CBSE Class 10 · Science

Class 10 Science – Chapter 10: The Human Eye and the Colourful World

The eye is a natural optical instrument that lets us see the world around us. Study its parts and working, the power of accommodation, common defects of vision (myopia, hypermetropia, presbyopia) and their correction, refraction through a prism, dispersion of white light into VIBGYOR, the rainbow, and atmospheric phenomena such as twinkling of stars and advanced sunrise — with labelled diagrams, solved numericals, tables and exam-ready practice.

1. The Human Eye

Human Eye

The human eye is a natural optical instrument that works like a camera — it collects light from an object and forms a sharp, real, inverted image on a light-sensitive screen (the retina) at its back.

The eye has a nearly spherical shape, about 2.3 cm in diameter. It uses a convex lens (the eye lens) of adjustable focal length to focus light. Its power and flexibility let us see objects placed both very close to us (up to the near point) and very far away (up to infinity).

Parts of the Eye and Their Functions

PartWhat it isFunction
CorneaThe transparent, curved front membrane of the eyeMost of the refraction of light happens at the cornea; it forms the protective transparent cover that lets light enter
IrisThe coloured, ring-like muscular structure behind the corneaControls the size of the pupil and therefore the amount of light entering the eye
PupilThe dark, circular central opening of the irisRegulates (adjusts) the amount of light that enters the eye
Eye lensA transparent, biconvex, flexible structure behind the pupilFocuses light to form a sharp image on the retina
Ciliary musclesMuscles attached to the eye lensChange the curvature (and hence the focal length) of the lens — this is called accommodation
RetinaThe light-sensitive screen at the back of the eyeConverts light into electrical signals; contains rod and cone cells
Optic nerveThe nerve that connects the eye to the brainCarries the electrical signals from the retina to the brain
Aqueous humourA clear watery fluid between the cornea and the lensProvides nutrients to the eye and maintains its shape; also helps in refraction
Vitreous humourA clear gel-like fluid filling the space between the lens and the retinaMaintains the spherical shape of the eye and keeps the retina in position
Fig. 1 – Horizontal section of the human eye Cornea Iris Pupil Eye lens Ciliary muscles Retina Optic nerve To brain → Aqueous humour – between cornea and lens Vitreous humour – gel between lens and retina

2. Functioning of the Human Eye

Seeing an object involves a step-by-step journey of light through the eye, ending with the brain interpreting the picture.

1
Light enters through the cornea. The transparent cornea refracts (bends) the incoming light the most, directing it towards the pupil.
2
Light passes through the pupil. The pupil is the central dark opening of the iris through which light travels.
3
Iris controls the amount of light. In bright light the iris contracts the pupil (narrows it), and in dim light it expands the pupil (widens it) so the right amount of light enters.
4
Eye lens focuses the light. The biconvex eye lens converges the rays and forms a sharp image.
5
Image is formed on the retina. The lens focuses light on the light-sensitive retina at the back of the eye.
6
Retina contains light-sensitive cells. Rods (for dim light) and cones (for colour) convert light into electrical signals.
7
Signals are transmitted through the optic nerve. These electrical impulses travel from the retina to the brain.
8
The brain interprets the information. The brain processes the inverted image and makes us see the object erect and in its true position.
Key Idea

The image formed on the retina is real and inverted. The brain interprets this inverted image and we perceive the object as erect.

Parallel rays from a distant object Retina Inverted image Fig. 2 – The eye lens focuses light to form a real, inverted image on the retina

3. Power of Accommodation

Accommodation

Accommodation is the ability of the eye lens to change its focal length by changing its curvature, so that we can focus clearly on objects at different distances.

The eye lens is flexible. When we look at a near object, the ciliary muscles contract, making the lens thicker (more curved). A thicker lens has a shorter focal length, so it can focus the diverging rays from a close object onto the retina. When we look at a distant object, the ciliary muscles relax, the lens becomes thinner (less curved), its focal length increases, and parallel rays are focused on the retina.

Understand

Change in curvature leads to change in focal length. This is why the eye can focus on objects at different distances even though the lens cannot move back and forth like a camera lens.

Near Point and Far Point of the Eye

PointDefinitionNormal value
Near pointThe closest distance at which a normal eye can see objects clearly25 cm
Far pointThe farthest distance at which a normal eye can see objects clearlyInfinity
CLEAR VISION RANGE OF A NORMAL EYE

A normal human eye can clearly see objects lying anywhere from the near point (25 cm) to the far point (infinity). Objects closer than 25 cm appear blurred because the lens cannot become curved enough.

Accommodation of the eye lens Near object – lens becomes thick Ciliary muscles contract, f decreases Distant object – lens becomes thin Muscles relax, f increases Fig. 3 – The eye lens changes its curvature during accommodation; near objects need a thicker lens

4. Defects of Vision (Overview)

Sometimes the eye loses its ability to focus clearly. The three common defects of vision studied in Class 10 are:

DefectWhat the person cannot see clearlyCorrective lens
Myopia (short-sightedness / near-sightedness)Distant objects appear blurredConcave lens of suitable power
Hypermetropia (long-sightedness / far-sightedness)Nearby objects appear blurredConvex lens of suitable power
Presbyopia (age-related)Nearby objects (and often distant objects too) appear blurredConvex lens, commonly bifocal or progressive lenses

Each defect is studied in detail in the next section, with its cause and the ray-diagram explanation.

5. Myopia (Short-Sightedness)

Myopia

Myopia is a defect of vision in which a person can see nearby objects clearly but distant objects appear blurred. It is also called near-sightedness or short-sightedness.

Cause and Ray-Diagram Explanation

  • The image of a distant object is formed in front of the retina instead of on it.
  • This happens because the eye lens is too curved (too convex) or the eyeball is too long, so the focal length of the eye lens is too small.
  • The far point of a myopic eye is not infinity — it is much closer. Objects beyond this far point cannot be seen clearly.
Myopic eye – image of a distant object forms in front of the retina Early focus – image in front of retina Retina Fig. 4 – In myopia the eye lens converges rays too strongly; distant objects are focused before the retina

Correction of Myopia

Myopia is corrected using a concave lens of suitable power. The concave lens diverges the incoming parallel rays slightly, so they are focused exactly on the retina.

Correction of myopia using a concave lens Concave lens Image now forms correctly on retina Retina Fig. 5 – A concave lens diverges parallel rays so the myopic eye focuses the image on the retina

6. Hypermetropia (Long-Sightedness)

Hypermetropia

Hypermetropia is a defect of vision in which a person can see distant objects clearly but nearby objects appear blurred. It is also called long-sightedness or far-sightedness.

Cause and Ray-Diagram Explanation

  • The image of a nearby object is formed behind the retina instead of on it.
  • This happens because the eye lens is too thin (less curved) or the eyeball is too short, so the focal length of the eye lens is too long.
  • The near point of a hypermetropic eye is farther away than 25 cm. The person cannot focus on objects closer than their near point.
Hypermetropic eye – image of a nearby object forms behind the retina Focus would be behind the retina Retina Fig. 6 – In hypermetropia the eye lens converges rays too weakly; nearby objects focus behind the retina

Correction of Hypermetropia

Hypermetropia is corrected using a convex lens of suitable power. The convex lens converges the rays from a nearby object slightly towards each other before they enter the eye, so the eye lens focuses them exactly on the retina.

Correction of hypermetropia using a convex lens Convex lens Image now forms correctly on retina Retina Fig. 7 – A convex lens converges rays so the hypermetropic eye focuses the image on the retina
THE LENS RULE

Myopia → concave lens (to diverge rays). Hypermetropia → convex lens (to converge rays). Remember: “can’t see far → concave; can’t see near → convex”.

7. Presbyopia

Presbyopia

Presbyopia is an age-related defect of vision in which the eye lens gradually loses its flexibility and hence its power of accommodation, so the person cannot see nearby objects clearly even if their far vision is normal.

  • It usually develops in old age because the ciliary muscles become weak and the lens hardens, so the lens cannot become thick enough to focus on near objects.
  • A presbyopic person needs a convex lens for reading (to focus near objects).
  • If a person suffers from myopia as well as presbyopia, they need bifocal lenses — the upper part is a concave lens (for distance) and the lower part is a convex lens (for reading).
Note

Presbyopia is not a normal defect like myopia or hypermetropia — it is a natural result of ageing. The symptom (trouble seeing close objects) is similar to hypermetropia, but the cause and age group are different.

Bifocal lens used in presbyopia Upper: concave lens (distance) Lower: convex lens (reading) Distant object Near (book) Fig. 8 – Bifocal lenses correct presbyopia: concave for distance, convex for reading

8. Defects of Vision – Quick Summary

DefectCauseImage positionCorrection
MyopiaEye lens too curved; eyeball too longIn front of the retinaConcave lens of suitable power
HypermetropiaEye lens too flat; eyeball too shortBehind the retinaConvex lens of suitable power
PresbyopiaAge-related loss of flexibility of the eye lensNear objects not focused (far vision may be normal or defective)Convex lens for reading; bifocal lenses if myopic too
Exam Tip

Remember the corrective lens by the shape: myopia → concave (thinner in the middle); hypermetropia / presbyopia → convex (thicker in the middle).

9. Refraction Through a Prism

Prism

A prism is a transparent optical object with polished, flat faces, usually made of glass, having a triangular cross-section. Light entering a prism is refracted twice — once at each of its two refracting faces.

Triangular prism terms

  • Angle of prism (A): the angle between the two refracting faces.
  • Angle of incidence (i): the angle between the incident ray and the normal at the first face.
  • Angle of refraction (r): the angle between the refracted ray and the normal inside the prism.
  • Angle of emergence (e): the angle between the emergent ray and the normal at the second face.
  • Angle of deviation (D): the angle between the emergent ray and the incident ray produced forwards.
Refraction of light through a glass prism D (deviation) Incident ray Emergent ray i e Glass prism Fig. 9 – A ray of light bends towards the base of the prism on each face; the total bend is the angle of deviation D
Key Behaviour

In a glass prism (denser than air) the ray bends towards the normal at the first face and away from the normal at the second face, and in both cases it bends towards the base of the prism. That is why the emergent ray is tilted towards the base and is deviated from the incident direction.

10. Dispersion of White Light

Dispersion

The splitting of white light into its constituent colours on passing through a prism is called dispersion of white light.

When white light (e.g. from the Sun) enters a glass prism, it splits into a band of seven colours. This band is called a spectrum. The colours, in order from the top of the band, are:

VIBGYOR

Violet – Indigo – Blue – Green – Yellow – Orange – Red.

Dispersion of white light into the VIBGYOR spectrum White light Violet Red Fig. 10 – A prism disperses white light into its seven constituent colours, violet deviating the most
Which Colour Deviates the Most?

Violet light deviates the most and red light deviates the least, because violet has the smallest wavelength and red has the largest wavelength of the visible spectrum. That is why violet appears at the upper (more deviated) end of the spectrum and red at the lower end.

11. Recombination of Colours

If the colours produced by one prism are passed through another prism placed upside down, they recombine to give white light again.

Recombination of the spectrum by a second inverted prism White light Prism (disperses) Inverted prism (recombines) White light Fig. 11 – A second prism, placed inverted, recombines the seven colours back into white light
Conclusion

The interference by the first prism is not the source of the colours — the colours were already present in white light. The prism merely separates them. Recombination removes any doubt that a prism creates colours.

12. The Rainbow

Rainbow

A rainbow is a natural spectrum of sunlight formed in the sky after rain, when sunlight passes through millions of tiny water droplets that act simultaneously as prisms.

1
Refraction: sunlight enters a raindrop and refracts (bends) into its coloured components.
2
Internal reflection: the light reflects once inside the droplet at its internal surface.
3
Dispersion and emergence: the light disperses further and emerges from the droplet, producing a band of colours.
4
Arc of colours: millions of droplets together produce a large arc — red on the outer edge and violet on the inner edge.
Remember

A rainbow is seen opposite the Sun, and it involves refraction, internal reflection and dispersion of sunlight by water droplets — not a single prism.

Formation of a rainbow by a water droplet Sunlight Refraction at entry Internal reflection inside the drop Emergent dispersed light to the observer Fig. 12 – A raindrop refracts, internally reflects and disperses sunlight to form a rainbow

13. Atmospheric Refraction

Atmospheric Refraction

The refraction of light caused by the non-uniform density of the Earth’s atmosphere is called atmospheric refraction. The air is denser near the ground and rarer higher up, so light bends gradually as it travels through it.

Why Light Bends

As light passes from a rarer (less dense) layer above to a denser layer below, it bends towards the normal layer by layer. Over many layers the path becomes a gentle curve instead of a straight line. The result is that astronomical objects appear slightly higher than their true position.

Atmospheric refraction – light bends layer by layer Air layers (increasing density) Earth surface Star (apparent position) True direction Bent path Fig. 13 – The atmosphere bends starlight, so a star appears slightly higher than its true position
Note

Atmospheric refraction is the reason the stars in the night sky appear to twinkle, and why we see the Sun before actual sunrise and after actual sunset (advanced sunrise / delayed sunset).

14. Twinkling of Stars

Twinkling

Stars appear to twinkle (their brightness and apparent position fluctuate) because their light is continuously refracted through moving layers of air of different densities in the atmosphere.

Twinkling of a star due to moving air of varying density Moving air layers of different densities Star Paths change as air moves Fig. 14 – Starlight follows a fluctuating path through unequal air layers, producing twinkling

The air above the Earth is not still — pockets of warm and cool air keep moving. As starlight passes through them, the apparent position and the amount of light reaching the eye change slightly from moment to moment. Since stars are point sources placed very far away, even a small fluctuation is noticeable, so they appear to twinkle.

Why Don’t Planets Twinkle?

Planets are extended sources — many point sources spread over a small disc. The fluctuations from different points average out, so the total light from a planet does not vary noticeably. That is why planets do not twinkle.

15. Advanced Sunrise and Delayed Sunset

Advanced Sunrise / Delayed Sunset

Because of atmospheric refraction, the Sun appears to rise about two minutes before actual sunrise and set about two minutes after actual sunset.

Advanced sunrise and delayed sunset by atmospheric refraction Apparent Sun Actual Sun (below horizon) Light bent towards Earth Fig. 15 – Atmospheric refraction bends sunlight so we see the Sun before it rises and after it sets

Even when the Sun is actually below the horizon, its light reaching the Earth passes through thicker layers of the atmosphere and is bent towards the Earth’s surface by atmospheric refraction, so the Sun appears higher in the sky than its true position. This makes us see the Sun a little before it actually rises and a little after it actually sets.

Same Principle

The apparent raised position of the Sun (or any star) is the same effect as a coin appearing raised when seen through a glass of water — it is atmospheric refraction.

16. Scattering of Light

Scattering

The process in which light is forced to deviate from its straight path and spread in different directions by small particles in the medium is called scattering of light.

Why the Sky is Blue

The air particles scatter light of shorter wavelengths (blue end of the spectrum) much more strongly than light of longer wavelengths (red end). Blue light is scattered in all directions by the atmosphere, so the sky appears blue when we look in any direction except directly at the Sun.

Why the Sun Appears Red at Sunrise and Sunset

At sunrise and sunset, sunlight travels through a much thicker layer of the atmosphere. Most of the blue and shorter wavelengths are scattered away, so the light that reaches us is dominated by the red (longer wavelength) component.

Summary

Day sky → blue (scattering of short wavelengths). Sun at noon → white (light travels a short path, little scattering). Sun at sunrise/sunset → reddish (long wavelengths reach us after scattering removes blue).

17. Important Definitions

TermDefinition
Near point (least distance of distinct vision)The closest distance at which a normal eye can see an object clearly — 25 cm.
Far pointThe farthest point up to which a normal eye can see objects clearly — infinity.
Power of accommodationThe ability of the eye lens to adjust its focal length by changing its curvature to focus objects at different distances.
DispersionThe splitting of white light into its constituent colours by a prism or water droplets.
SpectrumThe band of colours (VIBGYOR) obtained when white light is dispersed.
Atmospheric refractionThe bending of light due to the varying density of air in the Earth’s atmosphere.
ScatteringThe deflection of light into various directions by the particles of the medium.

18. Formulas and Must-Know Facts

The Lens Formula and Power

For a corrective spectacle lens we use the lens formula given by Newton’s lens equation. If a lens of focal length f produces an image at distance v of an object placed at distance u:

\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

The power of a lens is:

\[ P = \frac{1}{f\ \text{(in metres)}} \]

FormulaUsed for
\( P = \frac{100}{f\ \text{(in cm)}} \)Converting focal length (cm) to power (D)
\( f = \frac{100}{P} \) cmConverting power (D) to focal length (cm)
Myopia: \( f = -D \)Concave lens, D = distance of far point in metres
Power of myopic lens \( P = -\frac{1}{D} \)D in metres

Must-Know Facts

  • A normal eye focuses objects from the near point (25 cm) up to the far point (infinity).
  • The image formed on the retina is real and inverted; the brain makes it appear erect.
  • White light disperses into VIBGYOR; violet deviates the most, red the least.
  • A rainbow is formed by refraction + internal reflection + dispersion in water droplets.
  • Stars twinkle; planets do not twinkle (extended sources).
  • Sun is visible about 2 min before sunrise and 2 min after sunset.

19. Solved Examples

Example 1

A student cannot see objects clearly beyond 2 m. Identify the defect and find the power of the lens required to correct it.

Solution:

Condition: far point is 2 m (earlier than infinity) → Myopia. Use a concave lens with focal length \( f = -2\ \text{m} \). Power \( P = \frac{1}{f} = \frac{1}{-2} = -0.5\ \text{D} \).

Example 2

A person has a far point of 1 m. What lens is required?

Solution:

Myopia. \( f = -1\ \text{m} \), \( P = -1\ \text{D} \). A concave lens of power 1 D.

Example 3

A far point is 50 cm. Find the lens power.

Solution:

\( f = -0.5\ \text{m} \), \( P = -2\ \text{D} \), concave lens.

Example 4

Far point is 5 m. Find the lens power and type.

Solution:

\( f = -5\ \text{m} \), \( P = -0.2\ \text{D} \), concave lens.

Example 5

Far point is 3 m. Suggest the correction.

Solution:

\( P = -\frac{1}{3}\ \text{D} \approx -0.33\ \text{D} \), concave lens.

Example 6

A person’s near point is 1 m. Find the power of lens to correct hypermetropia.

Solution:

Object at \( u = -25\ \text{cm} \) must form image at \( v = -100\ \text{cm} \). Using \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \): \( \frac{1}{f} = \frac{1}{-100} - \frac{1}{-25} = \frac{3}{100} \), so \( f = \frac{100}{3}\ \text{cm} = 33.3\ \text{cm} \). In metres \( f = 0.333\ \text{m} \), \( P = +3\ \text{D} \) (convex lens).

Example 7

Near point is 40 cm. Find the correction power.

Solution:

\( u = -25\ \text{cm}, v = -40\ \text{cm} \). \( \frac{1}{f} = \frac{1}{-40} - \frac{1}{-25} = \frac{-5+8}{200} = \frac{3}{200} \), \( f = 66.7\ \text{cm} = 0.667\ \text{m} \), \( P = +1.5\ \text{D} \) (convex).

Example 8

Near point is 50 cm. Find the power.

Solution:

\( \frac{1}{f} = \frac{1}{-50} - \frac{1}{-25} = \frac{-1+2}{50} = \frac{1}{50} \), \( f = 50\ \text{cm} = 0.5\ \text{m} \), \( P = +2\ \text{D} \) (convex).

Example 9

Near point is 75 cm. Find the power.

Solution:

\( \frac{1}{f} = \frac{1}{-75} - \frac{1}{-25} = \frac{-1+3}{75} = \frac{2}{75} \), \( f = 37.5\ \text{cm} = 0.375\ \text{m} \), \( P \approx +2.67\ \text{D} \) (convex).

Example 10

A person with presbyopia needs a near point of 25 cm but has a near point of 125 cm. Find the reading power.

Solution:

\( u = -25\ \text{cm}, v = -125\ \text{cm} \). \( \frac{1}{f} = \frac{1}{-125} - \frac{1}{-25} = \frac{-1+5}{125} = \frac{4}{125} \), \( f = 31.25\ \text{cm} = 0.3125\ \text{m} \), \( P = +3.2\ \text{D} \) (convex).

Example 11

Which lens is used to correct myopia in a person whose far point is 10 m?

Solution:

Concave lens, \( P = -\frac{1}{10} = -0.1\ \text{D} \).

Example 12

A lens of power +3 D is prescribed. Is it convex or concave? What is its focal length?

Solution:

Positive power → convex. \( f = \frac{1}{3}\ \text{m} = 33.3\ \text{cm} \).

Example 13

A person needs a lens of power −2 D. What is the focal length and defect?

Solution:

Negative power → concave, myopia. \( f = -\frac{1}{2}\ \text{m} = -50\ \text{cm} \); far point = 50 cm.

Example 14

The far point of a myopic person is 80 cm. Find the lens power.

Solution:

\( f = -0.8\ \text{m} \), \( P = -1.25\ \text{D} \), concave lens.

Example 15

Explain why the pupil contracts in bright light.

Solution:

The iris narrows the pupil so that less light enters the eye, protecting the retina from excessive brightness.

Example 16

Why does the eye lens have a variable focal length?

Solution:

The ciliary muscles change the curvature (and hence focal length) of the lens — the larger the curvature, the smaller the focal length.

Example 17

A student sees clearly up to 4 m. Name the defect and the lens used.

Solution:

Myopia; concave lens of power \( -0.25\ \text{D} \).

Example 18

The spectrum of white light after a prism has red at one end and violet at the other. Which end deviates more?

Solution:

Violet deviates more; red deviates the least.

Example 19

Why do stars appear to twinkle but planets do not?

Solution:

Stars are point sources at a great distance, so atmospheric fluctuation is visible as twinkling. Planets are extended sources; the fluctuations average out.

Example 20

Why does the Sun appear red at sunrise?

Solution:

Sunlight travels through a thicker atmosphere layer; blue wavelengths scatter away, leaving red light to reach us.

Example 21

The sky appears dark from the surface of the Moon. Explain.

Solution:

There is no atmosphere on the Moon to scatter sunlight, so the sky appears dark.

Example 22

What is the function of the cornea?

Solution:

The transparent, curved cornea performs most of the refraction of light entering the eye and protects the eye.

Example 23

Where is the image of a distant object formed in a hypermetropic eye?

Solution:

The image of a near object forms behind the retina in a hypermetropic eye (its far vision is normal).

Example 24

A near point of 60 cm needs what power?

Solution:

\( \frac{1}{f} = \frac{1}{-60} - \frac{1}{-25} = \frac{-5+12}{300} = \frac{7}{300} \), \( f = 42.9\ \text{cm} = 0.429\ \text{m} \), \( P \approx +2.33\ \text{D} \) (convex).

Example 25

A person reads comfortably at 25 cm with a +2.5 D lens. What is their unaided near point?

Solution:

For hypermetropia, the lens must form the image of an object at 25 cm at the person’s near point N. \( f = \frac{1}{2.5}\ \text{m} = 40\ \text{cm} \). Using \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \) with \( u = -25\ \text{cm} \) and \( v = -N \): \( \frac{1}{40} = \frac{-1}{N} + \frac{1}{25} \), so \( \frac{1}{N} = \frac{1}{25} - \frac{1}{40} = \frac{3}{200} \), giving \( N = 66.7\ \text{cm} \). The near point is 66.7 cm.

20. Application-Based Questions

Q1.

Original — CBSE Exam-Style

Ravi reads on a moving bus. His eyes keep changing focus to avoid blurring. Name the property of the eye that enables this.

Answer:

The power of accommodation of the eye lens (ciliary muscles change the lens curvature).

Q2.

Original — CBSE Exam-Style

A doctor prescribes a reader +2.5 D for a 55-year-old. Explain why reading power, not distance vision, needed correction.

Answer:

With age the lens hardens and loses accommodation, so near objects (reading) blur — presbyopia. A convex lens (+ power) restores near focus. Far vision may still be normal.

Q3.

Original — CBSE Exam-Style

An optician gives a myopic boy −0.5 D lenses. Where is his far point?

Answer:

\( f = \frac{1}{0.5} = 2\ \text{m} \). Far point is 2 m.

Q4.

Original — CBSE Exam-Style

During a noisy, dust-filled construction site, a worker finds distant objects blurred but nearby ones fine. Identify the likely defect and corrective lens.

Answer:

Myopia; concave lens of suitable power.

Q5.

Original — CBSE Exam-Style

Why is the pupil black in colour though the iris is colourful?

Answer:

The pupil is an opening; light entering it is absorbed inside the eye, so very little light returns and it looks black.

Q6.

Original — CBSE Exam-Style

A chef sees oil films on water showing rainbow colours. Explain using concepts from this chapter.

Answer:

A thin oil film separates sunlight into colours by interference/thin-film effects; in essence white light gets split into colours when reflected, like dispersion.

Q7.

Original — CBSE Exam-Style

Suggest two reasons an astronaut in space sees the sky as black even on a sunny day.

Answer:

No atmosphere for scattering, and no atmosphere to refract light — so darkness in all directions except the disc of the Sun.

Q8.

Original — CBSE Exam-Style

At what time of day can you see your own shadow most clearly, and why?

Answer:

Near noon, because the Sun is overhead, light travels the shortest atmospheric path with least scattering, giving sharp shadows.

21. Competency-Based and Reasoning Questions

C1.

Original — CBSE Exam-Style

A person can see objects clearly only between 1 m and 4 m. Name the two likely defects and the combined lens needed.

Answer:

Far point 4 m → myopia (concave, \( -0.25\ \text{D} \)). Near point 1 m with normal accommodation decline → presbyopia/hypermetropia component (convex). Bifocal: upper concave for distance, lower convex for near.

C2.

Original — CBSE Exam-Style

Two students, one with normal vision and one wearing +2 D lenses, view a board. Which student’s eyes would focus a distant object with ciliary muscles relaxed?

Answer:

The student with normal vision — their lens is not accommodating for far objects. The +2 D wearer needs the lens (still accommodates up to infinity after correction). Far point corrected = infinity.

C3.

Original — CBSE Exam-Style

Given that red light deviates least in a prism, predict which colour would suffer maximum deviation if red were replaced by an even longer wavelength.

Answer:

Longer wavelength → even less deviation than red. (Wavelength inversely related to deviation.)

C4.

Original — CBSE Exam-Style

A rainbow is seen in the morning only in the west. Justify with a ray-level reasoning.

Answer:

We see a rainbow opposite the Sun. In the morning the Sun is in the east, so the bow appears in the west.

C5.

Original — CBSE Exam-Style

Compare the path of starlight through the atmosphere with the path of a ray passing through a water-filled glass slab. State one similarity and one difference.

Answer:

Similarity: both bend light (refraction). Difference: a glass slab causes a single lateral displacement with parallel emergent ray, while the atmosphere bends light gradually in a curve since density changes continuously.

C6.

Original — CBSE Exam-Style

If the Earth had no atmosphere, would the Sun appear white at noon and set abruptly with no red glow? Give reasoning.

Answer:

Yes. Without an atmosphere: no scattering (sky looks black, Sun stays white), and no atmospheric refraction (no advanced sunrise, no delayed sunset, no red glow at sunset).

22. Case-Based Learning

Case Study 1

Original — CBSE Exam-Style

Riya (14) playing cricket notices the ball blurs when it is very far. At school, the far board letters are clear, but far mountains are hazy.

Q. Name Riya’s defect and the type of lens she needs.

Answer:

Myopia; concave lens suitable to bring the far point to infinity.

Case Study 2

Original — CBSE Exam-Style

Grandfather (70) can’t read newspaper without glasses; he also needs distance glasses for driving.

Q. Which defect of vision is this, and which lens arrangement helps?

Answer:

Presbyopia (with existing myopia or hypermetropia); bifocal lenses (concave upper, convex lower).

Case Study 3

Original — CBSE Exam-Style

After a rainy evening, a student sees a full arc of colours across the sky while the Sun sets behind him.

Q. Name the phenomenon and the order of colours from the outer to inner edge.

Answer:

A rainbow (refraction + internal reflection + dispersion). Outer edge red, inner edge violet.

Case Study 4

Original — CBSE Exam-Style

At midnight a stargazer records stars flickering but Mars steady.

Q. Explain using atmospheric phenomena why Mars does not twinkle.

Answer:

Stars are point sources; fluctuations show as twinkling. Mars is an extended source; fluctuation from different points averages out, so no visible twinkling.

Case Study 5

Original — CBSE Exam-Style

A traveller sees the Sun’s upper edge above the horizon even while it is technically below the horizon.

Q. Name the phenomenon that lets this happen.

Answer:

Atmospheric refraction — light bending towards the Earth makes the Sun appear above the horizon. This gives advanced sunrise / delayed sunset.

23. Important Comparison Tables

1. Myopia vs. Hypermetropia

AspectMyopiaHypermetropia
Near objectsSeen clearlyBlurred
Distant objectsBlurredSeen clearly
CauseEye lens too curved / eyeball longEye lens too flat / eyeball short
Far pointCloser than infinityInfinity
Near point25 cm (normal)More than 25 cm
Image positionIn front of retinaBehind retina
Corrective lensConcaveConvex

2. Myopia vs. Presbyopia

AspectMyopiaPresbyopia
Age at onsetUsually in childhood/teenageOld age (natural ageing)
CauseAbnormal eye shape/lensHardening of lens, weak ciliary muscles
Near visionNormalBlurred
Corrective lensConcave (distance)Convex / bifocal

3. Concave vs. Convex Lens

AspectConcave lensConvex lens
ShapeThinner at centreThicker at centre
BehaviourDiverges parallel raysConverges parallel rays
Power signNegativePositive
CorrectsMyopiaHypermetropia, presbyopia

4. Near Point vs. Far Point

AspectNear pointFar point
MeaningClosest clearly visible distanceFarthest clearly visible distance
Normal value25 cmInfinity
Affected byHypermetropia, presbyopiaMyopia

5. Dispersion vs. Recombination

AspectDispersionRecombination
EffectWhite splits into coloursColours merge back to white
DeviceSingle prismSecond prism inverted
Result spectrumVIBGYOR bandWhite emergent light

6. Refraction vs. Scattering

AspectRefractionScattering
CauseChange in speed of light in a mediumDeflection by particles in a medium
ExamplePrism bending, coin raised in waterBlue sky, red sun at sunset
Wavelength dependenceSlightStrong (short wavelengths scatter more)

7. Twinkling Stars vs. Non-twinkling Planets

AspectStarsPlanets
Source typePoint sourceExtended source
Common distanceVery farNearer (share the Solar System)
TwinklingYesNo
ReasonAtmospheric fluctuation visibleFluctuations average out

24. Common Mistakes to Avoid

#Common mistakeWrite this instead
1Confusing which defect needs concave vs convexMyopia → concave; hypermetropia / presbyopia → convex
2Saying the eye forms an erect imageThe retina image is real and inverted
3Forgetting that near point is 25 cmNear point = 25 cm (least distance of distinct vision)
4Believing dispersion creates coloursColours were already present in white light
5Wrong order of spectrumViolet at the upper (more deviated) end, red at the lower
6Thinking a rainbow needs a single prismRainbow = water droplets (refraction + internal reflection + dispersion)
7Saying stars twinkle because they emit flickering lightThey twinkle due to atmospheric refraction of their light
8Thinking planets twinkle tooPlanets are extended sources; they do not visibly twinkle
9Mixing refraction and scattering in sunrise/sunsetRed-sun colour is scattering; early sunrise/late sunset timing is refraction
10Using power formula with cm instead of m\( P = 1/f \) only with f in metres

25. Exam Tips

  • Label the cornea, pupil, iris, eye lens, ciliary muscles, retina and optic nerve in ray diagrams — these get direct marks.
  • In numericals, write the sign convention (u negative, f positive for convex, negative for concave) before substituting.
  • Always answer defect questions in three parts: name the defect, cause, and corrective lens with power.
  • Quote textbook VALUES: near point 25 cm, far point infinity, prism disperses VIBGYOR, violet deviates most.
  • For 2-mark explanations about the sky/red sunset, connect wavelength to scattering rather than stating just the observation.
  • Practise at least 5 spectrum-order and 5 lens-power problems before the exam.
  • Write the rainbow formation in three clear steps (refraction → internal reflection → dispersion/emergence).

26. Quick Revision

5–10 Minute Revision

Go through these points just before the exam:

  • Human eye: natural optical instrument; image forms real and inverted on the retina; brain interprets it erect.
  • Accommodation: ciliary muscles change lens curvature/focal length to focus near and far objects.
  • Range of vision: near point 25 cm to far point infinity.
  • Myopia: distant blurred, image before retina, concave lens.
  • Hypermetropia: near blurred, image behind retina, convex lens.
  • Presbyopia: age-related loss of accommodation; convex/bifocal to read.
  • Prism: light bends towards base; emergent ray deviated from incident ray.
  • Dispersion: white → VIBGYOR; violet deviates most, red least.
  • Recombination: inverted second prism recombines colours to white.
  • Rainbow: droplets act as prisms; seen opposite to the Sun; red outer, violet inner.
  • Atmospheric refraction: stars twinkle; advanced sunrise; delayed sunset; Sun appears 2 min early/late.
  • Scattering: blue sky; red Sun at sunrise/sunset; dark sky on the Moon.

27. Self-Practice Questions

1.

MCQ

The image of an object formed on the retina of the human eye is (a) virtual and erect (b) real and inverted (c) virtual and inverted (d) real and erect.

2.

MCQ

A person having an axially long eyeball suffers from (a) hypermetropia (b) myopia (c) presbyopia (d) astigmatism.

3.

MCQ

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

4.

MCQ

Which colour deviates the most when white light passes through a glass prism? (a) Red (b) Orange (c) Violet (d) Green.

5.

MCQ

The phenomenon owing to which the sky appears blue is (a) refraction (b) dispersion (c) scattering (d) reflection.

6.

MCQ

A myopic person’s far point is 1 m. The power of the corrective lens is (a) +1 D (b) −1 D (c) +0.5 D (d) −2 D.

7.

MCQ

Which of the following does NOT involve refraction of light? (a) Twinkling of stars (b) Advanced sunrise (c) Blue colour of the sky (d) Coin appearing raised in water.

8.

MCQ

A hypermetropic person needs a lens of power (a) negative (b) positive (c) zero (d) any power.

9.

MCQ

The band of seven colours obtained from white light is called (a) reflection (b) spectrum (c) scattering (d) deviation.

10.

MCQ

Stars appear higher than their true position because of (a) dispersion (b) atmospheric refraction (c) scattering (d) total internal reflection.

11.

VSA

State the function of the iris.

12.

VSA

What is the far point of a normal human eye?

13.

VSA

Which part of the eye changes the focal length of the eye lens?

14.

VSA

Name the defect in which the image is formed in front of the retina.

15.

VSA

Why does the sky appear dark to an astronaut in space?

16.

SA I

State the cause and correction of hypermetropia.

17.

SA I

Why do stars twinkle but planets do not?

18.

SA I

Why does the Sun appear red at sunrise and sunset although it is white at noon?

19.

SA I

Draw the path of light refracted through a glass prism and label the angle of deviation.

20.

SA I

What is power of accommodation? State its normal range in the human eye.

21.

SA II

A person’s far point is 4 m. Find the power of the correcting lens.

22.

SA II

A person’s near point is 100 cm. Find the power and type of corrective lens.

23.

SA II

Describe with a diagram the formation of a rainbow.

24.

SA II

Distinguish between myopia and hypermetropia in a table (4 points).

25.

SA II

Explain advanced sunrise and delayed sunset with the help of atmospheric refraction.

26.

Reasoning

A concave lens of power −0.25 D is given. Locate the far point of the eye it corrects.

27.

Reasoning

If violet deviates the most, would a violet object placed at the base of a prism appear higher? Give a one-line reason.

28.

Reasoning

Why does the pupil contract in bright light?

29.

Reasoning

Two identical prisms are placed in a spectrometer: one upright, one inverted. Predict the emergent light.

30.

Reasoning

During a partial eclipse, why does the sky not turn completely dark in the shadow?

28. Answer Key and Hints

QAnswer / Hint
1(b) real and inverted
2(b) myopia
3(c) 25 cm
4(c) violet
5(c) scattering
6(b) −1 D
7(c) Blue colour of the sky (it is scattering)
8(b) positive
9(b) spectrum
10(b) atmospheric refraction
11Controls size of pupil / amount of light entering the eye
12Infinity
13Ciliary muscles
14Myopia
15No atmosphere in space to scatter light
16Cause: eye lens too flat or eyeball too short; correction: convex lens
17Stars are point sources; twinkling due to atmospheric refraction. Planets are extended sources; averages out.
18Longer atmospheric path; short wavelengths scatter away; red reaches us
19Refer Fig. 7; emergent ray deviated towards base; angle D between incident and emergent rays
20Ability of lens to change focal length by changing curvature; from near point 25 cm to infinity
21\( f = -4\ \text{m} \), \( P = -0.25\ \text{D} \), concave
22\( u = -25 \), \( v = -100\ \text{cm} \); \( f = 100/3\ \text{cm} \); \( P = +3\ \text{D} \), convex
23Refer section 12: refraction ← droplets split and reflect light (with diagram)
24See comparison table 1
25Atmosphere bends light towards surface; Sun appears above horizon ~2 min before/after
26\( f = -4\ \text{m} \); far point = 4 m
27Yes, higher apparent position corresponds to greater deviation
28Iris narrows pupil to reduce light entry, protecting the retina
29White light: the second inverted prism recombines the spectrum
30Scattering + refraction continue; partial light still reaches the shadowed region
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