Welcome to the NCERT Solutions for Class X Science Chapter 11 – Human Eye and Colourful World. This chapter is all about understanding the properties of light, how it interacts with different objects, and how the human eye perceives different colors.
In this chapter, you will learn about the structure and function of the human eye, the refraction of light, the different colors of light, and the phenomenon of dispersion. You will also learn about the formation of images by lenses and mirrors and the different defects of vision.
Our solutions are designed to help you understand the concepts thoroughly and prepare for your exams. We have provided step-by-step solutions to all the questions in the NCERT textbook, along with additional tips and tricks to help you score well in your exams.
At Swastik Classes, we believe in making learning fun and engaging. Our NCERT Solutions for Class X Science Chapter 11 – Human Eye and Colourful World are designed to make the learning process enjoyable and easy for you. So, let’s get started and explore the fascinating world of light and color together!
Download PDF of NCERT Solutions for Class 10 Science Chapter 11 Human Eye and Colourful World
Answers of Science NCERT solutions for class 10 Chapter 11 Human Eye and Colourful World
Chapter 11
Human Eye and Colorful World
In-Chapter Exercise
*Question 1:
What is meant by power of accommodation of the eye?
Answer:
When the ciliary muscles are relaxed, the eye lens becomes thin, the focal length increases, and the distant objects are clearly visible to the eyes. To see the nearby objects clearly, the ciliary muscles contract making the eye lens thicker. Thus, the focal length of the eye lens decreases and the nearby objects become visible to the eyes. Hence, the human eye lens is able to adjust its focal length to view both distant and nearby objects on the retina. This ability is called the power of accommodation of the eyes.
*Question 2:
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 is able to see nearby objects clearly, but he is unable to see objects beyond 1.2 m. This happens because the image of an object beyond 1.2 m is formed in front of the retina and not at the retina, as shown in the given figure.
To correct this defect of vision, he must use a concave lens. The concave lens will bring the image back to the retina as shown in the given figure.
Question 3:
What is the far point and near point of the human eye with normal vision?
Answer:
The near point of the eye is the minimum distance of the object from the eye, which can be seen distinctly without strain. For a normal human eye, this distance is 25 cm.
The far point of the eye is the maximum distance to which the eye can see the objects clearly. The far point of the normal human eye is infinity.
*Question 4:
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:
A student has difficulty in reading the blackboard while sitting in the last row. It shows that he is unable to see distant objects clearly. He is suffering from myopia. This defect can be corrected by using a concave lens.
Last Exercise
Question 1:
The human eye can focus objects at different distances by adjusting the focal length of the eye lens. This is due to
- presbyopia
- accommodation
- near-sightedness
- far-sightedness
Answer:
(b) Human eye can change the focal length of the eye lens to see the objects situated at various distances from the eye. This is possible due to the power of accommodation of the eye lens.
*Question 2:
The human eye forms the image of an object at its
(a) cornea (b) iris (c) pupil (d) retina
Answer:
(d) The human eye forms the image of an object at its retina.
Question 3:
The least distance of distinct vision for a young adult with normal vision is about
- 25 m
- 2.5 cm
- 25 cm
- 2.5 m
Answer:
(c) The least distance of distinct vision is the minimum distance of an object to see a clear and distinct image. It is 25 cm for a young adult with normal visions.
Question 4:
The change in focal length of an eye lens is caused by the action of the
- pupil
- retina
- ciliary muscles
- iris
Answer:
(c) The relaxation or contraction of ciliary muscles changes the curvature of the eye lens. The change in curvature of the eye lens changes the focal length of the eyes. Hence, the change in focal length of an eye lens is caused by the action of ciliary muscles.
*Question 5:
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?
Answer:
For distant vision P = −0.181 m, for near vision P = 0.667 m The power P of a lens of focal length f is given by the relation
- Power of the lens used for correcting distant vision = −5.5 D
Focal length of the required lens, f =
The focal length of the lens for correcting distant vision is −0.181 m.
- Power of the lens used for correcting near vision = +1.5 D
Focal length of the required lens, f =
The focal length of the lens for correcting near vision is 0.667 m.
*Question 6:
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?
Answer:
The person is suffering from an eye defect called myopia. In this defect, the image is formed in front of the retina. Hence, a concave lens is used to correct this defect of vision.
Object distance, u = infinity = Image distance, v = −80 cm Focal length = f
According to the lens formula,
We know,
A concave lens of power −1.25 D is required by the person to correct his defect.
*Question 7:
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.
Answer:
A person suffering from hypermetropia can see distinct objects clearly but faces difficulty in seeing nearby objects clearly. It happens because the eye lens focuses the incoming divergent rays beyond the retina. This defect of vision is corrected by using a convex lens. A convex lens of suitable power converges the incoming light in such a way that the image is formed on the retina, as shown in the following figure.
The convex lens actually creates a virtual image of a nearby object (N’ in the figure) at the near point of vision (N) of the person suffering from hypermetropia.
The given person will be able to clearly see the object kept at 25 cm (near point of the normal eye), if the image of the object is formed at his near point, which is given as 1m.
Object distance, u = −25 cm
Image distance, v = −1 m = −100 cm Focal length, f
Using the lens formula,
A convex lens of power +3.0 D is required to correct the defect.
Question 8:
Why is a normal eye not able to see clearly the objects placed closer than 25 cm
Answer:
A normal eye is not able to see clearly the objects placed closer than 25 cm because the ciliary muscles of the eyes are not able to contract beyond a certain limit.
Question 9:
What happens to the image distance in the eye when we increase the distance of an object from the eye?
Answer:
Since the size of eyes cannot increase or decrease, the image distance remains constant. When we increase the distance of an object from the eye, the image
distance in the eye does not change. The increase in the object distance is compensated by the change in the focal length of the eye lens. The focal length of the eyes changes in such a way that the image is always formed at the retina of the eye.
*Question 10:
Why do stars twinkle?
Answer:
Stars emit their own light and they twinkle due to the atmospheric refraction of light. Stars are very far away from the earth. Hence, they are considered as point sources of light. When the light coming from stars enters the earth’s atmosphere, it gets refracted at different levels because of the variation in the air density at different levels of the atmosphere. When the star light refracted by the atmosphere comes more towards us, it appears brighter than when it comes less towards us. Therefore, it appears as if the stars are twinkling at night.
*Question 11:
Explain why the planets do not twinkle?
Answer:
Planets do not twinkle because they appear larger in size than the stars as they are relatively closer to earth. Planets can be considered as a collection of a large number of point-size sources of light. The different parts of these planets produce either brighter or dimmer effect in such a way that the average of brighter and dimmer effect is zero. Hence, the twinkling effects of the planets are nullified and they do not twinkle.
*Question 12:
Why does the Sun appear reddish early in the morning?
Answer:
During sunrise, the light rays coming from the Sun have to travel a greater distance in the earth’s atmosphere before reaching our eyes. In this journey, the shorter wavelengths of lights are scattered out and only longer wavelengths are able to reach our eyes. Since blue colour has a shorter wavelength and red colour has a longer wavelength, the red colour is able to reach our eyes after the atmospheric scattering of light. Therefore, the Sun appears reddish early in the morning.
*Question 13:
Why does the sky appear dark instead of blue to an astronaut?
Answer:
The sky appears dark instead of blue to an astronaut because there is no atmosphere in the outer space that can scatter the sunlight. As the sunlight is not scattered, no scattered light reach the eyes of the astronauts and the sky appears black to them.
Access Answers to NCERT Class 10 Science Chapter 11 The Human Eye and Colorful World
Page No: 190
1. What is meant by power of accommodation of the eye?
Answer-
The ability of the lens of the eye to adjust its focal length to clearly focus rays coming from distant as well from a near objects on the retina, is known as the power of accommodation of the eye.
2. A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type of corrective lens used to restore proper vision?
Answer-
An individual with a myopic eye should use a concave lens of focal length 1.2 m so that he or she can restore proper vision.
3. What is the far point and near point of the human eye with normal vision?
Answer-
The minimum distance of the object from the eye, which can be seen distinctly without strain is called the near point of the eye. For a normal person’s eye, this distance is 25 cm.
The far point of the eye is the maximum distance to which the eye can see objects clearly. The far point of a normal person’s eye is infinity.
4. 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 student is suffering from short-sightedness or myopia. Myopia can be corrected by the use of concave or diverging lens of an appropriate power.
Page No: 197
Exercise
1. The human eye can focus 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
Answer-
(b) accommodation
Due to accommodation the human eye can focus objects at different distances by adjusting the focal length of the eye lens.
Page No: 198
2. The human eye forms an image of an object at its
(a) cornea
(b) iris
(c) pupil
(d) retina
Answer –
(d) retina
The retina is the layer of nerve cells lining the back wall inside the eye. This layer senses light and sends signals to the brain so you can see.
3. 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
Answer –
(c) 25 cm
25 cm is the least distance of distinct vision for a young adult with normal vision.
4. 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
Answer-
(c) ciliary muscles
The action of the ciliary muscles changes the focal length of an eye lens
5. 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?
Answer-
The power (P) of a lens of focal length f is given by the relation
Power (P) = 1/f
(i) Power of the lens (used for correcting distant vision) = – 5.5 D
Focal length of the lens (f) = 1/P
f = 1/-5.5
f = -0.181 m
The focal length of the lens (for correcting distant vision) is – 0.181 m.
(ii) Power of the lens (used for correcting near vision) = +1.5 D
Focal length of the required lens (f) = 1/P
f = 1/1.5 = +0.667 m
The focal length of the lens (for correcting near vision) is 0.667 m.
6. 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?
Answer-
The individual is suffering from myopia. In this defect, the image is formed in front of the retina. Therefore, a concave lens is used to correct this defect of vision.
Object distance (u) = infinity = ∞
Image distance (v) = – 80 cm
Focal length = f
According to the lens formula,
A concave lens of power – 1.25 D is required by the individual to correct his defect.
7. 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.
Answer-
An individual suffering from hypermetropia can see distinct objects clearly but he or she will face difficulty in clearly seeing objects nearby. This happens because the eye lens focuses the incoming divergent rays beyond the retina. This is corrected by using a convex lens. A convex lens of a suitable power converges the incoming light in such a way that the image is formed on the retina, as shown in the following figure.
The convex lens creates a virtual image of a nearby object (N’ in the above figure) at the near point of vision (N) of the individual suffering from hypermetropia.
The given individual will be able to clearly see the object kept at 25 cm (near point of the normal eye), if the image of the object is formed at his near point, which is given as 1 m.
Object distance, u= – 25 cm
Image distance, v= – 1 m = – 100 m
Focal length, f
Using the lens formula,
A convex lens of power +3.0 D is required to correct the defect.
8. Why is a normal eye not able to see clearly the objects placed closer than 25 cm?
Answer-
A normal eye is not able to see the objects placed closer than 25 cm clearly because the ciliary muscles of the eyes are unable to contract beyond a certain limit.
9. What happens to the image distance in the eye when we increase the distance of an object from the eye?
Answer-
The image is formed on the retina even on increasing the distance of an object from the eye. The eye lens becomes thinner and its focal length increases as the object is moved away from the eye.
10. Why do stars twinkle?
Answer-
The twinkling of a star is due to atmospheric refraction of starlight. The starlight, on entering the earth’s atmosphere, undergoes refraction continuously before it reaches the earth. The atmospheric refraction occurs in a medium of gradually changing refractive index.
11. Explain why the planets do not twinkle?
Answer-
Unlike stars, planets don’t twinkle. Stars are so distant that they appear as pinpoints of light in the night sky, even when viewed through a telescope. Since all the light is coming from a single point, its path is highly susceptible to atmospheric interference (i.e. their light is easily diffracted).
12. Why does the Sun appear reddish early in the morning?
Answer-
White light coming from the sun has to travel more distance in the atmosphere before reaching the observer. During this, the scattering of all colored lights except the light corresponding to red color takes place and so, only the red colored light reaches the observer. Therefore, the sun appears reddish at sunrise and sunset.
13. Why does the sky appear dark instead of blue to an astronaut?
Answer-
The sky appears dark instead of blue to an astronaut, as scattering of light does not take place outside the earth’s atmosphere.
NCERT Solutions for Class 10 Science Chapter 11: The Human Eye and Colourful World
NCERT Class 10 Science Chapter 11 describes the fine structure of the human eye. It does explain the reason behind the color of the sun during the time of sunrise and sunset. It explains the accommodation of the eye concept. Various defects that occur to the eye like refractive defects of vision which include hypermetropia, myopia, and presbyopia are discussed in NCERT Solutions. It describes the near point of the eye or the least distance of distinct vision. Topics included in this chapter are:
11.1 The Human eye (4 questions)
The topic discusses the importance of the Human Eye, its structure and how it functions. The human eye is one of the most valuable and sensitive sense organs. Of all the sense organs, the human eye is the most significant one as it enables us to see the beautiful, colourful world around us. The topic also explains the power of accommodation where it talks about the eye lens.
11.2 Defects of Vision and their Correction
The topic talks about defects of vision and their correction. There are mainly three common refractive defects of the eye. These are myopia or near-sightedness, hypermetropia or far-sightedness and Presbyopia. These defects can be corrected by the use of suitable spherical lenses.
11.3 Refraction of Light Through a Prism
You have learnt how the light gets refracted through a rectangular glass slab. For parallel refracting surfaces, as in a glass slab, the emergent ray is parallel to the incident ray. However, it is slightly displaced laterally. In this topic, you will get to learn how lights get refracted through a triangular glass prism by performing an activity.
11.4 Dispersion of White Light by a Glass Prism
You must have seen and appreciated the spectacular colours in a rainbow. In this topic, you will get to know how rainbow occurs by performing an activity related to the refraction of light through a prism. It will help you know the reason behind the dispersion of white light by a glass prism.
11.5 Atmospheric Refraction
The topic discusses Atmospheric Refraction and how it is related to the twinkling of stars and advance sunrise and delayed sunset. The twinkling of a star is due to atmospheric refraction of starlight.
11.6 Scattering of Light
In the previous class, you have learnt about the scattering of light by colloidal particles. The path of a beam of light passing through a true solution is not visible. However, its path becomes visible through a colloidal solution where the size of the particles is relatively larger. The topic further discusses the Tyndall Effect, why the colour of the clear sky blue? And the colour of the Sun at sunrise and sunset.
NCERT Solutions for Class 10 Science Chapter 11 The Human Eye and Colourful World
- NCERT Solutions for Class 10 explains the detailed structure of the human eye.
- Explains the various defects that happen for human and also describes the ways to correct it.
- Dispersion of light concept is made understandable in a simple way.
- It also explains the reason for the sky being blue along with the colour of the sun at sunset and sunrise events.
Key features of NCERT Solutions for Class 10 Science Chapter 11: The Human Eye and Colourful World
- The simple and easily understandable approach is followed in NCERT Solutions to make students aware of topics.
- NCERT Solutions offer detailed answers to all the questions to help students in their preparations.
- It is useful for successfully clearing the CBSE board exams in flying colours and is also the best guide for Science Olympiads and other competitive exams.
- Provides completely solved solutions to all the questions present in the respective NCERT textbooks.
Conclusion
The NCERT Solutions for Class X Science Chapter 11 – Human Eye and Colourful World provide a comprehensive understanding of the properties of light, how it interacts with different objects, and how the human eye perceives different colors. The chapter emphasizes the importance of light and color in our daily lives and highlights their various applications.
Our solutions provide a step-by-step explanation of all the questions in the NCERT textbook, making it easy for students to understand the concepts. We have also provided additional tips and tricks to help students score well in their exams.
By studying this chapter, students will not only gain knowledge about light and color but also develop an understanding of their significance in different fields like art, design, and photography. We need to appreciate the beauty of the colorful world around us and understand the science behind it.
At Swastik Classes, we believe in providing a holistic learning experience to our students, and our NCERT Solutions for Class X Science Chapter 11 – Human Eye and Colourful World are a testament to our commitment to quality education.
What will I get by learning the The Human Eye and Colourful World of the NCERT Solutions for Class 10 Science?
The NCERT Solutions for Class 10 Science The Human Eye and Colourful World provides a clear idea about the structure of the human eye. The various defects that happen in humans and the ways to correct it are also explained in this chapter. The concept of dispersion of light is explained in a simple language to help students score well in the board exams. The sunset and sunrise events are also covered in this chapter based on the CBSE syllabus and guidelines.
What are the defects of vision and ways to correct them covered in the Chapter 11 of NCERT Solutions for Class 10 Science?
Presbyopia, Hypermetropia and Myopia are the three defects of vision. A difficulty to see both nearby and far objects is called Presbyopia. A person will not be able to see nearby objects is called Hypermetropia and this defect can be solved using a convex lens. A person able to see only nearby objects is called Myopia and this defect can be solved using a concave lens. These concepts are explained with pictorial representation in the solutions PDF to help students score well in the board exams.
