KSEEB Solutions for Class 10 Science Chapter 11 Human Eye and Colourful World

Students can Download Science Chapter 11 Human Eye and Colourful World Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Science helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

Karnataka State Syllabus Class 10 Science Chapter 11 Human Eye and Colourful World

KSEEB Class 10 Science Human Eye and Colourful World Intext Questions and Answers

Question 1.
What is meant by power of accommodation of the eye?
Answer:
The ability of the eye lens to adjust its focal length is called accommodation.

Question 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:
Diverging or concave lens.

Question 3.
What is the far point and near point of the human eye with normal vision?
Answer:
For a human eye with normal vision, the far point is at infinity and near point is at 25cm from the eye.

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:
The child is. suffering from myopia. It can be corrected by using spectacles containing concave lens.

KSEEB Class 10 science Human Eye and Colourful World Textbook Exercise Questions and Answers

Question 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.

Question 2.
The human eye forms the image of an object at its.
(a) cornea
(b) iris
(c) pupil.
(d) retina.
Answer:
(d) retina.

Question 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.

Question 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.

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:
i) Power ,P = -5.5D,focal length t = ?

The – ve sign indicates that the lens is concave

ii) Power, P = + 1.5D, focal length = ?
\(P=\frac { 1 }{ f({ in\quad }m) } \)
\(+1.5=\frac{1}{f} \text { or } f=\frac{1}{+1.5}\)
f = +0.67 m
The +Ve sign indicates the lens is convex.

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 far point of a myopic person is 80 cm in front of the eye. This means that a person can see the distant object [ which is kept at infinity ] Clearly if the image of this object is formed at his far point
The negitive [-] sign indications that the lens is convave.

The concave lens of power – 1.25 D is required to correct the problem.

Question 7.
Make a diagram to show how hypermetropia is corrected. The near point of a hyperihetropic 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:
The hypermetropic eye is corrected by using a convex lens of appropriate focal length as spectacles in front of the eye. This convex lens makes a virtual image at the position of near point of the person so that the person is now able to see that image clearly.
Here, u = – 25cm [Normal hear point]
V = – 1m = – 100 cm [Near point of defective eye]
F = ?
Using lens formula,

Thus the convex lens of focal length + 33.3 cm is required to correct defect.
Here F = 33.3cm = 0.33m

Thus the convex lens of power + 3.0D is required

Question 8.
Why is a normal eye not able to see clearly the objects placed closer than 25 cm?
Answer:
Ciliary muscles in the eye are not contracted near. Because of this reason, normal eye is not able to see clearly the objects placed closer than 25 cm.

Question 9.
What happens to the image distance in the eye when we increase the distance of an object from the eye?
Answer:
Lens of our eye become thin and its focal length increases and no image is formed on the ratina. So we cannot see distant objects.

Question 10.
Why do stars twinkle?
Answer:
The twinkling of a star is due to the atmospheric refraction of starlight. The physical conditions of the earth’s atmosphere are not stationary. Since the stars are very distant, they approximate point-sized sources of light. As the path of rays of light coming from the star goes on varying slightly, the apparent position of the star fluctuates and the amount of starlight entering the eye flickers the star sometimes appears brighter, and at some other time, fainter, which is the twinking effect.

Question 11.
Explain why the planets do not twinkle.
Answer:
The planets are much closer to the earth and are thus seen as extended sources. If we consider a planet as a collection of a large number of point – sized sources of light, the total variation in the amount of light entering our eye from all the individual, point – sized sources will average out to zero, thereby nullifying the twinkling effect.

Question 12.
Why does the Sun appear reddish early in the morning?
Answer:
In the early morning (at the time of sunrise). When the sun in near the horizon, the sunlight has to travel the greatest distance through the atmosphere to reach us. During this long journey of sunlight, most of the blue colour and shorter wavelength present in it is scattered out and away from our line of sight. So, the light reaching us directly from the rising sun consists mainly of longer wavelength, red colour due to which 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 containing air in the outer space to scatter sunlight. As there is no scattered light to reach our eyes in outer space, so, the sky appears dark.

Choose the correct answer:-

Question 1.
The inability of a lens to form point image of an axial point object is called.
a) Chromatic aberration
b) Spherical aberration
c) Myopia
d) Distortion
Answer:
b) Spherical aberration

Question 2.
Spherical aberration can be corrected by using.
a) Anastigmat lens
b) Concave lens
c) Convex lens
d) Cylindrical lens
Answer:
a) Anastigmat lens

Question 3.
The distance of distinct vision for a normal eye is
a) 2.5cm
b) 25cm
c) 2.5m
d) 25m
Answer:
b) 25cm

Question 4.
A man is unable to see objects distinctly at a distance less than 1 metre. He is suffering from
a) astigmatism
b) Myopia
c) Distortion
d) Hypermetropia
Answer:
d) Hypermetropia

Question 5.
In a myopic eye, the rays from infinity are brought to focus at a point.
a) In front of retina
b) On the retina
c) Behind the retina
d) On the eye,lens
Answer:
a) In front of retina

Question 6.
Myopia occurs due to
a) Increase in the focal length of the eye lens
b) Decrease in focal length of the eye lens
c) Contraction of the eye ball
d) Decrease in distance between the ratina and the eye lens
Answer:
b) Decrease in focal length of the eye lens

Question 7.
Hypermetropia can be corrected by the use of
a) Convex lens
b) Concave lens
c) Cylinderical lens
d) Anastigmat lens
Answer:
a) Convex lens

Question 8.
The impact of an image on the retina persists for ……..
(a) \(\frac{1}{10} \mathrm{s}\)
(b) \(\frac{1}{20} \mathrm{s}\)
(c) 10s
(d) 20s
Answer:
(b) \(\frac{1}{20} \mathrm{s}\)

Question 9.
The person suffering from both myopia and hypermetropia requires the use of.
a) Bifocal lens
b) Concave lens
c) Convex lens
d) Cylinderical lenses
Answer:
a) Bifocal lens

Question 10.
The splitting of white light into its component colour is called.
a) Splitting
b) Dispresion
c) Diffraction
d) Interference
Answer:
b) Dispresion

Question 11.
Grass appears green because it
a) Reflects green light
b) absorbs green light
c) Refracts green light
d) None of these
Answer:
a) Reflects green light

Question 12.
A red rose, when seen in green light, appears
a) Yellow
b) Red
c) green
d) Black
Answer:
d) Black

KSEEB Class 10 Science Human Eye and Colourful World Additional Questions and Answers

Question 1.
A person is advised to wear spectacles with concave lenses. What type of defect of vision is he suffering from?
Answer:
Near – sightedness / Short – sightedness / myopia.

Question 2.
A person is advised to wear spectacles with convex lenses. What type of defect of vision is he suffering from?
Answer:
Far – sightedness / Long – sightedness / hypermetropia.

Question 3.
What is colour blindness?
Answer:
Colour blindness is the defect of the eye due to which a person is unable to distinguish certain colours, sometimes even the primary colours.

Question 4.
The sun can be seen about two minutes before actual sunrise. Give reason?
Answer:
The sun can be seen about two minutes before actual sunrise because of atmospheric refraction.

Question 5.
How do we sec colours?
Answer:
The retina of a human eye has a large number of light-sensitive cells. These cells are of two types, i.e, rod cells and cone cells. The. rod-shaped cells show responsibility towards the intensity of light rays, while the cone-shaped cells respond to colours. It is cone cells. Which makes it possible for a person to see different colours and distinguish between them.

Question 6.
Why are danger signal lights red in colour?
Answer:
Danger signal lights are red in color because the red coloured light having longer wave length is scattered the least by fog or smoke. There fore, it can be seen clearly from a distance.

Question 7.
What is meant by dispersion of white light? Name the various colours of spectrum of white light in proper sequence?
Answer:
The splitting of white light into its component colours is called dispersion ‘ of light. The band of the coloured components formed due to the dispersion of white light is called spectrum. Seven colours of spectrum are violet, indigo, blue, green, yellow orange and red also known as VIBGYOR.

Question 8.
What is the causes of dispersion?
Answer:
All colours of light travelat the same speed in a vacuum. When these enter a transparent subtance like prism, all slow down by different amounts depending on their wave length . As these slow down by different amounts, different colors are refracted through different angles.

Question 9.
What is presbyopia? What causes presbyopia? How is presbyopia corrected?
Answer:
The power of accommodation of the eye decreases with ageing. For most people, the near point gradually recedes and the far point comes closer. This defect is called presbyopia. It arises due to the gradual weakening of the ciliary muscles and diminishing flexibility of the eye lens. Such a person may suffer from myopia and hypermetropia. This defect is then corrected by using bifocal lenses of suitable focal lengths. The upper part of the lens is concave lens which corrects myopia to see the distant objects clearly, while the lower part of the lens has convex lens which corrects the hypermetropia to see the nearby objects clearly.

Question 10.
What is meant by persistence of vision? We are able to see the movie picture in a cinema hall. How does this happen?
Answer:
The ability of the human eye to continue to see the image of an object for a very short duration even after the removal of that object is called persistence of vision. It is due to persistence of vision that we are able to see movie pictures in a cinema hall. The pictures in the form of a long film are projected on the screen at a rate of about 24 pictures per second. Under these conditions, the image of one picture persists on the retina of the eye till the image of the next picture falls on the screen, and so on. Due to this, the slightly different images of the successive pictures present on the film merge smoothly with one another and give us the feeling of continuity and moving images.

Question 11.
What is meant by scattering of light? Use this phenomenon to explain why the clear sky appears blue or the sun appears reddish at sunrise.
Answer:
The phenomenon in which a part of the light incident on a particle is redirected in different directions is called the scattering of light. At the time of sunrise and sunset, when the sun is near the horizon, sunlight travels a greater distance through the atmosphere to reach us. During this most of the shorter wave lengths present in it are scattered away from our line of sight by the molecules of air and other fine particles in the atmosphere. So, light reaching us directly from the rising or setting sun consists mainly of the longer wavelength red colour because of which the sun appears red.

Question 12.
Why is the colour of sky blue?
Answer:
The molecules of air and other fine particles in the atmosphere have size smaller than the wavelength of visible light. These are more effective in scattering light of shorter wavelengths at the blue end than light of longer wavelengths at the red end. When sunlight passes through the atmosphere, the fine particles in air scatter the blue colour more strongly than red. The scattered blue light enters our eyes. Since we see the blue light from every where overhead, the sky appears blue,

Question 13.
What is tyndall effect? Explain with an example?
Answer:
The scattering of light by particles in its path is called tyndall effect. When a beam of light ‘enters a smoke-filled dark room through a small hole. Then its path becomes visible to us. The tiny dust particles present in the air of room scatter the beam of light all round the room. Thus, scattering of light makes the particles visible, tyndall effect can also be observed when sunlight passes through a canopy of a dense forest. Here, tiny water droplets in the mist scatter light.

Question 14.
How does atmospheric refraction affect sunrise and sunset?
Answer:
The layers of air nearer to earth are denser than those above it. At sunrise and sunset when the sun is below the horizon, the light rays starting from sun are incident on these layers. They pass through successively denser layers and thus get bent more and more towards the normal until they fall upon the eye of the observer O. To the observer O these rays appear to come from s which is above horizon, it is for this reason that the sun is visible to us a little before it rises above the horizon and so also till a little later it sets below the horizon. The difference of time is about 2 minutes each for early rise and late setting of the sun.

Question 15.
A person needs a lens of power -4.5 D for correction of her vision
(a) What kind, of a defect in vision is she suffering from?
(b) What is the focal length of the corrective lens?
(c) What is the nature of the corrective lens?
Answer:
(a) Myopia
(b) \(\mathrm{f}=\frac{1}{-4.5}=\frac{-2}{9} \mathrm{m}=-0.22 \mathrm{m}\)
(c) Concave lens.

Question 16.
How are we able to see nearby and also the distant objects clearly?
Answer:
This is due to the ability of the eye lens to adjust its focal length which is known as accommodation. When the ciliary muscles are relaxed the lens become thin. Thus, its focal length increases This enables us to see distant objects clearly. When we are looking at objects closer to the eye, the ciliary muscles contract. This increases the curvature of the eye lens. The eye lens become thicker. The focal length of the eye lens decreases. This enables us to see nearby objects. A concave lens of focal length 4m clearly.

Question 17.
A person cannot see objects nearer than 75 cm from his eyes while a person with normal vision can see objects up to 25cm from his eyes. Find the nature, the focal length and the power of the correcting lens used for the defective vision?
Answer:
The defect of the vision is hypermetropia
Given, that u = – 25 cm and v- = – 75cm

The + sign with for the power indicates a convergent lens.

Question 18.
A person can see clearly up to 3 meters, prescribe a lens for spectacles so that he can see clearly up to 12 metres.
Answer:
Here , v = 3m, u = -12m,f = ?
\(\text { Using } \frac{1}{\mathrm{u}} \frac{-1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \text { (or) } \frac{-1}{3} \frac{+1}{12}=\frac{1}{\mathrm{f}}\)
A concave lens of focal length4 cm should be used.