The Human Eye and the Colourful World — Class 10 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 10 Physics chapter "The Human Eye and the Colourful World" — 10 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 10 Physics chapter "The Human Eye and the Colourful World" — 10 important questions with detailed answers…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
NCERT Solutions for Class 10 Physics — The Human Eye and the Colourful World. Step-by-step answers to all 10 textbook questions from this chapter, written for CBSE board preparation and free to use.
The Human Eye and the Colourful World — All 10 Questions Solved
Q1. Define the 'power of accommodation' of the human eye. How does the human eye achieve it?
The 'power of accommodation' of the human eye is its ability to adjust the focal length of the eye lens to see objects clearly at varying distances. This is done by changing the curvature of the eye lens.
How it is achieved:
1. For distant objects: When the eye needs to see distant objects, the ciliary muscles relax. This makes the eye lens thinner and its focal length increases, allowing the light rays from distant objects to focus precisely on the retina.
2. For nearby objects: When the eye needs to see nearby objects, the ciliary muscles contract. This causes the eye lens to become thicker and more curved, which decreases its focal length. This increased converging power helps to focus the light rays from nearby objects accurately on the retina.
This adjustment of focal length by the ciliary muscles allows us to see both distant and nearby objects clearly.
Q2. A person cannot see objects distinctly beyond 2 meters. What kind of defect of vision is he suffering from? What type of corrective lens should be used to restore proper vision? Calculate the power of the required lens.
1. Defect of Vision: The person cannot see distant objects clearly, which means he is suffering from Myopia (nearsightedness).
2. Corrective Lens: Myopia is corrected using a concave lens (diverging lens) of appropriate focal length.
3. Calculation of Power of the lens:
* For a myopic eye, the far point is closer than infinity. Here, the person's far point is 2 meters.
* To correct this defect, the concave lens should form a virtual image of a distant object (at infinity) at the person's far point (2 m).
* Object distance, $u = -\infty$ (objects at infinity).
* Image distance, $v = -2$ m (image formed at the person's far point).
* Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
* $\frac{1}{f} = \frac{1}{-2} - \frac{1}{-\infty}$
* $\frac{1}{f} = -\frac{1}{2} - 0$
* $\frac{1}{f} = -\frac{1}{2}$ m$^{-1}$
* Focal length, $f = -2$ m
* Now, calculate the power of the lens:
Power, $P = \frac{1}{f}$ (where f is in meters)
$P = \frac{1}{-2}$ D
$P = -0.5$ Dioptre (D)
Conclusion: The person is suffering from Myopia. A concave lens with a power of -0.5 D is required to correct his vision.
Q3. The near point of a hypermetropic person is 75 cm from the eye. What is the power of the lens required to enable him to read clearly a book held at 25 cm from the eye?
1. Defect of Vision: The person's near point is 75 cm, which is further than the normal near point (25 cm). This indicates he is suffering from Hypermetropia (farsightedness).
2. Corrective Lens: Hypermetropia is corrected using a convex lens (converging lens).
3. Calculation of Power of the lens:
* The book needs to be held at 25 cm (normal near point).
* Object distance, $u = -25$ cm $= -0.25$ m.
* The convex lens should form a virtual image of the book at the person's near point (75 cm) so that the eye can see it clearly.
* Image distance, $v = -75$ cm $= -0.75$ m.
* Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
* $\frac{1}{f} = \frac{1}{-0.75} - \frac{1}{-0.25}$
* $\frac{1}{f} = -\frac{1}{0.75} + \frac{1}{0.25}$
* $\frac{1}{f} = -\frac{4}{3} + \frac{4}{1}$
* $\frac{1}{f} = -\frac{4}{3} + \frac{12}{3}$
* $\frac{1}{f} = \frac{8}{3}$ m$^{-1}$
* Focal length, $f = \frac{3}{8}$ m $= 0.375$ m
* Now, calculate the power of the lens:
Power, $P = \frac{1}{f}$ (where f is in meters)
$P = \frac{1}{3/8}$ D
$P = \frac{8}{3}$ D
$P \approx +2.67$ Dioptre (D)
Conclusion: A convex lens with a power of approximately +2.67 D is required to enable the person to read a book clearly at 25 cm.
Q4. What is meant by the dispersion of white light? Draw a neat labelled diagram to show the dispersion of white light by a glass prism. Name the seven colours of the spectrum.
Dispersion of white light is the phenomenon of splitting of white light into its constituent seven colours when it passes through a transparent medium like a glass prism. This happens because different colours of light have different wavelengths and thus travel at different speeds through the prism, causing them to bend (refract) at different angles.
Diagram showing dispersion of white light by a glass prism:
```
White Light
|\
| \ Incident Ray
| \
| \
| \______ Angle of Incidence (i)
| /\
| / \
| / \
| / \
| / \
(Prism) \ Refracted light (Spectrum)
/ \ /\
/ \ / \
/ \ / \
/ \ / \
/ \ / \
/ \/__________\ <--- Base of Prism
/
/
/ V (Violet)
/ I (Indigo)
/ B (Blue)
/ G (Green)
Y (Yellow)
O (Orange)
R (Red)
<--- Angle of Deviation (δ) between incident white light and emergent colour rays.
```
(Note: In a textual format, drawing a precise ray diagram is challenging. Imagine a triangular prism. A white light ray enters one face, bends and splits into seven distinct colours, and then emerges from the other face, bending further, with violet being most deviated and red being least deviated.)
The seven colours of the spectrum (VIBGYOR):
1. Violet
2. Indigo
3. Blue
4. Green
5. Yellow
6. Orange
7. Red
Q5. Explain why stars twinkle, but planets do not.
The twinkling of stars and the non-twinkling of planets can be explained by the phenomenon of atmospheric refraction and the difference in their apparent sizes from Earth.
1. Stars Twinkle:
* Point Sources: Stars are extremely far away from Earth, so they appear almost as point sources of light to us.
* Atmospheric Refraction: The Earth's atmosphere is made up of layers of varying densities and temperatures. This causes the refractive index of the atmosphere to keep fluctuating randomly. When light from a star enters the Earth's atmosphere, it undergoes continuous refraction in different directions before reaching our eyes.
* Fluctuating Light: As the path of light rays from the star changes continuously due to these atmospheric variations, the apparent position of the star fluctuates, and the amount of starlight reaching our eyes varies. Sometimes more light reaches, sometimes less. This continuous variation in the intensity of light makes the star appear to twinkle.
2. Planets Do Not Twinkle:
* Extended Sources: Planets are much closer to Earth than stars and appear as extended sources of light (a collection of many point sources) to us.
* Averaging Effect: Because a planet is an extended source, it can be considered as a large number of point sources of light. While the light from individual 'point sources' on the planet might fluctuate due to atmospheric refraction, the total amount of light reaching our eye from all these point sources averages out. The increase in brightness from one part of the planet is often compensated by a decrease from another part.
* Net Effect: As a result, the total amount of light entering our eye from a planet remains nearly constant, and thus, planets do not appear to twinkle.
Q6. Why does the sky appear blue on a clear day?
The blue colour of the sky on a clear day is primarily due to the phenomenon of scattering of light by the tiny particles (molecules of air, dust particles, etc.) present in the Earth's atmosphere. This phenomenon is known as Rayleigh scattering.
Here's the explanation:
1. Composition of White Light: Sunlight, which appears white to us, is composed of different colours, each having a different wavelength (from violet to red).
2. Scattering by Air Molecules: The molecules of air (like nitrogen and oxygen) and other fine particles in the atmosphere have sizes smaller than the wavelength of visible light. These particles are more effective at scattering light of shorter wavelengths (like blue and violet) than light of longer wavelengths (like red and orange).
3. Preferential Scattering of Blue Light: Blue light, having a shorter wavelength, is scattered much more strongly and in all directions by the atmospheric particles compared to red light, which has a longer wavelength. (According to Rayleigh's law of scattering, the intensity of scattered light is inversely proportional to the fourth power of the wavelength, I $\propto$ 1/$\lambda^4$).
4. Reaching Our Eyes: When we look at the sky, the blue light scattered from all directions by the atmospheric particles enters our eyes. This is why the sky appears blue.
Although violet light has an even shorter wavelength than blue and scatters more, our eyes are more sensitive to blue light. Also, some violet light is absorbed in the upper atmosphere. Hence, we perceive the sky as blue rather than violet.
Q7. Explain why the Sun appears reddish at sunrise and sunset.
The Sun appears reddish at sunrise and sunset due to the phenomenon of scattering of light by the Earth's atmosphere.
Here's the detailed explanation:
1. Longer Path Through Atmosphere: During sunrise and sunset, the Sun is near the horizon. At this time, sunlight has to travel a much longer distance through the Earth's atmosphere to reach our eyes compared to when the Sun is overhead (at noon).
2. More Scattering of Shorter Wavelengths: As sunlight travels this longer path, the shorter wavelength colours (like blue and violet) are scattered away much more effectively by the tiny air molecules and particles in the atmosphere. This is due to Rayleigh scattering, where shorter wavelengths scatter more intensely.
3. Remaining Longer Wavelengths: By the time the sunlight, depleted of most of its blue and violet components, reaches our eyes, the light that remains consists mainly of longer wavelength colours (like red, orange, and yellow). These colours are scattered least.
4. Reddish Appearance: As a result, the Sun and the sky around it appear reddish or orange-red. The more the light travels through the atmosphere, the more the blue light is scattered away, leaving behind the red components to reach our eyes.
Q8. What is an angle of deviation for a prism? Draw a ray diagram showing the path of light through a triangular glass prism and label the angle of deviation.
Angle of Deviation (δ) for a prism:
It is defined as the angle between the direction of the incident ray and the direction of the emergent ray when light passes through a prism.
Ray diagram showing the path of light through a triangular glass prism:
```
Incident Ray (PQ) E F
\ / \
\ R / \
\ / \
\ / \
\ / \
\ / \
\ / \
\ / \
A-----------------B
\ /
\ /
\ /
\ /
\ /
\ /
\ / Emergent Ray (RS)
\ /
In the diagram above:
* ABC is the cross-section of a triangular glass prism.
* PQ is the incident ray.
* QR is the refracted ray inside the prism.
* RS is the emergent ray.
* ∠i is the angle of incidence.
* ∠r1 is the angle of refraction at the first surface (AB).
* ∠r2 is the angle of incidence at the second surface (AC).
* ∠e is the angle of emergence.
* The angle between the direction of the incident ray (extended forward) and the emergent ray (extended backward) is the Angle of Deviation (δ).
(Note: In a textual format, drawing a precise ray diagram is challenging. Imagine a ray entering surface AB, bending towards the normal, then exiting surface AC, bending away from the normal. The angle formed by extending the incident ray and emergent ray to meet is the angle of deviation.)
```
Q9. Briefly describe the functions of the following parts of the human eye: (a) Pupil (b) Ciliary muscles (c) Retina.
Here are the brief functions of the specified parts of the human eye:
(a) Pupil:
* Function: The pupil acts like an aperture (opening) whose size is regulated by the iris. Its main function is to control and regulate the amount of light entering the eye. In bright light, the iris contracts the pupil to reduce the amount of light, preventing over-exposure. In dim light, the iris expands the pupil to allow more light to enter, helping us to see better.
(b) Ciliary muscles:
* Function: These muscles are attached to the eye lens. Their primary function is to change the curvature and therefore the focal length of the eye lens, a process called accommodation. By relaxing or contracting, they make the lens thinner or thicker, allowing the eye to focus on objects at different distances (near or far) clearly on the retina.
(c) Retina:
* Function: The retina is the light-sensitive screen or layer at the back of the eye. It contains millions of light-sensitive cells (rods and cones). Its main functions are:
1. To form a real and inverted image of the object being viewed.
2. To convert the light energy into electrical signals. These electrical signals are then sent to the brain via the optic nerves, where the brain interprets them, and we perceive the object as upright and complete.
Q10. Differentiate between Myopia and Hypermetropia based on: (a) cause, (b) near point/far point, and (c) corrective lens.
Here's a differentiation between Myopia and Hypermetropia:
| Feature | Myopia (Nearsightedness) | Hypermetropia (Farsightedness) |
| :---------------- | :-------------------------------------------------------- | :------------------------------------------------------------- |
| (a) Cause | 1. Excessive curvature of the eye lens (lens becomes too converging).<br>2. Elongation of the eyeball (eyeball becomes too long), so the retina is too far from the lens. | 1. Low converging power of the eye lens (focal length is too long).<br>2. Shortening of the eyeball (eyeball becomes too short), so the retina is too close to the lens. |
| (b) Near Point/Far Point | Far Point: The far point comes closer than infinity. The person can see nearby objects clearly but cannot see distant objects distinctly.<br>Near Point: Usually remains at 25 cm (normal). | Near Point: The near point recedes further away from 25 cm. The person can see distant objects clearly but cannot see nearby objects distinctly.<br>Far Point: Usually remains at infinity (normal). |
| (c) Corrective Lens | Corrected by using a concave lens (diverging lens) of appropriate power. This lens diverges the light rays before they enter the eye, making the image form correctly on the retina. | Corrected by using a convex lens (converging lens) of appropriate power. This lens converges the light rays before they enter the eye, making the image form correctly on the retina. |
The Human Eye and the Colourful World — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. Which part of the human eye controls the amount of light entering the eye?
- A. Cornea
- B. Iris
- C. Pupil
- D. Retina
Answer: B. Iris — The iris is a muscular diaphragm that controls the size of the pupil. By adjusting the pupil's size, it regulates the amount of light entering the eye.
MCQ 2. What is the ability of the eye lens to adjust its focal length to see objects at different distances called?
- A. Myopia
- B. Hypermetropia
- C. Accommodation
- D. Dispersion
Answer: C. Accommodation — Accommodation is the process by which the eye changes its optical power to maintain a clear image or focus on an object as its distance changes.
MCQ 3. A person cannot see distant objects clearly but can see nearby objects distinctly. Which defect of vision is this, and what kind of lens is required for correction?
- A. Myopia, Concave lens
- B. Hypermetropia, Convex lens
- C. Myopia, Convex lens
- D. Hypermetropia, Concave lens
Answer: A. Myopia, Concave lens — This defect is myopia, or nearsightedness, where distant objects appear blurry. It is corrected by using a concave lens of appropriate power, which diverges the light rays before they enter the eye.
MCQ 4. In hypermetropia, or farsightedness, the image of a nearby object is formed:
- A. In front of the retina
- B. Behind the retina
- C. On the retina
- D. On the blind spot
Answer: B. Behind the retina — Hypermetropia occurs when the eye lens cannot converge light rays sufficiently, or the eyeball is too short. As a result, the image of nearby objects forms behind the retina.
MCQ 5. The phenomenon of splitting of white light into its constituent colours when passing through a prism is called:
- A. Refraction
- B. Reflection
- C. Dispersion
- D. Scattering
Answer: C. Dispersion — Dispersion is the phenomenon where white light separates into its spectrum of colours (VIBGYOR) when passing through a transparent medium like a prism, due to different speeds of different colours in the medium.
MCQ 6. Which colour of light deviates the most when passing through a glass prism?
- A. Red
- B. Orange
- C. Green
- D. Violet
Answer: D. Violet — Violet light has the shortest wavelength among visible colours and therefore experiences the greatest refractive index and deviation when passing through a prism. Red light deviates the least.
MCQ 7. The twinkling of stars is primarily due to:
- A. Dispersion of light by the atmosphere
- B. Reflection of light by the atmosphere
- C. Refraction of light by the atmosphere
- D. Scattering of light by the atmosphere
Answer: C. Refraction of light by the atmosphere — The twinkling of stars is caused by atmospheric refraction. As starlight passes through varying layers of air with different refractive indices, it undergoes continuous changes in direction, making the stars appear to twinkle.
MCQ 8. The phenomenon responsible for the advanced sunrise and delayed sunset is:
- A. Reflection
- B. Dispersion
- C. Atmospheric refraction
- D. Scattering
Answer: C. Atmospheric refraction — Atmospheric refraction causes the sun to be visible for about two minutes before the actual sunrise and for two minutes after the actual sunset. This is because light rays from the sun bend as they pass through Earth's atmosphere.
MCQ 9. The blue colour of the sky is due to:
- A. Reflection of sunlight
- B. Refraction of sunlight
- C. Dispersion of sunlight
- D. Scattering of sunlight
Answer: D. Scattering of sunlight — The blue colour of the sky is due to the scattering of sunlight by tiny particles in the atmosphere. Blue light, having a shorter wavelength, is scattered more efficiently than red light (Rayleigh scattering).
MCQ 10. At sunrise and sunset, the sun appears reddish. This is because:
- A. Red light is scattered most
- B. Blue light is scattered most
- C. Red light is scattered least
- D. Blue light is scattered least
Answer: C. Red light is scattered least — During sunrise and sunset, sunlight travels a longer distance through the atmosphere. Most of the blue light is scattered away, leaving the red light, which is scattered least, to reach our eyes.
MCQ 11. The screen on which the image is formed in the human eye is called the:
- A. Cornea
- B. Pupil
- C. Iris
- D. Retina
Answer: D. Retina — The retina is the light-sensitive layer at the back of the eye. It contains photoreceptor cells that convert light into electrical signals, which are then sent to the brain.
MCQ 12. Presbyopia is a defect of vision which arises due to:
- A. Weakening of ciliary muscles and diminishing flexibility of the eye lens
- B. Irregular shape of the cornea
- C. Elongation of the eyeball
- D. Shortening of the eyeball
Answer: A. Weakening of ciliary muscles and diminishing flexibility of the eye lens — Presbyopia is an age-related condition where the eye's natural lens loses its flexibility and the ciliary muscles weaken, making it difficult to focus on nearby objects.
MCQ 13. A ray of light passes through a prism. If the angle of incidence is gradually increased from a small value, the angle of deviation will:
- A. Always increase
- B. Always decrease
- C. First decrease then increase
- D. First increase then decrease
Answer: C. First decrease then increase — For a prism, as the angle of incidence increases, the angle of deviation first decreases to a minimum value and then starts increasing. This behavior is represented by the 'i-δ' curve.
MCQ 14. Due to atmospheric refraction, the apparent position of a star is:
- A. Higher than its actual position
- B. Lower than its actual position
- C. At its actual position
- D. Not affected
Answer: A. Higher than its actual position — As light from a star passes from the rarer atmosphere to the denser atmosphere, it bends towards the normal. This makes the star appear slightly higher than its actual position.
MCQ 15. The phenomenon of scattering of light by colloidal particles is known as:
- A. Refraction
- B. Dispersion
- C. Tyndall effect
- D. Reflection
Answer: C. Tyndall effect — The Tyndall effect is the scattering of light by colloidal particles in a medium. It makes the path of a light beam visible, such as light passing through a dusty room.
MCQ 16. A person's vision becomes blurry and hazy due to the clouding of the eye lens, which can eventually lead to partial or complete loss of vision. This condition is known as:
- A. Myopia
- B. Hypermetropia
- C. Presbyopia
- D. Cataract
Answer: D. Cataract — Cataract is a medical condition where the lens of the eye becomes progressively opaque, resulting in blurred vision. It can be corrected through surgical removal of the opaque lens and implantation of an artificial lens.
MCQ 17. The pupil of the eye contracts in bright light and expands in dim light to:
- A. Change the focal length of the lens
- B. Regulate the amount of light entering the eye
- C. Change the curvature of the cornea
- D. Focus the image on the retina
Answer: B. Regulate the amount of light entering the eye — The pupil acts like an aperture, and its size is adjusted by the iris. In bright light, it contracts to reduce light entry, while in dim light, it expands to allow more light in, optimizing vision.
MCQ 18. A person has a far point of 80 cm. To correct this defect, the power of the required lens should be:
- A. +1.25 D
- B. -1.25 D
- C. +0.8 D
- D. -0.8 D
Answer: B. -1.25 D — A far point of 80 cm indicates myopia. For correction, a concave lens is needed that forms a virtual image of an object at infinity at 80 cm. Thus, focal length f = -80 cm = -0.8 m. Power P = 1/f = 1/(-0.8) = -1.25 D.
MCQ 19. Dispersion of white light occurs because:
- A. All colours of white light travel with the same speed in glass
- B. All colours of white light travel with different speeds in glass
- C. The prism absorbs certain colours
- D. The prism reflects certain colours
Answer: B. All colours of white light travel with different speeds in glass — Dispersion happens because the speed of light, and hence the refractive index, varies for different colours (wavelengths) in a transparent medium like glass. This causes each colour to bend at a slightly different angle.
MCQ 20. Danger signals are red in colour because:
- A. Red light is scattered the most
- B. Red light is scattered the least
- C. Red light travels fastest
- D. Red light is most comfortable for the eyes
Answer: B. Red light is scattered the least — Red light has the longest wavelength in the visible spectrum and is scattered the least by atmospheric particles. This allows it to travel the farthest through fog and smoke without losing intensity, making it ideal for danger signals.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.