CBSE Class 10 · Physics
CBSE Class 10 Physics The Human Eye And The Colourful World Previous Year Questions
Help your child ace CBSE Class 10 Physics The Human Eye and The Colourful World with this expert-curated collection of previous year board questions. Covering defects of vision, dispersion of light, scattering, and atmospheric refraction, every question comes with a detailed step-by-step solution to build your child’s confidence before the board exam.
CBSE Class 10 Physics The Human Eye And The Colourful World — Questions with Solutions
In the following diagram showing dispersion of white light by a glass prism, the colours ‘P’ and ‘Q’ respectively are:
Solution
Answer: Option (B) is correct.
Explanation:
When white light passes through a glass prism, it splits into seven colours — Violet, Indigo, Blue, Green, Yellow, Orange, and Red — due to dispersion of light. Each colour bends by a different amount because of its different wavelength.
Violet light bends the most, while Red light bends the least.
Hence, in the diagram: P represents Violet (more deviation) and Q represents Red (less deviation).
Explanation:
When white light passes through a glass prism, it splits into seven colours — Violet, Indigo, Blue, Green, Yellow, Orange, and Red — due to dispersion of light. Each colour bends by a different amount because of its different wavelength.
Violet light bends the most, while Red light bends the least.
Hence, in the diagram: P represents Violet (more deviation) and Q represents Red (less deviation).
In the human eye, the part which allows light to enter into the eye is:
Solution
Answer: Option (B) is correct.
Explanation: The pupil is the opening at the centre of the iris through which light enters the eye. Its size adjusts automatically to control the amount of light entering, depending on the surrounding brightness. The cornea helps in focusing light and the retina detects it, but the pupil is the actual entry point for light.
Explanation: The pupil is the opening at the centre of the iris through which light enters the eye. Its size adjusts automatically to control the amount of light entering, depending on the surrounding brightness. The cornea helps in focusing light and the retina detects it, but the pupil is the actual entry point for light.
The change in the focal length of an eye lens in human beings is caused by the action of:
Solution
Answer: Option (B) is correct.
Explanation: The ciliary muscles control the shape of the eye lens. When they contract, the lens becomes more curved, shortening the focal length to focus on nearby objects. When they relax, the lens flattens, increasing the focal length for distant objects. This ability to change focal length is called the power of accommodation.
Explanation: The ciliary muscles control the shape of the eye lens. When they contract, the lens becomes more curved, shortening the focal length to focus on nearby objects. When they relax, the lens flattens, increasing the focal length for distant objects. This ability to change focal length is called the power of accommodation.
When light enters the atmosphere, it strikes on extremely fine particles, which deflect the rays of light in all possible directions. This is due to:
Solution
Answer: Option (C) is correct.
Explanation: When light enters the Earth’s atmosphere, it interacts with extremely fine particles such as dust, water droplets, and gas molecules. These particles scatter the light in all directions — a phenomenon known as scattering of light. Scattering is responsible for the blue sky during the day and the reddening of the sun during sunrise and sunset.
Explanation: When light enters the Earth’s atmosphere, it interacts with extremely fine particles such as dust, water droplets, and gas molecules. These particles scatter the light in all directions — a phenomenon known as scattering of light. Scattering is responsible for the blue sky during the day and the reddening of the sun during sunrise and sunset.
The face of the moon that is visible to us is called the near side, and the face of the moon which is invisible to us is called the far side. What colour would the sky appear to an astronaut standing on the “far side” of the moon and why?
Solution
Answer: Option (C) is correct.
Explanation: The far side of the Moon does receive sunlight, but since the Moon has no atmosphere, there is no scattering of light. On Earth, the atmosphere scatters sunlight making the sky appear blue. Without any atmosphere on the Moon, there is no scattering, so the sky would appear black even during the day.
Explanation: The far side of the Moon does receive sunlight, but since the Moon has no atmosphere, there is no scattering of light. On Earth, the atmosphere scatters sunlight making the sky appear blue. Without any atmosphere on the Moon, there is no scattering, so the sky would appear black even during the day.
When we enter a dark room coming from outside, immediately the things inside the room do not appear clear to our eyes. This is because:
Solution
Answer: Option (B) is correct.
Explanation: When we move from a bright environment to a dark one, our pupils need time to dilate (widen) to allow more light into the eye. This process — known as dark adaptation — takes a few moments. In a bright room, pupils are small to limit light intake, but in darkness they must expand, which does not happen instantaneously.
Explanation: When we move from a bright environment to a dark one, our pupils need time to dilate (widen) to allow more light into the eye. This process — known as dark adaptation — takes a few moments. In a bright room, pupils are small to limit light intake, but in darkness they must expand, which does not happen instantaneously.
The phenomena of light responsible for the working of the human eye is:
Solution
Answer: Option (B) is correct.
Explanation: The working of the human eye primarily relies on refraction. Light passes through the cornea and the lens, bending to focus onto the retina. The power of accommodation (Option C) is a property of the eye, but it too relies on refraction as its underlying physical mechanism.
Explanation: The working of the human eye primarily relies on refraction. Light passes through the cornea and the lens, bending to focus onto the retina. The power of accommodation (Option C) is a property of the eye, but it too relies on refraction as its underlying physical mechanism.
Study the following ray diagram. In this diagram, the angle of incidence, the angle of emergence, and the angle of deviation respectively have been represented by:
Solution
Answer: Option (C) is correct.
Explanation:
• p — angle of incidence: formed between the incident ray and the normal at the first surface.
• y — angle of emergence: formed between the emergent ray and the normal at the second surface.
• z — angle of deviation: formed between the incident ray extended and the emergent ray.
Explanation:
• p — angle of incidence: formed between the incident ray and the normal at the first surface.
• y — angle of emergence: formed between the emergent ray and the normal at the second surface.
• z — angle of deviation: formed between the incident ray extended and the emergent ray.
In an experiment to trace the path of a ray of light through a triangular glass prism, a student would observe that the emergent ray:
Solution
Answer: Option (D) is correct.
Explanation: When light travels from air (rarer medium) into glass (denser medium), it bends towards the normal. As it exits through the second face of the prism, it moves from glass to air and bends away from the normal. Since light exits from the thicker base side of the prism, the emergent ray bends towards the thicker part (base) of the prism.
Explanation: When light travels from air (rarer medium) into glass (denser medium), it bends towards the normal. As it exits through the second face of the prism, it moves from glass to air and bends away from the normal. Since light exits from the thicker base side of the prism, the emergent ray bends towards the thicker part (base) of the prism.
Assertion (A): A hypermetropic person prefers to remove his spectacles while driving.
Reason (R): When a hypermetropic person wearing the spectacles looks at a distant object, the parallel rays from the distant object get converged in front of the retina. The image thus appears blurred.
(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
(B) Both (A) and (R) are true but R is NOT the correct explanation of (A).
(C) (A) is true but (R) is false.
(D) (A) is false and (R) is true.
Reason (R): When a hypermetropic person wearing the spectacles looks at a distant object, the parallel rays from the distant object get converged in front of the retina. The image thus appears blurred.
(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
(B) Both (A) and (R) are true but R is NOT the correct explanation of (A).
(C) (A) is true but (R) is false.
(D) (A) is false and (R) is true.
Solution
Answer: Option (A) is correct.
Explanation: A hypermetropic person uses a convex lens to converge light for near vision. However, when wearing convex spectacles and looking at distant objects, the parallel rays from that distant object get converged in front of the retina, blurring the image. To avoid this, the person prefers to remove their spectacles while driving (viewing distant objects). Both A and R are true, and R correctly explains A.
Explanation: A hypermetropic person uses a convex lens to converge light for near vision. However, when wearing convex spectacles and looking at distant objects, the parallel rays from that distant object get converged in front of the retina, blurring the image. To avoid this, the person prefers to remove their spectacles while driving (viewing distant objects). Both A and R are true, and R correctly explains A.
Assertion (A): A beam of white light gives a spectrum on passing through a glass prism.
Reason (R): Speed of light outside the prism is different from the speed of light inside the prism.
(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
(B) Both (A) and (R) are true but R is NOT the correct explanation of (A).
(C) (A) is true but (R) is false.
(D) (A) is false and (R) is true.
Reason (R): Speed of light outside the prism is different from the speed of light inside the prism.
(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
(B) Both (A) and (R) are true but R is NOT the correct explanation of (A).
(C) (A) is true but (R) is false.
(D) (A) is false and (R) is true.
Solution
Answer: Option (B) is correct.
Explanation: A beam of white light does give a spectrum (VIBGYOR) on passing through a glass prism — Assertion (A) is true. While it is also true that the speed of light outside and inside the prism are different (Reason R is true), this does not correctly explain the spectrum formation. The actual reason dispersion occurs is that the refractive index of the glass is different for each colour (each wavelength), causing them to bend by different amounts.
Explanation: A beam of white light does give a spectrum (VIBGYOR) on passing through a glass prism — Assertion (A) is true. While it is also true that the speed of light outside and inside the prism are different (Reason R is true), this does not correctly explain the spectrum formation. The actual reason dispersion occurs is that the refractive index of the glass is different for each colour (each wavelength), causing them to bend by different amounts.
Write the function of each of the following parts of the human eye:
(i) Cornea
(ii) Iris
(iii) Crystalline lens
(iv) Ciliary muscles
(i) Cornea
(ii) Iris
(iii) Crystalline lens
(iv) Ciliary muscles
Answer
(i) Cornea: It is the transparent front surface of the eye that allows light to enter and helps in focusing the incoming light rays.
(ii) Iris: It controls the amount of light entering the eye by adjusting the size of the pupil according to the brightness of the surroundings.
(iii) Crystalline lens: It converges the incoming light rays and focuses them onto the retina to form a clear image.
(iv) Ciliary muscles: They adjust the focal length of the eye lens by contracting or relaxing, enabling the eye to focus on objects at different distances (accommodation). They also help hold the lens in place.
(ii) Iris: It controls the amount of light entering the eye by adjusting the size of the pupil according to the brightness of the surroundings.
(iii) Crystalline lens: It converges the incoming light rays and focuses them onto the retina to form a clear image.
(iv) Ciliary muscles: They adjust the focal length of the eye lens by contracting or relaxing, enabling the eye to focus on objects at different distances (accommodation). They also help hold the lens in place.
(a) List two differences in the characteristic properties of the virtual images formed by the two types of spherical lenses (concave and convex).
(b) How are these characteristics of the two lenses used in the correction of the two common defects of vision namely myopia and hypermetropia?
(b) How are these characteristics of the two lenses used in the correction of the two common defects of vision namely myopia and hypermetropia?
Answer
(a) Two differences in virtual images formed by concave and convex lenses:
(i) Concave lenses always form virtual, erect, and smaller images, while convex lenses can form virtual, erect, and magnified images (depending on object position).
(ii) The virtual image formed by a concave lens is always situated behind the lens (on the same side as the object), while the virtual image formed by a convex lens is also behind the lens but is magnified compared to the object.
(b) Use in correction of vision defects:
Myopia: A concave lens is used to correct myopia. It diverges the incoming light rays before they enter the eye, shifting the focal point back onto the retina so that distant objects appear clear.
Hypermetropia: A convex lens is used to correct hypermetropia. It converges the incoming light rays before they enter the eye, bringing the focus forward onto the retina so that nearby objects appear clear.
(i) Concave lenses always form virtual, erect, and smaller images, while convex lenses can form virtual, erect, and magnified images (depending on object position).
(ii) The virtual image formed by a concave lens is always situated behind the lens (on the same side as the object), while the virtual image formed by a convex lens is also behind the lens but is magnified compared to the object.
(b) Use in correction of vision defects:
Myopia: A concave lens is used to correct myopia. It diverges the incoming light rays before they enter the eye, shifting the focal point back onto the retina so that distant objects appear clear.
Hypermetropia: A convex lens is used to correct hypermetropia. It converges the incoming light rays before they enter the eye, bringing the focus forward onto the retina so that nearby objects appear clear.
(a) A lens of focal length 5 cm is being used by Debashree in the laboratory as a magnifying glass. Her least distance of distinct vision is 25 cm.
(i) What is the magnification obtained by using the glass?
(ii) She keeps a book at a distance of 10 cm from her eyes and tries to read. She is, however, unable to read. What is the reason for this?
(b) Ravi kept a book at a distance of 10 cm from the eyes of his friend Hari. However, Hari is unable to read it.
(i) What is the magnification obtained by using the glass?
(ii) She keeps a book at a distance of 10 cm from her eyes and tries to read. She is, however, unable to read. What is the reason for this?
(b) Ravi kept a book at a distance of 10 cm from the eyes of his friend Hari. However, Hari is unable to read it.
Answer
(a)(i) Given: image distance $v = -25 \, \text{cm}$, focal length $f = 5 \, \text{cm}$
Using the lens formula: $$\frac{1}{f} = \frac{1}{v} – \frac{1}{u}$$ $$\frac{1}{5} = \frac{1}{-25} – \frac{1}{u}$$ $$\frac{1}{u} = \frac{1}{25} – \frac{1}{5} = \frac{1-5}{25} = \frac{-4}{25}$$ So $u = -6.25 \, \text{cm}$
Magnification: $m = \frac{v}{u} = \frac{-25}{-6.25} = 4$
The magnification obtained is 4.
(a)(ii) She was unable to read because 10 cm is less than her least distance of distinct vision (25 cm). The book is placed closer than her near point, so the eye lens cannot converge the rays to form a sharp image on the retina.
(b) Hari is unable to read because his least distance of distinct vision is more than 25 cm — the book is placed closer than his near point, suggesting he may be suffering from hypermetropia.
Using the lens formula: $$\frac{1}{f} = \frac{1}{v} – \frac{1}{u}$$ $$\frac{1}{5} = \frac{1}{-25} – \frac{1}{u}$$ $$\frac{1}{u} = \frac{1}{25} – \frac{1}{5} = \frac{1-5}{25} = \frac{-4}{25}$$ So $u = -6.25 \, \text{cm}$
Magnification: $m = \frac{v}{u} = \frac{-25}{-6.25} = 4$
The magnification obtained is 4.
(a)(ii) She was unable to read because 10 cm is less than her least distance of distinct vision (25 cm). The book is placed closer than her near point, so the eye lens cannot converge the rays to form a sharp image on the retina.
(b) Hari is unable to read because his least distance of distinct vision is more than 25 cm — the book is placed closer than his near point, suggesting he may be suffering from hypermetropia.
The near point of the eye of a person is 50 cm. Find the nature and power of the corrective lens required by the person to enable him to see clearly the objects placed at 25 cm from the eye.
Answer
Given:
Object distance, $u = -25 \, \text{cm}$
Image distance, $v = -50 \, \text{cm}$ (image must form at the person’s near point)
Using the lens formula: $$\frac{1}{f} = \frac{1}{v} – \frac{1}{u} = \frac{1}{-50} – \frac{1}{-25} = -\frac{1}{50} + \frac{1}{25} = \frac{-1+2}{50} = \frac{1}{50}$$ Thus $f = 50 \, \text{cm} = 0.5 \, \text{m}$
Power: $P = \dfrac{1}{f} = \dfrac{1}{0.5} = +2 \, \text{D}$
The corrective lens required is a convex lens with power +2 D. The positive power indicates the lens is convex (converging), used to correct hypermetropia (the person cannot see near objects clearly).
Object distance, $u = -25 \, \text{cm}$
Image distance, $v = -50 \, \text{cm}$ (image must form at the person’s near point)
Using the lens formula: $$\frac{1}{f} = \frac{1}{v} – \frac{1}{u} = \frac{1}{-50} – \frac{1}{-25} = -\frac{1}{50} + \frac{1}{25} = \frac{-1+2}{50} = \frac{1}{50}$$ Thus $f = 50 \, \text{cm} = 0.5 \, \text{m}$
Power: $P = \dfrac{1}{f} = \dfrac{1}{0.5} = +2 \, \text{D}$
The corrective lens required is a convex lens with power +2 D. The positive power indicates the lens is convex (converging), used to correct hypermetropia (the person cannot see near objects clearly).
(a) A person is suffering from both myopia and hypermetropia.
(i) What kind of lenses can correct this defect?
(ii) How are these lenses prepared?
(b) A person needs a lens of power +3D for correcting his near vision and –3D for correcting his distant vision. Calculate the focal lengths of the lenses required to correct these defects.
(i) What kind of lenses can correct this defect?
(ii) How are these lenses prepared?
(b) A person needs a lens of power +3D for correcting his near vision and –3D for correcting his distant vision. Calculate the focal lengths of the lenses required to correct these defects.
Answer
(a)(i) The type of lens required is a bifocal lens.
(a)(ii) A bifocal lens consists of both a concave lens and a convex lens in a single frame:
• The upper portion contains a concave lens to correct distant vision (myopia).
• The lower portion contains a convex lens to correct near vision (hypermetropia).
(b) Focal length for near vision ($P_1 = +3 \, \text{D}$): $$f_1 = \frac{1}{P_1} = \frac{1}{+3} = +0.33 \, \text{m} = +33.33 \, \text{cm}$$ Focal length for distant vision ($P_2 = -3 \, \text{D}$): $$f_2 = \frac{1}{P_2} = \frac{1}{-3} = -0.33 \, \text{m} = -33.33 \, \text{cm}$$ The convex lens has focal length +33.33 cm and the concave lens has focal length –33.33 cm.
(a)(ii) A bifocal lens consists of both a concave lens and a convex lens in a single frame:
• The upper portion contains a concave lens to correct distant vision (myopia).
• The lower portion contains a convex lens to correct near vision (hypermetropia).
(b) Focal length for near vision ($P_1 = +3 \, \text{D}$): $$f_1 = \frac{1}{P_1} = \frac{1}{+3} = +0.33 \, \text{m} = +33.33 \, \text{cm}$$ Focal length for distant vision ($P_2 = -3 \, \text{D}$): $$f_2 = \frac{1}{P_2} = \frac{1}{-3} = -0.33 \, \text{m} = -33.33 \, \text{cm}$$ The convex lens has focal length +33.33 cm and the concave lens has focal length –33.33 cm.
A student needs spectacles of power –0.5 D for the correction of his vision.
(i) Name the defect in vision the student is suffering from.
(ii) Find the nature and focal length of the corrective lens.
(iii) List two causes of this defect.
(i) Name the defect in vision the student is suffering from.
(ii) Find the nature and focal length of the corrective lens.
(iii) List two causes of this defect.
Answer
(i) The student is suffering from Myopia (nearsightedness / short-sightedness).
(ii) The corrective lens required is a concave (diverging) lens.
Focal length: $$f = \frac{1}{P} = \frac{1}{-0.5} = -2 \, \text{m}$$ The focal length of the corrective lens is $f = -2 \, \text{m}$.
(iii) Two causes of myopia:
1. Excessive curvature of the eye lens (making it too converging).
2. Elongation of the eyeball (increasing the distance between the lens and retina).
(ii) The corrective lens required is a concave (diverging) lens.
Focal length: $$f = \frac{1}{P} = \frac{1}{-0.5} = -2 \, \text{m}$$ The focal length of the corrective lens is $f = -2 \, \text{m}$.
(iii) Two causes of myopia:
1. Excessive curvature of the eye lens (making it too converging).
2. Elongation of the eyeball (increasing the distance between the lens and retina).
Due to gradual weakening of ciliary muscles and diminishing flexibility of the eye lens, a certain defect of vision arises.
(i) Write the name of this defect.
(ii) Name the type of lens required by such persons to improve the vision.
(iii) Explain the structure and function of such a lens.
(i) Write the name of this defect.
(ii) Name the type of lens required by such persons to improve the vision.
(iii) Explain the structure and function of such a lens.
Answer
(i) The defect is called Presbyopia.
(ii) The type of lens required to correct Presbyopia is a bifocal lens.
(iii) Structure and Function of the Bifocal Lens:
• Upper portion: Contains a concave (diverging) lens, which helps the person view far-off objects clearly (correcting the myopic component if present).
• Lower portion: Contains a convex (converging) lens, which helps the person view nearby objects clearly (correcting the hypermetropic component).
This dual structure allows the presbyopic person to see clearly at both near and far distances using a single pair of spectacles.
(ii) The type of lens required to correct Presbyopia is a bifocal lens.
(iii) Structure and Function of the Bifocal Lens:
• Upper portion: Contains a concave (diverging) lens, which helps the person view far-off objects clearly (correcting the myopic component if present).
• Lower portion: Contains a convex (converging) lens, which helps the person view nearby objects clearly (correcting the hypermetropic component).
This dual structure allows the presbyopic person to see clearly at both near and far distances using a single pair of spectacles.
The colour of clear sky from the Earth appears blue, but from space, it appears black. Why?
Answer
From Earth — Sky appears Blue: The Earth’s atmosphere contains fine particles (dust, gas molecules) that scatter light. Shorter wavelengths such as blue light are scattered much more than longer wavelengths by these particles. This scattered blue light reaches our eyes from all directions, making the sky appear blue.
From Space — Sky appears Black: Space has no atmosphere and therefore no particles to scatter light. Without scattering, light travels in straight lines and there is no diffused light reaching the observer from the surroundings. Hence, the sky appears dark and black from space.
From Space — Sky appears Black: Space has no atmosphere and therefore no particles to scatter light. Without scattering, light travels in straight lines and there is no diffused light reaching the observer from the surroundings. Hence, the sky appears dark and black from space.
Give reasons for the following:
(a) Red traffic signals can be seen from a very long distance.
(b) Stars appear to be slightly higher than their actual position.
(a) Red traffic signals can be seen from a very long distance.
(b) Stars appear to be slightly higher than their actual position.
Answer
(a) Red traffic signals visible from a long distance:
Red light has the highest wavelength among visible colours. Due to its long wavelength, red light is the least scattered by atmospheric particles. Since very little red light is scattered away, it can travel long distances without losing intensity, making it visible from very far away. This is why red is used for danger signals and traffic lights.
(b) Stars appear slightly higher than their actual position:
This phenomenon is due to atmospheric refraction. As light from a star enters the Earth’s atmosphere, it passes through layers of air with gradually increasing optical density. This causes the light to bend (refract) towards the normal as it moves from less dense to more dense layers. This bending makes the apparent position of the star slightly higher than its true geometric position in the sky.
Red light has the highest wavelength among visible colours. Due to its long wavelength, red light is the least scattered by atmospheric particles. Since very little red light is scattered away, it can travel long distances without losing intensity, making it visible from very far away. This is why red is used for danger signals and traffic lights.
(b) Stars appear slightly higher than their actual position:
This phenomenon is due to atmospheric refraction. As light from a star enters the Earth’s atmosphere, it passes through layers of air with gradually increasing optical density. This causes the light to bend (refract) towards the normal as it moves from less dense to more dense layers. This bending makes the apparent position of the star slightly higher than its true geometric position in the sky.
State the phenomena observed in the given diagram. Explain with reference to the diagram, which of the two lights mentioned below will have a higher wavelength?
Answer
Phenomenon: The phenomenon observed is dispersion of light — the splitting of white light into its seven constituent colours when passed through a prism.
Explanation of wavelength: The velocity of light in a medium is directly proportional to its wavelength (for a given frequency). Since yellow light travels faster through the prism than blue light (it is deviated less), yellow light has a greater wavelength than blue light.
Therefore, yellow light has the higher wavelength.
Explanation of wavelength: The velocity of light in a medium is directly proportional to its wavelength (for a given frequency). Since yellow light travels faster through the prism than blue light (it is deviated less), yellow light has a greater wavelength than blue light.
Therefore, yellow light has the higher wavelength.
How will you use two identical prisms so that a narrow beam of white light incident on one prism emerges out of the second prism as white light? Draw the diagram.
Answer
To recombine dispersed white light, the second prism is placed in an inverted position (apex pointing downward) relative to the first prism. The first prism disperses the white light into its constituent colours (VIBGYOR). When this dispersed spectrum enters the second inverted prism, each colour refracts in the opposite direction, and all the colours recombine to emerge as white light.
A student traces the path of a ray of light through a glass prism but leaves it incomplete and unlabelled. Complete and label the diagram with ∠i, ∠e, ∠r, and ∠D.
Answer
1. ∠i — Angle of incidence: The angle between the incident ray and the normal drawn at the first surface of the prism.
2. ∠r — Angle of refraction: The angle between the refracted ray inside the prism and the normal at the first surface.
3. ∠e — Angle of emergence: The angle between the emergent ray and the normal drawn at the second surface of the prism.
4. ∠D — Angle of deviation: The angle between the original direction of the incident ray (extended) and the direction of the emergent ray.
2. ∠r — Angle of refraction: The angle between the refracted ray inside the prism and the normal at the first surface.
3. ∠e — Angle of emergence: The angle between the emergent ray and the normal drawn at the second surface of the prism.
4. ∠D — Angle of deviation: The angle between the original direction of the incident ray (extended) and the direction of the emergent ray.
In the figure given below, a narrow beam of white light is shown to pass through a triangular glass prism. After passing through the prism, it produces a spectrum XY on the screen.
(a) Name the phenomenon.
(b) State the colours seen at X and Y.
(c) Why do different colours of white light bend at different angles through a prism?
(a) Name the phenomenon.
(b) State the colours seen at X and Y.
(c) Why do different colours of white light bend at different angles through a prism?
Answer
(a) The phenomenon is called Dispersion of light.
(b) X: Violet Y: Red
(Violet bends the most and is at the top of the spectrum; Red bends the least and is at the bottom.)
(c) Different colours of white light have different wavelengths and therefore different speeds inside the prism. Since the refractive index of the glass is different for each wavelength, each colour bends by a different angle when entering and exiting the prism. Violet light (shortest wavelength) bends the most and red light (longest wavelength) bends the least.
(b) X: Violet Y: Red
(Violet bends the most and is at the top of the spectrum; Red bends the least and is at the bottom.)
(c) Different colours of white light have different wavelengths and therefore different speeds inside the prism. Since the refractive index of the glass is different for each wavelength, each colour bends by a different angle when entering and exiting the prism. Violet light (shortest wavelength) bends the most and red light (longest wavelength) bends the least.
Why is Tyndall effect shown by colloidal particles? State four instances of observing the Tyndall effect.
Answer
Why colloidal particles show the Tyndall effect:
The size of colloidal particles is roughly comparable to the wavelength of visible light. When light passes through a colloid, these particles scatter the light in all directions, making the path of the light beam visible. This scattering of light by colloidal particles is called the Tyndall effect.
Four instances of the Tyndall effect:
1. Scattering of light by dust and smoke particles in fog, making car headlight beams visible.
2. When sunlight passes through a gap in the canopy of a forest, the beam of light is visible due to scattering by dust and water droplets.
3. The blue colour of the sky — fine atmospheric particles scatter blue light more than other colours.
4. When a beam of light passes through milk — milk is a colloid and its fat globules scatter the light.
The size of colloidal particles is roughly comparable to the wavelength of visible light. When light passes through a colloid, these particles scatter the light in all directions, making the path of the light beam visible. This scattering of light by colloidal particles is called the Tyndall effect.
Four instances of the Tyndall effect:
1. Scattering of light by dust and smoke particles in fog, making car headlight beams visible.
2. When sunlight passes through a gap in the canopy of a forest, the beam of light is visible due to scattering by dust and water droplets.
3. The blue colour of the sky — fine atmospheric particles scatter blue light more than other colours.
4. When a beam of light passes through milk — milk is a colloid and its fat globules scatter the light.
Observe the following diagram and answer the following questions:
(i) Identify the defect of vision shown.
(ii) List two causes.
(iii) Name the type of lens used for the correction of this defect.
(i) Identify the defect of vision shown.
(ii) List two causes.
(iii) Name the type of lens used for the correction of this defect.
Answer
(i) The defect of vision shown is Myopia (nearsightedness), as the image is formed in front of the retina for distant objects.
(ii) Two causes of Myopia:
1. Elongation of the eyeball.
2. Excessive curvature of the eye lens.
(iii) Myopia is corrected by using a concave lens of suitable focal length.
(ii) Two causes of Myopia:
1. Elongation of the eyeball.
2. Excessive curvature of the eye lens.
(iii) Myopia is corrected by using a concave lens of suitable focal length.
List two differences in the characteristic properties of the virtual images formed by the two types of spherical lenses (concave and convex). How are these characteristics of the two lenses used in the correction of the two common defects of vision, namely myopia and hypermetropia?
Answer
Two differences in virtual images:
1. Concave lenses always form virtual, erect, and smaller images, while convex lenses can form virtual, erect, and magnified images or real, inverted images depending on object position.
2. The virtual image formed by a concave lens is always behind the lens, closer to the lens than the object. The virtual image formed by a convex lens is also behind the lens but is magnified.
Correction of vision defects:
Myopia: A concave lens diverges the incoming light rays before they reach the eye, effectively shifting the focal point back onto the retina, allowing the person to see distant objects clearly.
Hypermetropia: A convex lens converges the light rays before they enter the eye, moving the focal point forward onto the retina, allowing the person to see nearby objects clearly.
1. Concave lenses always form virtual, erect, and smaller images, while convex lenses can form virtual, erect, and magnified images or real, inverted images depending on object position.
2. The virtual image formed by a concave lens is always behind the lens, closer to the lens than the object. The virtual image formed by a convex lens is also behind the lens but is magnified.
Correction of vision defects:
Myopia: A concave lens diverges the incoming light rays before they reach the eye, effectively shifting the focal point back onto the retina, allowing the person to see distant objects clearly.
Hypermetropia: A convex lens converges the light rays before they enter the eye, moving the focal point forward onto the retina, allowing the person to see nearby objects clearly.
State reasons for myopia. With the help of ray diagrams, show the:
(a) Image formation by a myopic eye.
(b) Correction of myopia using an appropriate lens.
(a) Image formation by a myopic eye.
(b) Correction of myopia using an appropriate lens.
Answer
Reasons for Myopia:
1. Excessive curvature of the eye lens — the lens becomes too converging.
2. Elongation of the eyeball — the retina is too far from the lens.
(a) Image formation by a myopic eye:
In a myopic eye, light rays from a distant object converge in front of the retina (before reaching it), forming a blurred image on the retina. This happens because the focal length of the eye lens is too short relative to the eyeball length.
(b) Correction of Myopia:
A concave lens of suitable focal length is placed in front of the eye. It diverges the incoming parallel rays before they enter the eye, so that after refraction by the eye lens, they converge exactly on the retina, producing a clear image.
1. Excessive curvature of the eye lens — the lens becomes too converging.
2. Elongation of the eyeball — the retina is too far from the lens.
(a) Image formation by a myopic eye:
In a myopic eye, light rays from a distant object converge in front of the retina (before reaching it), forming a blurred image on the retina. This happens because the focal length of the eye lens is too short relative to the eyeball length.
(b) Correction of Myopia:
A concave lens of suitable focal length is placed in front of the eye. It diverges the incoming parallel rays before they enter the eye, so that after refraction by the eye lens, they converge exactly on the retina, producing a clear image.
The near point of the eye of a person is 50 cm. Find the nature and power of the corrective lens required by the person to enable him to see clearly the objects placed at 25 cm from the eye.
Answer
Given:
Object distance, $u = -25 \, \text{cm}$
Image distance, $v = -50 \, \text{cm}$ (image must form at the near point of 50 cm)
Using the lens formula: $$\frac{1}{f} = \frac{1}{v} – \frac{1}{u} = \frac{1}{-50} – \frac{1}{-25} = -\frac{1}{50} + \frac{1}{25} = \frac{-1+2}{50} = \frac{1}{50}$$ Thus $f = 50 \, \text{cm} = 0.5 \, \text{m}$
Power: $P = \dfrac{1}{f} = \dfrac{1}{0.5} = +2 \, \text{D}$
The corrective lens is a convex lens with power +2 D. The positive power confirms it is a converging (convex) lens, used to correct hypermetropia.
Object distance, $u = -25 \, \text{cm}$
Image distance, $v = -50 \, \text{cm}$ (image must form at the near point of 50 cm)
Using the lens formula: $$\frac{1}{f} = \frac{1}{v} – \frac{1}{u} = \frac{1}{-50} – \frac{1}{-25} = -\frac{1}{50} + \frac{1}{25} = \frac{-1+2}{50} = \frac{1}{50}$$ Thus $f = 50 \, \text{cm} = 0.5 \, \text{m}$
Power: $P = \dfrac{1}{f} = \dfrac{1}{0.5} = +2 \, \text{D}$
The corrective lens is a convex lens with power +2 D. The positive power confirms it is a converging (convex) lens, used to correct hypermetropia.
A person is unable to see objects distinctly placed within 75 cm from his eyes.
(a) Name the defect of vision the person is suffering from.
(b) List its two possible causes.
(c) Calculate the power of the lens needed to correct this defect. Assume that the near point for the normal eye is 25 cm.
(a) Name the defect of vision the person is suffering from.
(b) List its two possible causes.
(c) Calculate the power of the lens needed to correct this defect. Assume that the near point for the normal eye is 25 cm.
Answer
(a) The person is suffering from Hypermetropia (long-sightedness / farsightedness). The person cannot see objects placed within 75 cm, meaning their near point is 75 cm instead of the normal 25 cm.
(b) Two possible causes:
1. Decrease in the curvature of the eye lens (lens becomes too flat).
2. Shortening of the eyeball.
(c) Calculation of power of corrective lens:
Object distance, $u = -25 \, \text{cm}$ (normal near point)
Image distance, $v = -75 \, \text{cm}$ (person’s near point)
Using the lens formula: $$\frac{1}{f} = \frac{1}{v} – \frac{1}{u} = \frac{1}{-75} – \frac{1}{-25} = -\frac{1}{75} + \frac{1}{25} = \frac{-1+3}{75} = \frac{2}{75}$$ Thus $f = \dfrac{75}{2} = 37.5 \, \text{cm} = 0.375 \, \text{m}$
Power: $P = \dfrac{1}{0.375} = +2.67 \, \text{D}$
A convex lens of power +2.67 D is required to correct the defect.
(b) Two possible causes:
1. Decrease in the curvature of the eye lens (lens becomes too flat).
2. Shortening of the eyeball.
(c) Calculation of power of corrective lens:
Object distance, $u = -25 \, \text{cm}$ (normal near point)
Image distance, $v = -75 \, \text{cm}$ (person’s near point)
Using the lens formula: $$\frac{1}{f} = \frac{1}{v} – \frac{1}{u} = \frac{1}{-75} – \frac{1}{-25} = -\frac{1}{75} + \frac{1}{25} = \frac{-1+3}{75} = \frac{2}{75}$$ Thus $f = \dfrac{75}{2} = 37.5 \, \text{cm} = 0.375 \, \text{m}$
Power: $P = \dfrac{1}{0.375} = +2.67 \, \text{D}$
A convex lens of power +2.67 D is required to correct the defect.
(a) A student is unable to see clearly the words written on the blackboard placed at a distance of approximately 3 m from him. Name the defect of vision the boy is suffering from. State the possible causes of this defect and explain the method of correcting it.
(b) Why do stars twinkle? Explain.
(b) Why do stars twinkle? Explain.
Answer
(a) Defect of Vision: The student is suffering from Myopia (nearsightedness).
Causes of Myopia:
1. Excessive curvature of the eye lens, making it too converging.
2. Elongation of the eyeball, causing the retina to lie too far from the lens.
Method of Correction: Myopia is corrected by using a concave lens of suitable focal length placed in front of the eye. The concave lens diverges the incoming light rays so that after refraction by the eye lens, they converge precisely on the retina, enabling the student to see distant objects clearly.
(b) Why do stars twinkle?
Stars twinkle due to atmospheric refraction. Starlight travels through the Earth’s atmosphere, which is composed of layers of air with varying temperatures and densities. As the light passes through these continuously moving layers, it undergoes refraction in different directions at different instants. This causes the apparent position and brightness of the star to fluctuate rapidly, producing the twinkling effect.
Planets do not twinkle because they are much closer to Earth and subtend a larger angle — they effectively act as extended sources, so the small fluctuations average out and appear steady.
Causes of Myopia:
1. Excessive curvature of the eye lens, making it too converging.
2. Elongation of the eyeball, causing the retina to lie too far from the lens.
Method of Correction: Myopia is corrected by using a concave lens of suitable focal length placed in front of the eye. The concave lens diverges the incoming light rays so that after refraction by the eye lens, they converge precisely on the retina, enabling the student to see distant objects clearly.
(b) Why do stars twinkle?
Stars twinkle due to atmospheric refraction. Starlight travels through the Earth’s atmosphere, which is composed of layers of air with varying temperatures and densities. As the light passes through these continuously moving layers, it undergoes refraction in different directions at different instants. This causes the apparent position and brightness of the star to fluctuate rapidly, producing the twinkling effect.
Planets do not twinkle because they are much closer to Earth and subtend a larger angle — they effectively act as extended sources, so the small fluctuations average out and appear steady.
(a) A person is suffering from both myopia and hypermetropia.
(i) What kind of lenses can correct this defect?
(ii) How are these lenses prepared?
(b) A person needs a lens of power +3 D for correcting his near vision and –3 D for correcting his distant vision. Calculate the focal lengths of the lenses required to correct these defects.
(i) What kind of lenses can correct this defect?
(ii) How are these lenses prepared?
(b) A person needs a lens of power +3 D for correcting his near vision and –3 D for correcting his distant vision. Calculate the focal lengths of the lenses required to correct these defects.
Answer
(a)(i) The lens required is a bifocal lens.
(a)(ii) A bifocal lens is prepared by combining two types of lenses in a single frame:
• The upper portion consists of a concave lens to help with distant vision (correcting myopia).
• The lower portion consists of a convex lens to help with near vision (correcting hypermetropia).
(b) Focal length calculations:
For near vision ($P_N = +3 \, \text{D}$): $$f_N = \frac{1}{P_N} = \frac{1}{+3} = +0.33 \, \text{m} = +33.33 \, \text{cm}$$ The convex lens has a focal length of +33.33 cm.
For distant vision ($P_D = -3 \, \text{D}$): $$f_D = \frac{1}{P_D} = \frac{1}{-3} = -0.33 \, \text{m} = -33.33 \, \text{cm}$$ The concave lens has a focal length of –33.33 cm.
(a)(ii) A bifocal lens is prepared by combining two types of lenses in a single frame:
• The upper portion consists of a concave lens to help with distant vision (correcting myopia).
• The lower portion consists of a convex lens to help with near vision (correcting hypermetropia).
(b) Focal length calculations:
For near vision ($P_N = +3 \, \text{D}$): $$f_N = \frac{1}{P_N} = \frac{1}{+3} = +0.33 \, \text{m} = +33.33 \, \text{cm}$$ The convex lens has a focal length of +33.33 cm.
For distant vision ($P_D = -3 \, \text{D}$): $$f_D = \frac{1}{P_D} = \frac{1}{-3} = -0.33 \, \text{m} = -33.33 \, \text{cm}$$ The concave lens has a focal length of –33.33 cm.
(a) What is meant by the term ‘power of accommodation’? Name the component of the eye that is responsible for the power of accommodation.
(b) A student sitting at the back bench in a class has difficulty in reading the blackboard. What could be his defect of vision? Draw ray diagrams to illustrate the image formation of the blackboard when he sits in the:
(i) back seat
(ii) front seat.
State two possible causes of this defect. Explain the method of correcting this defect with the help of a ray diagram.
(b) A student sitting at the back bench in a class has difficulty in reading the blackboard. What could be his defect of vision? Draw ray diagrams to illustrate the image formation of the blackboard when he sits in the:
(i) back seat
(ii) front seat.
State two possible causes of this defect. Explain the method of correcting this defect with the help of a ray diagram.
Answer
(a) Power of Accommodation:
The power of accommodation is the ability of the eye lens to adjust its focal length in order to focus clearly on objects at varying distances (both near and far). The ciliary muscles are responsible for this by contracting or relaxing to change the curvature of the eye lens.
(b) Defect of Vision: The student is suffering from Myopia (short-sightedness).
Causes of Myopia:
1. Excessive curvature of the eye lens.
2. Elongation of the eyeball.
Image Formation:
• Back seat: The blackboard is far away. In a myopic eye, the light from the distant board converges in front of the retina, so the image on the retina is blurred.
• Front seat: The blackboard is now at a short distance. The myopic eye can focus on nearby objects, so the image forms correctly on the retina and appears clear.
Correction: A concave lens of suitable focal length is used. It diverges the light rays before they enter the eye, shifting the point of convergence back onto the retina for distant objects.
The power of accommodation is the ability of the eye lens to adjust its focal length in order to focus clearly on objects at varying distances (both near and far). The ciliary muscles are responsible for this by contracting or relaxing to change the curvature of the eye lens.
(b) Defect of Vision: The student is suffering from Myopia (short-sightedness).
Causes of Myopia:
1. Excessive curvature of the eye lens.
2. Elongation of the eyeball.
Image Formation:
• Back seat: The blackboard is far away. In a myopic eye, the light from the distant board converges in front of the retina, so the image on the retina is blurred.
• Front seat: The blackboard is now at a short distance. The myopic eye can focus on nearby objects, so the image forms correctly on the retina and appears clear.
Correction: A concave lens of suitable focal length is used. It diverges the light rays before they enter the eye, shifting the point of convergence back onto the retina for distant objects.
Savera passed a beam of white light through a series of equilateral prisms as shown:
(a) What colour(s) will be seen on the screen?
(b) Copy the diagram above and draw the beam entering Prism 1 and emerging from Prism 3, falling on the screen.
(c) Name all the processes that take place when the beam of light enters Prism 1 and emerges from Prism 3.
(a) What colour(s) will be seen on the screen?
(b) Copy the diagram above and draw the beam entering Prism 1 and emerging from Prism 3, falling on the screen.
(c) Name all the processes that take place when the beam of light enters Prism 1 and emerges from Prism 3.
Answer
(a) Colours seen on the screen:
The colours seen on the screen will be the full spectrum of white light — VIBGYOR (Violet, Indigo, Blue, Green, Yellow, Orange, Red).
(b) Diagram:
The white light enters Prism 1 and is dispersed into its constituent colours. The dispersed beam passes through Prism 2 and Prism 3, with the spectrum appearing on the screen.
(c) Processes that take place:
1. Refraction — Light bends as it enters and exits each prism (moves between air and glass).
2. Dispersion — White light splits into its component colours (VIBGYOR) due to the different refractive indices of the glass for each wavelength. Violet bends the most and Red bends the least.
The colours seen on the screen will be the full spectrum of white light — VIBGYOR (Violet, Indigo, Blue, Green, Yellow, Orange, Red).
(b) Diagram:
The white light enters Prism 1 and is dispersed into its constituent colours. The dispersed beam passes through Prism 2 and Prism 3, with the spectrum appearing on the screen.
(c) Processes that take place:
1. Refraction — Light bends as it enters and exits each prism (moves between air and glass).
2. Dispersion — White light splits into its component colours (VIBGYOR) due to the different refractive indices of the glass for each wavelength. Violet bends the most and Red bends the least.
Define the term ‘power of accommodation’. Write the modification in the curvature of the eye lens which enables us to see the nearby objects clearly.
Answer
Power of accommodation: It is the ability of the eye lens to adjust its focal length so as to focus clearly on objects placed at different distances from the eye.
Modification in curvature to see nearby objects: To see nearby objects clearly, the ciliary muscles contract, causing the eye lens to become more curved (thicker / greater curvature). This shortens the focal length of the lens, enabling it to converge the diverging rays from a nearby object onto the retina.
Modification in curvature to see nearby objects: To see nearby objects clearly, the ciliary muscles contract, causing the eye lens to become more curved (thicker / greater curvature). This shortens the focal length of the lens, enabling it to converge the diverging rays from a nearby object onto the retina.
Explore More CBSE Class 10 Physics Chapters
Frequently Asked Questions
What is the chapter The Human Eye and The Colourful World about in CBSE Class 10 Physics? ⌄
This chapter covers the structure and functioning of the human eye, including defects of vision (myopia, hypermetropia, and presbyopia) and their correction using lenses. It also explains optical phenomena such as refraction, dispersion of light by a prism, scattering of light, atmospheric refraction, and the reasons behind natural observations like the blue sky, twinkling of stars, and the reddening of the sun at sunrise and sunset.
How many marks does this chapter carry in the CBSE Class 10 board exam? ⌄
The Human Eye and The Colourful World is part of the Light unit in CBSE Class 10 Physics. This unit typically carries around 12–14 marks in the Science board paper. Questions from this chapter appear as 1-mark MCQs, 2-mark and 3-mark short answers, and 5-mark long answer or case study questions, making it one of the higher-weightage chapters in the Science paper.
What are the most important topics your child must focus on in this chapter? ⌄
The most important topics are: the structure and functions of parts of the human eye, the three defects of vision (myopia, hypermetropia, presbyopia) along with their causes, ray diagrams, and lens corrections, power of accommodation, dispersion of white light through a prism (with VIBGYOR), scattering of light (Tyndall effect), atmospheric refraction, and numerical problems on lens formula and power of a lens. All of these appear regularly in previous year board papers.
What common mistakes do students make when answering questions from this chapter? ⌄
One very common error is confusing which lens corrects which defect — students sometimes write that myopia is corrected by a convex lens (it is corrected by a concave lens). Another frequent mistake is drawing ray diagrams incorrectly or forgetting to label angles such as ∠i, ∠e, and ∠D. In numerical problems, students often forget to convert focal lengths from centimetres to metres before calculating power. Practising previous year questions with Angle Belearn’s step-by-step solutions helps your child avoid all these errors.
How does Angle Belearn help students score well in this chapter? ⌄
Angle Belearn’s CBSE specialists curate chapter-wise previous year question banks with fully verified, step-by-step solutions — including all ray diagrams and numerical workings. Students who practise on Angle Belearn develop the habit of writing structured answers with correct diagram labelling, which earns full marks in board exams. Regular practice builds the speed and accuracy your child needs to walk into the exam with complete confidence.
