Retinaa thin layer of light-sensitive cells at the backthe screen: the image forms here; its cells turn light into electrical signals
Optic nervea bundle of nerve fibrescarries the signals from the retina to the brain

The cornea and the eye lens together act like one convex lens. For an object in front of the eye they form a real, inverted, diminished image on the retina — exactly the image a convex lens forms of an object beyond $2F$ (Section 1.4). The retina has two kinds of light-sensitive cells: rods, which respond to dim light, and cones, which respond to colour in bright light. The brain receives the signals and interprets the upside-down image as the right way up. Where the optic nerve leaves the eye there are no light-sensitive cells at all; this small patch is called the blind spot.

⚠️ Two facts students often get wrong

1. The eye lens does not do most of the focusing. About two-thirds of the bending happens at the cornea, where light passes from air into the eye. The eye lens provides the adjustable part.

2. The eye focuses by changing focal length, not distance. In a camera the lens moves; in the eye the distance from lens to retina is fixed and the lens changes its shape instead. When the pupil gets smaller in bright sunlight, that changes the amount of light, not the focus.

Accommodation: Focusing Near and Far

Because the lens-to-retina distance is fixed (about $2\ \text{cm}$), the eye can form a sharp image of objects at different distances only by changing the focal length of the eye lens. Light from a distant object arrives almost parallel and needs only a weak lens; light from a near object arrives spreading out and needs a stronger, more curved lens to bring it to a point on the retina.

📘 Definition 2.1 — Power of accommodation
The power of accommodation of the eye is the ability of the eye lens to adjust its focal length, so that both near and distant objects are focused sharply on the retina. The adjustment is made by the ciliary muscles changing the curvature of the eye lens.

There is a limit to how much the lens can bulge, and so a limit to how close an object can be and still look sharp. Try reading this line with your nose almost touching the screen: the letters blur, however hard you try.

📘 Definition 2.2 — Near point, far point and range of vision

The near point of the eye is the nearest point at which an object can be seen clearly without strain. For a normal young adult eye it is about $25\ \text{cm}$ from the eye; this distance is called the least distance of distinct vision.

The far point is the farthest point at which an object can be seen clearly. For a normal eye it is at infinity.

A normal eye therefore sees clearly everything from about $25\ \text{cm}$ to infinity — its range of vision.

Interactive 1 — The Accommodating Eye

Move the object with the slider (the strip at the top shows how far away it is) and watch the eye lens change shape so that the rays from the object meet on the retina. The cornea and lens are treated as one lens about $2\ \text{cm}$ from the retina; the eye needs $50\ \text{D}$ for a distant object and more for a near one. Bring the object closer than the near point and the lens is already as thick as it can get — the rays meet behind the retina and the image blurs. Switch to the older eye to see why reading glasses are needed after about 45. Press Play to sweep the object in and out. (The eye is drawn enlarged, and any focusing error inside it is drawn four times larger so that you can see it; the readouts give the true values.)

—
object—
power needed—
eye lens power—
ciliary muscles—
image—

Myopia and Hypermetropia

An eye whose range of vision is smaller than "$25\ \text{cm}$ to infinity" has a defect of vision, also called a refractive defect. The two most common ones are mirror images of each other: in one the eye bends light too much, in the other too little. Both are corrected with a spectacle (or contact) lens of the right power.

📘 Definition 2.3 — Myopia and hypermetropia

Myopia (near-sightedness): near objects are seen clearly but distant objects look blurred. The far point is nearer than infinity — perhaps only a few metres, or less. The image of a distant object forms in front of the retina.

Hypermetropia (far-sightedness): distant objects are seen clearly but near objects look blurred. The near point is farther than $25\ \text{cm}$. The image of a nearby object would form behind the retina.

MyopiaHypermetropia
Sees clearlynear objectsdistant objects
Point that has movedfar point comes nearer than infinitynear point moves beyond $25\ \text{cm}$
Image (uncorrected)in front of the retinabehind the retina
Causes(i) eye lens too curved — too powerful, focal length too short; (ii) eyeball too long(i) eye lens not curved enough — focal length too long; (ii) eyeball too short
Corrected withconcave (diverging) lens, negative powerconvex (converging) lens, positive power

Why a concave lens fixes myopia. The myopic eye, fully relaxed, can focus only light that spreads out from its far point. A concave lens takes the parallel rays from a distant object and makes them diverge as if they came from the far point. The eye then focuses them on the retina without trouble.

Why a convex lens fixes hypermetropia. The hypermetropic eye, straining as hard as it can, can focus only light that spreads out from its near point $N$ (farther than $25\ \text{cm}$). A convex lens takes the light from a book at $25\ \text{cm}$ and forms a virtual, erect image of it at the near point; the eye then sees that image clearly.

📐 Result 2.1 — The lens that corrects myopia

For a myopic eye with far point at distance $x$, take the corrective lens to be close to the eye and use the New Cartesian sign convention. The lens must form an image of an object at infinity ($u=-\infty$) at the far point ($v=-x$). The lens formula gives

$$\frac1f=\frac1v-\frac1u=-\frac1x-0,\qquad f=-x .$$

The lens is concave, its focal length equals the far-point distance, and its power is $P=\dfrac1f=-\dfrac1x$ with $x$ in metres.

📐 Result 2.2 — The lens that corrects hypermetropia

For a hypermetropic eye with near point at distance $N$ (greater than $25\ \text{cm}$), the lens must form an image of an object at $25\ \text{cm}$ ($u=-25\ \text{cm}$) at the near point ($v=-N$). The lens formula gives

$$\frac1f=\frac1v-\frac1u=\frac{1}{25\ \text{cm}}-\frac1N .$$

Since $N\gt25\ \text{cm}$, $f$ is positive: the lens is convex. Its power is $P=\dfrac1f$ with $f$ in metres.

🧭 Problem-solving strategy — spectacle numericals
  1. Decide the defect: a far point that is too near ⟹ myopia; a near point that is too far ⟹ hypermetropia.
  2. Write the sign convention: object on the left, distances measured from the lens (taken to be at the eye), light travels left to right. Both the object and the virtual image are in front: $u$ and $v$ are negative.
  3. Myopia: $u=-\infty$, $v=-(\text{far point})$, so $f=-(\text{far point})$. Hypermetropia: $u=-25\ \text{cm}$, $v=-(\text{near point})$, then $\dfrac1f=\dfrac1v-\dfrac1u$.
  4. Convert $f$ to metres and find $P=\dfrac1f$ in dioptres, with its sign.
  5. Check the sign: myopia must give a negative power (concave), hypermetropia a positive power (convex).
✏️ Example 1 — A student who cannot read the board
A student sees clearly only up to $80\ \text{cm}$ from her eyes. Name the defect and find the focal length and power of the lens she needs.
Defect Her far point is $80\ \text{cm}$ instead of infinity: myopia.
Signs New Cartesian convention, lens at the eye: an object at infinity, $u=-\infty$, must have its image at the far point, $v=-80\ \text{cm}$.
Formula $\dfrac1f=\dfrac1v-\dfrac1u=\dfrac{1}{-80}-0$, so $f=-80\ \text{cm}=-0.80\ \text{m}$, and $$P=\frac1f=\frac{1}{-0.80\ \text{m}}=-1.25\ \text{D}.$$
Meaning A concave lens of focal length $80\ \text{cm}$ (power $-1.25\ \text{D}$). Check: myopia needs a negative power. ✓
✏️ Example 2 — Glasses for reading
A man can see distant objects clearly, but his near point is $1.0\ \text{m}$ from his eyes. What lens does he need to read a book held at $25\ \text{cm}$?
Defect His near point is beyond $25\ \text{cm}$: hypermetropia.
Signs The book is the object, $u=-25\ \text{cm}$; the lens must place its (virtual) image at his near point, $v=-100\ \text{cm}$.
Formula $$\frac1f=\frac1v-\frac1u=\frac{1}{-100}-\frac{1}{-25}=\frac{-1+4}{100}=\frac{3}{100},$$ so $f=+33.3\ \text{cm}=+0.333\ \text{m}$ and $P=\dfrac{1}{0.333\ \text{m}}=+3.0\ \text{D}$.
Meaning A convex lens of focal length $33\ \text{cm}$ (power $+3.0\ \text{D}$). Check: positive power for hypermetropia ✓; and the image ($100\ \text{cm}$) is farther than the object ($25\ \text{cm}$), as it should be for a virtual image formed by a convex lens with the object inside $F$.
✏️ Example 3 — Reading a prescription
An optometrist's prescription reads "$-2.5\ \text{D}$" for both eyes. What kind of lens is this, what is its focal length, and how far can the patient see clearly without it?
Lens Negative power ⟹ concave (diverging) lens: the patient is myopic.
Focal length $f=\dfrac1P=\dfrac{1}{-2.5\ \text{D}}=-0.40\ \text{m}=-40\ \text{cm}$.
Far point For myopia $f=-(\text{far point})$, so the far point is $40\ \text{cm}$: without spectacles the patient sees clearly only up to about $40\ \text{cm}$ — enough to read, not enough to recognise a friend across the road.
✏️ Example 4 — Working backwards to the near point
A woman reads comfortably at $25\ \text{cm}$ when she wears convex lenses of power $+2.0\ \text{D}$. Where is her near point without the lenses?
Data $f=\dfrac1P=\dfrac{1}{2.0}\ \text{m}=+50\ \text{cm}$; the book is at $u=-25\ \text{cm}$.
Formula The lens places the image of the book at her near point: $$\frac1v=\frac1f+\frac1u=\frac1{50}+\frac{1}{-25}=\frac{1-2}{50}=-\frac1{50},\qquad v=-50\ \text{cm}.$$
Meaning The image is $50\ \text{cm}$ in front of the lens, so her near point is $50\ \text{cm}$. Check with Result 2.2: $\dfrac1{25}-\dfrac1{50}=\dfrac1{50}$, giving $f=+50\ \text{cm}$. ✓
Interactive 2 — Defects of Vision and the Corrective Lens

Choose a defect and set how bad it is with the slider. A myopic eye looks at a distant object (parallel rays); a hypermetropic eye tries to read a book $25\ \text{cm}$ away, straining as hard as it can. Without help the rays meet in front of, or behind, the retina. Tick wear the corrective lens: the lens power is worked out from Result 2.1 or 2.2 and the rays are traced through it — watch them land on the retina. Press Play to put the spectacles on and take them off. (The eye is drawn enlarged, and the focusing errors inside it are drawn four times larger so that you can see them; the readouts give the true values.)

40
far point—
without the lens—
lens needed—
power $P$—
image now—

Presbyopia, Bifocal Lenses and Caring for Your Eyes

With age the eye lens becomes stiffer and the ciliary muscles weaken, so the lens can no longer bulge as much. The power of accommodation falls and the near point creeps outwards — from about $10\ \text{cm}$ in a young child, to $25\ \text{cm}$ in a young adult, to a metre or more in many people over sixty. Try Interactive 1 with the older eye.

📘 Definition 2.4 — Presbyopia
Presbyopia is the loss of the power of accommodation with age: the near point moves farther away, so nearby objects cannot be seen clearly without spectacles. It is caused by the gradual weakening of the ciliary muscles and the loss of flexibility of the eye lens. It is corrected, like hypermetropia, with a convex lens for reading.

Some people suffer from myopia and presbyopia together: they need a concave lens to see far and a convex lens to read. A bifocal lens provides both in one frame — the upper part is concave, for looking ahead at distant objects, and the lower part is convex, for looking down at a book. (Modern "progressive" lenses change smoothly from one power to the other.) Defects can also be corrected with contact lenses or with laser surgery that reshapes the cornea.

Another common problem in old age is the cataract: the eye lens slowly becomes milky and cloudy, and vision dims. It is not corrected by spectacles; instead the cloudy lens is removed in a short operation and replaced with a clear artificial lens. Cataract surgery restores sight to millions of people every year and is one of the most successful operations in medicine.

✏️ Example 5 — Designing a bifocal lens
A 62-year-old man cannot see clearly beyond $2.0\ \text{m}$, and his near point is $60\ \text{cm}$. Find the powers of the upper and lower parts of the bifocal lens he needs.
Upper part For distance vision (myopia), $f=-(\text{far point})=-2.0\ \text{m}$, so $P=\dfrac{1}{-2.0\ \text{m}}=-0.50\ \text{D}$.
Lower part For reading at $25\ \text{cm}$: $u=-25\ \text{cm}$, $v=-60\ \text{cm}$, $$\frac1f=\frac1{-60}-\frac1{-25}=\frac{-5+12}{300}=\frac{7}{300},$$ so $f=+42.9\ \text{cm}=+0.429\ \text{m}$ and $P=+2.3\ \text{D}$.
Meaning Upper half: concave, $-0.50\ \text{D}$; lower half: convex, $+2.3\ \text{D}$. Check the signs: the distance part must be negative and the reading part positive. ✓
✏️ Example 6 — Diagnosis from the symptoms
Name the defect, or condition, in each case, and say how it is corrected. (a) Ravi, 14, reads his textbook easily but squints at the scoreboard at a cricket match. (b) Ms Fernandes, 50, has always had perfect eyesight but now holds her phone at arm's length to read messages. (c) Mr Okafor, 70, says everything looks as if he is seeing through frosted glass, with or without spectacles.
(a) Near is clear, far is blurred: myopia. Corrected with a concave lens whose focal length equals his far-point distance.
(b) Distance vision is fine but the near point has moved out with age: presbyopia. Corrected with convex reading glasses.
(c) A cloudy, milky view that no lens improves suggests a cataract. Treated by surgery: the clouded eye lens is replaced with an artificial one.
⚠️ Common mistakes

1. Power with $f$ in centimetres. $P=\dfrac{1}{f}$ gives dioptres only when $f$ is in metres: $f=-80\ \text{cm}$ gives $P=-1.25\ \text{D}$, not $-0.0125$. Equivalently, $P=\dfrac{100}{f\ \text{in cm}}$.

2. Mixing up the two defects. Myopia — the "my" eye sees things near me; the far point is too near; concave lens. Hypermetropia — the near point is too far; convex lens.

3. Using $u=-25\ \text{cm}$ for myopia. The myopic eye's problem is distant objects, so $u=-\infty$. The $25\ \text{cm}$ appears only in hypermetropia and presbyopia.

4. Making $v$ positive. The corrective lens forms a virtual image on the same side as the object (at the far point or near point), so $v$ is negative in both defects.

🚀 Beyond the board syllabus (enrichment only) — optional: how near can you see with your glasses on?

Not listed in JEE Main, NEET or JEE Advanced 2026 (none of them lists the human eye), and beyond the CBSE Class 10 syllabus, which stops at the power of the corrective lens — read this for understanding only.

Spectacles for myopia move the far point out to infinity — but they move the near point too. Suppose a myopic eye has far point $50\ \text{cm}$ and near point $15\ \text{cm}$, and wears its correct lens, $f=-50\ \text{cm}$ ($-2.0\ \text{D}$). The nearest object it can now see is the one whose image the lens throws to $v=-15\ \text{cm}$:

$$\frac1u=\frac1v-\frac1f=\frac{1}{-15}-\frac{1}{-50}=\frac{-10+3}{150}=-\frac{7}{150}.$$

So $u=-21\ \text{cm}$: with spectacles on, this person sees clearly from about $21\ \text{cm}$ to infinity. Working out this new "range of vision" with spectacles is a good test of the lens formula.

Exercises

I
The eye and accommodation.
  • 1 Which part of the eye does most of the bending of the light that enters it?
  • 2 What controls the size of the pupil? Why does the pupil become smaller in bright sunlight?
  • 3 Where is the image formed in the eye, and what are its nature and size?
  • 4 You look up from your book at a tree far away. What happens to the ciliary muscles, the shape of the eye lens and its focal length?
  • 5 Define the near point and far point of the eye and give their values for a normal young eye.
  • 6 Why can you not read a book held $8\ \text{cm}$ from your eyes, however hard you try?
  • 7 Walking from bright sunlight into a dark cinema, you see almost nothing at first and then gradually more. Explain.
  • 8 What is a cataract, and how is it treated?
1 The cornea, where light passes from air into the eye.
2 The iris. In bright light it contracts the pupil so that less light enters and the retina is not dazzled or damaged.
3 On the retina; the image is real, inverted and diminished.
4 The ciliary muscles relax, the lens becomes thinner, and its focal length increases (its power decreases).
5 Near point: the nearest point seen clearly without strain, about $25\ \text{cm}$. Far point: the farthest point seen clearly, at infinity.
6 $8\ \text{cm}$ is inside the near point. Even with the ciliary muscles fully contracted the lens cannot become any thicker, so its focal length cannot get short enough; the image would form behind the retina and looks blurred.
7 In sunlight the pupil was small. In the dark the iris slowly widens it so that more light enters; this takes some time, and the retina also takes time to adjust to dim light.
8 The eye lens becomes cloudy and milky, so vision dims. It is treated by surgery that replaces the cloudy lens with a clear artificial lens.
II
Myopia. Use the New Cartesian sign convention and take the lens to be close to the eye.
  • 1 A person cannot see clearly beyond $2.0\ \text{m}$. Find the power of the lens needed.
  • 2 Find the focal length and power of the lens for a far point of $50\ \text{cm}$.
  • 3 Find the power of the lens for a far point of $1.2\ \text{m}$.
  • 4 A myopic person wears lenses of power $-4.0\ \text{D}$. Where is his far point?
  • 5 A prescription reads $-0.40\ \text{D}$. Find the focal length of the lens and the patient's far point.
  • 6 State two causes of myopia.
  • 7 Describe ray diagrams showing (a) a myopic eye looking at a distant object, (b) the same eye with its corrective lens.
  • 8 A student in the back row is $6.0\ \text{m}$ from the board but sees clearly only up to $3.0\ \text{m}$. What lens should she wear?
1 $f=-2.0\ \text{m}$, $P=-0.50\ \text{D}$ (concave).
2 $f=-50\ \text{cm}=-0.50\ \text{m}$, $P=-2.0\ \text{D}$.
3 $f=-1.2\ \text{m}$, $P=-\tfrac{1}{1.2}=-0.83\ \text{D}$.
4 $f=\tfrac{1}{-4.0}=-0.25\ \text{m}$, so the far point is $25\ \text{cm}$ from the eye.
5 $f=\tfrac{1}{-0.40}=-2.5\ \text{m}$; far point $2.5\ \text{m}$.
6 (i) The eye lens is too curved (too powerful); (ii) the eyeball is too long.
7 (a) Parallel rays enter the eye and meet at a point in front of the retina, then spread out again, so a blurred patch falls on the retina. (b) A concave lens in front of the eye makes the parallel rays diverge as if from the far point; the eye bends them to meet exactly on the retina.
8 Myopia with far point $3.0\ \text{m}$: concave lens, $f=-3.0\ \text{m}$, $P=-0.33\ \text{D}$.
III
Hypermetropia and presbyopia. Take the normal reading distance as $25\ \text{cm}$ unless told otherwise.
  • 1 The near point of a hypermetropic eye is $50\ \text{cm}$. Find the focal length and power of the lens needed for reading.
  • 2 Repeat for a near point of $75\ \text{cm}$.
  • 3 Repeat for a near point of $2.0\ \text{m}$.
  • 4 A person reads at $25\ \text{cm}$ wearing $+1.5\ \text{D}$ lenses. Find his near point.
  • 5 A person reads at $25\ \text{cm}$ wearing $+2.5\ \text{D}$ lenses. Find her near point.
  • 6 State two causes of hypermetropia.
  • 7 Why do many people over about 45 need reading glasses even though they have never needed spectacles before?
  • 8 What is a bifocal lens? Which part is used for reading, and what kind of lens is it?
  • 9 A musician wants to read music on a stand $30\ \text{cm}$ away. His near point is $90\ \text{cm}$. Find the power of the lens he needs.
Check your spectacle answers: enter the far point or near point
$f$—
$P$—
lens—
1 $u=-25$, $v=-50$: $\tfrac1f=\tfrac1{25}-\tfrac1{50}=\tfrac1{50}$, $f=+50\ \text{cm}$, $P=+2.0\ \text{D}$.
2 $\tfrac1f=\tfrac1{25}-\tfrac1{75}=\tfrac2{75}$, $f=+37.5\ \text{cm}$, $P=+2.7\ \text{D}$.
3 $\tfrac1f=\tfrac1{25}-\tfrac1{200}=\tfrac7{200}$, $f=+28.6\ \text{cm}$, $P=+3.5\ \text{D}$.
4 $f=+66.7\ \text{cm}$: $\tfrac1v=\tfrac1{66.7}-\tfrac1{25}=0.0150-0.0400=-0.0250\ \text{cm}^{-1}$, $v=-40\ \text{cm}$: near point $40\ \text{cm}$.
5 $f=+40\ \text{cm}$: $\tfrac1v=\tfrac1{40}-\tfrac1{25}=-\tfrac{3}{200}$, $v=-66.7\ \text{cm}$: near point about $67\ \text{cm}$.
6 (i) The focal length of the eye lens is too long (the lens is not curved enough); (ii) the eyeball is too short.
7 Presbyopia: with age the ciliary muscles weaken and the lens stiffens, so the power of accommodation falls and the near point moves beyond $25\ \text{cm}$. Convex reading glasses are needed.
8 A lens with two parts of different power: the upper part (concave) for distant vision and the lower part (convex) for reading. It is used when a person has both myopia and presbyopia.
9 $u=-30\ \text{cm}$, $v=-90\ \text{cm}$: $\tfrac1f=\tfrac1{30}-\tfrac1{90}=\tfrac2{90}$, $f=+45\ \text{cm}$, $P=+2.2\ \text{D}$.
IV
Applications and challenges.
  • 1 Bifocals. A grandmother's near point is $1.25\ \text{m}$ and her far point is $2.5\ \text{m}$. Find the powers of the upper and lower parts of her bifocal lenses.
  • 2 A concave lens of focal length $60\ \text{cm}$ gives a patient clear distance vision. Name the defect, give the patient's far point and the power of the lens.
  • 3 Trial frame. An optometrist finds that two lenses of power $-1.0\ \text{D}$ and $-0.75\ \text{D}$ placed together give a patient sharp distance vision. Find the patient's far point.
  • 4 Explain why convex spectacles would make a myopic person's distance vision worse, not better.
  • 5 A man has to hold his newspaper at arm's length, but reads it at $25\ \text{cm}$ with $+3.0\ \text{D}$ glasses. Find his near point.
  • 6 Challenge. A camera and an eye both form a real image of a distant scene. Explain how each one adjusts to form a sharp image of a nearby object instead.
  • 7 Eye testing. A child's far point is $1.5\ \text{m}$. Find the power of the lens needed, and explain why distance-vision charts are placed about $6\ \text{m}$ from the patient.
1 Upper: $f=-2.5\ \text{m}$, $P=-0.40\ \text{D}$. Lower: $\tfrac1f=\tfrac1{25}-\tfrac1{125}=\tfrac4{125}$, $f=+31.25\ \text{cm}$, $P=+3.2\ \text{D}$.
2 Myopia; far point $60\ \text{cm}$; $P=\tfrac{1}{-0.60}=-1.7\ \text{D}$.
3 Powers of lenses in contact add: $P=-1.0+(-0.75)=-1.75\ \text{D}$, $f=\tfrac{1}{-1.75}=-0.571\ \text{m}$; far point about $57\ \text{cm}$.
4 A myopic eye already bends light too much, so distant images form in front of the retina. A convex lens adds more converging power and moves the image even farther in front of the retina. A concave lens is needed to reduce the total converging power.
5 $f=+33.3\ \text{cm}$: $\tfrac1v=\tfrac3{100}-\tfrac1{25}=-\tfrac1{100}$, $v=-100\ \text{cm}$: near point $1.0\ \text{m}$.
6 The camera keeps the focal length of its lens fixed and moves the lens farther from the sensor. The eye cannot change the lens-to-retina distance; instead the ciliary muscles contract, the lens becomes thicker and its focal length becomes shorter (accommodation).
7 $f=-1.5\ \text{m}$, $P=-0.67\ \text{D}$. Light from a chart $6\ \text{m}$ away enters the small pupil almost parallel, so it tests the eye as if the object were at infinity — while still fitting inside a clinic room.
🎯 Exam corner — board and entrance-exam practice BoardJEE Main

E1. Board · 3 marks A student cannot see objects clearly beyond $1.25\ \text{m}$. Name the defect, give two possible causes, and find the power of the lens that corrects it.

Idea Far point less than infinity ⟹ myopia. Causes: the eye lens is too curved (too powerful), or the eyeball is too long.
Convention Lens at the eye; $u=-\infty$, $v=-1.25\ \text{m}$, so $f=-1.25\ \text{m}$.
Answer $P=\dfrac{1}{-1.25\ \text{m}}=-0.80\ \text{D}$, a concave lens. Exam tip: state the sign of the power and the type of lens — "$0.8\ \text{D}$" alone loses a mark.

E2. Board · 3 marks The near point of a hypermetropic eye is $1.5\ \text{m}$. Find the power of the lens required to read at $25\ \text{cm}$, and describe the ray diagram for the corrected eye.

Convention $u=-25\ \text{cm}$; the image must be at the near point, $v=-150\ \text{cm}$.
Working $\dfrac1f=\dfrac1v-\dfrac1u=-\dfrac1{150}+\dfrac1{25}=\dfrac{-1+6}{150}=\dfrac1{30}$, so $f=+30\ \text{cm}=+0.30\ \text{m}$ and $P=+3.3\ \text{D}$.
Diagram Rays from a point on the book at $25\ \text{cm}$ pass through the convex lens and leave diverging less, as if from the near point $N$ (dashed back-extensions to $N$); the eye then focuses them on the retina.
Answer A convex lens of power about $+3.3\ \text{D}$. Exam tip: in hypermetropia problems the object is at $25\ \text{cm}$ and the image is at the near point — never the other way round.

E3. Board · Case study At a free eye camp, a 65-year-old woman tells the doctor she cannot read the newspaper unless it is $80\ \text{cm}$ away, and that faces across the street look blurred beyond $2.0\ \text{m}$. The doctor prescribes bifocal spectacles. (i) Find the powers of the upper and lower parts of the lenses. [2 marks] (ii) Why does her near point move away with age? [1 mark] (iii) Why is the reading part placed at the bottom of the lens? [1 mark]

(i) Upper (distance): $f=-2.0\ \text{m}$, $P=-0.50\ \text{D}$. Lower (reading): $\tfrac1f=\tfrac1{25}-\tfrac1{80}=\tfrac{16-5}{400}=\tfrac{11}{400}$, $f=+36.4\ \text{cm}$, $P=+2.75\ \text{D}\approx+2.8\ \text{D}$.
(ii) Presbyopia: the ciliary muscles weaken and the eye lens loses its flexibility, so the power of accommodation decreases.
(iii) We look downwards when reading and straight ahead when looking at distant things, so the convex reading part goes at the bottom and the concave distance part at the top. Exam tip: in a bifocal answer, always say which part is concave and which is convex.

E4. Board · 1 mark A person needs spectacles of power $-5.0\ \text{D}$ for distant vision. The far point of his eye is (A) $20\ \text{cm}$ (B) $50\ \text{cm}$ (C) $2.0\ \text{m}$ (D) $5.0\ \text{m}$

Working $f=\tfrac{1}{-5.0}\ \text{m}=-0.20\ \text{m}$; for myopia the far point equals $|f|$.
Answer (A) $20\ \text{cm}$. Exam tip: far point in cm $=\dfrac{100}{|P|}$ — a two-second calculation.

E5. JEE Main The near point of a person's eye is $40\ \text{cm}$. To read a book held at $25\ \text{cm}$ the power of the lens required is (A) $+1.5\ \text{D}$ (B) $-1.5\ \text{D}$ (C) $+2.5\ \text{D}$ (D) $+4.0\ \text{D}$

Working $u=-25\ \text{cm}$, $v=-40\ \text{cm}$: $\tfrac1f=\tfrac1{25}-\tfrac1{40}=\tfrac{8-5}{200}=\tfrac{3}{200}\ \text{cm}^{-1}$, so $f=+66.7\ \text{cm}$ and $P=\tfrac{100}{66.7}=+1.5\ \text{D}$.
Answer (A). Exam tip: working in dioptres directly is fastest: $P=\tfrac{100}{25}-\tfrac{100}{40}=4.0-2.5=+1.5\ \text{D}$. Option (C) is the trap of using $\tfrac{100}{40}$ alone, (D) of using $\tfrac{100}{25}$ alone.

E6. Board · 1 mark Assertion (A): A normal eye can see clearly both a distant star and a book held at $25\ \text{cm}$. Reason (R): The ciliary muscles can change the focal length of the eye lens. (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.

Idea A is true: the normal range of vision is $25\ \text{cm}$ to infinity. R is true, and it is exactly the reason: changing the focal length (accommodation) keeps both images on the retina.
Answer (a). Exam tip: ask "does R answer the question why A?" — here it does.
🔬 Research–Teaching NexusResearch-tutored · optional, about 45 minutes

Research question. How does the near point change with age, and at what age do people start to need reading glasses?

Why it matters. Eye-health researchers estimate that roughly 800 million people worldwide have near vision problems from presbyopia that is not corrected, most of them in low- and middle-income countries, where a simple pair of reading glasses can decide whether a tailor can thread a needle, a farmer can read a seed packet, or a shopkeeper can check a bill. At the other end of life, short sight (myopia) is rising fast among school students in many countries — especially in East Asian cities — and large studies predict that about half of the world's population could be myopic by 2050. Spending more time outdoors in childhood is one of the factors being studied.

What to do.

  1. Make a test card: a few lines of small print on a card. Hold a ruler against your cheekbone, and slowly bring the card towards one eye (cover the other) until the print just begins to blur. Record this distance — your near point. Repeat three times and take the mean.
  2. With permission, measure the near point of 6–10 family members or neighbours of different ages (for example 8, 15, 30, 45, 60 and 75 years). Anyone who wears reading glasses should take them off for the test.
  3. For each person whose near point is more than $25\ \text{cm}$, use Result 2.2 to work out the power of reading glasses they would need.
  4. Plot near point (vertical) against age (horizontal), and power needed against age on a second graph.

Tools: a 30 cm ruler or a measuring tape, a test card, a calculator or spreadsheet, graph paper.

Hand in: your table of measurements and powers, the two graphs, and a paragraph answering the research question. Compare your powers with the "+1.0", "+1.5", "+2.0" … reading glasses sold in pharmacies.

Going further. Find out why ready-made reading glasses are fine for many people with presbyopia, but not for people with myopia or with different powers in the two eyes.

Instructor note: typical near points are about $7$–$10\ \text{cm}$ at age 10, $12$–$15\ \text{cm}$ at 20, $25\ \text{cm}$ around 40, $50\ \text{cm}$ around 50 and $1\ \text{m}$ or more beyond 60 (wide variation). Corresponding reading powers from $P=\tfrac{100}{25}-\tfrac{100}{N}$ (N in cm): $N=50$ gives $+2.0\ \text{D}$, $N=100$ gives $+3.0\ \text{D}$, $N=200$ gives $+3.5\ \text{D}$. Measurement uncertainty of $\pm2\ \text{cm}$ matters little for large $N$. Remind students never to test anyone's eyes with sharp or pointed objects.

🧪 Self-check quiz

0/5 correct

Q1. The least distance of distinct vision for a normal young adult eye is about

Q2. When you shift your gaze from a distant hill to a book in your hands, the eye lens

Q3. In an uncorrected myopic eye, the image of a distant object is formed

Q4. A person's far point is $2.0\ \text{m}$. The power of the lens needed for clear distance vision is

Q5. Presbyopia is caused by

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