Class – 10 HSLC HUMAN EYE AND COLOURFUL WORLD INTEXT QUESTIONS AND ANSWERS

INTEXT QUESTIONS AND ANSWER of Page No – 190

1. 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

1. The human eye can focus objects at different distances by adjusting the focal length of the eye lens . This is due to

1. Presbyopia
2. accommodation
3. near- sightedness
4. far- sightedness

Ans : b. accommodation

2. The human eye forms the image of an object at its

1. Cornea
2. Iris
3. Pupil
4. Retina

Ans: d. Retina

3. The least distance of distinct vision for a young adult with normal vision is about

1. 25 m
2. 2.5 m
3. 25 cm
4. 2.5 cm

Ans : a. 25 m

4. The change in focal length of an eye lens is caused by the action of the

1. pupil
2. retina
3. ciliary muscles
4. 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