Class – 10 HSLC HUMAN EYE AND COLOURFUL WORLD INTEXT QUESTIONS AND ANSWERS
INTEXT QUESTIONS AND ANSWER of Page No – 190
- What is meant by power of accommodation of the eye ?
Ans : The power of accommodation of the eye is the maximum variation of its power for focusing on near and far (distant) objects . For a normal eye the power of accommodation is about 4 dioptres.
2. A person with myopic eye can not see objects beyond 1.2 m distinctly . What should be the type of the corrective lens used to restore proper vision ?
Ans: A person with a myopic eye can use concave or diverging lens to restore proper vision.
Class – 10 HSLC HUMAN EYE AND COLOURFUL WORLD INTEXT QUESTIONS AND ANSWERS
3. What is the far point and near point of the human eye with normal vision?
Ans: For a normal eye with normal vision the far point is at infinity and the near point is at 25 cm from the eye.
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 ?
Ans : The child is suffering from myopic . He should use concave lenses of suitable focal length.
Class – 10 HSLC HUMAN EYE AND COLOURFUL WORLD INTEXT QUESTIONS AND ANSWERS
Class 10 Science Chapter 11 The Human Eye and the Colourful World answer
TEXTBOOK EXERCISE
- The human eye can focus objects at different distances by adjusting the focal length of the eye lens . This is due to
- Presbyopia
- accommodation
- near- sightedness
- far- sightedness
Ans : b. accommodation
2. The human eye forms the image of an object at its
- Cornea
- Iris
- Pupil
- Retina
Ans: d. Retina
3. The least distance of distinct vision for a young adult with normal vision is about
- 25 m
- 2.5 m
- 25 cm
- 2.5 cm
Ans : a. 25 m
4. The change in focal length of an eye lens is caused by the action of the
- pupil
- retina
- ciliary muscles
- iris
Ans : c. ciliary muscles
Class – 10 HSLC HUMAN EYE AND COLOURFUL WORLD INTEXT QUESTIONS AND ANSWERS
5. A person needs a lens of power – 5.5 dioptres for corecting 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) distanct vision and (ii) near vision ?
Ans : (i) Power of distant viewing part of the lens, P1 = – 5.5 D
Focal length of this part , f1 = 1 / P1
= 1/ -5.5 m
= – 18.73 cm
(ii) As power of the near vision part is measured relative to the main part of the lens of power – 5.5 D , so we use
=P1 + P2 = P
=> -5.5 + P2 = +1.5
=> P2 = + 1.5 + 5.5
=> P2 = + 6.5 D
Focal length of near vision part, f2 = 1 / P2
=1/ + 6.5 m
= + 15.4 cm
Class – 10 HSLC HUMAN EYE AND COLOURFUL WORLD INTEXT QUESTIONS AND ANSWERS
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 ?
Ans : For the myopic eye
u = – a
v = – 80 cm
f = 1
1/ v – 1/ u = 1/f
=> (1 / – 80) – ( 1 / – a) = 1 / f
=> f = – 80 cm
therefore, f = – 0. 80 m
Power of the lens is P = 1 / f
=> P = 1 / – 0.80
=> P = – 1. 25 D
A concave lens , P = – 1.25 D
Class – 10 HSLC HUMAN EYE AND COLOURFUL WORLD INTEXT QUESTIONS AND ANSWERS
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 m.
Ans: The object placed at 25 cm from the correcting lens must produce a virtual image at 1 m or 100 cm.
v = – 25 cm
u = – 100 cm
by lens formula ,
1 / f = 1 / v – 1 / u
=> 1 / f = 1 / -100 – 1 / – 25
=> 1 / f = 1/ – 100 + 1 / 25
=> 1 / f = 3 / 100
=> f = 100 / 3 cm
f = 1 / 3 m
therefore, Power , P = 1 / f
= 1 / (1/3)
= + 3 D
Class – 10 HSLC HUMAN EYE AND COLOURFUL WORLD INTEXT QUESTIONS AND ANSWERS
8. Why a normal eye not able to see clearly the objects placed closer than 0.5 cm ?
Ans : A normal eye is unable to clearly see the objects place closer than 25 cm because the ciliary muscles of the eye unable to contract beyond a certain limit .
If the object is placed at the distance less than 25 cm from the eye than the object appears blured and produces strain in the eyes .
9. What happens to the image distance in the eye when we increase the distance of an object from the eye ?
Ans : The image distance remain same equal to the distance of the retina from the eye lens . They change in object distance is compensated by the change in the focal length of the eye lens due to the action of ciliary muscles so that a clear image is formed on the same retina .
Class – 10 HSLC HUMAN EYE AND COLOURFUL WORLD INTEXT QUESTIONS AND ANSWERS
10 . Why do stars twinkle ?
Ans : Twinkling of stars : The apparent position of star is slightly different from the actual position due to refraction of star light by the atmosphere . Further , this apparent position is not stationery but keeps on changing due to the change in atmospheric conditions like density , temperature etc. The path of the rays of light coming from the star goes on varying slightly . The amount of light entering our eyes from a particular star increases or decreases randomly with time . Sometimes , the star appears bright and other times , it appears fainter . This gives rise to the twinkling effect of the star.
Class – 10 HSLC HUMAN EYE AND COLOURFUL WORLD INTEXT QUESTIONS AND ANSWERS
11. Explain Why the planet do not twinkle ?
Ans: Planets do not twinkle because they appear larger in size than the stars as they are relatively closer to earth . Planets can be considered as a collection of a larger number of point – size sources of light . The different parts of these planets produce either brighter or dimmer effect is zero . Hence , the twinkling effect of the planets are nullified and they do not twinkle .
12. Why does the sun appear reddish early in the morning ?
Ans : During sunrise , the light rays coming from the sun have to travel a greater distance in the earth’s atmosphere before reaching our eyes . In this journey, the shorter wave length of lights are scattered out and only longer wave length are able to reach our eyes. Since blue colour has a shorter wave length and red colour has a longer wave length. The red colour is able to reach our eyes after the atmospheric scattering of light. Therefore , the sun appears reddish early in the morning .
Class – 10 HSLC HUMAN EYE AND COLOURFUL WORLD INTEXT QUESTIONS AND ANSWERS
13. Why does the sky appear dark instead of blue to an astronaut ?
Ans : The sky appears dark instead of blue to an astronaut because there is no atomsphere in the outerspace that can scatter the sunlight . As the sunlight is not scattered light reach the eyes of the astronaut and the sky appears black to them .
Class – 10 HSLC HUMAN EYE AND COLOURFUL WORLD INTEXT QUESTIONS AND ANSWERS
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Class – 10 HSLC HUMAN EYE AND COLOURFUL WORLD INTEXT QUESTIONS AND ANSWERS
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