Chapter 10 | The Human Eye and the Colourful World

See the World Through Science

Master how the eye lens focuses, learn the defect–correction pairs with bulletproof tables, and decode why the rainbow has 7 colours, stars twinkle and the sky is blue — with toggle answers, PYQs and 3-level tests.

🎯 Golden Board Rules: (1) Power of a lens: $P = \frac{1}{f(m)}$ — always convert focal length cm → m first. (2) Myopia → concave (−) lens; Hypermetropia → convex (+) lens. (3) For myopia correction, corrective lens focal length $f = -$ (far point distance). (4) Always state where the image is formed (in front of / beyond the retina) and write the lens name + its action in every diagram question.

Concepts & Visual Theory

Comprehensive Board Theory, Ray Diagrams, Defect Corrections & Atmospheric Optics

The Human Eye: Structure, Optical Parts & Physiological Functions

Direct Board Definition — Human Eye: The human eye is one of the most valuable and sensitive optical sense organs. It functions like a sophisticated photographic camera, using a natural convex lens system to form a real, inverted, and diminished image of an external object on a light-sensitive screen called the retina.

Anatomy of the Human Eye — Key Structural & Optical Components:

Part of Eye	Structural Nature	Physiological & Optical Function
Cornea	Thin, transparent, spherical membrane bulging outwards on the anterior surface.	Acts as the primary optical window; accounts for nearly 75% to 80% of total light refraction entering the eye.
Iris	Dark, muscular, pigmented diaphragm located directly behind the cornea.	Controls the color of the eye and regulates the size of the pupil to control the amount of light entering the eye.
Pupil	Central circular aperture/opening in the middle of the iris.	Acts as a variable light gateway: constricts in bright light (preventing retinal glare) and dilates in dim light (admitting maximum illumination).
Crystalline Eye Lens	Double convex lens composed of transparent, flexible, fibrous jelly-like protein material.	Provides fine optical adjustment and variable focal length to sharply focus objects at varying distances onto the retina.
Ciliary Muscles	Circular ring of smooth muscles holding and suspending the eye lens in position.	Modifies the curvature and thickness of the eye lens to alter its focal length during accommodation.
Retina	Delicate inner neuro-sensory lining composed of millions of light-sensitive photoreceptor cells.	Acts as the photographic screen. Houses two types of photoreceptors: • Rods: Sensitive to dim light illumination (twilight vision). • Cones: Sensitive to bright light and color discrimination.
Optic Nerve	Bundle of nerve fibers originating from the posterior pole of the eyeball.	Transmits visual photochemical signals as electrical impulses from the retina directly to the visual cortex of the brain.
Aqueous & Vitreous Humour	Clear fluids filling the anterior chamber ($n \approx 1.336$) and posterior chamber ($n \approx 1.337$).	Maintain intraocular hydrostatic pressure, nourish non-vascular structures, and preserve the spherical geometry of the eyeball (~$2.3\text{ cm}$ diameter).

Image Formation — The Eye vs Photographic Camera Analogy

Optical Nature of Retinal Image: The convex lens system of the eye converges incident light rays from an external object onto the retina, forming an inverted, real, and diminished image. When photoreceptors (rods and cones) absorb photons, they generate action potentials (electrical impulses) which travel via the optic nerve to the brain's cerebral visual cortex, where perception is processed upright.

Comparative Table — Human Eye vs Modern Photographic Camera:

Feature / Component	Human Eye	Photographic Camera
Light Entering Window	Cornea (curved transparent living tissue)	Front glass lens element / protective glass filter
Aperture / Light Control	Iris adjusts diameter of pupil automatically	Mechanical diaphragm adjusts circular aperture size ($f$-stops)
Refracting Lens System	Single crystalline organic double-convex lens of variable focal length	Multiple glass convex/concave lens assembly of fixed focal length
Method of Focusing	Ciliary muscles alter lens curvature (Accommodation) while distance to retina ($v$) stays constant	Lens is mechanically moved forward or backward to adjust distance ($v$) to the sensor
Light-Sensitive Screen	Retina (millions of living rods & cones generating nerve impulses)	Electronic CMOS/CCD digital sensor chip or photographic film
Nature of Real Image	Real, inverted, diminished on retina	Real, inverted, diminished on sensor / film

Power of Accommodation — Near Point & Far Point

The eye lens is made of a fibrous, jelly-like material. Its curvature can be modified by the ciliary muscles:

🔄 How Focal Length Changes While Seeing

Situation	Ciliary Muscles	Lens	Focal length
Seeing a distant object	Relaxed	Becomes thin	Increases
Seeing a nearby object	Contract	Becomes thicker (more curved)	Decreases

📖 Accommodation = the ability of the eye lens to adjust its focal length (change curvature) so that clear images of objects at different distances are formed on the retina.

Least distance of distinct vision (near point): the minimum distance at which objects can be seen most distinctly without strain — about 25 cm for a young adult with normal vision.

Far point: the farthest point up to which the eye can see objects clearly — infinity for a normal eye. A normal eye can see clearly from 25 cm to infinity.

⚠️ Try reading a page held very close to the eye — it gets blurred or strains the eye, because the focal length of the eye lens cannot decrease below a minimum limit. With age the near point recedes (moves farther away) — this is presbyopia.

🧠 Mnemonic "2-5-2-5": Normal near point = 25 cm; far point = infinity. A young adult sees everything between these two limits comfortably.

📌 In-Text Questions (NCERT Page 164)

Page 164 • Q1 What is meant by power of accommodation of the eye?

Answer

The power of accommodation is the ability of the eye lens to adjust its focal length (by changing its curvature through the ciliary muscles) so that the eye can focus clearly both nearby and distant objects, forming their images on the retina.

Page 164 • Q3 What is the far point and near point of the human eye with normal vision?

Answer

The far point of a normal eye is infinity and the near point (least distance of distinct vision) is about 25 cm from the eye.

Myopia (Near-Sightedness) & Its Correction

A person with myopia can see nearby objects clearly but cannot see distant objects distinctly. Such a person has the far point nearer than infinity (e.g., a few metres). The image of a distant object is formed in front of the retina, not on it.

🔭 Myopia at a Glance

Aspect	Detail
Also known as	Near-sightedness / short-sightedness
Can see clearly	Nearby objects
Cannot see clearly	Distant objects (far point becomes nearer than infinity)
Image of distant object	Formed in front of the retina
Causes	(i) Excessive curvature of the eye lens, or (ii) elongation of the eyeball
Correction	Concave (diverging) lens of suitable power — it diverges the rays slightly so they meet on the retina

🧲 Golden formula: For a myopic eye with far point distance $d$, the corrective lens must form the image of an object at infinity at the far point. So the focal length is $f = -d$, and $P = \dfrac{1}{f(m)} = -\dfrac{1}{d(m)}$ (concave ⇒ negative).

$$P_{\text{myopia}} = -\frac{1}{d_{\text{far point (m)}}} \qquad \text{(concave / diverging lens)}$$ Focal length $f$ = distance of far point in front of the eye, taken as negative.

Worked Example (NCERT Page 164 Q2): A person with a myopic eye cannot see objects beyond 1.2 m distinctly. Type & power of corrective lens?
Far point $d = 1.2m$, so $f = -1.2m$.
$P = \dfrac{1}{f} = \dfrac{1}{-1.2} = \mathbf{-0.83D}$.
Correction: a concave lens of power about −0.83 D (focal length −1.2 m).

📌 In-Text Questions (NCERT Page 164)

Page 164 • Q2 A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type of the corrective lens used to restore proper vision?

Answer

The person's far point is at 1.2 m. The corrective lens must form the image of an object placed at infinity at the far point, i.e. the focal length $f = -1.2m$. Hence a concave (diverging) lens of power $P = \dfrac{1}{-1.2} = \mathbf{-0.83D}$ must be used.

Page 164 • Q4 A student has difficulty reading the blackboard while sitting in the last row. What could be the defect the child is suffering from? How can it be corrected?

Answer

The child cannot see distant objects (the blackboard) clearly — this is myopia (near-sightedness). It can be corrected by using a concave (diverging) lens of suitable power, which brings the image of the distant object back on to the retina.

Hypermetropia (Far-Sightedness) & Its Correction

A person with hypermetropia can see distant objects clearly but cannot see nearby objects distinctly. The near point is farther away than the normal near point (25 cm), so such a person has to hold reading material much beyond 25 cm for comfortable reading. Light rays from a close object get focused at a point behind the retina.

🔭 Hypermetropia at a Glance

Aspect	Detail
Also known as	Far-sightedness / long-sightedness
Can see clearly	Distant objects
Cannot see clearly	Nearby objects (near point shifts farther than 25 cm)
Image of nearby object	Formed behind the retina
Causes	(i) Focal length of the eye lens is too long, or (ii) eyeball has become too small
Correction	Convex (converging) lens of appropriate power — it provides the additional converging power needed to form the image on the retina

🧲 Golden formula: For a hypermetropic eye whose near point is $N'$, the corrective convex lens must make an object placed at the normal near point $(u = -25cm)$ appear at $v = -N'$. Use the lens formula $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$, then $P = \dfrac{100}{f(cm)}$.

$$\frac1f = \frac1v - \frac1u \quad with \quad u = -25cm,\ \ v = -N' $$ $v$ is negative because the virtual image is formed on the same side as the object.

Worked Example (NCERT Exercise Q7): Near point of a hypermetropic eye is 1 m (100 cm). Assume normal near point 25 cm.
$u = -25cm$, $v = -100cm$.
$\dfrac{1}{f} = \dfrac{1}{-100} - \dfrac{1}{-25} = \dfrac{-1+4}{100} = \dfrac{3}{100} \implies f = +33.3cm$.
$P = \dfrac{100}{f(cm)} = \dfrac{100}{100/3} = \mathbf{+3.0D}$ (convex lens).

Presbyopia, Bifocal Lenses & Cataract

With ageing, the power of accommodation of the eye usually decreases — the near point gradually recedes and people find it difficult to see nearby objects comfortably. This defect is called Presbyopia.

🏷️ Presbyopia at a Glance

Aspect	Detail
Causes	Gradual weakening of ciliary muscles and diminishing flexibility of the eye lens with age
Effect	Near point gradually recedes; difficulty in seeing nearby objects distinctly
Correction	Convex lenses for near vision; bifocal lenses if the person also has myopia
Bifocal lens	Upper part = concave lens (for distant vision); lower part = convex lens (for near vision)

Think! Some people suffer from both myopia and hypermetropia — they need bi-focal lenses (upper concave for distance, lower convex for near). Today these refractive defects can also be corrected by contact lenses or surgical intervention (e.g., LASIK).

📋 COMPARISON TABLE — The 3 Defects of Vision (Board Favourite)

Defect	Can't see	Where image forms	Cause	Corrective lens
Myopia	Distant objects	In front of retina	Excess curvature of lens / elongated eyeball	Concave (−P)
Hypermetropia	Nearby objects	Behind retina	Focal length too long / eyeball too small	Convex (+P)
Presbyopia	Nearby objects (old age)	Behind retina (near objects)	Weak ciliary muscles / inflexible lens	Convex or bifocal

📌 Extra fact: When the crystalline lens at old age becomes milky and cloudy, the condition is cataract, which causes partial/complete loss of vision — it can be restored by cataract surgery.

Refraction of Light Through a Prism

A triangular glass prism has two triangular bases and three rectangular lateral surfaces inclined to each other. The angle between its two lateral (refracting) faces is called the angle of the prism ($\angle A$). Unlike a glass slab (emergent ray parallel to incident ray), the prism bends the emergent ray towards its base, so the emergent ray is not parallel to the incident ray.

🔺 Key Terms for the Prism Diagram

Term	Meaning
PE	Incident ray
EF	Refracted ray (inside the prism)
FS	Emergent ray
∠i	Angle of incidence (at face AB)
∠r	Angle of refraction (inside the prism)
∠e	Angle of emergence (at face AC)
∠D	Angle of deviation — angle between the incident ray and the emergent ray

$$\angle D = i + e - A$$ The emergent ray bends towards the base of the prism; the deviation angle depends on the prism angle and i, e.

Why does the emergent ray bend at an angle? At face AB light goes air→glass, bending towards the normal; at face AC light goes glass→air, bending away from the normal. Because the two faces are inclined (not parallel), the two bends add together and produce a net deviation instead of cancelling as in a glass slab.

🧪 (Activity 10.1) Fix pins P, Q on an incident line PE; look through face AC; align pins R, S so all four appear in a straight line; trace the outline and join E–F to get the refracted ray. Mark $i$, $r$, $e$ and $D$.

Dispersion of White Light — The VIBGYOR Spectrum

When a narrow beam of white light is passed through a glass prism, it splits into a band of seven colours: Violet, Indigo, Blue, Green, Yellow, Orange, Red. The band of coloured components is called its spectrum, and the splitting of white light into its component colours is called dispersion.

🌈 VIBGYOR — The Splitting of White Light

Violet → Indigo → Blue → Green → Yellow → Orange → Red (from most to least deviated)

⚖️ How Different Colours Behave in a Prism

Property	Red light	Violet light
Wavelength	Longest (largest λ)	Shortest (smallest λ)
Speed in glass	Fastest among the colours	Slowest among the colours
Bending in prism	Bends the least	Bends the most
Position in spectrum	Lower/outer end	Upper/inner end

🧪 Newton's experiment: A first prism splits white light into a spectrum; a second identical prism kept inverted recombines all the colours into a beam of white light. This proved that sunlight is made up of seven colours. Any light giving a spectrum similar to sunlight is called white light.

🧠 Mnemonic "VIBGYOR": Violet Indigo Blue Green Yellow Orange Red. Trick: "Very Intense Blue Gets Yellow On Rainbows".

Rainbow Formation — Nature's Spectrum

A rainbow is a natural spectrum appearing in the sky after a rain shower. It is caused by dispersion of sunlight by tiny water droplets present in the atmosphere. Each tiny droplet acts like a small prism.

🌈 3 Steps Inside a Raindrop (Board Definition)

Step	What happens	Optical phenomenon
1	Sunlight enters the droplet	Refraction + dispersion
2	Light reflects off the inner back surface of the droplet	Total internal reflection
3	Colours come out of the droplet toward the observer	Refraction again (dispersed colours emerge)

Key facts:
• A rainbow is always formed in a direction opposite to the Sun.
• You can also see a rainbow on a sunny day near a waterfall or water fountain, with the Sun behind you.
• Every droplet disperses light and this internally-reflected, dispersed light reaches the observer's eye — different colours appear at different directions.

Atmospheric Refraction — Twinkling Stars & Delayed Sunset

Light coming from celestial objects passes through the Earth's atmosphere, whose refractive index changes gradually (denser near the ground, rarer higher up). The continuous bending of light through the atmosphere is called atmospheric refraction.

🌌 Three Famous Effects of Atmospheric Refraction

Effect	Reason	Extra detail
Twinkling of stars	Atmospheric refraction; the path of starlight keeps fluctuating	Stars are point-sized sources, so the light entering the eye flickers → brightness varies
Stars appear higher	Atmosphere bends starlight towards the normal	Apparent position is slightly above the actual position, especially near the horizon
Advance sunrise & delayed sunset	Atmospheric refraction of sunlight	Sun is visible about 2 minutes before actual sunrise and 2 minutes after actual sunset; its disc appears flattened near the horizon

🪐 Why don't planets twinkle? Planets are very close compared to stars, so they appear as extended sources (a collection of many point sources). The total variation in light entering our eye from all the points averages out to zero, nullifying the twinkling effect.

Hot-air wavering: The same small-scale effect is seen when objects flicker above a fire/radiator — hot air above is less dense (lower refractive index) than cool air above it, so the apparent position of the object fluctuates.

🧠 Mnemonic "P-P-E": Point sources → Produce twinkle; Extended sources → no twinkle. Stars ≈ points (twinkle), planets ≈ extended (no twinkle).

Scattering of Light — Blue Sky, Red Sun & Tyndall Effect

When light strikes tiny particles (smoke, dust, water droplets, air molecules), it bounces off them in different directions — this is scattering of light. The colour of scattered light depends on the size of the scattering particles and on the wavelength of light.

🎨 Colourful Phenomena — Reason Table (Memorise)

Phenomenon	Scientific reason
Clear sky appears blue	Air molecules and fine particles scatter blue light (shorter wavelength) more strongly than red light (λ of red ≈ 1.8× blue). Scattered blue light enters our eyes.
Sky looks dark (to an astronaut at high altitude)	There is no atmosphere at very high altitudes → no scattering → sky appears dark
Danger signals are red	Red is least scattered by fog/smoke, so it is seen in the same colour from far away
Sun looks reddish at sunrise/sunset	Sunlight travels through a thicker layer of atmosphere; blue light is scattered away, red (least scattered) reaches our eyes
Path of light beam visible in a dark room	Tyndall effect — scattering of light by colloidal particles (dust, smoke, water droplets) makes the beam visible

🧫 Tyndall effect is the scattering of light by colloidal particles. You see it when a fine beam of sunlight enters a smoke-filled room, or when sunlight passes through the canopy of a dense forest (mist droplets scatter light).

Size rule: Very fine particles scatter mainly blue light; larger particles scatter light of longer wavelengths; if particles are large enough, scattered light may even appear white.

🧠 Mnemonic "BLUE = Bottom Bounces Best": Shorter (blue) wavelengths are scattered the most, longer (red) wavelengths the least. So blue takes a scenic route to your eyes; red travels straight through fog and smog.

NCERT Chapter-End Exercises (Q1 to Q12)

Complete official textbook questions with interactive click-to-reveal step-by-step solutions

NCERT Official In-Text Questions — Page 190

In-Text Q1. What is meant by power of accommodation of the eye?

In-Text Q2. A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type of the corrective lens used to restore proper vision?

In-Text Q3. What is the normal vision of a human eye with respect to the near point and the far point?

In-Text Q4. A student has difficulty reading the blackboard while sitting in the last row. What could be the defect the child is suffering from? How can it be corrected?

NCERT Chapter-End Exercises (Q1 to Q12)

Q1. The human eye can focus on objects at different distances by adjusting the focal length of the eye lens. This is due to
(a) presbyopia. (b) accommodation.
(c) near-sightedness. (d) far-sightedness.

Solution

Correct Option: (b) accommodation

Explanation: The ability of the eye lens to adjust its focal length (by changing its curvature through the ciliary muscles) so as to focus objects at different distances clearly is called accommodation.

Q2. The human eye forms the image of an object at its
(a) cornea. (b) iris. (c) pupil. (d) retina.

Solution

Correct Option: (d) retina

Explanation: The eye lens forms a real, inverted image of the object on the light-sensitive screen called the retina, which has light-sensitive cells that convert light into electrical signals.

Q3. The least distance of distinct vision for a young adult with normal vision is about
(a) 25 m. (b) 2.5 cm. (c) 25 cm. (d) 2.5 m.

Solution

Correct Option: (c) 25 cm

Explanation: The minimum distance at which objects can be seen most distinctly without strain is called the least distance of distinct vision (near point). For a young adult with normal vision it is about 25 cm.

Q4. The change in focal length of an eye lens is caused by the action of the
(a) pupil. (b) retina. (c) ciliary muscles. (d) iris.

Solution

Correct Option: (c) ciliary muscles

Explanation: The ciliary muscles change the curvature of the jelly-like eye lens, thereby changing its focal length. When they contract the lens becomes thicker (shorter $f$ for near objects); when relaxed the lens becomes thin (longer $f$ for distant objects).

Q5. A person needs a lens of power –5.5 dioptres for correcting his distant vision. For correcting his near vision he needs a lens of power +1.5 dioptre. What is the focal length of the lens required for correcting (i) distant vision, and (ii) near vision?

Solution

(i) Distant vision: $P = -5.5D$

$$f = \frac1P = \frac1{-5.5} = -0.1818m = \mathbf{-18.18cm}$$

(ii) Near vision: $P = +1.5D$

$$f = \frac1P = \frac1{+1.5} = +0.667m = \mathbf{+66.7cm}$$

Answer: (i) Focal length = $-\mathbf{18.2cm}$ (concave/diverging lens), (ii) Focal length = $+\mathbf{66.7cm}$ (convex/converging lens).

Q6. The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?

Solution

Given: Far point $= 80cm = 0.8m$, so the corrective lens must form the image of an object at infinity at the far point: $v = -80cm$, $f = -80cm = -0.8m$.

$$P = \frac1{f(m)} = \frac1{-0.8} = \mathbf{-1.25D}$$

Since the power is negative, the lens is a concave (diverging) lens of power −1.25 D.

Q7. Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.

Solution

Diagram note: In a hypermetropic eye the image of a nearby object forms behind the retina; a convex lens in front of the eye converges the rays so the image is formed on the retina.

Given: Near point of the eye $N' = 1m = 100cm$; object is at the normal near point $u = -25cm$. The corrective lens must form a virtual image at the eye's near point: $v = -100cm$.

$$\frac1f = \frac1v - \frac1u = \frac1{-100} - \frac1{-25} = -\frac1100 + \frac125 = \frac{-1 + 4}100 = \frac3100$$

$$f = +\frac1003cm = +33.3cm = +0.333m, \quad P = \frac1f = \frac1{0.333} = \mathbf{+3D}$$

A convex lens of power +3 D and focal length +33.3 cm corrects the defect.

Q8. Why is a normal eye not able to see clearly the objects placed closer than 25 cm?

Solution

To see objects closer than 25 cm, the eye lens would have to become even thicker so that its focal length decreases further. However, the focal length of the eye lens cannot be decreased below a certain minimum limit (maximum accommodation is limited). Hence the image cannot be focused on the retina and appears blurred, causing strain. Thus 25 cm is the least distance of distinct vision for a normal eye.

Q9. What happens to the image distance in the eye when we increase the distance of an object from the eye?

Solution

The image distance in the eye is always constant because the eye lens and the retina are at fixed positions. When the object distance increases, the eye lens becomes thin and its focal length increases (ciliary muscles relax), so that the image continues to be formed exactly on the retina. So the image distance does not change.

Q10. Why do stars twinkle?

Solution

Stars twinkle due to atmospheric refraction of starlight. Stars are very distant and appear as point-sized sources. As starlight travels through the Earth's atmosphere (whose refractive index changes gradually and whose physical conditions are not stationary), the path of the rays keeps varying slightly. Hence the apparent position fluctuates and the amount of starlight entering the eye flickers — the star appears sometimes brighter and sometimes fainter. This is the twinkling effect.

Q11. Explain why the planets do not twinkle.

Solution

Planets are much closer to the Earth than stars and are seen as extended sources (a collection of a large number of point-sized sources). If we consider a planet as a collection of point sources, the total variation in the amount of light entering the eye from all the individual point-sized sources averages out to zero. This nullifies the twinkling effect, so planets shine steadily.

Q12. Why does the sky appear dark instead of blue to an astronaut?

Solution

The blue colour of the sky is due to scattering of sunlight by the fine particles and air molecules present in the Earth's atmosphere. At very high altitudes, there is no atmosphere for scattering to take place. For an astronaut, no light is therefore scattered toward the eye from the air, and the sky appears dark.

CBSE Class 10 Science • Chapter 10 PYQ Bank (2011–2026)

Q1–Q92 | Aligned with 2026-27 Syllabus | Verified Board PYQs | MCQ • A-R • SA • Case Study • LA

⚠️ CBSE Examiner Common Deductions (Mark Scheme Warnings):

Drawing ray diagram without labelling retina, lens, cornea → −0.5M
Saying "convex lens corrects myopia" instead of concave → −1M
Forgetting sign while calculating power (negative for myopia) → −1M
Writing ROYGBIV instead of VIBGYOR for order of deviation → −0.5M
Saying "red is most scattered" instead of "least scattered" → −1M
Not naming all three phenomena for rainbow → −0.5M each
Defining dispersion without mentioning different speeds → −0.5M

📌 Topic 1 — Functioning of a Lens in the Human Eye (Q1 to Q11)

MCQ — 1M | CBSE Recurring 2018–2026

Q1. Which part of the human eye controls the amount of light entering the eye?

Explanation

Correct Option: (B) Pupil

The pupil is the circular opening in the centre of the iris that regulates and controls the amount of light entering the eye. The iris (dark muscular diaphragm) controls the size of the pupil by contraction or relaxation.

MCQ — 1M | CBSE Recurring

Q2. The lens in the human eye is:

Explanation

Correct Option: (C) Convex

The crystalline eye lens is a fibrous, jelly-like, biconvex (convex) structure. It provides the finer adjustment of focal length by changing curvature via the ciliary muscles.

MCQ — 1M | CBSE Recurring 2015–2026

Q3. The least distance of distinct vision for a normal human eye is:

Explanation

Correct Option: (C) 25 cm

The minimum distance at which objects can be seen most distinctly without strain is 25 cm for a young adult with normal vision — the near point (least distance of distinct vision).

MCQ — 1M | CBSE Recurring

Q4. The far point of a normal human eye is:

Explanation

Correct Option: (D) Infinity

The far point is the farthest point that the normal eye can see clearly. For a normal eye, it is at infinity.

MCQ A-R — 1M | CBSE 2023; 2024; 2025

Q5. Assertion (A): The human eye can change its focal length to focus on objects at different distances.
Reason (R): The ciliary muscles change the curvature (and hence the focal length) of the eye lens — this ability is called power of accommodation.

Explanation

Correct Option: (A) — Both A and R are true, and R correctly explains A. Ciliary muscles adjust the curvature of the eye lens to change its focal length — this is the power of accommodation.

SA — 2M | CBSE 2016; 2024

Q6. Define the term 'power of accommodation' of the human eye. What is the function of the iris and pupil of the eye?

Answer

Power of accommodation: The ability of the eye lens to adjust its focal length (by the action of ciliary muscles changing its curvature) so that objects at different distances form clear images on the retina.

Iris: A dark muscular diaphragm that controls the size of the pupil. Pupil: The circular opening in the centre of the iris that regulates the amount of light entering the eye.

SA — 3M | CBSE 2024

Q7. Define power of accommodation of the human eye. What happens to the image distance in the eye when we increase the distance of an object from the eye? Name and explain the role of the part of the human eye responsible for this.

Answer

Power of accommodation: The ability of the eye lens to adjust its focal length so that objects at different distances form clear, sharp images on the retina.

Image distance: When the object is moved farther, the eye lens must increase its focal length (become thinner/flatter) to maintain the image on the retina. The image distance in the eye remains essentially constant (image always on retina) while the focal length changes.

Ciliary muscles: For distant objects → ciliary muscles relax → lens becomes thin (longer focal length). For nearby objects → ciliary muscles contract → lens becomes more curved (shorter focal length).

SA — 3M | CBSE 2016; 2019; 2022; Recurring

Q8. Draw a neat labelled diagram of the human eye. Name the part that: (i) controls the amount of light entering the eye, (ii) is responsible for converting light signals to electrical signals, (iii) focuses the image on the retina.

Answer

Light enters through the cornea → pupil → crystalline lens → forms real, inverted image on retina → signals sent to brain via optic nerve.

Diagram: Draw the spherical eyeball; label cornea, iris, pupil, crystalline lens, ciliary muscles, retina, optic nerve with arrows showing path of light.

(i) Pupil (controlled by iris) — controls light entering the eye.
(ii) Retina — light-sensitive cells convert light into electrical signals sent to brain via optic nerve.
(iii) Crystalline lens (eye lens) — focuses image on the retina via ciliary muscles.

SA — 2M | CBSE Recurring 2015–2025

Q9. What is the range of vision of a normal adult eye? Why is a normal eye unable to see objects placed closer than 25 cm?

Answer

Range of vision: 25 cm (near point) to infinity (far point).

Why not closer than 25 cm? At 25 cm the ciliary muscles are fully contracted (maximum curvature, minimum focal length). The eye cannot increase converging power further. Objects closer than 25 cm produce a highly diverging beam that the eye lens cannot converge on to the retina → blurred image + eye strain.

SA — 3M | CBSE 2017; 2020; 2022; Recurring

Q10. Explain the working of a human eye as an optical instrument. How does it adjust its focal length to see objects at different distances?

Answer

Light enters through the cornea (most refraction here) → pupil (controlled by iris) → crystalline lens → real, inverted image on retina → electrical signals via optic nerve → brain interprets the image.

Accommodation: Distant objects: ciliary muscles relax → lens thin → focal length increases. Near objects: ciliary muscles contract → lens more curved → focal length decreases. This ensures a sharp image always forms on the retina.

Case Study — 4M | CBSE 2023; 2024; 2025; 2026

Q11. Case Study — The Human Eye as an Optical Instrument

The human eye works like a camera. Light enters through the cornea, passes through the pupil (controlled by the iris), and is focused by the flexible crystalline lens onto the retina. The retina converts the light image into electrical signals sent to the brain via the optic nerve. The eye lens can change its curvature using ciliary muscles — this is called accommodation. Near point = 25 cm; Far point = infinity (normal eye).

(i) Name the part of the eye that acts like a camera diaphragm. What is its function?
(ii) Define power of accommodation.
(iii) A person can see clearly only between 50 cm and 500 cm. Find the power of lenses for: (a) distant vision, (b) near vision.
(iv) What is the function of the retina?

Answer

(i) Iris acts like a camera diaphragm — it controls the size of the pupil, regulating the amount of light entering the eye.

(ii) Power of accommodation: The ability of the eye lens to adjust its focal length using ciliary muscles so that clear images of objects at different distances form on the retina.

(iii) Far point = 500 cm = 5 m; Near point = 50 cm = 0.5 m.
(a) Distant: $f = -5$ m → $P = \mathbf{-0.2D}$ (concave)
(b) Near: $u = -25$ cm, $v = -50$ cm → $\frac{1}{f}=\frac{1}{-50}-\frac{1}{-25}=\frac{1}{50}$ → $f=50$ cm = 0.5 m → $P = \mathbf{+2D}$ (convex)

(iv) Retina: Light-sensitive screen containing rods and cones that convert the light image into electrical signals transmitted to the brain via the optic nerve.

📌 Topic 2 — Defects of Vision and Their Corrections (Q12 to Q30)

MCQ — 1M | CBSE Recurring 2016–2026

Q12. Which lens is used to correct myopia?

Explanation

Correct Option: (B) Concave lens — In myopia image forms in front of retina. A concave (diverging) lens shifts the image back to fall on the retina.

MCQ — 1M | CBSE Recurring 2016–2026

Q13. Which lens is used to correct hypermetropia?

Explanation

Correct Option: (B) Convex lens — In hypermetropia image forms behind retina. A convex (converging) lens brings it forward onto the retina.

MCQ — 1M | CBSE Recurring 2018–2025

Q14. A man suffering from myopia cannot see objects distinctly at a distance greater than 2 m. Which lens is required to correct this defect?
(A) −2 D (B) +2 D (C) −0.5 D (D) +0.5 D

Answer

Far point = 2 m → $f = -2$ m → $P = \frac{1}{-2} = \mathbf{-0.5D}$ (Option C). A concave lens of power −0.5 D.

MCQ — 1M | CBSE Recurring 2019–2025

Q15. The near point of a hypermetropic person lies at 75 cm. Which lens is required to correct this defect? (Normal near point = 25 cm)
(A) +3 D (B) −3 D (C) +2 D (D) −2 D

Answer

$u = -25$ cm, $v = -75$ cm → $\frac{1}{f}=\frac{1}{-75}-\frac{1}{-25}=\frac{2}{75}$ → $f=37.5$ cm = $0.375$ m → $P \approx +2.67$ D ≈ +3 D (Option A). Convex lens.

MCQ — 1M | CBSE Recurring 2020–2025

Q16. Cataract is caused due to:

Explanation

Correct Option: (C) — Cataract is caused by clouding/opacification of the crystalline lens. Not a refractive defect; treated by surgical lens replacement.

MCQ — 1M | CBSE 2017; Recurring

Q17. A person uses spectacles with convex lenses. The name of the defect of vision he is suffering from is:

Explanation

Correct Option: (C) Hypermetropia — Convex (converging) lenses correct hypermetropia where nearby object images form behind the retina.

MCQ A-R — 1M | CBSE 2023; 2024; 2025; 2026

Q18. Assertion (A): Myopia is corrected using a concave lens.
Reason (R): In a myopic eye, the image is formed in front of the retina because the eyeball is elongated or the lens is too converging. A concave lens diverges the light rays to shift the image back onto the retina.

Explanation

Correct Option: (A) — Both A and R are true, and R correctly explains A.

MCQ A-R — 1M | CBSE 2024; 2025

Q19. Assertion (A): A person with presbyopia needs bifocal lenses.
Reason (R): Presbyopia is an age-related condition where both near and far vision are affected due to weakening of ciliary muscles and hardening of the eye lens. Bifocal lenses have a concave upper half (for distance) and a convex lower half (for reading).

Explanation

Correct Option: (A) — Both A and R are true and R correctly explains A. Presbyopia affects both near and far vision; bifocal lenses have concave upper (distant) and convex lower (reading) halves.

VSA — 1M | CBSE 2016

Q20. Priya prefers to sit in the front row as she finds it difficult to read the blackboard from the last desk in her classroom. Name the defect of vision she is suffering from.

Answer

Myopia (short-sightedness) — cannot see distant objects (blackboard) clearly but sees near objects well.

SA — 2M | CBSE 2017

Q21. Name the type of defect of vision a person is suffering from if he uses convex lenses in his spectacles. Give two possible causes for this defect.

Answer

Defect: Hypermetropia (far-sightedness)

Two causes: (i) The eyeball is too short; (ii) The focal length of the eye lens is too large (lens too weak/flat — insufficient converging power).

Numerical — 2M | CBSE Recurring 2019–2025

Q22. Sheela cannot read a newspaper held closer than 100 cm. Find (i) the name of the defect, (ii) the required power of lens with sign and calculation. (Normal near point = 25 cm)

Solution

(i) Cannot see closer than 100 cm → Hypermetropia.

(ii) $u = -25$ cm, $v = -100$ cm.
$\frac{1}{f}=\frac{1}{-100}-\frac{1}{-25}=\frac{-1+4}{100}=\frac{3}{100}$
$f = 33.3$ cm = $0.333$ m → $P = \mathbf{+3D}$ (convex lens)

Numerical — 2M | CBSE Recurring 2018–2025

Q23. A short-sighted person has a far point of 1.2 m. Find the power of the concave lens required to correct this defect for distant vision.

Solution

$f = -1.2$ m (concave) → $P = \frac{1}{-1.2} = \mathbf{-0.83D}$

LA — 5M | CBSE 2012; 2015; 2017; Recurring

Q24. What is myopia? List two causes. How can it be corrected using a lens? Draw ray diagrams: (i) defective eye, (ii) corrected eye.

Solution

Myopia: A defect where nearby objects are seen clearly but distant objects appear blurred. Image of distant object forms in front of the retina.

Two causes: (i) Eye lens is too converging (excessive curvature); (ii) Eyeball is elongated.

Correction: Concave (diverging) lens — diverges parallel incoming rays so the eye lens focuses them on the retina.

Diagrams: (i) Show parallel rays from infinity converging in front of retina. (ii) Show concave lens diverging rays; eye lens then converges them on retina.

LA — 5M | CBSE 2011; 2017; Recurring

Q25. What is hypermetropia? Describe with a ray diagram how this defect can be corrected. List two possible causes.

Solution

Hypermetropia: Distant objects seen clearly but nearby objects appear blurred. Image of nearby object forms behind the retina.

Two causes: (i) Eyeball too short; (ii) Focal length of eye lens too large (weak converging power).

Correction: Convex (converging) lens — converges diverging rays from nearby objects before they enter the eye, so the image falls on the retina.

Diagrams: (i) Diverging rays from nearby object focused behind retina. (ii) Convex lens converges rays; eye then focuses on retina.

LA — 5M | CBSE 2016; 2019; Recurring

Q26. A student is unable to see the blackboard placed at ~4 m. (i) Name the defect of vision. (ii) With labelled ray diagrams explain: (a) the defective eye, and (b) the corrected eye after using a suitable lens.

Solution

(i) Myopia — cannot see distant objects (4 m).

(ii)(a) Defective eye: Parallel rays from blackboard converge to a point in front of the retina → blurred image on retina.
(b) Corrected eye: Concave lens placed in front — diverges parallel rays so the effective virtual source is within the far point; eye lens then focuses on retina.

SA — 3M | CBSE 2017

Q27. An old person has difficulty in reading a book as well as seeing distant objects. (i) Name the defect. (ii) What type of lens corrects this? Explain its structure. (iii) State the cause of this defect.

Answer

(i) Presbyopia.

(ii) Bifocal lens: Upper part = concave (for distant vision); Lower part = convex (for near/reading vision).

(iii) Cause: With age, ciliary muscles weaken and the eye lens loses flexibility (hardens), reducing the power of accommodation. Both near and far vision are affected.

LA — 5M | CBSE 2014

Q28. (a) List three common refractive defects of vision. Suggest the way of correcting each. (b) About 45 lakh people in developing countries suffer from corneal blindness. About 30 lakh children below age 12 can be cured by replacing the defective cornea with a donated eye. How and why can students of your age create awareness about this?

Solution

(a)
1. Myopia → concave lens.
2. Hypermetropia → convex lens.
3. Presbyopia → bifocal lens.

(b) Students can: (i) organise eye donation camps and talks in schools; (ii) put up posters/banners encouraging eye donation; (iii) use social media to spread awareness; (iv) encourage family and neighbours to pledge eye donation. This is vital as a single donation can restore sight to children suffering from corneal blindness.

SA — 3M | CBSE 2024; Recurring

Q29. [Ray diagram shows image forming behind retina for a nearby object.] (i) Name the defect, stating the part of eye responsible. (ii) List two causes. (iii) Name the type of lens used to correct it and state its role.

Answer

(i) Hypermetropia — the eye lens (too weak/long focal length) or the overall converging power of the eye is insufficient.

(ii) (a) Focal length of eye lens too large; (b) Eyeball too short.

(iii) Convex (converging) lens — it converges diverging rays from the nearby object before entering the eye, providing additional converging power so the image falls exactly on the retina.

Case Study — 4M | CBSE 2023; 2024; 2025; 2026

Q30. Case Study — Defects of Vision and Their Correction

Myopia: image forms in front of retina; corrected by concave lens. Hypermetropia: image forms behind retina; corrected by convex lens. Presbyopia: both near and far vision affected in old age; corrected by bifocal lens. Cataract: clouding of crystalline lens; treated by surgery (not a refractive defect).

(i) A person reads at 30 cm clearly but cannot see a blackboard at 10 m. What is the defect? What lens corrects it?
(ii) Another person cannot read closer than 50 cm. What is the defect? Find the power of lens required (near point = 25 cm).
(iii) Distinguish between myopia and hypermetropia w.r.t.: (a) image position, (b) eyeball shape, (c) corrective lens.
(iv) Why does presbyopia require bifocal lenses? What does each half do?

Answer

(i) Cannot see distant → Myopia. Corrected by concave lens.

(ii) Cannot read closer than 50 cm → Hypermetropia.
$u=-25$ cm, $v=-50$ cm → $\frac{1}{f}=\frac{1}{50}$ → $f=50$ cm = 0.5 m → $P=\mathbf{+2D}$

(iii)

Feature	Myopia	Hypermetropia
Image position	In front of retina	Behind retina
Eyeball	Elongated	Shortened
Corrective lens	Concave (−)	Convex (+)

(iv) Presbyopia affects both near and far vision. Upper concave half → distant vision; Lower convex half → near/reading vision.

📌 Topic 3 — Applications of Spherical Mirrors and Lenses (Q31 to Q39)

MCQ — 1M | CBSE Recurring 2015–2026

Q31. Which type of mirror is used as a rear-view mirror in vehicles?

Explanation

Correct Option: (B) Convex — A convex mirror always forms a virtual, erect, diminished image regardless of object position, giving a wider field of view.

MCQ — 1M | CBSE Recurring 2016–2025

Q32. A dentist uses which type of mirror to see an enlarged image of a patient's tooth?

Explanation

Correct Option: (C) Concave — When tooth is within focal length, a concave mirror forms a virtual, erect, magnified image.

MCQ — 1M | CBSE Recurring 2016–2025

Q33. Concave mirrors are used in solar furnaces because:

Explanation

Correct Option: (B) — Concave mirrors converge parallel solar rays to the focus, concentrating energy to generate very high temperatures.

MCQ A-R — 1M | CBSE 2024; 2025

Q34. Assertion (A): Large concave mirrors are used in reflecting telescopes to collect light from distant stars.
Reason (R): Concave mirrors converge a parallel beam of incoming light to its focal point, enabling collection and focus of light from distant sources.

Explanation

Correct Option: (A) — Both A and R are true and R correctly explains A.

VSA — 1M | CBSE 2016

Q35. Name the type of mirror used in solar furnaces. Explain how a high temperature is achieved.

Answer

Large concave mirror. It converges parallel rays of sunlight to a small focal point, concentrating huge solar energy in a tiny area → extremely high temperature generated.

SA — 2M | CBSE Recurring 2015–2025

Q36. Why do we prefer a convex mirror as a rear-view mirror in vehicles? Draw a ray diagram to show image formation.

Answer

A convex mirror is preferred because: (i) It always gives a virtual, erect, diminished image for all positions of the object. (ii) It has a wider field of view compared to a plane or concave mirror of the same size, helping the driver see more of the traffic behind.

Virtual, erect, diminished image forms behind the mirror for all object positions — giving a wider field of view.

Ray diagram: Draw a convex mirror. Two incident rays from an object — one parallel to the principal axis (reflects as if diverging from focus behind mirror), one towards the centre of curvature (reflects back along itself). Trace back to find the virtual, erect, diminished image behind the mirror.

SA — 3M | CBSE Recurring 2015–2024

Q37. Give two uses each of: (i) concave mirrors, and (ii) convex mirrors in daily life. Give reasons for each use.

Answer

(i) Concave mirrors:
1. Shaving/make-up mirror: Object within focal length → virtual, erect, magnified image for close-up inspection.
2. Headlight/searchlight reflector: Source at focus → parallel beam of reflected light → strong directed beam.

(ii) Convex mirrors:
1. Rear-view mirror: Always erect, diminished → wider field of view for driver.
2. Security/shop mirrors (corner mirrors): Wide-angle diminished image allows surveillance of a large area.

SA — 3M | CBSE Recurring 2017–2025

Q38. Where should an object be placed in front of a convex lens to use it as: (i) a magnifying glass, (ii) a projector lens, (iii) a camera lens? State the characteristics of the image in each case.

Answer

(i) Magnifying glass: Object between O and F (u < f) → Virtual, erect, magnified image on same side as object.

(ii) Projector lens: Object between F and 2F (f < u < 2f) → Real, inverted, magnified image beyond 2F on other side.

(iii) Camera lens: Object beyond 2F (u > 2f) → Real, inverted, diminished image between F and 2F on other side.

SA — 3M | CBSE 2018; 2021; Recurring

Q39. List four uses of concave mirrors. For each use, state the position of the object and the nature of image required.

Answer

Use	Object position	Image required
Shaving/make-up mirror	Between pole and F	Virtual, erect, magnified
Dentist's mirror	Between pole and F	Virtual, erect, magnified
Headlight/searchlight reflector	At focus (F)	Parallel beam (image at infinity)
Solar furnace/cooker	Parallel rays (sun at infinity)	Real, concentrated at F

📌 Topic 4 — Refraction of Light Through a Prism (Q40 to Q49)

MCQ — 1M | CBSE Recurring 2017–2026

Q40. When a ray of light passes through a glass prism, the emergent ray:

Explanation

Correct Option: (C) — A ray always bends towards the base of the prism due to the geometry of refraction at the two inclined surfaces.

MCQ — 1M | CBSE Recurring 2018–2026

Q41. The angle between the incident ray and the emergent ray when a ray of light passes through a glass prism is called the:

Explanation

Correct Option: (C) Angle of deviation (∠D) — given by $D = i + e - A$.

VSA — 1M | CBSE 2011; 2012; 2014; 2015; 2016

Q42. Name the component of white light that deviates: (i) the least, and (ii) the most, while passing through a glass prism.

Answer

(i) Red — least deviation (longest λ, minimum refractive index).
(ii) Violet — most deviation (shortest λ, maximum refractive index).

VSA — 1M | CBSE 2011

Q43. Light of two colours A and B passes through a glass prism. Colour A deviates more than colour B. Which colour has a higher speed in the prism?

Answer

Greater deviation → higher refractive index → lower speed. A deviates more → A has lower speed → Colour B has a higher speed in the prism.

MCQ A-R — 1M | CBSE 2023; 2024; 2025; 2026

Q44. Assertion (A): When white light passes through a glass prism, violet light deviates the most and red light deviates the least.
Reason (R): The refractive index of glass is maximum for violet light (shortest wavelength) and minimum for red light (longest wavelength). Greater refractive index means greater bending.

Explanation

Correct Option: (A) — Both A and R are true and R correctly explains A.

MCQ A-R — 1M | CBSE 2024; 2025

Q45. Assertion (A): A glass prism produces a spectrum of white light but a glass slab does not.
Reason (R): In a glass slab, the two refracting surfaces are parallel, so the dispersion produced at the first surface is cancelled at the second surface. In a prism, the surfaces are inclined, so dispersion is not cancelled.

Explanation

Correct Option: (A) — Both A and R are true and R correctly explains A.

SA — 3M | CBSE 2015; 2016; Recurring

Q46. What is the angle of deviation in a prism? Draw a labelled diagram showing the refraction of a monochromatic ray through a glass prism. Mark the angle of incidence, angle of refraction, angle of emergence, and angle of deviation.

Answer

i = angle of incidence, e = angle of emergence, r₁ and r₂ = angles of refraction at the two faces, A = prism angle, D = angle of deviation.

Angle of deviation (∠D): The angle between the direction of the incident ray (ended) and the direction of the emergent ray after passing through the prism. $D = i + e - A$.

Diagram labels: PE (incident ray at surface 1), EF (refracted ray inside prism), FS (emergent ray at surface 2). Mark: ∠i (at surface 1), ∠r₁ and ∠r₂ (inside prism), ∠e (at surface 2), ∠D (between extended PE and emergent FS), prism angle A.

SA — 2M | CBSE 2024; Recurring

Q47. A student traces the path of a ray of light through a rectangular glass slab and a triangular glass prism. In which case does the emergent ray appear shifted/deviated? Give reason.

Answer

Glass prism: The emergent ray is deviated (bent) from the direction of the incident ray.
Reason: In a glass slab, the two surfaces are parallel — deviation at the first surface is exactly reversed at the second → emergent ray is parallel to incident ray (only laterally shifted). In a prism, the surfaces are inclined → deviations at both surfaces add up → the emergent ray is bent towards the base by the angle of deviation.

SA — 2M | CBSE 2017; 2020; Recurring

Q48. Why does a glass prism cause bending of light towards its base? Draw a ray diagram showing the path of a narrow beam of white light through a glass prism.

Answer

Violet (shortest wavelength, highest refractive index) deviates most; Red (longest wavelength, lowest refractive index) deviates least.

At each inclined face of the prism, the ray bends towards the normal on entering and away from the normal on exiting. Due to the geometry of the inclined surfaces, both refractions bend the ray in the same direction (towards the thicker base). The effects add up, so the net bending is always towards the base.

Diagram: Draw a triangular prism (apex up). Narrow white beam enters one face → splits into VIBGYOR inside → all colours emerge bending towards the base, with violet bending most and red least.

LA — 5M | CBSE 2017

Q49. State the cause of dispersion of white light by a glass prism. How did Newton, using two identical glass prisms, show that white light is made of seven colours? Draw a ray diagram to show the path of a narrow beam of white light through two identical prisms arranged in an inverted position with respect to each other.

Solution

Newton's two-prism experiment: Prism 1 (apex up) disperses → VIBGYOR; Prism 2 (inverted/apex down) recombines → White light.

Cause of dispersion: Different colours of white light have different wavelengths and hence travel at different speeds in glass → different refractive indices → different amounts of bending at the prism faces.

Newton's two-prism experiment:
1. Prism 1 (apex up) disperses white light into VIBGYOR spectrum.
2. Prism 2 (identical, placed inverted — apex down) is placed next to Prism 1 such that all seven colours enter it.
3. Prism 2 recombines all the colours back into a single beam of white light.
Conclusion: White light is a mixture of seven colours; the prism only separates them, does not create them; dispersion is reversible.

Ray diagram: Prism 1 (apex up) → VIBGYOR dispersed out. Prism 2 (inverted, apex down) → all colours recombine → white light out. Mark the path of each colour.

📌 Topic 5 — Dispersion of Light (Q50 to Q64)

MCQ — 1M | CBSE Recurring 2017–2026

Q50. The phenomenon of splitting white light into its component colours is called:

Explanation

Correct Option: (C) Dispersion — The splitting of white light into VIBGYOR by a prism.

MCQ — 1M | CBSE Recurring

Q51. The correct sequence of colours in the visible spectrum (from most deviated to least deviated) is:

Explanation

Correct Option: (B) VIBGYOR — Violet deviates most → Red deviates least. Mnemonic: "Very Intense Blue Gets Yellow On Rainbows."

MCQ — 1M | CBSE Recurring 2018–2025

Q52. Which colour of white light has the highest frequency?

Explanation

Correct Option: (D) Violet — Shortest wavelength → highest frequency (f = c/λ).

MCQ — 1M | CBSE Recurring 2019–2026

Q53. Why does the refractive index of glass differ for different colours of light?

Explanation

Correct Option: (B) — Refractive index $n = c/v$. Different colours travel at different speeds (v) in glass → different n values. Violet: slowest speed, highest n; Red: fastest speed, lowest n.

MCQ — 1M | CBSE Recurring

Q54. The intensity of scattered light by particles in the atmosphere varies as the nth power of the wavelength of light, where n =

Explanation

Correct Option: (C) −4 — Rayleigh scattering: Intensity $\propto \frac{1}{\lambda^4} = \lambda^{-4}$, so n = −4.

MCQ A-R — 1M | CBSE 2023; 2024; 2025

Q55. Assertion (A): A rainbow is always observed opposite to the direction of the sun.
Reason (R): For the dispersed light from water droplets to reach the observer's eyes, the sun must be behind the observer. The droplets act like small prisms, refracting and internally reflecting sunlight back toward the observer.

Explanation

Correct Option: (A) — Both A and R are true and R correctly explains A.

VSA — 1M | CBSE 2012; 2015; Foreign 2016

Q56. Define dispersion of light. What is the cause of dispersion?

Answer

Dispersion: The splitting of white light into its constituent seven colours (VIBGYOR) on passing through a prism.

Cause: Different colours have different wavelengths → travel at different speeds in glass → different refractive indices → different amounts of bending.

VSA — 1M | CBSE Recurring

Q57. What is a spectrum? Write the colours of the visible spectrum in order from least deviated to most deviated.

Answer

Spectrum: The band of coloured components obtained when white light undergoes dispersion.

Least to most deviated: Red → Orange → Yellow → Green → Blue → Indigo → Violet (ROYGBIV).

VSA — 1M | CBSE 2014

Q58. Name the colours that suffer: (i) maximum deviation, and (ii) minimum deviation while passing through a glass prism.

Answer

(i) Violet — maximum deviation (shortest λ, highest n).
(ii) Red — minimum deviation (longest λ, lowest n).

SA — 3M | CBSE 2009; 2011; 2012; 2014; 2017; Recurring

Q59. What is dispersion of light? What is the cause of dispersion? Draw a neat diagram to show the dispersion of white light by a glass prism.

Answer

Dispersion: Splitting of white light into VIBGYOR (seven component colours) on passing through a glass prism.

Cause: Each colour has a different wavelength → different speed in glass → different refractive index → different angle of bending → colours separate.

Violet (shortest λ, highest n) deviates most; Red (longest λ, lowest n) deviates least. The band of colours is the spectrum.

Diagram: Triangular prism (apex up). Narrow white beam enters one face → splits into VIBGYOR emerging from the second face. Label violet (most bent, near base) and red (least bent).

SA — 2M | CBSE Recurring 2015–2025

Q60. Name the phenomenon occurring in nature that is due to dispersion of light. Name the three optical phenomena responsible for the formation of a rainbow.

Answer

Natural phenomenon due to dispersion: Rainbow.

Three phenomena for rainbow: (1) Refraction (entering the droplet), (2) Dispersion (splitting into VIBGYOR), (3) Total internal reflection (at the back of the droplet) + refraction again on exit.

SA — 2M | CBSE 2015; Recurring

Q61. A glass prism is able to produce a spectrum when white light passes through it, but a glass slab does not produce any spectrum. Explain why.

Answer

Prism: The two refracting surfaces are inclined. Dispersion at surface 1 is enhanced at the inclined surface 2 — different colours emerge at different angles forming a spectrum.

Glass slab: The two surfaces are parallel. Dispersion at surface 1 is exactly cancelled at surface 2 (because the surfaces are parallel and the deviation is equal and opposite). All colours emerge parallel to the incident ray — no spectrum.

SA — 3M | CBSE 2022; Recurring

Q62. Savera passed a beam of white light through three equilateral prisms — Prism 1, Prism 2 (inverted relative to 1), Prism 3 (same orientation as 1). (a) Name the colours she would see on the screen. (b) Draw the path of the beam from Prism 1 to Prism 3. (c) Name all the processes that take place.

Answer

(a) Prism 1 → VIBGYOR. Prism 2 (inverted) → recombines → white. Prism 3 → VIBGYOR again. Screen shows VIBGYOR (seven colours).

Newton's two-prism experiment: Prism 1 disperses white light into VIBGYOR; Prism 2 (inverted) recombines them back into white light.

(b) White → Prism 1 → VIBGYOR → Prism 2 (inverted) → white → Prism 3 → VIBGYOR → screen.

(c) Refraction at every glass-air interface; Dispersion in Prism 1 and Prism 3 (splitting into colours); Recombination/Recomposition in Prism 2 (inverted prism reverses dispersion).

SA — 3M | CBSE 2019; 2022; Recurring

Q63. Explain why different colours of white light undergo different amounts of bending while passing through a prism. How does this lead to the formation of a spectrum?

Answer

White light consists of seven colours, each with a different wavelength. In glass, different colours travel at different speeds (violet slowest, red fastest). Since $n = c/v$, each colour has a different refractive index. As bending (deviation) depends on refractive index, violet (highest n, shortest λ) bends most and red (lowest n, longest λ) bends least.

When these differently-bent colours emerge from the second prism face at different angles, they spread out spatially, forming a visible band of colours — the spectrum (VIBGYOR).

Case Study — 4M | CBSE 2023; 2024; 2025; 2026

Q64. Case Study — Dispersion of Light through a Prism

When a narrow beam of white light is passed through a glass prism, it splits into seven colours — VIBGYOR. This is called dispersion. The cause is that different colours have different speeds (and hence different refractive indices) in glass. Violet has the highest refractive index (bends most); red has the lowest (bends least). Newton showed that the spectrum could be recombined into white light by a second inverted prism, proving dispersion is reversible.

(i) Define dispersion of light. State its cause.
(ii) In a glass prism, red is at the top and violet at the bottom. Is the apex up or down? Explain.
(iii) Which colour emerges with: (a) maximum angle of deviation, (b) minimum angle of deviation?
(iv) How did Newton prove that white light is a mixture of seven colours using two prisms?

Answer

(i) Dispersion: Splitting of white light into VIBGYOR on passing through a prism. Cause: Different colours travel at different speeds in glass → different refractive indices → different bending.

(ii) Red at top = least deviated at top; violet at bottom = most deviated at bottom. Light bends towards the base. If violet (most deviated) is at the bottom, the base is at the bottom → apex is up (pointing upward).

(iii) (a) Violet — maximum deviation. (b) Red — minimum deviation.

(iv) Prism 1 (apex up) dispersed white light into VIBGYOR. Prism 2 (identical, inverted/apex down) placed next to Prism 1 recombined all colours back into a single white beam. This proved white light is a mixture of seven colours and that a prism only separates them.

📌 Topic 6 — Scattering of Light (Q65 to Q78)

MCQ — 1M | CBSE Recurring 2015–2026

Q65. The blue colour of the clear sky is due to:

Explanation

Correct Option: (C) — Fine particles (air molecules) scatter shorter wavelengths (blue/violet) much more than longer wavelengths. Scattered blue light reaches our eyes from all parts of the sky → blue sky.

MCQ — 1M | CBSE 2024

Q66. When a beam of white light passes through a region having very fine dust particles, the colour of light mainly scattered in that region is:

Explanation

Correct Option: (C) Blue — Very fine dust particles scatter shorter wavelengths (blue) preferentially (Rayleigh scattering).

MCQ — 1M | CBSE Recurring 2019–2025

Q67. Clouds appear white because:

Explanation

Correct Option: (B) — Large water droplets scatter all wavelengths equally → mixture of all colours → appears white.

MCQ — 1M | CBSE 2016

Q68. Why is red colour selected for danger signal lights?

Explanation

Correct Option: (B) — Red has the longest wavelength → least scattered by fog/smoke → travels farthest → visible from greatest distance.

MCQ — 1M | CBSE Recurring 2018–2025

Q69. The Tyndall effect is observed when:

Explanation

Correct Option: (B) — Tyndall effect: scattering of light by colloidal/fine particles makes the beam path visible.

MCQ A-R — 1M | CBSE 2023; 2024; 2025; 2026

Q70. Assertion (A): The sky appears dark (black) to astronauts in outer space.
Reason (R): There is no atmosphere in outer space, so there are no particles to scatter sunlight. Without scattering, no light reaches the eye from the sky, making it appear black.

Explanation

Correct Option: (A) — Both A and R are true and R correctly explains A. No atmosphere = no scattering = dark sky.

MCQ A-R — 1M | CBSE 2024; 2025

Q71. Assertion (A): Smoke from a chimney appears bluish when viewed sideways (reflected light) but reddish when viewed in the direction of the light source (transmitted light).
Reason (R): Very fine smoke particles scatter shorter wavelengths (blue) more, so blue-scattered light reaches the eye from the side; in the forward direction (transmitted), longer wavelengths (red-enriched) dominate.

Explanation

Correct Option: (A) — Both A and R are true and R correctly explains A.

VSA — 1M | CBSE 2016

Q72. What will be the colour of the sky if it is observed from a place in the complete absence of any atmosphere?

Answer

The sky would appear dark / black — no atmosphere means no particles to scatter sunlight.

SA — 2M | CBSE 2012; 2015

Q73. Define scattering of light. List the factors on which the scattering of light depends.

Answer

Scattering of light: When light falls on particles in the medium, the particles absorb the incident light energy and re-emit it in all directions — this is called scattering of light.

Factors:
1. Wavelength of light (λ): Shorter wavelengths scattered much more (intensity ∝ 1/λ⁴).
2. Size of scattering particles: Fine particles (smaller than λ) scatter shorter wavelengths preferentially; large particles scatter all wavelengths equally.

SA — 2M | CBSE Recurring 2015–2024

Q74. What is the Tyndall effect? Give one example where this effect can be observed in daily life.

Answer

Tyndall effect: When a beam of light passes through a colloidal solution or a medium containing fine suspended particles, the path of the beam becomes visible due to scattering of light by the particles.

Example: A beam of sunlight entering a dusty room through a small opening — the path of the beam is clearly visible as fine dust particles scatter the light.

SA — 2M | CBSE Recurring 2015–2025

Q75. Why is the colour of the clear sky blue? Explain using the concept of scattering of light.

Answer

The atmosphere contains fine particles (air molecules, dust). When sunlight passes through, these particles scatter shorter wavelengths (blue/violet) much more strongly than longer wavelengths — intensity of scattering $\propto \frac{1}{\lambda^4}$ (Rayleigh scattering). The scattered blue light reaches our eyes from all parts of the sky (the eye is more sensitive to blue than to violet), making the clear sky appear blue.

SA — 2M | CBSE Recurring 2016–2025

Q76. Why do clouds appear white even though they are made of water droplets and dust? Explain on the basis of scattering.

Answer

Clouds consist of large water droplets and large dust particles — much bigger than the wavelengths of visible light. Such large particles scatter all wavelengths (colours) approximately equally. When all colours are equally scattered and mixed, the result appears white.

SA — 2M | CBSE 2024

Q77. State the dependence of colour of light we receive on the size of particles of the medium through which the beam passes.

Answer

Fine particles (smaller than λ of light): Scatter shorter wavelengths (blue/violet) preferentially. Light scattered sideways appears bluish; light transmitted straight through appears reddish/orange (blue removed).

Large particles (bigger than λ of light): Scatter all wavelengths equally. Both scattered and transmitted light appear white (e.g., clouds).

Case Study — 4M | CBSE 2023; 2024; 2025; 2026

Q78. Case Study — Scattering of Light

Scattering depends on: (i) wavelength — shorter λ scattered much more (intensity ∝ 1/λ⁴, Rayleigh scattering); (ii) particle size — fine particles scatter blue preferentially; large particles scatter all colours equally (white). Tyndall effect: scattering by colloidal particles makes the light path visible.

(i) State the colour of the sky on a clear day and explain using scattering.
(ii) Why does the sky appear black when viewed from the moon?
(iii) A glass of milk appears slightly bluish when a beam of white light is shone through it. Why?
(iv) Why is red used for danger signals and not blue, even though blue is more easily scattered?

Answer

(i) Blue sky — fine air molecules scatter shorter wavelengths (blue) much more than longer wavelengths ($\propto 1/\lambda^4$). Scattered blue light reaches our eyes from all directions → blue sky.

(ii) The moon has no atmosphere → no particles → no scattering → sky appears dark/black.

(iii) Milk contains fine colloidal particles (fat globules, proteins) smaller than visible light wavelengths. They scatter shorter wavelengths (blue) preferentially (Tyndall effect) → milk appears slightly bluish when viewed sideways.

(iv) Red has the longest wavelength → least scattered by fog/smoke ($\propto 1/\lambda^4$) → travels farthest with minimum intensity loss → visible from greater distance. Blue is easily scattered and lost quickly in foggy conditions → not reliable for long-range danger signals.

📌 Topic 7 — Applications in Daily Life (Q79 to Q86)

⚠️ Colour of the sun at sunrise/sunset is explicitly excluded from 2026-27 CBSE syllabus.

MCQ — 1M | CBSE Recurring 2017–2026

Q79. The formation of a rainbow after rain is caused by:

Explanation

Correct Option: (A) — Rainbow is formed by refraction + dispersion + total internal reflection of sunlight inside water droplets.

MCQ — 1M | CBSE Recurring 2020–2025

Q80. Which of the following is NOT caused by scattering of light?

Explanation

Correct Option: (D) Formation of a rainbow — Rainbow is caused by refraction, dispersion, and internal reflection — NOT scattering.

MCQ A-R — 1M | CBSE 2023; 2024; 2025; 2026

Q81. Assertion (A): A rainbow is a natural spectrum of sunlight formed in the sky.
Reason (R): Water droplets in the atmosphere act as small prisms, refracting and dispersing sunlight; in addition, internal reflection occurs inside each droplet, directing the dispersed colours toward the observer.

Explanation

Correct Option: (A) — Both A and R are true and R correctly explains A.

SA — 3M | CBSE Recurring 2015–2025

Q82. Explain how a rainbow is formed in the sky after a rain shower. Name the three optical phenomena responsible for the formation of a rainbow.

Answer

Red (least deviated) appears on the outer arc; Violet (most deviated) appears on the inner arc. The sun must be behind the observer.

Three phenomena: (1) Refraction, (2) Dispersion, (3) Total internal reflection.

Formation: Tiny water droplets suspended in the air after rain act as tiny prisms:
(1) Sunlight enters a droplet and is refracted and dispersed (split into VIBGYOR).
(2) The dispersed colours undergo total internal reflection at the back surface of the droplet.
(3) On exiting the droplet, the colours are refracted again, further separating them.
An arc of VIBGYOR colours is seen with the sun behind the observer: red on the outside, violet on the inside.

SA — 2M | CBSE 2019; 2022; Recurring

Q83. When is a rainbow observed? An observer can see a rainbow only when the sun is behind him/her. Explain why.

Answer

When observed: After/during a rain shower, when sunlight falls on water droplets present in the sky on the side opposite to the sun.

Why sun must be behind: Inside the water droplet, light undergoes internal reflection which sends the dispersed colours back roughly in the direction the light came from. For this reflected dispersed light to reach the observer's eye, the droplets must be in front and the sun behind the observer — so the internally reflected light travels from the droplets toward the observer, forming the rainbow in the sky opposite to the sun.

SA — 2M | CBSE 2016

Q84. The sun appears white at noon. Explain why, using the concept of scattering of light.

Answer

At noon, the sun is overhead and sunlight travels through the shortest possible path in the atmosphere. Since the path length is short, only a very small amount of blue light is scattered away. The sunlight reaching our eyes directly still contains all colours in nearly equal proportions → the sun appears white (or slightly yellowish-white) at noon.

SA — 3M | CBSE 2018; 2021; 2023; Recurring

Q85. Give a scientific reason for each of the following: (i) The sky appears blue on a clear day. (ii) Danger signals use red colour. (iii) Clouds appear white.

Answer

(i) Blue sky: Fine air molecules and particles scatter shorter wavelengths (blue) much more than longer ones (Rayleigh scattering: $\propto 1/\lambda^4$). Scattered blue light reaches us from all directions → sky appears blue.

(ii) Red danger signals: Red has the longest visible wavelength → least scattered by fog/smoke/dust → travels farthest without significant loss → visible from greater distances.

(iii) White clouds: Large water droplets scatter all wavelengths approximately equally. The equal mixture of all scattered colours appears white.

SA — 2M | CBSE 2019; Recurring

Q86. Explain why a spectroscope uses a prism to identify elements. What property of prisms makes this possible?

Answer

A spectroscope uses a prism because the prism disperses light into its component wavelengths. Each element emits light of specific characteristic wavelengths when heated/excited — producing a unique line emission spectrum.

Property used: Dispersion — different wavelengths (colours) have different refractive indices in glass and are therefore bent by different amounts. This separates the wavelengths emitted by an element spatially, allowing identification from the pattern and positions of spectral lines.

📌 Topic 8 — Mixed Long Answer Questions (5M) (Q87 to Q92)

LA — 5M | CBSE 2011; 2017; 2022; Recurring

Q87. (a) What is hypermetropia? List two causes. (b) Draw ray diagrams showing (i) a hypermetropic eye and (ii) corrected eye. (c) A person with hypermetropia has near point at 75 cm. What is the power of corrective lens? (Normal near point = 25 cm)

Solution

(a) Hypermetropia: Defect where nearby objects appear blurred; image forms behind the retina.
Two causes: (i) Eyeball too short; (ii) Eye lens focal length too large (weak converging power).

(b) Diagrams:
(i) Rays from nearby object → focused behind retina (defective).
(ii) Convex lens before eye → converges rays → image on retina (corrected).

(c) $u = -25$ cm, $v = -75$ cm.
$\frac{1}{f}=\frac{1}{-75}-\frac{1}{-25}=\frac{-1+3}{75}=\frac{2}{75}$
$f = 37.5$ cm = $0.375$ m → $P = +2.67 \approx \mathbf{+3D}$ (convex lens)

LA — 5M | CBSE 2012; 2015; 2019; Recurring

Q88. (a) What is myopia? List two causes. (b) Draw ray diagrams showing (i) myopic eye and (ii) corrected eye. (c) A person cannot see beyond 1.5 m. What is the power of corrective lens?

Solution

(a) Myopia: Defect where distant objects appear blurred; image forms in front of the retina.
Two causes: (i) Eye lens too converging (excessive curvature); (ii) Eyeball elongated.

(b) Diagrams:
(i) Parallel rays from infinity → focused in front of retina (defective).
(ii) Concave lens diverges rays → eye then focuses on retina (corrected).

(c) Far point = 1.5 m → $f = -1.5$ m.
$P = \frac{1}{-1.5} = \mathbf{-0.67D}$ (concave lens)

LA — 5M | CBSE 2009; 2011; 2012; 2014; 2017; Recurring

Q89. (a) What is dispersion of white light? State its cause. (b) Draw a labelled diagram showing dispersion by a glass prism. (c) A glass prism produces a spectrum but a glass slab does not. Explain why. (d) Name the colours on the two extreme ends of the spectrum.

Solution

(a) Dispersion: Splitting of white light into VIBGYOR on passing through a prism. Cause: Different colours travel at different speeds in glass → different refractive indices → different bending.

Violet (shortest λ, highest n) deviates most; Red (longest λ, lowest n) deviates least. The band of colours is the spectrum.

(b) Diagram: Triangular prism (apex up). White beam enters → VIBGYOR spectrum on screen. Label: prism, incident ray (white), VIBGYOR emerging (violet most bent near base, red least bent).

(c) Prism: Inclined faces → deviations add up → spectrum formed. Slab: Parallel faces → dispersion at face 1 is reversed at face 2 → no spectrum (only lateral shift).

(d) Violet (most deviated end) and Red (least deviated end).

LA — 5M | CBSE 2012; 2015; 2018; 2021; Recurring

Q90. (a) Define scattering of light. On what factors does it depend? (b) Explain why: (i) sky appears blue, (ii) clouds appear white, (iii) red light is used for danger signals. (c) State the Tyndall effect with one daily-life example.

Solution

(a) Scattering: Re-emission of absorbed light energy by particles of the medium in all directions. Depends on: (i) wavelength — shorter λ scattered more (∝ 1/λ⁴); (ii) particle size — fine particles scatter blue preferentially; large particles scatter all wavelengths equally.

(b)(i) Blue sky: Fine particles scatter shorter λ (blue) more → scattered blue from all directions → blue sky.
(ii) White clouds: Large water droplets scatter all wavelengths equally → white appearance.
(iii) Red signals: Longest λ → least scattered → visible from greatest distance in fog/smoke.

(c) Tyndall effect: Scattering by colloidal particles making the light path visible. Example: Sunbeam visible in a dusty room through a small window.

LA — 5M | CBSE 2019; 2022; 2025; Recurring

Q91. (a) List three common refractive defects of vision. Suggest the way of correcting each. (b) A person needs a lens of power −5.5 D for distant vision and +1.5 D for reading. (i) What is the focal length of each lens? (ii) What defect does each lens correct?

Solution

(a)
1. Myopia → concave lens.
2. Hypermetropia → convex lens.
3. Presbyopia → bifocal lens.

(b)(i)
$P = -5.5$ D → $f = \frac{1}{-5.5} = -0.182$ m = −18.2 cm (concave).
$P = +1.5$ D → $f = \frac{1}{1.5} = 0.667$ m = +66.7 cm (convex).

(ii)
−5.5 D (concave) → corrects myopia (for distant vision).
+1.5 D (convex) → corrects hypermetropia/presbyopia (for near/reading vision).

LA — 5M | CBSE 2018; 2021; 2023; Recurring

Q92. (a) Explain with a labelled diagram the formation of a rainbow. Name the two optical phenomena responsible. (b) Why is a rainbow always seen opposite to the sun? (c) When fine dust particles are present in a sunlit room, the beam of light is clearly visible. Name this phenomenon and explain why it occurs.

Solution

Red (least deviated) appears on the outer arc; Violet (most deviated) appears on the inner arc. The sun must be behind the observer.

(a) Rainbow formation — Two phenomena: Dispersion and Total internal reflection (also refraction).
Sunlight enters a water droplet → refracted and dispersed at entry → totally internally reflected at back surface → refracted again at exit → each colour exits at a different angle → VIBGYOR arc seen.
Diagram: Show a water droplet with white ray entering, dispersion inside, internal reflection at back, coloured rays emerging at different angles. Red at top of arc, violet at bottom.

(b) The internal reflection inside droplets directs dispersed light back toward the direction of the incoming sunlight. For this reflected light to reach the observer's eye, the sun must be behind the observer and droplets in front — so the rainbow always appears opposite to the sun.

(c) Tyndall effect — Fine dust particles in the room are colloidal in size. They scatter the beam of sunlight, re-emitting it in all directions. This makes the path of the light beam visible to an observer viewing from the side.

📊 Topic-Wise Summary & CBSE Exam Blueprint (Q1–Q92)

Topic-Wise Question Breakdown (Q.1 to Q.92 | 2026-27 Syllabus)

Topic	MCQ (1M)	A-R (1M)	SA/VSA (2–3M)	Case Study (4M)	LA (5M)	Total Qs
1. Functioning of Lens in Human Eye	4	1	5	1	—	Q1–Q11 (11)
2. Defects of Vision & Corrections	6	2	7	1	3	Q12–Q30 (19)
3. Applications of Mirrors & Lenses	3	1	5	—	—	Q31–Q39 (9)
4. Refraction through a Prism	2	2	4	—	1	Q40–Q49 (10)
5. Dispersion of Light	4	1	9	1	—	Q50–Q64 (15)
6. Scattering of Light	5	2	6	1	—	Q65–Q78 (14)
7. Applications in Daily Life	2	1	5	—	—	Q79–Q86 (8)
8. Mixed Long Answers (5M)	—	—	—	—	6	Q87–Q92 (6)
TOTAL PYQ BANK	26	10	41	4	10	92 Questions

⚠️ CBSE Examiner Common Deductions (2022–2026 Mark Schemes)

Mistake	Marks Lost
Drawing ray diagram without labelling retina, lens, cornea	−0.5M
Not drawing image at correct position for defective vs corrected eye	−1M
Saying "convex lens corrects myopia" instead of concave	−1M
Forgetting sign while calculating power (negative for myopia)	−1M
Writing ROYGBIV instead of VIBGYOR for order of deviation	−0.5M
Saying "red is most scattered" instead of "least scattered"	−1M
Not naming all three phenomena for rainbow (refraction, dispersion, internal reflection)	−0.5M each
Confusing Tyndall effect with dispersion	−1M
Not writing "towards the base of the prism" for bending direction	−0.5M
Defining dispersion without mentioning "different colours/wavelengths have different speeds"	−0.5M

Complete Chapter Revision Notes

Definition cheat-sheet • Defect master tables • VIBGYOR & colourful-world mnemonics • Last-minute board checklist

📖 One-Line Definitions (Board Favourites)

Accommodation: The ability of the eye lens to adjust its focal length so that objects at different distances form clear images on the retina.
Near point (least distance of distinct vision): The minimum distance at which objects can be seen most distinctly without strain — 25 cm for a young adult.
Far point: The farthest point up to which the eye can see objects clearly — infinity for a normal eye.
Myopia: Defect in which the far point shifts closer than infinity; image of distant objects forms in front of the retina; corrected by a concave lens.
Hypermetropia: Defect in which the near point recedes beyond 25 cm; image of nearby objects forms behind the retina; corrected by a convex lens.
Presbyopia: Age-related decrease of accommodation (weak ciliary muscles, inflexible lens); corrected by convex or bifocal lenses.
Angle of deviation (∠D): The angle between the incident ray and the emergent ray on passing through a prism, $D = i + e - A$.
Dispersion: Splitting of white light into its component colours (VIBGYOR) by a prism.
Spectrum: The band of coloured components of a light beam.
Tyndall effect: Scattering of light by colloidal particles which makes the path of a light beam visible.
Atmospheric refraction: Refraction of light by the Earth's atmosphere, whose refractive index changes gradually.

👁️ Defects of Vision — Master Table (High-Yield)

Defect	Why it happens	Image position	Corrective lens	Formula link
Myopia	Excess curvature of eye lens / elongated eyeball	In front of retina	Concave (−)	$P = -\dfrac{1}{d_{far} \text{ (m)}}$
Hypermetropia	Focal length too long / eyeball too small	Behind retina	Convex (+)	$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$; $u=-25$, $v=-N'$
Presbyopia	Weakening of ciliary muscles & inflexible lens (age)	Behind retina (near objects)	Convex or bifocal	Upper concave − lower convex

🌈 The Colourful World — Phenomenon → Reason Table

Phenomenon	Cause	Key line for the answer
Blue sky	Scattering	Shorter wavelengths (blue) are scattered more strongly than red
Dark sky for astronaut	No atmosphere	No scattering without atmosphere
Red Sun at sunrise/sunset	Scattering	Longer atmospheric path; blue scattered away, red reaches eye
Red danger signals	Scattering	Red is least scattered by fog/smoke → visible far away
Rainbow	Dispersion + internal reflection	Droplets act like prisms; opposite to the Sun
Spectrum by prism	Dispersion	Different colours deviate by different angles
Twinkling of stars	Atmospheric refraction	Point sources; fluctuating path of starlight
Advance sunrise / delayed sunset	Atmospheric refraction	Sun appears ~2 min early and sets ~2 min late; disc flattened
Visible light beam in a dark room	Tyndall effect	Colloidal particles scatter the beam

⚡ Master Formula Sheet

Concept	Formula	Key Sign Rule / Note
Power of a lens	$P = \dfrac{1}{f(m)} = \dfrac{100}{f(cm)}$	Unit = dioptre (D). Convex = $+$, Concave = $-$
Focal length from power	$f = \dfrac{1}{P}$ (metres)	Convert to cm × 100
Myopia correction	$f = -d_{\text{far point}}$	$P = -\dfrac{1}{d \text{ (m)}}$ → concave lens
Hypermetropia correction	$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$	$u = -25cm$, $v = -N'$ → convex lens
Angle of deviation	$\angle D = i + e - A$	Prism deviates emergent ray towards the base

🧠 High-Yield Mnemonics & Memory Tricks

VIBGYOR: Violet → Indigo → Blue → Green → Yellow → Orange → Red. Trick: "Very Intense Blue Gets Yellow On Rainbows".
M-C = In front: Myopia-Concave → image in front of retina; H-Cx = Beyond: Hypermetropia-Convex → image beyond retina.
P-P-E: Point sources → Produce twinkle; Extended sources → no twinkle (planets).
Blue Belongs to Short: Blue is Bent (scattered) Best because it has the shortest wavelength — so the sky is blue and setting sun is red.
R-R = Retina Role: Retina is the recording surface like a camera's film/sensor.
Cave = Inward: ConCAVE lens bulges inward (diverges rays, power negative). ConVEX bulges outward (converges, power positive).

🎯 Last-Minute Board Exam Checklist

Always convert focal length to metres before computing power in dioptres.
State the type of lens (concave/convex) + its power + where the image now forms (on the retina).
For myopia: $f = -$ far point distance. For hypermetropia: solve lens formula with $u = -25$ cm and $v = -N'$.
Mention wavelength in scattering answers: shorter λ scattered more strongly.
In the prism diagram label PE (incident), EF (refracted), FS (emergent), ∠i, ∠r, ∠e, ∠A, ∠D.
In the eye diagram, add arrows on the light rays and label cornea, iris, pupil, lens, retina and optic nerve.
Recall: a normal eye sees clearly from 25 cm to infinity; far point of a myopic eye is closer than infinity.

Chapter 10 Mastery Test

3 progressive difficulty levels • 10 MCQs each • Instant scoring and answer review

Concepts & Visual Theory

🔄 How Focal Length Changes While Seeing

🔭 Myopia at a Glance

🔭 Hypermetropia at a Glance

🏷️ Presbyopia at a Glance

📋 COMPARISON TABLE — The 3 Defects of Vision (Board Favourite)

🔺 Key Terms for the Prism Diagram

🌈 VIBGYOR — The Splitting of White Light

⚖️ How Different Colours Behave in a Prism

🌈 3 Steps Inside a Raindrop (Board Definition)

🌌 Three Famous Effects of Atmospheric Refraction

🎨 Colourful Phenomena — Reason Table (Memorise)

NCERT Chapter-End Exercises (Q1 to Q12)

CBSE Class 10 Science • Chapter 10 PYQ Bank (2011–2026)

📌 Topic 1 — Functioning of a Lens in the Human Eye (Q1 to Q11)

Q1. Which part of the human eye controls the amount of light entering the eye?

Q2. The lens in the human eye is:

Q3. The least distance of distinct vision for a normal human eye is:

Q4. The far point of a normal human eye is:

Q5. Assertion (A): The human eye can change its focal length to focus on objects at different distances.Reason (R): The ciliary muscles change the curvature (and hence the focal length) of the eye lens — this ability is called power of accommodation.

Q6. Define the term 'power of accommodation' of the human eye. What is the function of the iris and pupil of the eye?

Q7. Define power of accommodation of the human eye. What happens to the image distance in the eye when we increase the distance of an object from the eye? Name and explain the role of the part of the human eye responsible for this.

Q8. Draw a neat labelled diagram of the human eye. Name the part that: (i) controls the amount of light entering the eye, (ii) is responsible for converting light signals to electrical signals, (iii) focuses the image on the retina.

Q9. What is the range of vision of a normal adult eye? Why is a normal eye unable to see objects placed closer than 25 cm?

Q10. Explain the working of a human eye as an optical instrument. How does it adjust its focal length to see objects at different distances?

Q11. Case Study — The Human Eye as an Optical Instrument

📌 Topic 2 — Defects of Vision and Their Corrections (Q12 to Q30)

Q12. Which lens is used to correct myopia?

Q13. Which lens is used to correct hypermetropia?

Q14. A man suffering from myopia cannot see objects distinctly at a distance greater than 2 m. Which lens is required to correct this defect?(A) −2 D (B) +2 D (C) −0.5 D (D) +0.5 D

Q15. The near point of a hypermetropic person lies at 75 cm. Which lens is required to correct this defect? (Normal near point = 25 cm)(A) +3 D (B) −3 D (C) +2 D (D) −2 D

Q16. Cataract is caused due to:

Q17. A person uses spectacles with convex lenses. The name of the defect of vision he is suffering from is:

Q18. Assertion (A): Myopia is corrected using a concave lens.Reason (R): In a myopic eye, the image is formed in front of the retina because the eyeball is elongated or the lens is too converging. A concave lens diverges the light rays to shift the image back onto the retina.

Q20. Priya prefers to sit in the front row as she finds it difficult to read the blackboard from the last desk in her classroom. Name the defect of vision she is suffering from.

Q21. Name the type of defect of vision a person is suffering from if he uses convex lenses in his spectacles. Give two possible causes for this defect.

Q22. Sheela cannot read a newspaper held closer than 100 cm. Find (i) the name of the defect, (ii) the required power of lens with sign and calculation. (Normal near point = 25 cm)

Q23. A short-sighted person has a far point of 1.2 m. Find the power of the concave lens required to correct this defect for distant vision.

Q24. What is myopia? List two causes. How can it be corrected using a lens? Draw ray diagrams: (i) defective eye, (ii) corrected eye.

Q25. What is hypermetropia? Describe with a ray diagram how this defect can be corrected. List two possible causes.

Q26. A student is unable to see the blackboard placed at ~4 m. (i) Name the defect of vision. (ii) With labelled ray diagrams explain: (a) the defective eye, and (b) the corrected eye after using a suitable lens.

Q27. An old person has difficulty in reading a book as well as seeing distant objects. (i) Name the defect. (ii) What type of lens corrects this? Explain its structure. (iii) State the cause of this defect.

Q29. [Ray diagram shows image forming behind retina for a nearby object.] (i) Name the defect, stating the part of eye responsible. (ii) List two causes. (iii) Name the type of lens used to correct it and state its role.

Q30. Case Study — Defects of Vision and Their Correction

📌 Topic 3 — Applications of Spherical Mirrors and Lenses (Q31 to Q39)

Q31. Which type of mirror is used as a rear-view mirror in vehicles?

Q32. A dentist uses which type of mirror to see an enlarged image of a patient's tooth?

Q33. Concave mirrors are used in solar furnaces because:

Q34. Assertion (A): Large concave mirrors are used in reflecting telescopes to collect light from distant stars.Reason (R): Concave mirrors converge a parallel beam of incoming light to its focal point, enabling collection and focus of light from distant sources.

Q35. Name the type of mirror used in solar furnaces. Explain how a high temperature is achieved.

Q36. Why do we prefer a convex mirror as a rear-view mirror in vehicles? Draw a ray diagram to show image formation.

Q37. Give two uses each of: (i) concave mirrors, and (ii) convex mirrors in daily life. Give reasons for each use.

Q38. Where should an object be placed in front of a convex lens to use it as: (i) a magnifying glass, (ii) a projector lens, (iii) a camera lens? State the characteristics of the image in each case.

Q39. List four uses of concave mirrors. For each use, state the position of the object and the nature of image required.

📌 Topic 4 — Refraction of Light Through a Prism (Q40 to Q49)

Q40. When a ray of light passes through a glass prism, the emergent ray:

Q41. The angle between the incident ray and the emergent ray when a ray of light passes through a glass prism is called the:

Q42. Name the component of white light that deviates: (i) the least, and (ii) the most, while passing through a glass prism.

Q43. Light of two colours A and B passes through a glass prism. Colour A deviates more than colour B. Which colour has a higher speed in the prism?

Q46. What is the angle of deviation in a prism? Draw a labelled diagram showing the refraction of a monochromatic ray through a glass prism. Mark the angle of incidence, angle of refraction, angle of emergence, and angle of deviation.

Q47. A student traces the path of a ray of light through a rectangular glass slab and a triangular glass prism. In which case does the emergent ray appear shifted/deviated? Give reason.

Q48. Why does a glass prism cause bending of light towards its base? Draw a ray diagram showing the path of a narrow beam of white light through a glass prism.

📌 Topic 5 — Dispersion of Light (Q50 to Q64)

Q50. The phenomenon of splitting white light into its component colours is called:

Q51. The correct sequence of colours in the visible spectrum (from most deviated to least deviated) is:

Q52. Which colour of white light has the highest frequency?

Q53. Why does the refractive index of glass differ for different colours of light?

Q54. The intensity of scattered light by particles in the atmosphere varies as the nth power of the wavelength of light, where n =

Q56. Define dispersion of light. What is the cause of dispersion?

Q57. What is a spectrum? Write the colours of the visible spectrum in order from least deviated to most deviated.

Q58. Name the colours that suffer: (i) maximum deviation, and (ii) minimum deviation while passing through a glass prism.

Q59. What is dispersion of light? What is the cause of dispersion? Draw a neat diagram to show the dispersion of white light by a glass prism.

Q60. Name the phenomenon occurring in nature that is due to dispersion of light. Name the three optical phenomena responsible for the formation of a rainbow.

Q61. A glass prism is able to produce a spectrum when white light passes through it, but a glass slab does not produce any spectrum. Explain why.

Q63. Explain why different colours of white light undergo different amounts of bending while passing through a prism. How does this lead to the formation of a spectrum?

Q64. Case Study — Dispersion of Light through a Prism

📌 Topic 6 — Scattering of Light (Q65 to Q78)

Q65. The blue colour of the clear sky is due to:

Q66. When a beam of white light passes through a region having very fine dust particles, the colour of light mainly scattered in that region is:

Q67. Clouds appear white because:

Q5. Assertion (A): The human eye can change its focal length to focus on objects at different distances.
Reason (R): The ciliary muscles change the curvature (and hence the focal length) of the eye lens — this ability is called power of accommodation.

Q14. A man suffering from myopia cannot see objects distinctly at a distance greater than 2 m. Which lens is required to correct this defect?
(A) −2 D (B) +2 D (C) −0.5 D (D) +0.5 D

Q15. The near point of a hypermetropic person lies at 75 cm. Which lens is required to correct this defect? (Normal near point = 25 cm)
(A) +3 D (B) −3 D (C) +2 D (D) −2 D

Q18. Assertion (A): Myopia is corrected using a concave lens.
Reason (R): In a myopic eye, the image is formed in front of the retina because the eyeball is elongated or the lens is too converging. A concave lens diverges the light rays to shift the image back onto the retina.

Q34. Assertion (A): Large concave mirrors are used in reflecting telescopes to collect light from distant stars.
Reason (R): Concave mirrors converge a parallel beam of incoming light to its focal point, enabling collection and focus of light from distant sources.

Q70. Assertion (A): The sky appears dark (black) to astronauts in outer space.
Reason (R): There is no atmosphere in outer space, so there are no particles to scatter sunlight. Without scattering, no light reaches the eye from the sky, making it appear black.

Q81. Assertion (A): A rainbow is a natural spectrum of sunlight formed in the sky.
Reason (R): Water droplets in the atmosphere act as small prisms, refracting and dispersing sunlight; in addition, internal reflection occurs inside each droplet, directing the dispersed colours toward the observer.