By the end of this chapter you'll be able to…

  • 1State the least distance of distinct vision and how it varies with age, and describe Activity 1
  • 2Define the angle of vision and explain why only part of a very close object is visible
  • 3Label the parts of the human eye and state the function of cornea, iris, pupil, lens, ciliary muscles and retina
  • 4Explain accommodation, and why a fixed image distance of 2.5 cm requires a variable focal length
  • 5Calculate f max and f min for the eye lens using the lens formula
  • 6Define far point and near point, and explain myopia and hypermetropia in terms of the focal length range
  • 7Derive the focal length of the correcting lens for myopia and for hypermetropia
  • 8Explain presbyopia and why a bi-focal lens corrects it
  • 9Define the power of a lens and convert between power and focal length
  • 10Define the four prism angles and describe the pin-and-chart lab activity including the deviation graph
  • 11Derive n = sin[(A+D)/2] / sin(A/2) and apply it numerically
  • 12Explain dispersion using wave theory and say why ray theory fails
  • 13Explain the rainbow, including why it is a cone and why one drop gives one colour
  • 14Explain scattering and apply it to the blue sky, the white sky on hot days, and red sunrise and sunset
💡
Why this chapter matters
This chapter takes the abstract lens formula of Chapter 4 and makes it concrete by applying it to the reader's own eye, deriving the 2.5 cm to 2.27 cm working range and then showing that every defect of vision is a failure to stay inside it. That single framing turns myopia, hypermetropia and presbyopia from three memorised definitions into one idea. The second half explains things every student has seen — a rainbow, a blue sky, a red sunset — and does so by admitting that ray theory fails and wave theory is needed. Written from the SCERT Telangana official 2026 Class 10 Physical Science textbook, pages 172-222.

Human Eye and Colourful World

1. What This Chapter Covers

This chapter takes the lens theory of Chapter 4 and turns it on the eye. It is allotted 10 periods in August, the most of any chapter in this volume, and runs from textbook page 172 to page 222.

The questions it sets out to answer are listed at the start:

  • What is the function of the lens in the human eye?
  • How does it help us see objects both far away and close by?
  • How is it possible to get the image at the same distance on the retina every time?
  • Are we able to see all objects in front of the eye clearly?
  • How do spectacle lenses correct defects of vision?

The second half turns outward, to the prism, the rainbow and the colour of the sky. The connecting idea is that a single physical quantity — refractive index, and how it varies with wavelength — explains all of it.

2. Least Distance of Distinct Vision (Textbook 5.1)

Activity 1

Hold a textbook in front of you and read it. Slowly bring it towards your eyes until it is very close. The printed letters blur, or you feel strain in the eye.

Now move the book slowly back to the position where you can read clearly without straining. Have a friend measure that distance. Repeat with several friends and average the values.

You will get about 25 cm. This is the least distance of distinct vision.

The chapter is careful that this is not a universal constant:

Age groupLeast distance of distinct vision
Below about 10 years7 to 8 cm, since the eye muscles are strong and flexible
A healthy adultAbout 25 cm
Old ageAbout 1 to 2 m or even more, as muscles cannot sustain strain

Activity 2 — the angle of vision

Prepare sticks or PVC pipes of 20, 30, 35, 40 and 50 cm. Place a retort stand on a table and stand so that your head is beside the vertical rod. Fix the clamp on the horizontal rod 25 cm from your eyes, and have a friend fix a 30 cm stick vertically in the clamp.

Keeping your vision parallel to the horizontal rod, try to see the top and bottom of the stick simultaneously, without moving your eyes. Then swap in the other sticks one by one without moving the clamp.

What the activity shows

Rays from the extreme ends of an object form an angle at the eye. If that angle is below 60 degrees you can see the whole object. If it is above 60 degrees you see only part of it.

That maximum angle is the angle of vision, about 60 degrees for a healthy human being, and like the least distance it varies with person and age.

So when an object AB is moved closer, to A′B′, you no longer see all of it — only the portion EF whose rays can still enter your eye.

3. Structure of the Human Eye (Textbook 5.2)

The eyeball is nearly spherical. Working inward from the front:

PartFunction
CorneaTransparent protective membrane over the sharply curved front; the part visible from outside
Aqueous humourLiquid filling the space behind the cornea
IrisMuscular diaphragm; the coloured part of the eye; controls the light entering
PupilThe small hole in the iris, acting as a variable aperture
Crystalline lensResponsible for image formation; hard in the middle, softer towards the edge
Ciliary musclesAttached to the lens; change its radii of curvature
RetinaRear part of the eyeball, where the image forms
Optic nerveCarries signals to the brain

The pupil appears black because light falling on it goes into the eye and has almost no chance of coming back out.

In low light the iris makes the pupil expand to let more light in; in bright light it makes the pupil contract to keep excess light out.

The fixed image distance

The distance between the lens and the retina is about 2.5 cm, and it is fixed. Whatever the object distance, the image distance cannot change.

From Chapter 4 you know that for different object positions the image distance stays constant only if the focal length changes. So the eye lens must be able to change its shape — and the ciliary muscles do exactly that, by changing the lens's radii of curvature.

  • Focusing a distant object: the ciliary muscles relax, the focal length takes its maximum value, equal to the lens-to-retina distance.
  • Focusing a closer object: the ciliary muscles strain, and the focal length decreases.

This process is called accommodation. The muscles cannot strain beyond a limit, which is precisely why there is a least distance of distinct vision.

What the retina does

The eye lens forms a real and inverted image on the retina. The retina is a delicate membrane containing about 125 million receptors called rods and cones — cones identify colour, rods identify the intensity of light. Signals travel to the brain through about 1 million optic nerve fibres, and the brain interprets them so that we perceive shape, size and colour.

4. The Range of the Eye Lens

The chapter now does something neat: it applies the lens formula to the eye itself.

Maximum focal length. For an object at infinity, u = −infinity and v = 2.5 cm:

1/f = 1/v − 1/u = 1/2.5 + 0, so f_max = 2.5 cm

Minimum focal length. For an object at the least distance of distinct vision, u = −25 cm and v = 2.5 cm:

1/f = 1/2.5 + 1/25 = 11/25, so f_min = 25/11 = 2.27 cm

So for any object between infinity and 25 cm, the eye lens adjusts its focal length between 2.5 cm and 2.27 cm. That narrow band is the eye's whole working range, and the next section is simply what happens when an eye cannot stay inside it.

5. Defects of Vision (Textbook 5.3)

There are three common defects.

Myopia, or near sightedness (5.3.1)

A person can see nearby objects clearly but not distant ones. Here the maximum focal length is less than 2.5 cm, so rays from a distant object form an image before the retina.

The farthest point from which an object appears clear is the far point (M), defined as the point of maximum distance at which the eye lens can form an image on the retina. Myopia is the defect in which people cannot see objects beyond the far point.

Correction. Bring the image of a distant object to lie between the far point and the least distance of distinct vision, so that it can act as an object for the eye lens. This requires a bi-concave lens.

To find its focal length, the lens must form an image at the far point for an object at infinity. With u = −infinity and v = −D where D is the far-point distance:

1/f = 1/(−D), so f = −D

The negative sign confirms it is a concave lens.

Hypermetropia, or far sightedness (5.3.2)

A person can see distant objects clearly but not nearby ones, because the minimum focal length is greater than 2.27 cm. Rays from a nearby object form an image beyond the retina.

The closest point at which the eye can form an image on the retina is the near point (H). Someone with hypermetropia cannot see objects between the near point and the least distance of distinct vision.

Correction. Use a lens that takes an object at L, the least distance of distinct vision, and forms its image at the near point. This requires a bi-convex lens.

With u = −25 cm and v = −d where d is the near-point distance:

1/f = 1/(−d) − 1/(−25) = 1/25 − 1/d

f = 25d/(d − 25) in centimetres

Since d > 25 cm, f comes out positive, confirming a biconvex lens is needed.

Presbyopia (5.3.3)

Here the power of accommodation decreases with ageing and the near point gradually recedes. The cause is the gradual weakening of the ciliary muscles and the diminishing flexibility of the eye lens.

With age a person may suffer from both myopia and hypermetropia. The correction is a bi-focal lens, with the upper portion concave and the lower portion convex.

Power of a lens (5.3.4)

The degree of convergence or divergence a lens can achieve is its power, defined as the reciprocal of focal length:

P = 1/f with f in metres, or P = 100/f with f in centimetres

The unit is the dioptre, denoted D.

Example 1. A doctor advises a 2D lens. What is its focal length? Using P = 100/f, 2 = 100/f, so f = 50 cm.

Every defect of vision is the eye lens leaving its working range Healthy eye: f adjusts between 2.27 cm and 2.50 cm Myopia f max is LESS than 2.5 cm Image of a distant object falls BEFORE retina Cannot see beyond the far point M Hypermetropia f min is MORE than 2.27 cm Image of a near object falls BEYOND retina Cannot see inside the near point H Corrected by a bi-concave lens, f = -D Corrected by a bi-convex lens, f = 25d/(d-25) Presbyopia: accommodation fails with age Bi-focal lens, concave above and convex below Power P = 1/f in metres, or 100/f in centimetres, measured in dioptres (D)

The three defects read as one idea: the eye lens can no longer keep its focal length inside the 2.27 to 2.50 cm band that a fixed 2.5 cm image distance demands.

6. The Prism (Textbook 5.4.1)

A prism is a transparent medium separated from its surroundings by at least two plane surfaces inclined at an angle, such that light entering one surface emerges from the other. A triangular glass prism has two triangular bases and three rectangular lateral surfaces.

The terms, using triangle PQR with the ray entering at M and leaving at N:

TermDefinition
Angle of incidence (i₁)Between the incident ray and the normal at M
Angle of emergence (i₂)Between the emergent ray and the normal at N
Angle of the prism (A)Between the plane surfaces PQ and PR; also called the refracting angle
Angle of deviation (d)Between the incident ray and the emergent ray

Lab activity — finding the refractive index of a prism

Materials. Prism, a 20 by 20 cm white chart, pencil, pins, scale and protractor.

Procedure. Place the prism on the chart with its triangular base down and draw its outline. Remove it and name the vertices P, Q and R — for many prisms the triangle is equilateral. Measure the angle between PQ and PR, which is the angle of the prism A.

Mark M on PQ, draw the normal there, and using the protractor mark an angle of 30 degrees to give the incident ray. Replace the prism and fix two pins vertically at A and B on that line. Looking through PR from the other side, fix two more pins at C and D so that all four appear to lie in a straight line.

Remove the prism and pins, and join the pin holes to PR to get the emergent ray at N. Measure the angle of emergence. Join M and N; the path through A, B, M, N, C and D is the complete path of the light.

To get the angle of deviation, extend both the incident and emergent rays until they meet at O and measure the angle between them.

Repeat for angles of incidence of 40 degrees, 50 degrees and so on.

The graph and minimum deviation

Plotting angle of incidence on the x-axis against angle of deviation on the y-axis gives a curve that first decreases and then increases.

Draw a tangent to the lowest point of the curve, parallel to the x-axis. Where it cuts the y-axis is the angle of minimum deviation, D. Dropping a line from the point of tangency to the x-axis gives the corresponding angle of incidence — and if you repeat the experiment at that angle, you find the angle of emergence equals the angle of incidence.

7. Refractive Index of a Prism (Textbook 5.4.2)

Step 1. From triangle OMN, the deviation is the sum of the two bends:

d = (i₁ − r₁) + (i₂ − r₂) = (i₁ + i₂) − (r₁ + r₂) ..... (1)

Step 2. From triangle PMN, A + (90 − r₁) + (90 − r₂) = 180, which simplifies to

r₁ + r₂ = A ..... (2)

Step 3. Combining (1) and (2):

A + d = i₁ + i₂ ..... (3)

This is the general relation between the four angles.

Step 4. Snell's law at M, with air outside, gives sin i₁ = n sin r₁ ..... (4), and at N gives n sin r₂ = sin i₂ ..... (5).

Step 5. At minimum deviation, i₁ = i₂, and the ray MN runs parallel to the base QR. Equation (3) becomes A + D = 2i₁, so

i₁ = (A + D)/2

Since i₁ = i₂, it follows that r₁ = r₂, so equation (2) gives 2r₁ = A, and

r₁ = A/2

Step 6. Substituting both into equation (4):

n = sin[(A + D)/2] / sin(A/2)

Example 2. A prism with A = 60 degrees gives a minimum deviation of 30 degrees. Then

n = sin(45) / sin(30) = (1/root2) / (1/2) = root2

So the refractive index is root 2.

8. Dispersion of Light (Textbook 5.4.3)

Activities 3 and 4

Activity 3. In a dark room, fix a thin wooden plank with a small hole vertically on a table, and place a prism between the plank and a white wall. A white light source behind the hole gives a narrow beam. Adjust the prism until you get an image on the wall. You see a band of colours.

Activity 4. Fill a metal tray with water and place a mirror in it at an angle to the water surface. Focus white light on the mirror through the water, and catch the result on a white card held above the surface. Again, colours.

The splitting of white light into different colours — VIBGYOR — is called dispersion.

The angle of deviation is minimum for red and maximum for violet.

Why ray theory cannot explain it

Here the chapter makes an unusually honest move. For a given refractive index there can be only one angle of minimum deviation, and by Fermat's principle light always takes the path of least time. Yet in activity 3 the light clearly took several different paths at once.

So ray theory is ruled out. White light must instead be treated as a collection of waves of different wavelengths — violet the shortest, red the longest. Light is an electromagnetic wave in which no particle physically oscillates; instead the electric and magnetic field magnitudes vary periodically at every point.

The actual explanation

The speed of light in vacuum is the same for all colours, but in a medium it depends on wavelength. Since refractive index is the ratio of the speed in vacuum to the speed in the medium, refractive index depends on wavelength. Each colour then takes its own least-time path, refracting to a different extent, and the colours separate.

Experimentally, refractive index decreases as wavelength increases. Red has the longest wavelength, so the lowest refractive index, so the least deviation — which is exactly what was observed.

Why a single colour does not split further

Frequency is a property of the source — the number of waves leaving it per second — and no medium can change it. At an interface, the number of waves arriving per second must equal the number passing any point beyond it, so frequency is unaltered by refraction while wavelength changes with the medium.

Since v = (frequency) × (wavelength) and frequency is fixed, v is proportional to wavelength. A coloured light therefore retains its colour through any transparent medium, and cannot split further.

9. The Rainbow

Activity 5 — making one

Stand facing a white wall lit by the sun, with the sun on your back. Hold a tube with water flowing through it and partly block the opening with your finger so the water sprays out like a fountain. Colours appear on the wall.

Inside a single drop

Sunlight enters a water drop near its top surface. At this first refraction the white light disperses, violet deviating most and red least. Reaching the far side of the drop, each colour is reflected back into the drop by total internal reflection. Arriving at the surface again, each colour is refracted into the air — and at this second refraction the angle between red and violet increases further.

The angle between the incoming and outgoing rays can be anything from 0 up to about 42 degrees, and a bright rainbow is seen near that maximum.

Why you see only one colour per drop

Although every drop disperses a full spectrum, an observer sees only a single colour from any one drop. If violet from a drop reaches your eye, the red from that same drop passes below your eye. To see red you must look at a drop higher in the sky.

  • Red is seen when the angle between the sunbeam and the returned light is 42 degrees.
  • Violet is seen at 40 degrees.
  • Between 40 and 42 degrees you see the remaining colours of VIBGYOR.

Why it is a bow

A rainbow is not a flat two-dimensional arc. It is a three-dimensional cone with its tip at your eye, made of layers. The drops sending red to your eye lie on the outermost layer of the cone, orange on the layer beneath, and so on inwards, with violet innermost.

10. Scattering of Light (Textbook 5.4.4)

The mechanism

Atoms and molecules exposed to light absorb light energy and re-emit part of it in different directions. That is scattering.

An atom or molecule responds to light when its size is comparable to the wavelength. Small particles are affected by higher frequency, shorter wavelength light, and larger ones by longer wavelengths. When the condition is met, the atom absorbs the light and vibrates, re-emitting a fraction of the energy in all directions with different intensities. The re-emitted light is scattered light and the atoms are scattering centres.

The angle between the incident light and the direction of observation is the angle of scattering, and the intensity of scattered light is maximum at 90 degrees.

Why the sky is blue

The atmosphere contains N₂ and O₂ molecules whose sizes are comparable to the wavelength of blue light, so they act as scattering centres for blue.

This also explains why the sky looks clearest blue when you look perpendicular to the direction of the sun's rays — that is the 90 degree maximum. Change your angle of view and the intensity of the blue changes with it.

Why the sky sometimes looks white

On a hot day, water vapour enters the atmosphere, so water molecules become abundant. A water molecule is larger than N₂ or O₂, so it scatters lower frequencies than blue. All those other colours reach your eye together, and the sky appears white.

Activity 6 — watching it happen

Take sodium thiosulphate (hypo) and sulphuric acid in a glass beaker and place it in abundant sunlight. Sulphur precipitates as the reaction proceeds, and the grains grow with time.

At first the grains are small, comparable to the wavelength of blue light, and appear blue. As they grow their size becomes comparable to the wavelengths of other colours too, so they scatter those as well, and the combination appears white.

Why sunrise and sunset are red

Molecules of a size comparable to the wavelength of red light are few in the atmosphere, so red is scattered least.

At sunrise and sunset, sunlight must travel a longer distance through the atmosphere to reach you. Every colour except red scatters away before arriving; red survives the journey, so the sun appears red.

At noon the path through the atmosphere is much shorter, so all colours reach your eye without much scattering and the sun appears white.

The Raman effect

Sir C.V. Raman, a Nobel Prize winner, explained light scattering in solids, gases and liquids. He found experimentally that the frequency of light scattered by liquids differs from the frequency of the incident light. This is the Raman effect, and scientists use it to determine the shapes of molecules.

Key words from the chapter

Least distance of distinct vision, angle of vision, accommodation of eye lens, myopia, hypermetropia, presbyopia, power of lens, prism, angle of prism or refracting angle of prism, angle of minimum deviation, dispersion, scattering.

11. Summary

The least distance of distinct vision is about 25 cm and the angle of vision about 60 degrees, both varying with person and age. The eye forms a real, inverted image on the retina, whose distance from the lens is fixed at about 2.5 cm, so focusing must be done by changing focal length. The ciliary muscles do this by altering the lens's curvature, a process called accommodation.

Applying the lens formula gives the working range: f_max = 2.5 cm for an object at infinity and f_min = 2.27 cm for an object at 25 cm. Each defect of vision is a failure to stay within that range.

Myopia has f_max below 2.5 cm, images of distant objects forming before the retina, corrected by a bi-concave lens of focal length −D. Hypermetropia has f_min above 2.27 cm, images of near objects forming beyond the retina, corrected by a bi-convex lens of focal length 25d/(d − 25). Presbyopia is the loss of accommodation with age, corrected by a bi-focal lens. Lens power is 1/f, in dioptres.

For a prism, A + d = i₁ + i₂ in general, and at minimum deviation, where the emergent ray is parallel to the base and i₁ = i₂, the refractive index is n = sin[(A + D)/2] / sin(A/2).

Dispersion cannot be explained by ray theory, since a single refractive index would give a single path. Treating light as waves of different wavelengths resolves it: refractive index depends on wavelength and decreases as wavelength increases, so red deviates least and violet most.

Frequency is fixed by the source and unchanged by refraction, which is why a single colour never splits further. The rainbow results from refraction, total internal reflection and a second refraction inside each drop, seen as a cone with red at 42 degrees on the outside and violet at 40 degrees inside.

Scattering is absorption and re-emission by particles whose size is comparable to the wavelength. N₂ and O₂ scatter blue, making the sky blue, with maximum intensity at 90 degrees to the sunlight. Larger water molecules on hot days scatter other colours and make the sky look white. At sunrise and sunset the long atmospheric path scatters everything but red away, so the sun appears red, while at noon the short path lets all colours through and it appears white.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Working range of the eye lens
f max = 2.5 cm for an object at infinity; f min = 25/11 = 2.27 cm for an object at 25 cm
Both follow from the lens formula with the fixed image distance v = 2.5 cm
Correcting lens for myopia
f = -D, where D is the distance of the far point
Object at infinity, image at the far point; the negative sign confirms a bi-concave lens
Correcting lens for hypermetropia
f = 25d/(d - 25) in centimetres, where d is the distance of the near point
Object at 25 cm, image at the near point; f is positive since d is greater than 25, so a bi-convex lens
Power of a lens
P = 1/f with f in metres, or P = 100/f with f in centimetres
Measured in dioptres, denoted D
General prism relation
A + d = i1 + i2, and r1 + r2 = A
From triangles OMN and PMN respectively
Refractive index of a prism
n = sin[(A + D)/2] / sin(A/2)
Valid at minimum deviation, where i1 = i2, r1 = r2 and the ray inside is parallel to the base
Dispersion rule
refractive index decreases as wavelength increases
Red has the longest wavelength, so the lowest refractive index and the least deviation; violet the most
Rainbow angles
red at 42 degrees, violet at 40 degrees between the sunbeam and the returned light
Red on the outer layer of the cone, violet innermost
⚠️

Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT
✗ Thinking the image distance in the eye changes with object distance
✓ The lens-to-retina distance is fixed at about 2.5 cm. It is the focal length that changes, through accommodation by the ciliary muscles altering the lens curvature. Getting this the wrong way round makes the whole 2.5 to 2.27 cm calculation impossible.
WATCH OUT
✗ Confusing far point with the least distance of distinct vision
✓ The far point is the maximum distance at which the eye can still form an image on the retina, and it is what myopia limits. The least distance of distinct vision, about 25 cm, is the nearest comfortable distance. Hypermetropia is described using the near point, a third and separate quantity.
WATCH OUT
✗ Assigning the wrong corrective lens to each defect
✓ Myopia, where the image falls before the retina, needs a diverging bi-concave lens. Hypermetropia, where the image falls beyond the retina, needs a converging bi-convex lens. Check with the sign of f: the myopia derivation gives f = -D, which is negative and therefore concave.
WATCH OUT
✗ Describing presbyopia as just a severe form of hypermetropia
✓ Presbyopia is the loss of the power of accommodation itself, from weakening ciliary muscles and a less flexible lens, and it may occur together with myopia in the same person. That is exactly why a bi-focal lens is needed, concave above and convex below, rather than a single convex lens.
WATCH OUT
✗ Using the prism refractive index formula away from minimum deviation
✓ n = sin[(A+D)/2] / sin(A/2) is derived on the assumptions i1 = i2 and r1 = r2, which only hold at minimum deviation. Away from it, only the general relation A + d = i1 + i2 applies.
WATCH OUT
✗ Saying refractive index increases with wavelength
✓ It decreases. Red has the longest wavelength and the lowest refractive index, so it deviates least; violet has the shortest wavelength and deviates most. Getting this backwards reverses the whole colour order of the spectrum and the rainbow.
WATCH OUT
✗ Claiming a single colour splits further on passing through another prism
✓ It does not. Frequency is a property of the source and is unchanged by refraction, so a coloured light retains its colour through any transparent medium. Only the wavelength and speed change with the medium.
WATCH OUT
✗ Thinking each raindrop sends you the whole spectrum
✓ Each drop does disperse a full spectrum, but only one colour from any given drop reaches your eye. If violet from a drop reaches you, its red passes below your eye, so you must look at a higher drop to see red. This is why the rainbow is spread out across the sky rather than appearing in each drop.
WATCH OUT
✗ Explaining a white sky by saying there is no scattering
✓ There is more scattering, not less. On hot days abundant water molecules, being larger than N2 and O2, scatter the longer wavelengths as well. All those colours arriving together make the sky look white.

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Human Eye and Colourful World?

10 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

10 questions~7 min

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

  • •Least distance of distinct vision about 25 cm; 7 to 8 cm in young children, 1 to 2 m in old age
  • •Angle of vision about 60 degrees; beyond it only part of the object is seen
  • •Cornea, aqueous humour, iris, pupil, crystalline lens, ciliary muscles, retina, optic nerve
  • •Pupil appears black because light entering it does not come back out
  • •Iris makes the pupil a variable aperture: expands in low light, contracts in bright light
  • •Lens-to-retina distance fixed at about 2.5 cm
  • •Accommodation: ciliary muscles change the lens curvature and hence focal length
  • •Retina has about 125 million rods and cones; cones for colour, rods for intensity; about 1 million optic nerve fibres
  • •f max = 2.5 cm, f min = 2.27 cm
  • •Myopia: f max below 2.5 cm, image before retina, cannot see beyond far point, corrected by bi-concave lens with f = -D
  • •Hypermetropia: f min above 2.27 cm, image beyond retina, cannot see inside near point, corrected by bi-convex lens with f = 25d/(d-25)
  • •Presbyopia: accommodation weakens with age; corrected by a bi-focal lens, concave above and convex below
  • •Power P = 1/f in metres, unit dioptre
  • •Prism terms: angle of incidence, angle of emergence, angle of the prism A, angle of deviation d
  • •A + d = i1 + i2 and r1 + r2 = A
  • •At minimum deviation i1 = i2, r1 = r2, and the internal ray is parallel to the base
  • •n = sin[(A + D)/2] / sin(A/2)
  • •Dispersion splits white light into VIBGYOR; red deviates least, violet most
  • •Refractive index decreases as wavelength increases
  • •Frequency is fixed by the source and unchanged by refraction, so a colour never splits further
  • •Rainbow: refraction, total internal reflection, second refraction; red at 42 degrees, violet at 40 degrees; a cone, not a flat arc
  • •Scattering needs particle size comparable to wavelength; intensity maximum at 90 degrees
  • •N2 and O2 scatter blue, giving a blue sky; water molecules on hot days make it white
  • •Red scatters least, so sunrise and sunset look red while noon looks white
  • •Raman effect: frequency of light scattered by liquids differs from the incident frequency

Telangana (TSBIE) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: No marks distribution is printed in the textbook for this chapter, so no total is claimed. The index gives 10 periods in August, the largest allotment in this volume. The categories below follow the book's own question structure; the marks column indicates question size rather than official weightage.

Question typeMarks eachTypical countWhat it tests
Reflections on concepts25Definitions of least distance of distinct vision, angle of vision and accommodation, parts of the eye, and the deviation graph
Application of concepts24Numerical work with the lens formula on the eye, power and focal length conversions, and the prism refractive index
Higher Order Thinking Questions44Why ray theory fails for dispersion, rainbow geometry, and scattering applied to sky and sun colours
Prep strategy
  • Learn the 2.5 cm and 2.27 cm calculation first, then define each defect as a departure from that range
  • Draw one labelled diagram of the eye and use it to answer every structure question
  • Practise the prism derivation, paying attention to where i1 = i2 is assumed
  • Keep a single table of red versus violet: wavelength, refractive index, deviation and rainbow angle
  • For scattering questions, always start from the rule that particle size must be comparable to wavelength

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Reading a spectacle prescription

Reading a spectacle prescription, where the power in dioptres tells you both the lens type and its focal length

Understanding why older people hold reading material furt…

Understanding why older people hold reading material further away, and why bi-focal lenses exist

Recognising why a doctor tests distance vision and near v…

Recognising why a doctor tests distance vision and near vision separately, since myopia and hypermetropia affect different ranges

Explaining rainbows

Explaining rainbows, and predicting where to stand to see one with the sun behind you

Explaining the blue sky

Explaining the blue sky, the white sky on humid days, and red sunrises and sunsets

The Raman effect

The Raman effect, used by scientists to determine the shapes of molecules

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

1
Start any defect-of-vision answer by naming which end of the 2.27 to 2.5 cm range has failed, then the lens that fixes it
2
Draw and label the eye diagram even when the question does not explicitly ask for one, as it supports most structure answers
3
In the prism derivation, state the minimum deviation condition before using it, since that is where the marks sit
4
For dispersion and scattering questions, name the physical rule first, refractive index versus wavelength or particle size versus wavelength, then apply it
5
Quote the rainbow angles 42 for red and 40 for violet precisely, as these are frequently the marked detail

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH
Derive the condition for minimum deviation formally by differentiating the deviation with respect to the angle of incidence
STRETCH
Work out why a secondary rainbow appears above the primary one with the colours reversed, using two internal reflections instead of one
STRETCH
Investigate the Rayleigh scattering dependence on the inverse fourth power of wavelength and check it against the chapter's qualitative account
STRETCH
Calculate the dispersive power of a prism from the refractive indices for red and violet, and relate it to achromatic prism combinations
STRETCH
Research how the Raman effect distinguishes molecular shapes, and why the scattered frequency differs from the incident one

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

Telangana SSC public examination — Physical Science paper, eye defects, prism numericals and scattering questions
Polytechnic and residential-school entrance tests in Telangana
NTSE and science olympiad screening papers, where prism and dispersion problems are standard

Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Because the image distance cannot change. The retina sits at a fixed distance of about 2.5 cm behind the lens, so the only free variable is the focal length. The ciliary muscles change the lens's radii of curvature to adjust it, which is what accommodation means.

For a healthy eye they coincide at about 25 cm. For a person with hypermetropia the near point has moved further out, so there is a gap between the near point and 25 cm in which objects cannot be seen clearly. The correcting lens is chosen precisely to bridge that gap, taking an object at 25 cm and imaging it at the near point.

The derivation assumes i1 = i2, which is true only at minimum deviation, where the ray inside the prism runs parallel to the base. That assumption then forces r1 = r2, which lets r1 be written as A/2 and i1 as (A + D)/2. At any other angle of incidence only the general relation A + d = i1 + i2 holds.

Geometry. Red leaves a drop at 42 degrees to the incoming sunlight and violet at 40 degrees, so only one of them can be aimed at your eye from a given drop. The red from a drop that sends you violet passes below your eye, which is why you must look higher in the sky to see red.

Yes — from the SCERT Telangana official Class 10 Physical Science eTextbook (x_physics_part-1_2026-27.pdf), pages 172 to 222, read directly, including all six activities, the lab activity and both worked examples.
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Last reviewed on 21 September 2026. Written and reviewed by subject-matter experts — read about our process.
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