# 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: Do you know that human eye has a lens in its structure? What is the function of this lens? Such interesting questions are answered with examples in solutions of NCERT for class 10 chapter 11 the human eye and the colorful world. This chapter talks about the human eye which is one of the most valuable and sensitive sense organs. Some of the topics which are covered in this chapter are the human eye, defects of vision and their correction, refraction of light through a prism, dispersion of white light by a glass prism, atmospheric refraction and scattering of light. All these topics have certain questions and in order to make learning easier, CBSE NCERT solutions for class 10 chapter 11 the human eye and the colorful world are given below in this article. You can also check out NCERT solutions for other chapters in science or for any other subject by clicking on the links given in this article.

## Types of questions asked from chapter 6 life processes board exams:

CBSE Class 10 science board exam will have the following types of questions:

• Practical based questions

## CBSE NCERT class 10 science chapter 11 the human eye and the colorful world weightage in board exams:

You should understand the reason why NCERT solutions for class 10 science chapter 11 the human eye and the colorful world are important because questions from this chapter carry weightage of around 8 marks in the CBSE board exams. Solutions of NCERT for class 10 science chapter 11 the human eye and the colorful world will help you crack these questions so that you can score well in your board exams. If you have any problem in answering these questions or you are not getting the correct answers then don't worry, NCERT solutions for class 10 science chapter 11 the human eye and the colorful world is there to help. For any doubt or query, you can always ask our experts for the solutions.

## NCERT solutions for class 10 science chapter 11 the human eye and the colorful world topic 11.2 Defects of vision and their correction:

The power of accommodation is the ability of an eye to focus near and far objects clearly and making 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.

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

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

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 unable to see the far objects properly.

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

## Solutions of NCERT for class 10 science chapter 11 the human eye and the colorful world Exercise:

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

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.

(a) cornea.   (b) iris.   (c) pupil.   (d) retina.

The human eye forms the image of an object at its $(d)\ retina.$

(a) 25 m.   (b) 2.5 cm.   (c) 25 cm.   (d) 2.5 m.

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

(a) pupil.                            (b) retina.
(c) ciliary muscles.            (d) iris.

The change in focal length of an eye - lens is caused by the action of the :

$(c)\ ciliary\ muscles$.

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).

The myopic person is suffering from the near-sightedness and 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 concave lens required is $-1.25\ D$.

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 point).

The image distance, $v = -1\ m$ (Near 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 $+0.3\ D$

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.

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.

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.

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 to the eye. Hence, twinkling is not observed in planets.

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 the sunrise and at the sunset.

The dark color is nothing but the absence of the light.

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 the sky appears black to them.

## CBSE NCERT solutions for class 10 science (chapter-wise)

 Chapter No. Chapter Name Chapter 1 NCERT solutions for class 10 science chapter 1 Chemical Reactions and Equations Chapter 2 Solutions of NCERT for class 10 science chapter 2 Acids, Bases, and Salts Chapter 3 CBSE NCERT solutions for class 10 science chapter 3 Metals and Non-metals Chapter 4 NCERT solutions for class 10 science chapter 3 Carbon and Its Compounds Chapter 5 Solutions of NCERT for class 10 science chapter 5 Periodic Classification of Elements Chapter 6 CBSE NCERT solutions for class 10 science chapter 6 Life Processes Chapter 7 NCERT solutions for class 10 science chapter 7 Control and Coordination Chapter 8 Solutions of NCERT for class 10 science chapter 8 How do Organisms Reproduce? Chapter 9 CBSE NCERT solutions for class 10 science chapter 9 Heredity and Evolution Chapter 10 NCERT solutions for class 10 science chapter 10 Light Reflection and Refraction Chapter 11 CBSE NCERT solutions for class 10 science chapter 11 The human eye and the colorful world Chapter 12 Solutions of NCERT for class 10 science chapter 12 Electricity Chapter 13 CBSE NCERT solutions for class 10 science chapter 13 Magnetic Effects of Electric Current Chapter 14 NCERT solutions for class 10 science chapter 14 Sources of Energy Chapter 15 Solutions of NCERT for class 10 science chapter 15 Our Environment Chapter 16 CBSE NCERT solutions for class 10 science chapter 16 Sustainable Management of Natural Resources

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