NCERT Solutions for Class 10 Science Chapter 11 The Human Eye and The Colorful World

NCERT Solutions for Class 10 Science Chapter 11 The Human Eye and The Colorful World

Edited By Vishal kumar | Updated on Jul 21, 2023 10:01 AM IST | #CBSE Class 10th

NCERT Solutions for Class 10 Science Chapter 11 – CBSE Free PDF Download

Class 10 science chapter 11 question answer- Welcome, Class 10th students, to the perfect destination for all your ch 11 science class 10 solution needs! This page has been curated with the sole purpose of providing you with comprehensive NCERT solutions for Chapter 11 - "The Human Eye and Colourful World." Whether you're seeking clarity on difficult concepts or looking for assistance with homework assignments, you'll find all the human eye and the colourful world question answer you need right here. Our NCERT class 10 science chapter 11 exercise solutions cover all the questions mentioned in your NCERT textbook, ensuring you have a thorough understanding of the chapter.

Do you know that the human eye has a lens in its structure? What is the function of this lens? Such interesting questions are answered with examples in NCERT Class 10 Science solutions chapter 11. The human eye is the most valuable and sensitive sense organ. Some of the topics which are covered in the Human Eye and Colourful World Class 10 chapter are the human eye, defects of vision and their correction, and refraction of light through a prism.

chapter 11 science class 10 contain the solution and answers to the questions asked in 11th lesson of NCERT class 10 science book. The detailed NCERT Class 10 Science solutions chapter 11 are helpful to understand the topics properly.
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NCERT Solutions for Class 10 Science: Important Formulas and Diagrams + eBook link

To score good marks in this chapter 11 class 10 science, important formulas and diagrams are crucial. Experts from Careers360 have created comprehensive formula notes and diagrams for this chapter, which are provided below. You can also download the PDF for all the chapters by clicking on the link below.

Human Eye

1689909476249

Dispersion of White Light in A Prism

1689909535631

Rainbow Formation

1689909596458

Download Ebook - NCERT Class 10 Science: Chapterwise Important Formulas, Diagrams, And Points

This ebook contains essential equations, diagrams, and key concepts that will frequently appear in your exams and monthly quizzes. By using this resource, you can swiftly review important formulas and familiarize yourself with relevant examples, aiding your exam preparation.

NCERT Solutions for Class 10 Science Chapter 11 The Human Eye and The Colorful World - Important Topics

Let's look at the important topics from human eye and colourful world class 10 ncert solutions that will help you revise the chapter easily in less time and prove useful during class tests, exams, and other assessments:

  • The anatomy of the human eye
  • The process of vision
  • Defects of vision and their correction
  • Refraction of light through a prism
  • Dispersion of white light by a glass prism
  • Rainbow formation
  • Scattering of light

Download Class 10 science chapter 11 question answer human eye and colourful world pdf for free.

This chapter 11 science class 10 question answer has been renumbered as Chapter 10 in accordance with the CBSE Syllabus 2023–24.

Chapter 11 Science Class 10 HumanEye and Colourful World

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Access to Class 10 human eye and the colourful world solutions: The Human Eye and Colourful World

Topic 11.2 - Defects of vision and their correction :

Q.1. What is meant by power of accommodation of the eye?

Answer:

The power of accommodation is the ability of an eye to focus near and far objects clearly and make the image on the retina by adjusting its focal length. However, the focal length of the eye lens cannot be decreased below a certain minimum limit.

Q.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 correction for a myopic eye should be the concave lens (negative power) to restore proper vision.

Q.3. What is the far point and near point of the human eye with normal vision?

Answer:

The far point of the human eye with normal vision is at infinity and the near point is 25cm distance from the eye.

Q.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:

Since the student has difficulty reading the blackboard while sitting in the last row, he is most likely to suffer from the disease of myopia (short-sightedness) i.e., he can see the closer objects clearly but is unable to see the far objects properly.

Hence it can be corrected by placing the spectacles with concave lenses of appropriate power.

The Human Eye and The Colorful World Exercise Solutions:

Q.1. The human eye can focus on objects at different distances by adjusting the focal length of the eye lens. This is due to

(a) presbyopia
(b) accommodation
(c) near-sightedness
(d) far-sightedness

Answer:

It is due to the accommodation of the eye. The human eye can adjust its focal length according to the different distances. Therefore, the option (b) is correct.

Q.2. The human eye forms the image of an object at its

(a) cornea

(b) iris

(c) pupil

(d) retina

Answer:

The human eye forms the image of an object at its retina. Therefore,the correct option is(d)\ retina.

Q.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:

The least distance of distinct vision for a young adult with normal vision is about (c) 25 cm

Q.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:

The change in focal length of an eye-lens is caused by the action of the :(c)\ ciliary\ muscles

Q.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 the distant vision:

The power of the lens, given P = -5.5\ D

Power(in\ D) = \frac{1}{f(m)}

Hence the focal length will be:

f= \frac{100}{-5.5} = -18.2\ cm

Therefore, the focal length of the lens required for correcting distant vision will be: -18.2\ cm (Here, the negative sign of focal length tells us that it is a concave lens).

For near vision:

The power of the lens, given P = +1.5\ D

Power(in\ D) = \frac{1}{f(m)}

Hence the focal length will be:

f= \frac{100}{+1.5} = 66.66\ cm

Therefore, the focal length of the lens required for correcting near vision will be: +66.7\ cm (Here, the positive sign of focal length tells us that it is a convex lens).

Q.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 myopic person is suffering from near-sightedness which can be corrected by putting the concave lens in spectacles.

Given the far point up to which a myopic person can see is 80cm in front of the eye.

Here, this person can see the distant object (kept at infinity) clearly if the image of this distant object is formed at his far point.

So, we have

The object distance, u = \infty \ (infinity)

The image distance, v = -80\ cm (far point, in front of the lens)

and the focal length, f = to find.

Hence substituting the values in the lens formula: we obtain

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

\Rightarrow \frac{1}{-80} - \frac{1}{\infty} = \frac{1}{f}

\Rightarrow \frac{1}{f} = \frac{1}{-80}

Or, \Rightarrow f = -80\ cm

Therefore, the focal length of the required concave lens will be 80\ cm

Now, Power,

Power = \frac{1}{f(m)} = \frac{1}{-0.8} = -1.25\ D

Hence the power of the concave lens required is -1.25\ D

Q.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:

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Given the near point for the hypermetropic eye is 1m.

Assume the near point of the normal eye is 25 cm.

So,

The object distance will be, u = -25\ cm (Normal near the point)

The image distance, v = -1\ m (Near the point of this defective eye) or -100\ cm

Then the focal length can be found from the lens formula:

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

Substituting the values in the equation, we obtain

\Rightarrow \frac{1}{-100} - \frac{1}{-25} = \frac{1}{f}

\Rightarrow \frac{-1+4}{100} = \frac{1}{f}

\Rightarrow \frac{3}{100} = \frac{1}{f}

\Rightarrow f = \frac{100}{3} = 33.3\ cm\ or\ 0.333\ m

Hence the power of the lens will be:

P =\frac{1}{f(m)} = \frac{1}{+0.333m} = +3.0\ D

Thus, the power of the convex lens required will be +0.3\ D

Q.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 have a limit of contraction and relaxation, to see closer objects, ciliary muscles, must decrease the focal length of the eye lens. However, the focal length of the eye lens cannot be decreased below a certain minimum limit. Due to this, our eyes are not able to see objects clearly when placed closer than 25cm.

Q.9. What happens to the image distance in the eye when we increase the distance of an object from the eye?

Answer:

There is no change to the image distance in the eye when we increase the distance of an object from the eye. The image is always formed in the retina of our eye.

Q.10. Why do stars twinkle?

Answer:

Stars twinkle because of turbulence in the atmosphere of the Earth.

The light from the star is refracted in different directions because of the variable refractive index. This seems to us like the position of stars to be changed slightly and also brightness and position, Hence, the twinkling of stars happens.

Q.11. Explain why the planets do not twinkle.

Answer:

Planets do not twinkle the way stars do, the reason is that stars are so far away that they appear to be a point source of light on the sky, while planets have finite size and the size of a planet "averages out" the turbulent effects of the atmosphere.

Therefore, we see a relatively stable image of the eye. Hence, twinkling is not observed on planets.

Q.12. Why does the Sun appear reddish early in the morning?

Answer:

At sunrise and sunset, the sun is closer to the horizon. The sunlight near the horizon passes through denser layers of the air and covers a larger distance before reaching our eyes. Most of the blue light gets scattered. The light that reaches our eyes is of longer wavelengths, mainly orange and red. That is why the sun appears red at sunrise and at sunset.

Q.13. Why does the sky appear dark instead of blue to an astronaut?

Answer:

The dark colour is nothing but the absence of light.

The sky appears dark instead of blue to an astronaut because there is no atmosphere in outer space that can scatter the sunlight. As the sunlight is not scattered, no scattered light reaches the eyes of the astronauts the sky appears black to them.

ch 11 Science Class 10- Types of Questions Asked

Before moving on to the exam, students must familiarize themselves with the types of questions that may be asked from class 10 science chapter 11 question answer. The following types of questions are commonly found in the exam:

  • Very short answer type

  • Short answer type

  • Long answer type

  • Practical based questions

Also, check - NCERT Solutions for Class 10 Maths

NCERT Class 10 Science Solutions Chapter 11 - Weightage in Board Exams

You should understand the reason why class 10 science chapter 11 question answer are important because questions from this chapter carry a weightage of around 8 marks in the CBSE 10th board exams. To score full marks, students should have to focus on the solution and the NCERT class 10 science notes that will help to score 8/8 marks from this chapter.

Key features of NCERT Solutions for Class 10 Science Chapter 11: The Human Eye and Colourful World

Comprehensive Understanding: The ch 11 science class 10 provide a thorough explanation of the chapter's topics helping students in better understanding the concepts.

Exam-Centric Approach: The class 10 chapter 11 science solution are designed with the exam pattern in mind, ensuring that students are well-prepared for their board exams.

Clarity on Questions: The class 10 science ch 11 question answer help students in solving various types of questions, be it short answers, long answers, or diagram-based questions.

Correct Answers: If students face difficulties or cannot find the correct answers, the chapter 11 science class 10 question answer act as a reliable source to clear their doubts and provide accurate responses.

Expert Guidance: The option to seek help from experts for any doubts or queries ensures that students receive the right guidance to excel in the subject.

Self-Assessment: Practicing with ncert class 10 science chapter 11 exercise solutions allows students to self-assess their understanding and identify areas that need further attention

How to Access NCERT Solutions for Class 10 Science Chapter 11?

  • Human Eye and Colourful World class 10 human eye and the colourful world solutions are easy to find as you just need to scroll down the article for solutions.

  • In order to navigate from other articles, click on the NCERT solutions from where you can land on to class article such as NCERT solutions for Class 10 and then click on a subject article such as NCERT solutions for class 10 science and you can access the article where solutions are given.

  • Not only you will find exercise solutions in ch 11 science class 10, but also solutions to the topic-wise questions.

  • Human Eye and Colourful World Class 10 science solutions can help you crack some of the questions asked in the CBSE board exam for science so that you can take a step forward.

The summary of the chapter 11 science class 10 human eye and the colourful world as follows:

  • The human eye has the ability to focus on both near and far objects, known as accommodation.
  • The near point of clear vision without strain is 25 cm for a person with normal vision.
  • Three common vision defects are myopia (nearsightedness), hypermetropia (farsightedness), and presbyopia.
  • Myopia is corrected with a concave lens, and hypermetropia is corrected with a convex lens.
  • The power of accommodation decreases with age.
  • Dispersion splits white light into its component colours.
  • Scattered light causes the blue colour of the sky and the red hue of the sun.

Also, Check NCERT Books and NCERT Syllabus here:

NCERT Solutions for Class 10 Science- Chapter Wise

NCERT Science Exemplar Solutions Class 10 - Chapter Wise

Frequently Asked Questions (FAQs)

1. What are the important topics of NCERT Class 10 Science solutions chapter 11?

Here is the list of some important topics:

  •  Functioning of a Lens in Human Eye

  • Applications of Spherical Mirrors & Lenses

  • Refraction of Light through a Prism

  • Dispersion & Scattering of Light

  • Power of Accommodation

  • Defects of Vision and Their Corrections

2. What benefits can be gained from studying chapter 11 class 10 science, specifically the topics on the human eye and the colourful world?

By studying the human eye and the colourful world solutions you will gain a comprehensive understanding of the following topics:

  • The anatomy and function of the human eye, including its ability to accommodate or focus on both near and distant objects.

  • The common vision defects such as myopia, hypermetropia, and presbyopia, and how they can be corrected.

  • The near point of clear vision and how it varies with age.

  • The principle of dispersion and how it splits white light into its component colours.

3. How can I download NCERT solutions for class 10 Science chapter 11 pdf?

Through NCERT solutions Class 10 science chapter 11 pdf download facility download this solution as a webpage to access offline. To get more problems on The Human Eye and The Colourful World refer to NCERT exemplar.

4. What is atmospheric refraction according to class 10 chapter 11 science?

 Atmospheric refraction is the bending of light as it travels through the Earth's atmosphere due to variations in temperature and density. This bending of light causes objects to appear displaced or distorted from their true position, leading to phenomena such as elevated sun or star positions and mirages.



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Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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