![\"\"](\"https:\/\/cbsencertanswers.com\/wp-content\/uploads\/2020\/12\/image.png\")
\r\n3.What is the far point and the near point of the human eye with normal vision?<\/h2>\r\n\r\n\r\n\r\n
Ans:- i)the far point of the human eye with normal vision is the location of the farthest object on which the fully released eye can focus (infinity).<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 ii) the the near point of the human eye with normal vision\u00a0 is 25cm (9.8 in)<\/p>\r\n\r\n\r\n\r\n
4. A student has difficulty in far point and the near point of the human eye with normal vision the blackboard while sitting in the last row. What could be the defect the child is suffering from? How could it be corrected?<\/h2>\r\n\r\n\r\n\r\n
Ans:- It means that the student cannot see clearly the distant objects. Thus, he is suffering from myopia.<\/p>\r\n\r\n\r\n\r\n
In order to correct his problem, he should be provided with a concave lens.<\/p>\r\n\r\n\r\n\r\n
YOU ARE READING -TEXTBOOK ANSWERS OF The Human Eye and the Colourful World<\/h3>\r\n\r\n\r\n- Arithmetic Progressions Chapter 5 NCERT Solutions For Class 10 CBSE Mathematics (Exercise 3 and 4)<\/a><\/li>\n
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Exercises of TEXTBOOK (NCERT) ANSWERS OF The Human Eye and the Colourful World<\/h2>\r\n\r\n\r\n\r\n1.The human eye can focus objects at different distances by adjusting the focal length of the eye lens. This is due to<\/h2>\r\n\r\n\r\n\r\n\r\n- Presbyopia<\/li>\r\n
- Accommodation<\/li>\r\n
- Near-sightedness<\/li>\r\n
- Far-sightedness<\/li>\r\n<\/ol>\r\n\r\n\r\n\r\n
Ans:- b) accommodation<\/p>\r\n\r\n\r\n\r\n
2.The human eye forms the image of an object at its<\/h2>\r\n\r\n\r\n\r\n\r\n- Cornea<\/li>\r\n
- Iris<\/li>\r\n
- Pupil<\/li>\r\n
- Retina<\/li>\r\n<\/ol>\r\n\r\n\r\n\r\n
Ans:- d) retina<\/p>\r\n\r\n\r\n\r\n
3.The least distance of vision for a young adult with normal vision is about<\/h2>\r\n\r\n\r\n\r\n\r\n- 25m<\/li>\r\n
- 2.5cm<\/li>\r\n
- 25cm<\/li>\r\n
- 2.5m<\/li>\r\n<\/ol>\r\n\r\n\r\n\r\n
\u00a0c) 25 cm<\/p>\r\n\r\n\r\n\r\n
4.The change in focal length of an eye is caused by the action of the<\/h2>\r\n\r\n\r\n\r\n\r\n- Pupil<\/li>\r\n
- Retina<\/li>\r\n
- Ciliary muscles<\/li>\r\n
- Iris<\/li>\r\n<\/ol>\r\n\r\n\r\n\r\n
Ans:- c) ciliary muscles<\/p>\r\n\r\n\r\n\r\n
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\u00a0<\/h2>\r\n\r\n\r\n\r\n\r\n- Distant vision<\/li>\r\n
- Near vision<\/li>\r\n<\/ol>\r\n\r\n\r\n\r\n
Ans:- a) distant\u00a0 vision =-0.181m<\/p>\r\n\r\n\r\n\r\n
\r\n- near vision= 0.667m<\/li>\r\n<\/ul>\r\n\r\n\r\n\r\n
6. The far point of a myopic person is 80cm in front of the eye. What is the nature and power of the lens required to correct the problem?<\/h2>\r\n\r\n\r\n\r\n
Ans:- As the person is suffering from myopia, the image formed in his eye, ia at the retina. This, a concave lens is used to correct this defect.<\/p>\r\n\r\n\r\n\r\n
As we know that,<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Object distance, u= infinity<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Image distance, v=-80 cm<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Focal length =f<\/p>\r\n\r\n\r\n\r\n
According to lens formula,<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1\/V -1\/N =1\/f<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -1\/80 -1\/u =1\/f<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1\/f= -1\/80<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 f=-80cm Or, -0. 8m<\/p>\r\n\r\n\r\n\r\n
We know that,<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Power (P) =1\/f (m)<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 P=1\/-0.8 Or, -1. 25 D<\/p>\r\n\r\n\r\n\r\n
Thus, the person needs a concave lens, if Power=-1.25 D, to correct his defect.<\/p>\r\n\r\n\r\n\r\n
7.Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25cm.<\/h2>\r\n\r\n\r\n\r\n
Ans:- When a person suffers from hypermetropia, he can see distinct objects clearly, but he faces difficulties while seeing nearly objects. This is due to the lens focuses the incoming divergent rays beyond the retina of the eye.<\/p>\r\n\r\n\r\n\r\n
If the person uses a concave lens there will be more problems as the image will form at the retina, as shown in the figure, below\u2014<\/p>\r\n\r\n\r\n\r\n![\"TEXTBOOK](\"https:\/\/cbsencertanswers.com\/wp-content\/uploads\/2021\/01\/11.png\")
<\/figure>\r\n\r\n\r\n\r\nIf the image us formed at his near point (1m), the person will be able to see the object kept at 25 m(which is the near point of normal eye).<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Object distance, u=-25cm<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Image distance, v=-1m=100cm<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Focal length, f<\/p>\r\n\r\n\r\n\r\n
According to the lens formula,<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1\/v-1\/u=1\/f<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -1\/100-1\/-25=1\/f<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1\/25-1\/100=1\/f<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1\/f=(4-1) \/100<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 F=-100\/3 or, -33. 3cm Or, -0. 33 m<\/p>\r\n\r\n\r\n\r\n
Power, P= 1\/f(in metres)<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =1\/0.33m Or, 3.0D<\/p>\r\n\r\n\r\n\r\n
Thus, a concave lens of power +3.0 D should be provided to the person.<\/p>\r\n\r\n\r\n\r\n
8.Why is a normal eye not able to see clearly the objects placed closer than 25cm?<\/h2>\r\n\r\n\r\n\r\n
Ans:- A normal eye not able to see clearly the objects placed closer than 25cm because, the ciliary muscles are unable to contract beyond certain limit.<\/p>\r\n\r\n\r\n\r\n
9.What happens to the image distance in the eye when we increase the distance of an object from the eye?<\/h2>\r\n\r\n\r\n\r\n
Ans:- The image distance will\u00a0 be constant, as we know that, the size of the eyes cannot be increased or decreased.<\/p>\r\n\r\n\r\n\r\n
If we increase the object’s distance from our eye, the image distance in the eye does not change, but only the focal length of the eye lens, changes.<\/p>\r\n\r\n\r\n\r\n
As we know that, the focal length of the eye always changes, as the image to be formed at the retina of the eye.<\/p>\r\n\r\n\r\n\r\n
10.Why do stars twinkle?<\/h2>\r\n\r\n\r\n\r\n
Ans:- Stars twinkle in the night sky. Even though they are far away from us, but we can see them twinkling. This is because, they emit their own light which is being refracted but the atmosphere, and thus, it causes them to shine. The rays of the light, coming from the stars, it enters the earth’s atmosphere and gets refracted at various levels, which is due to the variation in the density of the air at different levels of the atmosphere. Thus, the stars appear brighter.<\/p>\r\n\r\n\r\n\r\n
11.Explain why the planets do not twinkle?<\/h2>\r\n\r\n\r\n\r\n
Ans:- As the planets are nearer to the earth than that of the stars, thus the planets sometimes appear larger than the stars in size. The planets sometimes, produce either bright or dimmer affects of light in such a way that, the average result for the light is zero. Thus, we cannot see the planets twinkling in the night sky.<\/p>\r\n\r\n\r\n\r\n
12.Why does the Sun appear reddish early in the morning?<\/h2>\r\n\r\n\r\n\r\n
Ans:- As the light rays of the Sun has to travel a greater distance in the earth’s atmosphere, before entering our eyes, during the rise of the sun, the shorter wavelength of the lights become less prominent than that of the longer wavelengths. And as we know that, blue colour is having a shorter wavelength than that of the red colour, the red colour scatters all over the sky, by the atmosphere. Thus, the Sun appears reddish to us, early in the morning.<\/p>\r\n\r\n\r\n\r\n
13.Why does the sky appear dark instead of blue to an astronaut?<\/h2>\r\n\r\n\r\n\r\n
Ans:- As we know that, space does not have its own atmosphere, which could scatter light rays of the Sun, so, the sky appears dark, instead of blue, to the astronauts.<\/p>\r\n\r\n\r\n\r\n
The experts have prepared the solutions of TEXTBOOK (NCERT) ANSWERS OF The Human Eye and the Colourful World CHAPTER-11 CBSE CLASS 10 SCIENCE(PHYSICS) on the best standard. If you have any query or suggestion ,reach us and we shall surely see to that attended. Hopefully the chapter TEXTBOOK (NCERT) ANSWERS OF The Human Eye and the Colourful World CHAPTER-11 CBSE CLASS 10 SCIENCE(PHYSICS) would serve to your purpose.<\/p>\r\n\r\n\r\n\r\n\r\n\r\n
Exercises of TEXTBOOK (NCERT) ANSWERS OF The Human Eye and the Colourful World<\/h2>\r\n\r\n\r\n\r\n1.The human eye can focus objects at different distances by adjusting the focal length of the eye lens. This is due to<\/h2>\r\n\r\n\r\n\r\n\r\n- Presbyopia<\/li>\r\n
- Accommodation<\/li>\r\n
- Near-sightedness<\/li>\r\n
- Far-sightedness<\/li>\r\n<\/ol>\r\n\r\n\r\n\r\n
Ans:- b) accommodation<\/p>\r\n\r\n\r\n\r\n
2.The human eye forms the image of an object at its<\/h2>\r\n\r\n\r\n\r\n\r\n- Cornea<\/li>\r\n
- Iris<\/li>\r\n
- Pupil<\/li>\r\n
- Retina<\/li>\r\n<\/ol>\r\n\r\n\r\n\r\n
Ans:- d) retina<\/p>\r\n\r\n\r\n\r\n
3.The least distance of vision for a young adult with normal vision is about<\/h2>\r\n\r\n\r\n\r\n\r\n- 25m<\/li>\r\n
- 2.5cm<\/li>\r\n
- 25cm<\/li>\r\n
- 2.5m<\/li>\r\n<\/ol>\r\n\r\n\r\n\r\n
\u00a0c) 25 cm<\/p>\r\n\r\n\r\n\r\n
4.The change in focal length of an eye is caused by the action of the<\/h2>\r\n\r\n\r\n\r\n\r\n- Pupil<\/li>\r\n
- Retina<\/li>\r\n
- Ciliary muscles<\/li>\r\n
- Iris<\/li>\r\n<\/ol>\r\n\r\n\r\n\r\n
Ans:- c) ciliary muscles<\/p>\r\n\r\n\r\n\r\n
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\u00a0<\/h2>\r\n\r\n\r\n\r\n\r\n- Distant vision<\/li>\r\n
- Near vision<\/li>\r\n<\/ol>\r\n\r\n\r\n\r\n
Ans:- a) distant\u00a0 vision =-0.181m<\/p>\r\n\r\n\r\n\r\n
\r\n- near vision= 0.667m<\/li>\r\n<\/ul>\r\n\r\n\r\n\r\n
6. The far point of a myopic person is 80cm in front of the eye. What is the nature and power of the lens required to correct the problem?<\/h2>\r\n\r\n\r\n\r\n
Ans:- As the person is suffering from myopia, the image formed in his eye, ia at the retina. This, a concave lens is used to correct this defect.<\/p>\r\n\r\n\r\n\r\n
As we know that,<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Object distance, u= infinity<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Image distance, v=-80 cm<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Focal length =f<\/p>\r\n\r\n\r\n\r\n
According to lens formula,<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1\/V -1\/N =1\/f<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -1\/80 -1\/u =1\/f<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1\/f= -1\/80<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 f=-80cm Or, -0. 8m<\/p>\r\n\r\n\r\n\r\n
We know that,<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Power (P) =1\/f (m)<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 P=1\/-0.8 Or, -1. 25 D<\/p>\r\n\r\n\r\n\r\n
Thus, the person needs a concave lens, if Power=-1.25 D, to correct his defect.<\/p>\r\n\r\n\r\n\r\n
7.Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25cm.<\/h2>\r\n\r\n\r\n\r\n
Ans:- When a person suffers from hypermetropia, he can see distinct objects clearly, but he faces difficulties while seeing nearly objects. This is due to the lens focuses the incoming divergent rays beyond the retina of the eye.<\/p>\r\n\r\n\r\n\r\n
If the person uses a concave lens there will be more problems as the image will form at the retina, as shown in the figure, below\u2014<\/p>\r\n\r\n\r\n\r\n<\/figure>\r\n\r\n\r\n\r\nIf the image us formed at his near point (1m), the person will be able to see the object kept at 25 m(which is the near point of normal eye).<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Object distance, u=-25cm<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Image distance, v=-1m=100cm<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Focal length, f<\/p>\r\n\r\n\r\n\r\n
According to the lens formula,<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1\/v-1\/u=1\/f<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -1\/100-1\/-25=1\/f<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1\/25-1\/100=1\/f<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1\/f=(4-1) \/100<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 F=-100\/3 or, -33. 3cm Or, -0. 33 m<\/p>\r\n\r\n\r\n\r\n
Power, P= 1\/f(in metres)<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =1\/0.33m Or, 3.0D<\/p>\r\n\r\n\r\n\r\n
Thus, a concave lens of power +3.0 D should be provided to the person.<\/p>\r\n\r\n\r\n\r\n
8.Why is a normal eye not able to see clearly the objects placed closer than 25cm?<\/h2>\r\n\r\n\r\n\r\n
Ans:- A normal eye not able to see clearly the objects placed closer than 25cm because, the ciliary muscles are unable to contract beyond certain limit.<\/p>\r\n\r\n\r\n\r\n
9.What happens to the image distance in the eye when we increase the distance of an object from the eye?<\/h2>\r\n\r\n\r\n\r\n
Ans:- The image distance will\u00a0 be constant, as we know that, the size of the eyes cannot be increased or decreased.<\/p>\r\n\r\n\r\n\r\n
If we increase the object’s distance from our eye, the image distance in the eye does not change, but only the focal length of the eye lens, changes.<\/p>\r\n\r\n\r\n\r\n
As we know that, the focal length of the eye always changes, as the image to be formed at the retina of the eye.<\/p>\r\n\r\n\r\n\r\n
10.Why do stars twinkle?<\/h2>\r\n\r\n\r\n\r\n
Ans:- Stars twinkle in the night sky. Even though they are far away from us, but we can see them twinkling. This is because, they emit their own light which is being refracted but the atmosphere, and thus, it causes them to shine. The rays of the light, coming from the stars, it enters the earth’s atmosphere and gets refracted at various levels, which is due to the variation in the density of the air at different levels of the atmosphere. Thus, the stars appear brighter.<\/p>\r\n\r\n\r\n\r\n
11.Explain why the planets do not twinkle?<\/h2>\r\n\r\n\r\n\r\n
Ans:- As the planets are nearer to the earth than that of the stars, thus the planets sometimes appear larger than the stars in size. The planets sometimes, produce either bright or dimmer affects of light in such a way that, the average result for the light is zero. Thus, we cannot see the planets twinkling in the night sky.<\/p>\r\n\r\n\r\n\r\n
12.Why does the Sun appear reddish early in the morning?<\/h2>\r\n\r\n\r\n\r\n
Ans:- As the light rays of the Sun has to travel a greater distance in the earth’s atmosphere, before entering our eyes, during the rise of the sun, the shorter wavelength of the lights become less prominent than that of the longer wavelengths. And as we know that, blue colour is having a shorter wavelength than that of the red colour, the red colour scatters all over the sky, by the atmosphere. Thus, the Sun appears reddish to us, early in the morning.<\/p>\r\n\r\n\r\n\r\n
13.Why does the sky appear dark instead of blue to an astronaut?<\/h2>\r\n\r\n\r\n\r\n
Ans:- As we know that, space does not have its own atmosphere, which could scatter light rays of the Sun, so, the sky appears dark, instead of blue, to the astronauts.<\/p>\r\n\r\n\r\n\r\n
The experts have prepared the solutions of TEXTBOOK (NCERT) ANSWERS OF The Human Eye and the Colourful World CHAPTER-11 CBSE CLASS 10 SCIENCE(PHYSICS) on the best standard. If you have any query or suggestion ,reach us and we shall surely see to that attended. Hopefully the chapter TEXTBOOK (NCERT) ANSWERS OF The Human Eye and the Colourful World CHAPTER-11 CBSE CLASS 10 SCIENCE(PHYSICS) would serve to your purpose.<\/p>\r\n\r\n\r\n\r\n\r\n\r\n
- \r\n
- Presbyopia<\/li>\r\n
- Accommodation<\/li>\r\n
- Near-sightedness<\/li>\r\n
- Far-sightedness<\/li>\r\n<\/ol>\r\n\r\n\r\n\r\n
Ans:- b) accommodation<\/p>\r\n\r\n\r\n\r\n
2.The human eye forms the image of an object at its<\/h2>\r\n\r\n\r\n\r\n
- \r\n
- Cornea<\/li>\r\n
- Iris<\/li>\r\n
- Pupil<\/li>\r\n
- Retina<\/li>\r\n<\/ol>\r\n\r\n\r\n\r\n
Ans:- d) retina<\/p>\r\n\r\n\r\n\r\n
3.The least distance of vision for a young adult with normal vision is about<\/h2>\r\n\r\n\r\n\r\n
- \r\n
- 25m<\/li>\r\n
- 2.5cm<\/li>\r\n
- 25cm<\/li>\r\n
- 2.5m<\/li>\r\n<\/ol>\r\n\r\n\r\n\r\n
\u00a0c) 25 cm<\/p>\r\n\r\n\r\n\r\n
4.The change in focal length of an eye is caused by the action of the<\/h2>\r\n\r\n\r\n\r\n
- \r\n
- Pupil<\/li>\r\n
- Retina<\/li>\r\n
- Ciliary muscles<\/li>\r\n
- Iris<\/li>\r\n<\/ol>\r\n\r\n\r\n\r\n
Ans:- c) ciliary muscles<\/p>\r\n\r\n\r\n\r\n
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\u00a0<\/h2>\r\n\r\n\r\n\r\n
- \r\n
- Distant vision<\/li>\r\n
- Near vision<\/li>\r\n<\/ol>\r\n\r\n\r\n\r\n
Ans:- a) distant\u00a0 vision =-0.181m<\/p>\r\n\r\n\r\n\r\n
- \r\n
- near vision= 0.667m<\/li>\r\n<\/ul>\r\n\r\n\r\n\r\n
6. The far point of a myopic person is 80cm in front of the eye. What is the nature and power of the lens required to correct the problem?<\/h2>\r\n\r\n\r\n\r\n
Ans:- As the person is suffering from myopia, the image formed in his eye, ia at the retina. This, a concave lens is used to correct this defect.<\/p>\r\n\r\n\r\n\r\n
As we know that,<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Object distance, u= infinity<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Image distance, v=-80 cm<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Focal length =f<\/p>\r\n\r\n\r\n\r\n
According to lens formula,<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1\/V -1\/N =1\/f<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -1\/80 -1\/u =1\/f<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1\/f= -1\/80<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 f=-80cm Or, -0. 8m<\/p>\r\n\r\n\r\n\r\n
We know that,<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Power (P) =1\/f (m)<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 P=1\/-0.8 Or, -1. 25 D<\/p>\r\n\r\n\r\n\r\n
Thus, the person needs a concave lens, if Power=-1.25 D, to correct his defect.<\/p>\r\n\r\n\r\n\r\n
7.Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25cm.<\/h2>\r\n\r\n\r\n\r\n
Ans:- When a person suffers from hypermetropia, he can see distinct objects clearly, but he faces difficulties while seeing nearly objects. This is due to the lens focuses the incoming divergent rays beyond the retina of the eye.<\/p>\r\n\r\n\r\n\r\n
If the person uses a concave lens there will be more problems as the image will form at the retina, as shown in the figure, below\u2014<\/p>\r\n\r\n\r\n\r\n
<\/figure>\r\n\r\n\r\n\r\nIf the image us formed at his near point (1m), the person will be able to see the object kept at 25 m(which is the near point of normal eye).<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Object distance, u=-25cm<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Image distance, v=-1m=100cm<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Focal length, f<\/p>\r\n\r\n\r\n\r\n
According to the lens formula,<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1\/v-1\/u=1\/f<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -1\/100-1\/-25=1\/f<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1\/25-1\/100=1\/f<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1\/f=(4-1) \/100<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 F=-100\/3 or, -33. 3cm Or, -0. 33 m<\/p>\r\n\r\n\r\n\r\n
Power, P= 1\/f(in metres)<\/p>\r\n\r\n\r\n\r\n
\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =1\/0.33m Or, 3.0D<\/p>\r\n\r\n\r\n\r\n
Thus, a concave lens of power +3.0 D should be provided to the person.<\/p>\r\n\r\n\r\n\r\n
8.Why is a normal eye not able to see clearly the objects placed closer than 25cm?<\/h2>\r\n\r\n\r\n\r\n
Ans:- A normal eye not able to see clearly the objects placed closer than 25cm because, the ciliary muscles are unable to contract beyond certain limit.<\/p>\r\n\r\n\r\n\r\n
9.What happens to the image distance in the eye when we increase the distance of an object from the eye?<\/h2>\r\n\r\n\r\n\r\n
Ans:- The image distance will\u00a0 be constant, as we know that, the size of the eyes cannot be increased or decreased.<\/p>\r\n\r\n\r\n\r\n
If we increase the object’s distance from our eye, the image distance in the eye does not change, but only the focal length of the eye lens, changes.<\/p>\r\n\r\n\r\n\r\n
As we know that, the focal length of the eye always changes, as the image to be formed at the retina of the eye.<\/p>\r\n\r\n\r\n\r\n
10.Why do stars twinkle?<\/h2>\r\n\r\n\r\n\r\n
Ans:- Stars twinkle in the night sky. Even though they are far away from us, but we can see them twinkling. This is because, they emit their own light which is being refracted but the atmosphere, and thus, it causes them to shine. The rays of the light, coming from the stars, it enters the earth’s atmosphere and gets refracted at various levels, which is due to the variation in the density of the air at different levels of the atmosphere. Thus, the stars appear brighter.<\/p>\r\n\r\n\r\n\r\n
11.Explain why the planets do not twinkle?<\/h2>\r\n\r\n\r\n\r\n
Ans:- As the planets are nearer to the earth than that of the stars, thus the planets sometimes appear larger than the stars in size. The planets sometimes, produce either bright or dimmer affects of light in such a way that, the average result for the light is zero. Thus, we cannot see the planets twinkling in the night sky.<\/p>\r\n\r\n\r\n\r\n
12.Why does the Sun appear reddish early in the morning?<\/h2>\r\n\r\n\r\n\r\n
Ans:- As the light rays of the Sun has to travel a greater distance in the earth’s atmosphere, before entering our eyes, during the rise of the sun, the shorter wavelength of the lights become less prominent than that of the longer wavelengths. And as we know that, blue colour is having a shorter wavelength than that of the red colour, the red colour scatters all over the sky, by the atmosphere. Thus, the Sun appears reddish to us, early in the morning.<\/p>\r\n\r\n\r\n\r\n
13.Why does the sky appear dark instead of blue to an astronaut?<\/h2>\r\n\r\n\r\n\r\n
Ans:- As we know that, space does not have its own atmosphere, which could scatter light rays of the Sun, so, the sky appears dark, instead of blue, to the astronauts.<\/p>\r\n\r\n\r\n\r\n
The experts have prepared the solutions of TEXTBOOK (NCERT) ANSWERS OF The Human Eye and the Colourful World CHAPTER-11 CBSE CLASS 10 SCIENCE(PHYSICS) on the best standard. If you have any query or suggestion ,reach us and we shall surely see to that attended. Hopefully the chapter TEXTBOOK (NCERT) ANSWERS OF The Human Eye and the Colourful World CHAPTER-11 CBSE CLASS 10 SCIENCE(PHYSICS) would serve to your purpose.<\/p>\r\n\r\n\r\n\r\n\r\n\r\n
- near vision= 0.667m<\/li>\r\n<\/ul>\r\n\r\n\r\n\r\n