CBSE Class 10 Science · 2026–27
Your eye keeps a nearby word sharp, a prism spreads a white beam into colours, and air sends blue light across the sky. Follow these connected stories with short explanations, light-path models and checks that ask you to explain what happens.
- Describe the eye’s light path and distinguish pupil control from accommodation.
- Explain myopia, hypermetropia and age-related accommodation changes using lens models.
- Solve simple corrective-lens exercises with signs and units.
- Trace light through a prism and explain dispersion and a primary rainbow.
- Explain scattering, the clear blue sky and the Tyndall effect.
01From an object to the retina
Look from a printed word to a distant tree. Both can look sharp even though they are at very different distances. Your eye is an optical system that adjusts the way it focuses light.
Follow the light: object → cornea → pupil → eye lens → retina. The cornea is the transparent curved front surface; it provides most of the eye’s refraction. The pupil is an opening in the coloured iris. The flexible convex eye lens makes the fine focusing adjustment.
A real, inverted image forms on the retina. Its light-sensitive cells produce electrical signals, carried by the optic nerve to the brain. These signals are processed into vision; the eye does not send seeing rays out to the object.
In the simple school model, the distance from the eye lens to the retina stays approximately fixed while focusing. A useful eye diagram therefore has a fixed receiving surface and an adjustable lens, rather than a screen that moves back and forth.
Pause. Think. Try.
02Brightness and focus are different jobs
In bright surroundings, the iris makes the pupil smaller. In dim surroundings, it makes the pupil larger so more light can enter. This controls the amount of light, not the eye lens’s focusing power.
The ciliary muscles help change lens shape and focal length. They perform a different job from the iris. “The pupil gets bigger to focus on a near object” confuses these two adjustments.
The retina contains rods, important for dim-light vision, and cones, important for colour vision and fine detail in suitable light. Seeing in a dim place improves through several adjustments, including pupil changes and the retina’s adaptation; it is not explained by pupil size alone.
For this topic, compare labelled drawings or ordinary room lighting. Do not shine a bright light into anyone’s eyes or use the Sun for an observation.
Pause. Think. Try.
03One retina, different object distances
Accommodation is the eye’s ability to adjust its lens focal length to focus objects at different distances on the retina.
| Viewing | Ciliary action and lens shape | Optical effect |
|---|---|---|
| Distant object | Muscles relax; lens becomes less curved/thinner | Longer focal length, lower converging power |
| Nearby object | Muscles contract; lens becomes more curved/thicker | Shorter focal length, higher converging power |
Rays from a nearby point arrive more divergent, so more convergence is needed. The lens-to-retina distance does not increase to accommodate them.
The far point is the farthest distance seen clearly without correction; for the standard normal-eye model it is infinity. The near point is the nearest distance seen clearly and comfortably. A conventional value for a young adult with normal vision is about 25 cm. It varies between people and with age; it is not a test for labelling a classmate’s eyesight.
Pause. Think. Try.
04Myopia: the distant image forms too soon
In the school model of myopia, or short-sightedness, distant objects are blurred while nearer objects can be clear. The far point lies at a finite distance rather than infinity.
For an uncorrected eye looking at a distant object, light comes to a focus in front of the retina. This can result from too much optical convergence for the eye’s length, or an eyeball that is too long.
A suitable concave, diverging lens spreads the incoming parallel rays before they reach the eye. Their backward extensions locate a virtual image at the eye’s far point. The eye can then focus those rays on the retina.
The spectacle lens does not move the retina or permanently change the eye lens. In a diagram, first show the uncorrected focus in front of the retina, then show the diverging corrective lens and the corrected focus on it.
Pause. Think. Try.
05Hypermetropia: nearby rays need more help
In the school model of hypermetropia, or long-sightedness, a nearby object at the usual reading distance is blurred. The near point is farther away than the conventional 25 cm.
The eye’s optical system has insufficient convergence for that near object, or the eyeball is too short. The rays would meet behind the retina. The retina intercepts them before they come to a sharp focus.
A suitable convex, converging lens increases the convergence before the light enters the eye. For an object at 25 cm, it can form an upright virtual image farther away, at the eye’s own near point. The eye then focuses that light on the retina.
Keep the two images separate in your explanation: the spectacle lens’s virtual image is outside the eye; the final real image is on the retina. The corrective convex lens is not forming a real screen image behind the person.
Pause. Think. Try.
06Age-related focus change is not lens clouding
With age, the eye lens becomes less flexible and accommodation decreases. Presbyopia makes close focusing harder and the near point recedes. It can occur alongside another refractive error. It is different from hypermetropia, even though both may make close work difficult.
Reading correction supplies suitable additional convergence. Some spectacles have different near and distance zones; the required powers depend on the person’s prescription. Do not assume every bifocal has one universal combination of positive and negative powers.
Cataract is a different idea: clouding within the lens, not a membrane growing across it and not simply failure to accommodate. Changing lens power does not remove that clouding.
For physics, identify the optical distinction: focus position, ability to change focus, or transparency. A classroom model explains light paths; actual eyesight concerns belong with an eye-care professional.
Pause. Think. Try.
07Work out a corrective lens in a model
Use the thin-lens relation 1/f=1/v −1/u. Assume the spectacle lens is close to the eye; ignore the small lens-eye separation. With light left to right, a real object in front has u<0, and a virtual image in front has v<0.
Distant-vision example
A model myopic eye has a far point 1.25 m away. For an object at infinity, the corrective lens should make a virtual image there: 1/u≈0 and v=−1.25 m. Thus f=−1.25 m and P=1/f=−0.80 D. The negative sign agrees with a diverging lens.
Near-vision example
A model eye has a near point 75 cm away. To view an object 25 cm away, use u=−0.25 m and v=−0.75 m. Then 1/f=−1/0.75−(−1/0.25)=−4/3+4=8/3 m−1. Hence P≈+2.67 D and f=+0.375 m. The positive sign agrees with a converging lens.
These are idealised exercises with stated near/far points. They do not determine a real person’s prescription from a reading observation.
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08A prism changes the outgoing direction
A triangular glass prism has two refracting faces inclined to one another. A ray entering from air bends towards the first normal. When it leaves for air, it bends away from the normal to the second face.
In the usual prism arrangement, the net deviation is towards the base, the thicker part of the prism. Unlike a parallel-sided slab, its two refracting-face normals are not parallel, so the emergent ray is not parallel to the original incident direction.
The angle of deviation is the angle between the original direction of the incident ray and the emergent direction. Extend the incident direction as a reference line to show it. Do not label the angle between a ray and a prism face as the angle of incidence: incidence is measured from the local normal.
For a ray-tracing practical, a teacher-controlled ray box and a prism outline let you mark the entrance/exit points and measure the deviation. No sunlight focusing is needed.
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09The prism separates colours already present
Dispersion is the separation of white light into its component colours. Ordinary glass has different refractive indices for different visible wavelengths, so the colours emerge with different deviations.
For an ordinary glass prism in the usual arrangement, red deviates least and violet most. Red has a longer wavelength than violet. The prism does not manufacture the colours; it separates components already in the incoming light.
The visible spectrum is continuous. The familiar sequence violet, indigo, blue, green, yellow, orange, red gives seven conventional colour names, not seven sharply separated physical strips.
A second matching prism placed appropriately in the opposite orientation can recombine the spread colours into white light. This supports the idea that white light contains a mixture of colours. Do not confuse this recombination of light with mixing classroom paints.
Pause. Think. Try.
10A raindrop is a tiny optical journey
A primary rainbow combines refraction, dispersion and internal reflection. Sunlight enters a water drop, changes direction and separates by colour. Part of that light reflects inside the drop, then refracts again on leaving.
You see a primary rainbow in the direction opposite the Sun, with sunlight behind you and suitable drops ahead. Red forms the outer edge and violet the inner edge. Different drops send the different colours towards your eye; it is not one drop painting a whole rainbow across the sky.
Use the phrase internal reflection here. The principal rainbow ray need not undergo total internal reflection. A careful diagram shows entry, one reflection and exit, with arrows along the actual light path.
A diagram or photograph is enough to study this process. Never look towards the Sun to search for a rainbow.
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11Why light arrives from the blue sky
Scattering redirects some light in different directions when it interacts with particles or molecules. It differs from a prism’s orderly separation of colours by refraction.
Molecules in the atmosphere scatter shorter visible wavelengths much more strongly than longer wavelengths. Some of that scattered light reaches your eye from directions across the sky, giving a clear daytime sky its blue appearance. The colour is not a reflection of an ocean beneath it.
Much larger particles, such as many cloud droplets, scatter visible colours more similarly, which helps explain a cloud’s white appearance in suitable illumination. Particle size and the incident light matter; not every scattered beam must look blue.
Far from an atmosphere, the sky can look dark because there is very little atmospheric scattering to send sunlight into your eye from those directions. An illuminated object can still be bright there.
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12Make a beam visible without making a new colour
The Tyndall effect is the visible scattering of a beam by sufficiently small dispersed particles. A torch beam may be hard to see from the side in clean air, but visible through mist or a suitable colloid because particles redirect some light towards the observer.
Teacher-led comparison: put clean water in one transparent container and very dilute milk in water in another. Shine an ordinary torch through each, directed away from eyes. Compare the beam viewed from the side against a dark background. Keep the containers away from electricity and do not drink the mixtures.
Record a simple table: liquid, side-view beam visibility, and what stayed the same (torch, container and viewing position). The purpose is to test whether dispersed particles make a light path more visible. This is a model of scattering, not an exact scale model of the atmosphere.
A bright patch on the far wall alone does not show the Tyndall effect: the key observation is light reaching you from the side of the beam’s path.
Pause. Think. Try.
13Explore further: a restless atmosphere
Air’s refractive index varies with conditions. As light travels through changing layers, its path can bend. This is atmospheric refraction.
A distant star acts nearly like a point source. Changing air can vary the light reaching your eye, so the star appears to twinkle. A planet presents a small apparent disc containing many contributing points; their fluctuations tend to average out, so planets usually twinkle less. “Never” is too strong, especially near the horizon.
Atmospheric refraction can also shift an apparent position upward, making the Sun visible when it is geometrically just below the horizon. The familiar few-minute sunrise/sunset estimate is approximate and depends on conditions and location.
This topic connects the ray model to observations. It is supplementary context here; the current assessed list explicitly names eye, prism, dispersion and scattering without a separate atmospheric-refraction item.
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14Curiosity: why the low Sun can look red
When sunlight travels a long oblique path through the atmosphere, more of its shorter-wavelength light is scattered out of the direct beam. The remaining direct light can be richer in longer wavelengths and look orange or red. Weather, dust and clouds alter the appearance.
The blue sky and reddish low Sun are complementary scattering ideas: one asks about light scattered towards you from the side; the other asks about the direct beam after a long path.
This is optional curiosity reading. The CBSE 2026–27 assessed topic list explicitly excludes the colour of the Sun at sunrise and sunset. A diagram is the right way to explore it; never stare at the Sun.
Pause. Think. Try.
Your formula shelf
| Idea | Relation | Remember |
|---|---|---|
| Corrective thin lens | 1/f=1/v −1/u | Signed distances in consistent units. Model assumes lens close to the eye. |
| Lens power | P=1/f | f in metres, P in dioptres. Diverging negative, converging positive. |
| Distant object in model myopia correction | 1/u≈0, so v≈f | Required virtual image at the finite far point; v and f negative. |
| Conventional normal near point | D≈25 cm | A standard young-adult model value, not a universal personal measurement. |
Work through the numbers
A model myopic eye has far point 1.25 m. Find the correcting lens power for a distant object, ignoring spectacle-eye separation.
For a distant object, 1/u≈0. The virtual image must be at v=−1.25 m. Hence f=v=−1.25 m and P=1/f=−0.80 D, a diverging lens.
A model eye has near point 75 cm. Find the lens power to view an object 25 cm away, ignoring spectacle-eye separation.
u=−0.25 m, v=−0.75 m. 1/f=1/v −1/u=−4/3+4=8/3 m−1. Thus P≈+2.67 D and f=+0.375 m. The positive power adds convergence.
A model correcting lens has power −1.25 D. Find its focal length and the distant virtual-image position.
f=1/P=−0.80 m. For an object at infinity, v=f=−0.80 m. The lens makes a virtual image 80 cm in front, as required for a model eye whose far point is there.
Bring it all together
Mix the ideas. Try each question before opening its explanation.
Write it. Explain it.
Use paper for a calculation, a labelled diagram or a short explanation. Then compare the reasoning. Suggested marks help you check how complete your answer is.
Two model eyes are drawn with the same retina position. In A, parallel rays focus before the retina. In B, a young eye has an eyeball shorter than needed for its optical power, so rays from an object 25 cm away would focus behind the retina. Name each refractive error, choose its corrective lens and explain one light-path change. · 4 marks
A represents myopia; a suitable concave lens first diverges distant parallel rays so the eye focuses them on the retina. B represents hypermetropia; a suitable convex lens adds convergence for the near rays so the final image falls on the retina. Neither correction moves the retina.
- Specify in front/behind retina
- Describe the final real image on retina, not a virtual retinal image
A model eye’s far point is 2 m. Find the nature, focal length and power of the lens needed for distant viewing. Neglect lens-eye separation. · 3 marks
The eye needs a virtual image at its far point. For a distant object 1/u≈0 and v=−2 m. Therefore f=−2 m and P=1/(−2)=−0.50 D. A suitable diverging/concave lens is required.
- 1: virtual image at v=−2 m; 1/u≈0
- 1: f=−2 m
- 1: P=−0.50 D and concave nature
Explain why a prism gives a spectrum of white light. State the least- and most-deviated colours for an ordinary glass prism, and explain what a reversed second prism can show. · 3 marks
The glass refracts different visible wavelengths by different amounts because its refractive index depends on wavelength. Red deviates least and violet most. A matching oppositely oriented second prism can recombine the separated colours into white light, showing they were already components of the incoming beam.
- 1: wavelength-dependent index/different refraction
- 1: red least/violet most
- 1: recombination and meaning
A torch beam is hard to see from the side in clean water but becomes visible from the side in very dilute milk. A learner says, “The milk makes blue paint, so this proves the sky is blue because of clouds.” Give a better explanation, identify a controlled variable and distinguish a model observation from the clear-sky explanation. · 4 marks
Dispersed milk particles scatter some torch light towards a side observer, making the beam visible; they do not create paint. Keep the torch, containers and viewing position alike while changing the liquid. The comparison models scattering. For a clear blue sky, air molecules preferentially scatter shorter visible wavelengths; large cloud droplets often scatter visible colours more similarly.
- 1: side-scattering/Tyndall explanation
- 1: valid controlled variable
- 1: no paint creation/model-limitation
- 1: clear-sky air molecules and shorter wavelengths