The Human Eye and the Colourful World: Class 10 Notes

What The Human Eye and the Colourful World is about

Weightage: This chapter is part of Unit III, Natural Phenomena, which carries 12 of the 80 theory marks together with Light – Reflection and Refraction.Source: CBSE Science syllabus 2026-27, Unit III Natural Phenomena

In real board papers: asked in 21 of the 21 board papers we hold (2023 to 2026), on average 4.4 marks per paper. See the questions

Key concepts

Structure of the human eye

NCERT section 10.1 The Human Eye

Think of your eye as a camera. Its lens focuses light onto the retina, a layer at the back of the eye packed with cells that respond to light.

Light gets in through the cornea, a thin, clear dome at the front of the eyeball. The cornea does most of the bending (refraction) of light, at its outer surface. The crystalline eye lens does much less. It only fine-tunes the focal length, so that objects both near and far can be brought into focus.

Just behind the cornea is the iris, a dark ring of muscle that opens and closes the pupil. The pupil is the opening that lets light in, so changing its size changes how much light enters.

The image the eye lens makes on the retina is real and inverted. When light falls on the retina's cells, they send electrical signals along the optic nerve to the brain. The brain makes sense of these signals, so you see objects as they really are.

The eyeball is roughly a sphere, about 2.3 cm across. The lens-to-retina distance is fixed, so the image distance cannot change. That is why the lens has to change its focal length instead.

Worked example

Name the part of the human eye that (i) does most of the refraction of light entering the eye, (ii) controls the size of the pupil, (iii) acts as the screen on which the image forms, (iv) carries signals from the eye to the brain. State the nature of the image formed on the screen. (3 marks)

Try it first, then show the solutionSolution

1. (i) Outer surface of the cornea. (½)
2. (ii) Iris. (½)
3. (iii) Retina. (½)
4. (iv) Optic nerve. (½)
5. The image on the retina is real and inverted. (1)

Answer: Cornea, iris, retina, optic nerve; the image on the retina is real and inverted.

Power of accommodation, near point and far point

NCERT section 10.1.1 Power of Accommodation

The eye lens is soft and jelly-like and made of fibres, so its shape can change. The ciliary muscles change how curved it is.

• Distant object: ciliary muscles relax, the lens becomes thin, its focal length increases.
• Near object: ciliary muscles contract, curvature increases, the lens becomes thick, its focal length decreases.

Accommodation is the eye's way of changing the focal length of its lens to bring objects at different distances into focus. Why is it needed? The lens-to-retina distance is fixed. So the only way to keep a sharp image on the retina for objects at different distances is to change the focal length.

There is a limit. The lens can become only so thick, so it cannot focus objects that are too close. Bring a page right up to your nose and the words go fuzzy, or your eyes start to ache. The closest distance at which you can see clearly and comfortably is the near point (least distance of distinct vision), about 25 cm for a young adult. The farthest point you can see clearly is the far point, which is infinity for a normal eye. So a normal eye sees clearly from 25 cm to infinity.

In old age the lens may turn milky and cloudy (cataract), causing partial or complete loss of vision. Cataract surgery can restore vision.

Worked example

Riya reads a book held at 30 cm and then looks up at a tree far away. (a) Describe what happens to her ciliary muscles, the thickness of her eye lens and its focal length in each case. (b) Why can she not read the book clearly when it is held 8 cm from her eyes? (3 marks)

Try it first, then show the solutionSolution

1. Reading at 30 cm: ciliary muscles contract, the eye lens becomes thicker, its focal length decreases, so the image of the near book forms on the retina. (1)
2. Looking at the distant tree: ciliary muscles relax, the lens becomes thinner, its focal length increases, so the image of the far tree forms on the retina. (1)
3. At 8 cm the book is closer than her near point (about 25 cm). Her lens cannot become thick enough to shorten its focal length that much, so the image cannot be focused on the retina; she sees a blur or feels strain. (1)

Answer: Near: muscles contract, lens thick, focal length short. Far: muscles relax, lens thin, focal length long. At 8 cm the lens cannot reduce its focal length enough, so the image is blurred.

Myopia (near-sightedness) and its correction

NCERT section 10.2 Defects of Vision and their Correction

A person with myopia sees nearby objects clearly but distant objects look blurred. The far point has moved closer than infinity, perhaps to a few metres.

Why? The image of a distant object forms in front of the retina, not on it. This happens for one of two reasons:

• the eye lens has excessive curvature, so it bends light too strongly, or
• the eyeball is elongated, so the retina is too far back.

The fix is a concave (diverging) lens of suitable power. It spreads the rays slightly before they enter the eye, so the image is pushed back onto the retina.

In ray terms, the concave lens takes parallel rays from a distant object and makes them seem to come from the person's own far point. The eye can see that point clearly.

This gives the rule for numericals. Take the object at infinity and the image at the far point, on the same side as the object. Then the focal length of the lens equals the far point distance with a negative sign. Convert to metres and use $P=\frac{1}{f}$. A myopia correction always has negative power.

Worked example

A student cannot see objects beyond 40 cm distinctly. Name the defect, state the type of corrective lens, and find its focal length and power using the New Cartesian sign convention. (3 marks)

Try it first, then show the solutionSolution

1. Defect: myopia. Correction: concave lens. (½)
2. The lens must form the image of an object at infinity at the far point: $u=-\infty$, $v=-40$ cm. (½)
3. Lens formula: $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{-40}-0$, so $f=-40$ cm $=-0.4$ m. (1)
4. Power: $P=\frac{1}{f}=\frac{1}{-0.4\text{ m}}=-2.5$ D. (1)

Answer: Myopia; concave lens of focal length $-40$ cm and power $-2.5$ D.

Hypermetropia (far-sightedness) and its correction

NCERT section 10.2 Defects of Vision and their Correction

A person with hypermetropia sees distant objects clearly but cannot see nearby objects distinctly. The near point has moved farther than the normal 25 cm. Such a person must hold reading material well beyond 25 cm.

Why? Light from a nearby object is focused at a point behind the retina. There are two causes:

• the focal length of the eye lens is too long, or
• the eyeball has become too small.

The fix is a convex (converging) lens of appropriate power. It supplies the extra focusing power the eye lacks, so the image forms on the retina.

In ray terms, the convex lens takes an object at 25 cm and forms a virtual, upright image at the person's own near point. The eye can see that image clearly.

Rule for numericals: take $u=-25$ cm (normal near point) and $v=-$(the defective near point). Both are on the object side, so both are negative. Put them into $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, find f, change it to metres and use $P=\frac{1}{f}$. The power always comes out positive.

Worked example

A person cannot read a book held closer than 50 cm. (a) Name the defect and give two possible causes. (b) Find the power of the lens needed so that he can read at 25 cm. (3 marks)

Try it first, then show the solutionSolution

1. Defect: hypermetropia (far-sightedness). (½)
2. Causes: focal length of the eye lens too long; eyeball too small. (½ + ½)
3. Lens must form the image of a book at 25 cm at his near point: $u=-25$ cm, $v=-50$ cm. (½)
4. $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=-\frac{1}{50}+\frac{1}{25}=\frac{1}{50}$, so $f=+50$ cm $=+0.5$ m. (½)
5. $P=\frac{1}{f}=\frac{1}{0.5\text{ m}}=+2$ D, a convex lens. (½)

Answer: Hypermetropia; caused by a long focal length of the eye lens or a small eyeball; a convex lens of power $+2$ D.

Presbyopia and bifocal lenses

NCERT section 10.2 Defects of Vision and their Correction

The power of accommodation usually decreases with age. For most people the near point slowly moves away, so nearby objects become hard to see without glasses. This defect is presbyopia.

The symptom is like hypermetropia, but the cause is different, and examiners like this contrast.

• Presbyopia is caused by the gradual weakening of the ciliary muscles and the diminishing flexibility of the eye lens. The lens can no longer be made thick enough for near objects.
• Hypermetropia is caused by a long focal length of the eye lens or a small eyeball.

Some people have both myopia and hypermetropia. They often need bifocal lenses. A common type has a concave lens in the upper part, which helps distant vision. It has a convex lens in the lower part, which helps near vision, such as reading while looking down.

Refractive defects can also be corrected with contact lenses or by surgery.

In numericals, a negative power means the concave (distance) part and a positive power means the convex (reading) part.

Worked example

An elderly person's prescription reads: distant vision $-2$ D, near vision $+2.5$ D. (a) Find the focal length of each lens. (b) Which type of spectacles will he need? State which lens goes in the upper part and which in the lower part, and why. (3 marks)

Try it first, then show the solutionSolution

1. Distant vision: $f=\frac{1}{P}=\frac{1}{-2}$ m $=-0.5$ m $=-50$ cm, a concave lens. (1)
2. Near vision: $f=\frac{1}{+2.5}$ m $=+0.4$ m $=+40$ cm, a convex lens. (1)
3. He needs bifocal lenses: concave lens in the upper part for distant vision, convex lens in the lower part for near vision such as reading. (1)

Answer: Distance lens $f=-50$ cm (concave), reading lens $f=+40$ cm (convex); bifocals with the concave part on top and the convex part below.

Refraction of light through a glass prism

NCERT section 10.3 Refraction of Light through a Prism

A prism is a glass block with triangle-shaped ends and three rectangular side faces that slope towards each other. The angle between the two faces the light passes through is the angle of the prism (∠A).

In Activity 10.1 you trace a ray through a prism using pins. PE is the incident ray, EF the refracted ray and FS the emergent ray. You mark ∠i, ∠r and ∠e, the angles of incidence, refraction and emergence.

• At the first face AB, light goes from air into glass and bends towards the normal.
• At the second face AC, light goes from glass into air and bends away from the normal.

Compare this with a rectangular glass slab (also a CBSE practical). A slab has parallel faces, so the bending at the second face undoes the first. The emergent ray is parallel to the incident ray, only shifted sideways (laterally displaced).

A prism's faces are inclined, so the two bendings do not cancel. The emergent ray leaves at an angle to the direction of the incident ray. This angle is the angle of deviation (∠D). Different colours deviating by different amounts is what causes dispersion.

Worked example

In the prism-tracing practical, a student marks PE (incident ray), EF and FS on the outline of prism ABC. (a) At which face does the ray bend towards the normal, and why? (b) Define the angle of deviation. (c) In the glass-slab practical the emergent ray was parallel to the incident ray, but here it is not. Explain why. (3 marks)

Try it first, then show the solutionSolution

1. (a) At face AB, because light passes from air (rarer) into glass (denser), so it bends towards the normal. (1)
2. (b) The angle between the direction of the incident ray and the emergent ray is the angle of deviation, ∠D. (1)
3. (c) The faces of a slab are parallel, so the bending at the second face cancels the first. The faces of a prism are inclined at the angle of the prism, so the bendings do not cancel and the emergent ray is deviated. (1)

Answer: Towards the normal at AB (air to glass); ∠D is the angle between incident and emergent rays; inclined faces stop the bendings cancelling, unlike the parallel faces of a slab.

Dispersion of white light and the rainbow

NCERT section 10.4 Dispersion of White Light by a Glass Prism

In Activity 10.2, a narrow beam of sunlight from a slit falls on a prism. A band of colours appears on a screen. This splitting of white light into its component colours is dispersion, and the band is a spectrum. The order is Violet, Indigo, Blue, Green, Yellow, Orange, Red (VIBGYOR).

Why does light split? Each colour bends through a different angle in the prism. Red bends the least and violet the most, so each colour leaves along its own path. This is refraction, not scattering.

Isaac Newton was the first to get the spectrum of sunlight with a prism. A second similar prism gave no new colours. When he placed an identical prism inverted behind the first, white light came out. This showed that sunlight is made of seven colours.

A rainbow is a natural spectrum seen after rain. Tiny water droplets act like small prisms:

• they refract and disperse sunlight as it enters,
• reflect it internally,
• and refract it again as it leaves.

A rainbow always forms opposite the Sun, so the Sun must be behind you.

Worked example

(a) Why does a glass prism split white light into colours? (b) What did Newton observe when he placed a second identical prism in an inverted position after the first, and what did he conclude? (c) Name, in order, the three processes light undergoes inside a raindrop to form a rainbow. (d) State two conditions needed to see a rainbow. (4 marks)

Try it first, then show the solutionSolution

1. (a) Different colours bend through different angles in the prism; red bends least and violet most, so the colours emerge along different paths and separate. (1)
2. (b) The colours recombined and white light emerged from the second prism; he concluded that sunlight is made up of seven colours. (1)
3. (c) Refraction with dispersion on entering the drop, internal reflection inside it, refraction again on leaving. (1)
4. (d) Tiny water droplets must be present in the air (after a shower, or at a waterfall or fountain), and the Sun must be behind the observer, since a rainbow forms opposite the Sun. (½ + ½)

Answer: Unequal bending of colours; recombination to white light showing sunlight has seven colours; refraction-dispersion, internal reflection, refraction; water droplets present and Sun behind the observer.

Atmospheric refraction: twinkling, planets and advance sunrise

NCERT section 10.5 Atmospheric Refraction

Air is not uniform. Hot air is less dense than cooler air and has a slightly lower refractive index. Above a fire, the hot air keeps moving, so objects seen through it appear to waver. This is atmospheric refraction on a small scale.

Twinkling of stars. Starlight passes through air whose refractive index changes gradually. It bends continuously towards the normal, so a star near the horizon appears slightly higher than it really is. The atmosphere is never still, so this apparent position keeps shifting. A star is so far away that it acts as a point source. So the light entering your eye flickers, sometimes brighter and sometimes fainter.

Planets are much closer and appear as extended sources, like many point sources together. Their variations average out to zero, so planets do not twinkle.

Advance sunrise and delayed sunset. Because of refraction, the Sun is seen about 2 minutes before it actually crosses the horizon and about 2 minutes after it actually sets. The apparent flattening of the Sun's disc then has the same cause.

Worked example

On a clear night, a star close to the horizon appears to flicker, while a bright planet nearby shines steadily. (a) Why does the star appear slightly higher than its actual position? (b) Why does it twinkle? (c) Why does the planet not twinkle? (3 marks)

Try it first, then show the solutionSolution

1. (a) Starlight passes through air of gradually changing refractive index and bends towards the normal, so the star's apparent position is slightly above its actual position. (1)
2. (b) The physical conditions of the atmosphere keep changing, so the path of starlight varies slightly; since the star is a point source, the amount of light entering the eye flickers, making it look brighter and fainter by turns. (1)
3. (c) The planet is much closer and acts as an extended source, a collection of many point sources; the variations from all of them average out to zero, so there is no twinkling. (1)

Answer: Refraction in air of changing refractive index raises the star's apparent position; changing conditions make the point-source star flicker; the planet is an extended source whose variations cancel.

Scattering of light, Tyndall effect and the blue sky

NCERT section 10.6 Scattering of Light

The atmosphere is a mixture of tiny particles: smoke, tiny water droplets, dust and air molecules. When a light beam hits such fine particles, they reflect light diffusely in all directions, so the beam's path becomes visible. Scattering by colloidal particles causes the Tyndall effect. You see it when a thin beam of sunlight enters a smoky room through a small hole, or passes through a dense, misty forest canopy.

The colour scattered depends on particle size. Very fine particles scatter mainly blue light. Larger particles scatter longer wavelengths. Large enough particles may scatter light that looks white.

Blue sky. Air molecules and fine particles are smaller than the wavelength of visible light. They scatter shorter wavelengths (blue) more strongly than longer ones (red), and red's wavelength is about 1.8 times blue's. This scattered blue light reaches your eyes. With no atmosphere there would be no scattering, and the sky would look dark. That is how it looks to passengers at very high altitudes.

Danger signals are red because red is least scattered by fog or smoke, so it keeps its colour over a distance.

Worked example

(a) Why does the clear sky appear blue from the earth's surface? (b) Why would the sky appear dark to an astronaut far above the atmosphere? (c) Why are danger signal lights red? (3 marks)

Try it first, then show the solutionSolution

1. (a) Air molecules and fine particles, smaller than the wavelength of visible light, scatter shorter wavelengths (blue) much more strongly than red; this scattered blue light enters our eyes. (1)
2. (b) Far above the atmosphere there are hardly any particles to scatter sunlight, so no scattered light reaches the eye and the sky looks dark. (1)
3. (c) Red light has the longest wavelength and is least scattered by fog or smoke, so it can be seen in the same colour from a distance. (1)

Answer: Blue is scattered most by fine air particles; no atmosphere means no scattering, so the sky is dark; red is least scattered, so it is visible from far away.

Science in real life: The Human Eye and the Colourful World

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Human Eye and Power of Accommodation
• Defects of Vision and Their Correction
• Refraction of Light Through a PrismComing soon
• Dispersion of White Light and RainbowComing soon
• Atmospheric RefractionComing soon
• Scattering of Light and Tyndall Effect

The Human Eye and the Colourful World: formulas and key facts

Power of a lens

$P=\frac{1}{f}$

f must be in metres; unit is the dioptre (D). Concave lens: negative power. Convex lens: positive power.

Lens formula

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

Use with the New Cartesian sign convention for corrective-lens numericals; distances on the object side are negative.

Normal eye range

Near point $\approx 25$ cm; far point $=\infty$

A normal young adult eye sees clearly anywhere from 25 cm to infinity.

Myopia correction rule

$u=-\infty ,v=-(\text{far point}),f=-(\text{far point})$

Concave lens; the focal length equals the far point distance with a negative sign.

Hypermetropia correction rule

$u=-25\text{ cm},v=-(\text{defective near point})$

Convex lens; substitute both values into the lens formula, then find P.

Accommodation key fact

Near object: ciliary muscles contract, lens thick, f decreases. Far object: muscles relax, lens thin, f increases.

Used in MCQs and 2-mark 'how does focal length change' questions.

Bifocal lens

Upper part concave (distant vision); lower part convex (near vision)

For people who suffer from both myopia and hypermetropia.

Presbyopia cause

Gradual weakening of ciliary muscles and diminishing flexibility of the eye lens

Keep this separate from the causes of hypermetropia.

Refraction through a prism

Air to glass at the first face: towards the normal. Glass to air at the second face: away from the normal.

The emergent ray is deviated by ∠D; in a glass slab the emergent ray is parallel to the incident ray, only laterally displaced.

Angle of deviation

$\angle D=anglebetweentheincidentrayandtheemergentray$

Mark it by extending the incident ray forward and the emergent ray backward.

Dispersion and spectrum order

VIBGYOR: violet bends most, red bends least

Dispersion is caused by different colours bending through different angles, not by scattering.

Rainbow formation

Refraction + dispersion, internal reflection, refraction

Always forms opposite the Sun; needs water droplets in the air with the Sun behind the observer.

Atmospheric refraction timings

Sun seen about 2 minutes before actual sunrise and about 2 minutes after actual sunset

Also explains the apparent flattening of the Sun's disc at sunrise and sunset.

Scattering and wavelength

Very fine particles scatter blue (short wavelength) most; red wavelength $\approx 1.8\times$ blue wavelength

Explains the blue sky, the dark sky at great height and red danger signals.

Tyndall effect

Scattering of light by colloidal particles makes the path of a light beam visible

Examples: a sunbeam in a smoke-filled room, sunlight through a misty forest canopy.

The Human Eye and the Colourful World: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks The Human Eye and the Colourful World

The question patterns to expect, and the method that works for each.

Tricks to remember The Human Eye and the Colourful World

Short memory aids. Say each one aloud once, then use it on a question the same day.

• VIBGYOR order

  Violet, Indigo, Blue, Green, Yellow, Orange, Red. Violet bends the most (it is at the 'very bent' end) and red bends the least.

  Use it when: Spectrum order, labelling a dispersion diagram, and 'which colour deviates least' MCQs.

• Myopia means Minus

  M for Myopia, M for Minus: a concave lens with negative power. Hypermetropia (far-sighted) gets the plus, a convex lens.

  Use it when: A quick sign check on every corrective-lens numerical.

• Close, Contract, thiCK

  For a Close object the ciliary muscles Contract and the lens becomes thiCK, so its focal length gets shorter. Far object: relax, thin, long.

  Use it when: Accommodation MCQs and 2-mark 'how does focal length change' questions.

• Rainbow recipe R-R-R

  Refract (with dispersion) in, Reflect inside, Refract out. The Sun stays behind you.

  Use it when: Rainbow diagram labelling and 'state the conditions' questions.

• Point twinkles, planet steady

  A star is a point, so its flicker shows. A planet is many points, so the flickers cancel.

  Use it when: 'Why don't planets twinkle' reasons, which need the words extended source.

• Small particle, small wavelength

  Tiny air particles scatter the short (blue) wavelength most; red, the longest, escapes, so it is used for danger.

  Use it when: Blue sky, dark sky for astronauts and red signals in one line of logic.

The Human Eye and the Colourful World: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in The Human Eye and the Colourful World

The slips examiners see most often, and what to write instead.

• ✗Mistake: Using focal length in centimetres in P = 1/f

  Students carry the value from the lens formula straight into the power formula without converting units.

  Wrong working

  $f=-50$ cm, so $P=\frac{1}{-50}=-0.02$ D

  Right working

  $f=-50$ cm $=-0.5$ m, so $P=\frac{1}{-0.5\text{ m}}=-2$ D

  ✓Do this: Write 'f in metres' as a separate line before every power calculation.

  Usually costs: Usually the final 1 mark of the numerical.

• ✗Mistake: Giving a positive power for myopia, or dropping the sign

  Students remember 'power' as a size and forget that the sign tells the lens type.

  Wrong working

  Far point 40 cm, so $P=2.5$ D, convex lens.

  Right working

  $u=-\infty$, $v=-40$ cm, $f=-0.4$ m, so $P=-2.5$ D, concave lens.

  ✓Do this: State the lens type first (myopia: concave), then check that the sign of your answer matches it.

  Usually costs: ½ to 1 mark for the sign and the nature of the lens.

• ✗Mistake: Swapping the ciliary muscle action

  It feels natural to think the muscles 'work harder' for distant objects.

  Wrong working

  For a near object the ciliary muscles relax and the lens becomes thin.

  Right working

  For a near object the ciliary muscles contract, the lens becomes thicker and its focal length decreases.

  ✓Do this: Use Close, Contract, thiCK and link it to 'shorter focal length for near objects'.

  Usually costs: The whole mark in an MCQ; 1 mark in a 2-mark explanation.

• ✗Mistake: Explaining dispersion by scattering

  Both topics are about colours, so students mix up the causes.

  Wrong working

  The prism scatters white light, so red bends least and violet most.

  Right working

  Different colours bend through different angles on refraction in the prism; red bends least and violet most. Scattering is not involved.

  ✓Do this: Dispersion = refraction by different amounts. Scattering = particles throwing light in all directions.

  Usually costs: The 1-mark assertion-reason, or 1 mark in a 3-mark answer.

• ✗Mistake: Wrong conditions for a rainbow

  Students imagine the Sun 'shining on' the rain from in front or from overhead.

  Wrong working

  A rainbow forms when the Sun is overhead and droplets are in the air.

  Right working

  A rainbow forms opposite the Sun: the observer must have the Sun behind them, with tiny water droplets in the air in front.

  ✓Do this: Always write 'Sun behind the observer' as one of the two conditions.

  Usually costs: ½ to 1 mark in a conditions question, or the full mark in assertion-reason.

• ✗Mistake: Incomplete reason for planets not twinkling

  Students stop at 'planets are closer', which does not explain anything by itself.

  Wrong working

  Planets do not twinkle because they are near the earth.

  Right working

  Planets are closer, so they are seen as extended sources, a collection of many point sources. The variations in light from all of them average out to zero, so there is no twinkling.

  ✓Do this: Always include the chain: closer, extended source, variations average out.

  Usually costs: 1 mark out of 2.

• ✗Mistake: Drawing a prism like a glass slab

  Students reuse the slab diagram from the previous chapter.

  Wrong working

  Emergent ray drawn parallel to the incident ray, or bent towards the normal at the second face.

  Right working

  Bend the ray towards the normal at the first face and away from the normal at the second face; the emergent ray is deviated by ∠D from the incident direction.

  ✓Do this: Draw dotted normals at both faces before drawing the rays, then check the bending at each face.

  Usually costs: 1 to 2 marks in a diagram question.

NCERT exercise guide: The Human Eye and the Colourful World

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

The Human Eye and the Colourful World: previous year questions

• Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 4.4 marks per paper; 60 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from The Human Eye and the Colourful World

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

Every The Human Eye and the Colourful World question, by year

Practice questions on The Human Eye and the Colourful World

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   When white light passes through a glass prism, which colour is deviated the most?
   (a) Red
   (b) Yellow
   (c) Green
   (d) Violet

   Hint

   Think of which end of VIBGYOR lies nearest the base of the prism.

   Answer

   (d) Violet. Violet bends the most and red the least, so violet lies nearest the base.

2. Q2 · 2 marks · Very short answerBoard level

   What is the Tyndall effect? Give two everyday situations where it can be seen.

   Hint

   Think of a beam of light passing through fine particles.

   Answer

   The Tyndall effect is the scattering of light by fine (colloidal) particles, which makes the path of a light beam visible. Examples: a thin beam of sunlight entering a smoke-filled room through a small hole, and sunlight passing through the canopy of a dense forest, where mist droplets scatter it.

3. Q3 · 2 marks · Very short answerBoard level

   A doctor prescribes a lens of power $-1.6$ D. Find its focal length, name the lens and state the defect it corrects.

   Hint

   Use $f=\frac{1}{P}$ and give the answer in metres first.

   Answer

   $f=\frac{1}{-1.6}=-0.625$ m $=-62.5$ cm. It is a concave lens, which corrects myopia (here, a far point of 62.5 cm).

4. Q4 · 3 marks · Short answerBoard level

   How does the colour of scattered light depend on the size of the scattering particles? Use this to explain (i) the blue colour of the clear sky and (ii) why light scattered by large particles may look white.

   Hint

   Compare particle size with the wavelength of visible light.

   Answer

   Very fine particles scatter mainly blue (shorter wavelength) light, larger particles scatter longer wavelengths, and large enough particles scatter all colours so the light looks white. (i) Air molecules are smaller than the wavelength of visible light, so they scatter blue much more strongly than red, and this scattered blue reaches our eyes. (ii) Large particles scatter all colours roughly equally, which together look white.

5. Q5 · 3 marks · Short answerBoard level

   Explain why the Sun is seen about 2 minutes before it actually crosses the horizon at sunrise. By about how much does this effect lengthen the time from apparent sunrise to apparent sunset? Name one more effect of the same phenomenon seen at the horizon.

   Hint

   Add the gain in the morning to the gain in the evening.

   Answer

   Sunlight is bent by atmospheric refraction as it passes through air of gradually changing refractive index, so the Sun appears above the horizon while it is actually just below it. The Sun is seen about 2 minutes early at sunrise and about 2 minutes late at sunset, so the day is lengthened by about 4 minutes. The apparent flattening of the Sun's disc at sunrise and sunset is due to the same phenomenon.

6. Q6 · 5 marks · Long answerHarder board level

   (a) What is hypermetropia? (b) Give two causes. (c) Describe in words the ray diagrams of the defect and its correction. (d) A person's near point is 125 cm. Find the power of the lens he needs to read at 25 cm.

   Hint

   For part (d), $u=-25$ cm and $v=-125$ cm.

   Answer

   (a) A defect in which distant objects are seen clearly but nearby objects are not, because the near point is beyond 25 cm. (b) The focal length of the eye lens is too long, or the eyeball has become too small. (c) Rays from an object at 25 cm focus behind the retina. A convex lens bends them so they appear to come from the near point and then focus on the retina. (d) $\frac{1}{f}=-\frac{1}{125}+\frac{1}{25}=\frac{4}{125}$, so $f=31.25$ cm. $P=\frac{100}{31.25}=+3.2$ D (convex lens).

7. Q7 · 1 mark · Assertion-reasonBoard level

   Assertion (A): A normal eye cannot see clearly an object held 10 cm away from it.
   Reason (R): The eye lens cannot become thick enough to reduce its focal length beyond a certain limit.
   (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.

   Hint

   Think about why the near point is about 25 cm.

   Answer

   (a) Both A and R are true and R is the correct explanation of A. The lens cannot be made thick enough to focus an object at 10 cm on the retina, so the image is blurred.

8. Q8 · 3 marks · Short answerHarder board level

   A student cannot see objects beyond 1 m clearly. (a) Name the defect and find the nature and power of the correct lens. (b) A friend suggests she should wear a convex lens of power $+1$ D instead. Explain why this would make her distant vision worse. (c) Give a 'draw and label' checklist for the ray diagram of her eye corrected with the right lens.

   Hint

   First decide where the image of a distant object forms in her eye, then ask whether the suggested lens moves it towards or away from the retina.

   Answer

   (a) Myopia. For the correcting lens, $u=-\infty$ and $v=-1$ m, so $f=-1$ m and $P=\frac{1}{-1}=-1$ D, a concave lens. (1)

   (b) In her eye, rays from a distant object already meet in front of the retina. A convex lens converges the rays further before they enter the eye, so they meet even farther in front of the retina and the blur gets worse. Only a diverging (concave) lens can move the image back onto the retina. (1)

   (c) Draw and label: the eyeball with the eye lens and retina; a concave lens (thinner in the middle) in front of the eye; parallel rays from a distant object diverging after the lens as if coming from the far point F, 1 m in front of the eye (show this with dotted lines); the rays then meeting on the retina. Labels: concave lens, far point F, eye lens, retina, image on the retina. Put arrows on every ray. Common slip: drawing the rays converging after the concave lens. (1)

9. Q9 · 4 marks · Short answerHOTS / competency

   Suppose two friends compare their spectacles. Meera's lens has power $-2$ D. Kabir's lens has power $+2$ D and lets him read a book held at 25 cm.
   (a) Name each person's defect. Find Meera's far point and Kabir's near point.
   (b) For fun, Meera puts on Kabir's spectacles and looks at a distant hill. Predict what happens to the image in her eye, and why.
   (c) Kabir's grandmother later needs a lens like Meera's for distance and one like Kabir's for reading. Suggest one pair of spectacles for her and say which part goes on top.

   Hint

   Use the sign of each power to find the lens type, then use the lens formula for Kabir with $u=-25$ cm.

   Answer

   (a) Meera: negative power, so a concave lens and myopia. $f=\frac{1}{-2}=-0.5$ m $=-50$ cm, and for myopia $f=-$(far point), so her far point is 50 cm. Kabir: positive power, so a convex lens and hypermetropia. $f=+0.5$ m $=+50$ cm. With $u=-25$ cm, $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{50}-\frac{1}{25}=-\frac{1}{50}$, so $v=-50$ cm: his near point is 50 cm. (2)

   (b) In Meera's myopic eye the image of a distant object already forms in front of the retina. Kabir's convex lens adds converging power, so the rays meet even earlier and the hill looks more blurred, not less. (1)

   (c) Bifocal lenses: a concave part of $-2$ D in the upper portion for distant vision, and a convex part of $+2$ D in the lower portion for reading. (1)

10. Q10 · 4 marks · Short answerHOTS / competency

    A myopic man's far point is 80 cm.
    (a) Find the power of the lens he needs.
    (b) Suppose he is given a lens of power $-2$ D by mistake. Find where this lens forms the image of a very distant object, and explain why it over-corrects his defect.
    (c) Some years later he also needs a lens of $+1$ D for reading. Find its focal length and name the type of spectacles he should now use, stating what goes in each part.

    Hint

    For a distant object, a lens forms its image at its own focus. Compare that distance with his far point.

    Answer

    (a) $f=-80$ cm $=-0.8$ m, so $P=\frac{1}{-0.8}=-1.25$ D, a concave lens. (1)

    (b) The wrong lens has $f=\frac{1}{-2}=-0.5$ m $=-50$ cm. For an object at $u=-\infty$, the image forms at $v=f=-50$ cm, that is 50 cm in front of the eye. The right lens would place it at his far point, 80 cm. So the given lens is stronger than needed: it diverges the rays more than necessary, and his eye must keep its ciliary muscles working (lens thicker) even to see distant objects, which causes strain. (2)

    (c) $f=\frac{1}{+1}=+1$ m $=+100$ cm, a convex lens. He should use bifocal lenses: the concave $-1.25$ D part in the upper portion for distant vision and the convex $+1$ D part in the lower portion for reading. (1)

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Where does most of the refraction of light entering the eye take place?Show answer

   At the outer surface of the cornea.

2. 2.Which part controls the size of the pupil?Show answer

   The iris, a dark ring of muscle around the pupil.

3. 3.What is the near point of a normal young adult eye?Show answer

   About 25 cm (least distance of distinct vision).

4. 4.What is the far point of a normal eye?Show answer

   Infinity.

5. 5.Which lens corrects myopia, and what is the sign of its power?Show answer

   A concave lens, with negative power.

6. 6.Give the two causes of hypermetropia.Show answer

   The focal length of the eye lens is too long, or the eyeball has become too small.

7. 7.Which colour deviates the least in a prism?Show answer

   Red (violet deviates the most).

8. 8.Why is the clear sky blue?Show answer

   Fine air particles scatter short-wavelength blue light more strongly than red, and this scattered blue light reaches our eyes.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you name the function of the cornea, iris, pupil, lens, ciliary muscles, retina and optic nerve?
• ☐Can you say what the ciliary muscles and lens do for a near and a far object?
• ☐Can you state the near point and far point of a normal eye?
• ☐Can you give the causes and the correcting lens for myopia and hypermetropia?
• ☐Can you explain how presbyopia differs from hypermetropia, and which part of a bifocal is concave?
• ☐Can you find the power of the lens for a far point of 50 cm (answer $-2$ D) without looking?
• ☐Can you draw a prism ray diagram with ∠i, ∠r, ∠e, ∠A and ∠D correctly marked?
• ☐Can you list VIBGYOR and say why red bends least?
• ☐Can you write the three steps of rainbow formation and the two conditions to see one?
• ☐Can you explain twinkling, steady planets, the blue sky, the dark sky at height and red danger signals?

The Human Eye and the Colourful World: frequently asked questions