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NCERT solutions for class 10 science chapter 11-The Human Eye and Colourful world

solutions of chapter 11 class 10 science

NCERT solutions for class 10 science chapter 11-The Human Eye and Colourful world

NCERT solutions for class 10 science chapter 11-The Human Eye and Colourful world are the solutions of unsolved questions of class 10 NCERT science text book chapter 11  prescribed by CBSE for the students of class 10. All these NCERT solutions for class 10 science chapter 11-The Human Eye and Colourful world is the best inputs for achieving excellent marks, therefore every student is compulsorily required to study these NCERT solutions periodically with a proper understanding of solutions of each question for upgrading their science knowledge so that they wouldn’t face any kinds of difficulty in studying science subject of higher classes.

Class 10 Science NCERT textbook chapter 11-The Human Eye and Colourful world is based on the structure and function of the human eye. You will study in this chapter defects of eyes like myopia(nearsightedness) , hypermetropia(farsightedness) and presbyopia along with their remedies, therefore to study chapter 11 – The Human Eye and Colourful world is needed for the awareness about one of the most important organs of our body due to which we become able to see the colourful world.

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solutions of chapter 11 class 10 science

 

NCERT solutions for class 10 science chapter 11-The Human Eye and Colourful world

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Q1. What is meant by the power of accommodation of the eye?

Ans. When we see distant objects the ciliary muscles in the crystalline lens dilates which causes an increase in focal length of the eye lens and we see the distant objects, these muscles contract causing a decrease in the focal length of the eye that enables eyes to see nearby objects. The ability of the lens to adjust its focal length to clearly focus rays coming from distant as well from a near object on the retina is known as the power of accommodation of the eye.

Q2. 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?

Ans. The person with a myopic eye who can’t see objects beyond 1.2 m should use a concave lens (diverging lens) of a focal length of 1.2 m so that the image of the objects formed near the eye lens could be shifted to the retina.

Q3. What are the far point and near point of the human eye with normal vision?

Ans. The minimum point 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, and the far point is the maximum distance at which the human eye is capable to see clearly, as example, we can see the stars which are too much far from us, therefore far point is infinity.

Q4. 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?

Ans. A student has difficulty in reading the blackboard while sitting in the last row, which means he is unable to see the objects after a certain distance, therefore the student is suffering from short-sightedness or myopia. Myopia can be corrected by the use of a concave or diverging lens of appropriate power.

NCERT solutions for class 10 science chapter 11-The Human Eye and Colourful world

Exercise

Q1. 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

Ans. (b) accommodation

Due to the accommodation of the human eye can focus objects at different distances by adjusting the focal length of the eye lens.

Q2. The human eye forms an image if an object at its

(a) cornea

(b) iris

(c) pupil

(c) retina

Ans(d) Retina
The retina is the layer of nerve cells lining the black wall inside the eye. This layer senses light and sends signals to the brain so you can see.

Q3.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

Ans(c) 25 cm
25 cm is the least distance of distinct vision for a young adult with normal vision.

Q4.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

Ans(c) ciliary muscles
The action of the ciliary muscles changes the focal length of an eye lens.

NCERT solutions for class 10 science chapter 11-The Human Eye and Colourful world

Q5. 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

Ans. The power (P) of the lens of focal length as is given by the relation power (P) = 1/f ⇒f =1/P
(i)The person needs a lens of power -5.5 dioptres for correcting his distance vision

∴ He would need a lens with the focal length = 1/P = 1/-5.5D = 0.1818 m ≈0.182 m=18.2 cm

(ii) The person needs a lens of power -5.5 dioptres for correcting his near vision

∴ He would need a lens with the focal length = 1/P = 1/+1.5D = 0. 6666 m ≈0.667 m=66.7 cm

Q6. The far point of a myopic person is 80 cm in front of the eye. What is the power of the lens required to correct the problem?

Ans. The far point of a myopic person is 80 cm means the image formed of the object is near the eye lens between the eye lens and retina when it is placed at a distance of more than 80 cm(i.e ∞ ). The eye lens is a convex lens, therefore we are required power of lens that could focus image at a distance,v  = -80 cm

The object is placed at a distance of more than 80 cm i.e u = ∞

According to lens formula, we can find the required focal length f of the lens

f = -80

The  focal length f of the lens is = -80 cm = -0.80 m

The power of the lense is = 1/f(m) = -1/0.80 m =-1.25 D

Negative sign shows that lens required to correct given myopic eye is concave lens.

NCERT solutions for class 10 science chapter 11-The Human Eye and Colourful world

Q7.Make a diagram of how hypermetropia is corrected. The near point of the hypermetropic eye is 1 m. What is the power of the lens required to correct the defect? Assume that the near point of the normal eye is 25 cm.

Ans. Hypermetropia is the defect of the eye in which a person can not see the objects lying nearer to a certain distance because its image is formed beyond the retina

The distance nearer to which a person can not see known as the near point of the hypermetropic eye, here in the diagram it is shown by N and the near point of the normal eye is N'(25 cm)

hypermetropic eye

 

For correcting a hypermetropic eye, a convex lens is used so that the net focal length of the eye lens and lens enables the ray of the nearby object to focus its image at the retina.

correction of hypermetropic eye

The near point of the hypermetropic eye is 100 cm, which means the person with such a defect can’t see the object nearer to 100 cm.

If the object is placed at 25 cm then we are needed a lens that is able to focus the virtual image of the object at 100 cm, for this, we are needed a convex lens that enables us to shift of the image formation behind the retina to the retina.

Applying the lens formula

1/f = 1/v – 1/u

u = -25 cm,v = -100 cm

1/f = 1/-100 – 1/-25

1/f = -1/100 +1/25

= (-1 +4)/100 = 3/100

f = 100/3 =33.33

Focal length of the lens is 33.33 cm ≈0.33 m

Power of the lens ,P = 1/f =1/.33 m ≈ +3 D

Q8. Why is normal eye not able to see clearly the objects placed closer than 25 cm?

Ans. 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 and as a result, the image of the object formed beyond the retina and we couldn’t see the object properly.

Q9. What happens to the image distance in the eye when we increase the distance of an object from the eye?

Ans. When we increase the distance of an object from the eye, the ciliary muscles of the crystalline lens expand causing an increase in focal length of the eye lens and thus image distance is unchanged, the image is formed at the retina as before.

NCERT solutions for class 10 science chapter 11-The Human Eye and Colourful world

Q10. Why do stars twinkle?

Ans. Stars are very far from us, the light ray from star continually refracted through different layers of the atmosphere, the temperatures of these atmospheric layers varies and thus their densities also vary due to which the path of starlight to our eyes changes that changes the virtual image of the star within little span of time,this effect makes our eyes to see the star twinkle.

Q11. Explain why the planets do not twinkle?

Ans. 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, but planets are very near to us as compared to star they look like a disc,a bunch of rays reaches to our eye through little atmospheric refraction,it is that’s why planet do not twinkle.

Q12. Why does the sun appear reddish early in the morning?

Ans. 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 to the observer. Therefore the sun appears reddish at sunrise and sunset.

Q13. Why does the sky appear dark instead of blue to an astronaut?

Ans. The sky appears dark instead of blue to an astronaut, as the scattering of light does not take place outside the earth’s atmosphere due to the absence of gas particles.

Summary of the chapter 11-The Human Eye and Colourful world

Structure of the human eye

Cornea: It is a transparent layer that covers the pupil and iris of the eye and contributes maximum refraction of the total refraction of the incident rays into the eye.

Aqueous humor: It is watery fluids that contain nutrients that regulate the pressure on the eye and repairs the cells of the cornea, pupil and iris after the cornea light is refracted through the aqueous humor and focused to the iris and pupil.

Pupil: This is a black dot at the center of the eye,it is an opening to the eyes which allows light rays to enter the eye.

Iris: Iris controls the amount of light entering the eye by controlling the size of the pupil.

Crystalline lens: Crystalline lens is made of ciliary muscles which vary the focal length of the eye by their expansion and contraction and enable us to see the distant and nearby objects.

Vitreous humor: Vitreous humor is jelly-like fluids that fill the space between crystalline lens and retina, it gives a shape to the eyeball and helps in the refraction of light ray focussing it to the retina.

Retina: After refraction, the light ray falls at the retina which converts the light signal to an electrical signal then these electrical signals are transported to the brain through the optic nerve. The electric signals are interpreted by the visual cortex of the brain we see the object.

Defects of the eye

The eye has three defects (i) myopia (ii) hypermetropia (iii) presbyopia

Myopia: Myopia is also known as nearsightedness, in this defect a person is unable to see after a certain distance. The distance at which a person with the myopic eye is able to see known as the far point of the myopic eye, it is treated by applying a convex lens to the eye. This defect occurs due to the bulged cornea that decreases the focal length of the eye lens and the image of objects forms near to the eye lens instead of at the retina,so the image of the distant objects looks blurred

Hypermetropia: Hypermetropia is also known as farsightedness, in this defect a person is unable to see nearby objects. The minimum distance to which a person with the hypermetropic eye is able to see known as the near point of the hypermetropic eye, it is treated by applying a concave lens to the eye. This defect occurs due to the non-contraction of the ciliary muscles of the crystalline lens that makes the eye lens thinner, in this defect eye become unable to accommodate to see nearby objects  and the image of nearby objects forms beyond the retina that is far from the eye lens, so the image of the nearby objects looks blurred

Presbyopia: Presbyopia generally occurs in old age, in this defect, the human eye is unable to see the nearby object as well as far objects, the eye becomes unable to accommodate to see the objects,it occurs because of weakening of ciliary muscles. It is remedied by applying a bifocal lens a combination of the concave and convex lens.

Cataract: Cataract occurs due to injury to the eye or diabetes, in this defect the eye lens turns cloudy due to the ruptured tissues in the eye lens, anybody suffering from cataract can’t see the objects clearly, it is treated by surgery and then by applying glasses.

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Chapter 1- Chemical reactions and equations Chapter 9- Heredity and Evolution
Chapter 2- Acid, Base and Salt Chapter 10- Light reflection and refraction
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