CBSE Class 10th Exam Date:17 Feb' 26 - 17 Feb' 26
Did you ever wonder why the sky is blue in the daytime but reddish-orange during sunrise and sunset or why stars sparkle only at night? These are explained by Class 10 Science Chapter 10, The Human Eye and the Colorful World. Significant phenomena such as how our eyes work and defects of vision that occur commonly, how light is deviated (refraction), how white light spreads into its colours (dispersion), why stars sparkle, how light gets scattered, and the Tyndall effect are all described in this chapter of NCERT solutions.
This Story also Contains
NCERT Solutions help students to learn these natural phenomena in more details, dispel some of the most widespread misconceptions and organize their studies properly so that they are able to pass their exams easily and successfully with the help of clear explanations and examples. The NCERT solutions for class 10 Science Chapter 10 contains: the PDF in form that can be downloaded, answers to intext questions, solutions to end of chapter exercises, Higher Order Thinking Skills (HOTS) questions, covers all the significant topics of this chapter.
With step-wise answers to all questions in the textbook, the NCERT Solutions of Class 10 Science Chapter 10 The Human Eye and the Colourful World will make students understand the main points. These solutions are provided in form of a free downloadable PDF which is convenient and easy to revise.
NCERT Solutions Class 10 Science Chapter 10: Human Eye and the Colourful World (Intext Questions) offers precise explanations by illustrated diagrams to enable students to study certain principles involved in eye functioning, defects of the eyes, atmospheric refraction, and dispersion of light. These step-by-step solutions enhance clarity of concepts and helps in exam preparation.
Q. 1. What is meant by the 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.
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.
Answer:
Since the student has difficulty reading the blackboard while sitting in the last row, he is most likely to suffer from myopia (short-sightedness), i.e., he can see 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.
NCERT Solutions for Class 10 Science Chapter 10: The Human Eye and the Colourful World (Exercise Questions) offer well-structured answers to all textbook questions. These solutions simplify complex topics like vision, refraction, rainbow formation, and scattering of light, making them easier to understand and revise for exams.
(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
Answer:
For the distant vision:
The power of the lens, given $P=-5.5 \mathrm{D}$
$
\operatorname{Power}(\text { in } D)=\frac{1}{f(m)}
$
Hence, the focal length will be:
$
f=\frac{100}{-5.5}=-18.2 \mathrm{~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 $P=+1.5 \mathrm{D}$
$
\operatorname{Power}(\text { in } D)=\frac{1}{f(m)}
$
Hence the focal length will be:
$
f=\frac{100}{+1.5}=66.66 \mathrm{~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).
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, $u=\infty$ (in finity $)$
The image distance, $v=-80 \mathrm{~cm}$ (far point, in front of the lens).
Hence substituting the values in the lens formula: we obtain
$
\begin{aligned}
& \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}
\end{aligned}
$
$ \Rightarrow f=-80 \mathrm{~cm}$
Therefore, the focal length of the required concave lens will be 80 cm
Now, Power,
$
\text { Power }=\frac{1}{f(m)}=\frac{1}{-0.8}=-1.25 \mathrm{D}
$
Hence the power of the concave lens required is -1.25 D
Answer:
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, $u=-25 \mathrm{~cm}$ (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
$
\begin{aligned}
& \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 \mathrm{~cm} \text { or } 0.333 \mathrm{~m}
\end{aligned}
$
Hence the power of the lens will be:
$
P=\frac{1}{f(m)}=\frac{1}{+0.333 \mathrm{~m}}=+3.0 \mathrm{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.
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.
Class 10 Science Chapter 10 The Human Eye and the Colourful World (HOTS Questions) will assess in-depth knowledge and concept application. The questions are not basic ones because they make students develop their own way of thinking about the real-life phenomena, such as myopia, dispersion, and atmospheric refraction.
Q1: The angle of a prism is $30^{\circ}$ and its refractive index is $\sqrt{2}$ and one of the surface is silvered. At what angle of incidence, a ray should be incident on one surface so that after reflection from the silvered surface, it retraces its path?
Answer:
$\begin{aligned} & \text { so } r_1+r_2=A \\ & \text { and } r_2=0 \\ & \text { Hence } r_1=30^{\circ} \\ & n_1 \sin i=n_2 \sin r \\ & \Rightarrow \sin i=(\sqrt{2})(1 / 2) \\ & i=45^{\circ}\end{aligned}$
Q2: One ray of light suffers minimum deviation in an equilateral prism P additional prism Q and R of identical shape and made up of same material shown in figure. The ray will now suffer ________
Answer:
Atmospheric refraction causes the sun to appear two minutes before it is actually above the horizon (advanced sunrise). Also, we can see the sun for about two minutes even after it is actually below the horizon (delayed sunset). Therefore, the time from sunrise to sunset is increased by about four minutes because of the refraction of light through the earth's atmosphere. If the earth had no atmosphere, the length of the day would have been shorter by about four minutes.
Q3: Which of the following are conditions necessary for observing a rainbow?
Answer:
A rainbow occurs due to reflection and refraction occurring from water droplets present in the air. So, to see a rainbow, it should have rained and the Sun should be present.
At the same time, the Sun should be behind the observer. If it is otherwise, only the glare of the Sun will be felt and no rainbow will be observed.
Q4: A beam of white light falls on a glass prism. The colour of light which undergoes the least bending on passing through the glass prism is :
Answer:
$\mu=\frac{\text { velocity of light in air or vacuum }}{\text { velocity of light in a given medium }}$
The refractive index of violet colour in the glass is greater than the refractive index of red, in other words, the speed of violet is the least in glass and the speed of red is the highest. So, the red light bends the least while the violet the most.
Q5: By how much time the day would have been shorter if the earth had no atmosphere?
Answer:
Atmospheric refraction causes the sun to appear two minutes before it is actually above the horizon (advanced sunrise). Also, we can see the sun for about two minutes even after it is actually below the horizon (delayed sunset). Therefore, the time from sunrise to sunset is increased by about four minutes because of the refraction of light through the earth's atmosphere. If the earth had no atmosphere, the length of the day would have been shorter by about four minutes.
NCERT Solutions Class 10 Science Chapter 10: The Human eye and the Colourful World have all the important topics like structure of the human eye, problems in vision, refraction in air, light scattering and formation of a rainbow. Students can learn the major concepts and successfully pass the exams with the utilization of these solutions.
10.1 The Human Eye
10.1.1 Power Of Accommodation
10.2 Defects Of Vision And Their Correction
10.3 Refraction Of Light Through A Prism
10.4 Dispersion Of White Light By A Glass Prism
10.5 Atmospheric Refraction
10.6 Scattering Of Light
10.6.1 Tyndall Effect
10.6.2 Why Is The Colour Of The Clear Sky Blue?
The Human Eye and the Colourful World NCERT Solutions Class 10 Science Chapter 10 also comprise a list of useful formulae on lens and refraction and vision correction. These solutions enable students to better solve numerical questions in less time and practice more in examination preparation.
$
\frac{1}{f}=\frac{1}{v}-\frac{1}{u}
$
Where:
$f=$ focal length of the lens
$v=$ image distance
$u=$ object distance
(Sign convention applies)
$
m=\frac{h^{\prime}}{h}=\frac{v}{u}
$
Where:
$m=$ magnification
$h^{\prime}=$ height of the image
$h=$ height of the object
$
n=\frac{\sin i}{\sin r}
$
Or,
$
n=\frac{c}{v}
$
Where:
$i=$ angle of incidence
$r=$ angle of refraction
$c=$ speed of light in vacuum
$v=$ speed of light in the medium
Learn accommodation – The capacity of the eye to accommodate lens shape to focus near and far objects.
Understand refraction using prism – Light is bent and separated into spectrum (dispersion).
Understand the effects of atmospheric refraction – Twinkling of stars, advanced sunrise, delayed sunset.
Understand scattering of light – Blue sky, red sunset phenomenon explained by scattering of shorter and longer wavelengths.
Label diagrams clearly – Eye, prism, spectrum diagrams assist in scoring marks.
Practice NCERT questions – In-text and exercise questions often appear in exams.
Conceptual Clarity: Complex concepts such as accommodation of the eye, scattering of light and dispersion have been explained in a language that anyone can understand making learning very effective.
Easy Exam Preparation: The in-text and exercise questions are all answered on the basis of the latest CBSE guidelines, which also can help students how questions actually appear in the board examination.
Step-by-Step Solutions: Examples of numerical problems on lens formula, magnification and power of lens are clearly provided step-to-step giving accuracy and confidence.
Higher Order Thinking Skills (HOTS) Questions: The chapter contains additional application-oriented questions that develop analytical abilities and aids in resolving complex-based questions in exams and competitive tests.
Revision Friendly: The short and organized solutions can be used in a last-minute revision before tests.
Real-Life Connections: Solutions bridges the gap between the knowledge the students read in the textbook and the real world by providing explanations to how rainbows form or why the sky is blue and why the sun appears to have a red color during sunset allowing the students to relate science to real life.
Also, check
NCERT Solutions for Class 10 Science provide detailed explanations of every chapter, covering both Physics, Chemistry, and Biology topics. With step-by-step answers, important formulas, and extra practice questions, these solutions help students build strong concepts and prepare effectively for their CBSE board exams.
Frequently Asked Questions (FAQs)
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
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.
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.
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.
On Question asked by student community
If you're looking for directions or steps to reach Sadhu Ashram on Ramgart Road in Aligarh, here’s how you can get there:
Starting Point:
Use Google Maps:
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
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
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
