{"id":6895,"date":"2024-01-23T11:45:56","date_gmt":"2024-01-23T11:45:56","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6895"},"modified":"2024-01-23T11:45:56","modified_gmt":"2024-01-23T11:45:56","slug":"ncert-solutions-class-10-science-chapter-11-human-eye-and-colourful-world","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-science-chapter-11-human-eye-and-colourful-world\/","title":{"rendered":"NCERT Solutions Class 10 (Science) Chapter 11 (Human Eye and Colourful World)"},"content":{"rendered":"

NCERT Solutions Class 10 Science\u00a0<\/span><\/strong><\/h2>\n

The NCERT Solutions in English Language for Class 10 Science Chapter – 11 (Human Eye and Colourful World) <\/strong>have been provided here to help the students solve the questions from this exercise.\u00a0<\/span><\/p>\n

Chapter – 11 (Human Eye and Colourful World)\u00a0<\/span><\/strong><\/h2>\n

Questions<\/strong><\/span><\/p>\n

1. What is meant by power of accommodation of the eye ?
\n<\/strong>Answer – <\/strong>The ability of the eye lens to adjust its focal length to see nearby and distant objects clearly.<\/span><\/p>\n

2. A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type of the corrective lens used to restore proper vision ?<\/strong><\/span>
\nAnswer –\u00a0<\/strong>A person with a myopic eye should use a concave lens of focal length 1.2 m so as to restore proper vision.<\/span><\/p>\n

3. What is the far point and near point of the human eye with normal vision ?\u00a0<\/strong><\/span>
\nAnswer – <\/strong>For a human eye with normal vision the far point is at infinity and the near point is at 25 cm from the eye.<\/span><\/p>\n

4. <\/strong>A student has difficulty in reading the black board while sitting in the last row. What could be the defect the child is suffering from ? How can it be corrected ?
\n<\/strong>Answer – <\/strong>Near sightedness or myopia. This defect can be corrected by using a concave lens of suitable focal length.<\/span><\/p>\n

Exercises<\/span><\/strong><\/span><\/p>\n

1. <\/strong>The human eye can focus objects at different distances by adjusting the focal length of the eye lens. This is due to<\/span>
\n(a)\u00a0presbyopia<\/span>
\n(b)\u00a0near sightedness<\/span>
\n(c)\u00a0accommodation<\/span>
\n(d)\u00a0far sightedness.<\/span>
\nAnswer –\u00a0<\/strong>(c) accommodation<\/span><\/p>\n

2. Human eye forms the image of an object at its<\/strong><\/span>
\n(a)\u00a0cornea<\/span>
\n(b)\u00a0iris<\/span>
\n(c)\u00a0pupil<\/span>
\n(d) retina\u00a0<\/span>
\nAnswer –\u00a0<\/strong>(d) retina <\/span><\/p>\n

3. The least distance of distinct vision for a young adult with normal vision is about\u00a0<\/strong><\/span>
\n(a)\u00a025 m<\/span>
\n(b)\u00a02.5 cm<\/span>
\n(c)\u00a025 cm<\/span>
\n(d) 50 cm.\u00a0<\/span>
\nAnswer – <\/strong>(c) 25 cm<\/span><\/p>\n

4. The change in focal length of an eye lens is caused by the action of<\/strong><\/span>
\n(a)\u00a0pupil<\/span>
\n(b)\u00a0retina<\/span>
\n(c)\u00a0ciliary muscles<\/span>
\n(d) iris.\u00a0<\/span>
\nAnswer –\u00a0<\/strong>(c)\u00a0ciliary muscles<\/span><\/p>\n

5. A person needs a lens of power \u2013 5.5 diopters for correcting distant vision. For correcting his near vision, he needs a lens of power +1.5 diopter. What is the focal length of the lens required for correcting<\/strong><\/span>
\n(i) distant vision and<\/span>
\n(ii) near vision ?<\/span>
\nAnswer –
\n<\/strong>(i) Power of lens needed for correction distant vision of the person (P) = -5.5 D <\/span>
\nFocal length of lens required for correcting distant vision (f) = 1\/P <\/span>
\n= 1\/-5.5 m = – 0.18 m = – 18 cm. <\/span><\/p>\n

(ii) For correcting near vision the power of lens required (P) = +1.5 D <\/span>
\nFocal length of lens required for correcting near vision (f) = 1\/P <\/span>
\n= 1\/1.5 m = 0.67 m = 67.0 cm.<\/span><\/p>\n

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?
\n<\/b><\/strong>Answer –\u00a0 <\/strong>To correct the myopia the person concerned should use concave lens of focal length (f) = -80 cm = -0.80 m <\/span>
\nPower of lens (P) = 1\/f(m) = 1\/-0.80 = 100\/-80 = -1.25 D.<\/span><\/p>\n

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 cm.<\/strong><\/span>
\nAnswer –\u00a0<\/strong>Diagram representing the correction of hypermetropia is a follows:
\n\"NCERT
\nNear point of defective eye is 1 m and that of normal eye is 25 cm. <\/span>
\nHere, u = -25 cm, v = -1m = 100 cm. <\/span>
\nUsing lens formula <\/span>
\n1\/f = 1\/v \u2013 1\/u <\/span>
\n1\/f = 1\/-100 + 1\/25 = 3\/100 <\/span>
\nf = 100\/3 cm = 1\/3m. <\/span>
\nP = 1\/f(m) <\/span>
\n= 1\/0.33 <\/span>
\n= +3.0 D.<\/span><\/p>\n

8. Why is a normal eye not able to see clearly the objects placed closer than 25 cm ?\u00a0<\/strong><\/span>
\nAnswer – <\/strong>The nearest position of an object from a normal human eye so that its sharp image is formed on retina is 25 cm. If the object is placed at a distance less than 25 cm, then the blurred image of the object is formed on retina as the focal length of eye lens cannot be decreased below a certain limit. Hence, eye cannot see it clearly.<\/span><\/p>\n

9. What happens to the image distance in the eye when we increase the distance of an object from the eye ?
\n<\/strong>Answer – <\/strong>The image distance remains the same in the eye because the eye has the ability to change the focal length of its lens to make the image always on the retina when the object distance increases from the eye.<\/span><\/p>\n

10. Why do stars twinkle ?
\n<\/strong>Answer –\u00a0<\/strong>Stars twinkle due to atmospheric refraction of starlight. As the stars are very away they behave as almost point sources of light. As on account of atmospheric refraction, 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 light entering the eye flickers, so sometimes the star appear brighter and at some other time, fainter. Thus the stars twinkle.<\/span><\/p>\n

11. Explain why the planets do not twinkle.\u00a0<\/strong><\/span>
\nAnswer –
\nPlanets do not twinkle:<\/strong>Planets are much closer to the earth and are seen as extended source. So, a planet may be considered as a collection of a large number of point-sized light sources. Although light coming from individual point-sized sources flickers but the total amount of light entering our eye from all the individual point-sized sources average out to be constant. Thereby, planets appear equally brighter and there is no twinkling of planets.<\/span><\/p>\n

12. Why does the Sun appear reddish early in the morning ?\u00a0<\/strong><\/span>
\nAnswer – <\/strong>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 wavelengths of lights are scattered out and only longer wavelengths are able to reach our eyes. Since blue colour has a\u00a0 shorter wavelength and red colour has a longer wavelength, the red colour is able to reach our eyes after the atmospheric scattering of light. Therefore, the Sun appears reddish early in the morning.<\/span><\/p>\n

13. Why does the sky appear dark instead of blue to an astronaut ?
\n<\/strong>Answer –\u00a0<\/strong>Blue colour of the sky is on account of scattering of light of shorter wavelength by particles in the atmosphere of earth. If the earth had no atmosphere, there would not have been any scattering and sky would have looked dark. When astronaut in his spacecraft goes above the atmosphere of earth, sky appears dark to him because there is no scattering of light.<\/span><\/p>\n

 <\/p>\n

Go Back To Chapters<\/i><\/b><\/a><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

NCERT Solutions Class 10 Science\u00a0 The NCERT Solutions in English Language for Class 10 Science Chapter – 11 (Human Eye and Colourful World) have been provided here to help the students solve the questions from this exercise.\u00a0 Chapter – 11 (Human Eye and Colourful World)\u00a0 Questions 1. What is meant by power of accommodation of […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1125],"tags":[1385,1133,1408,1407,1388,5],"class_list":["post-6895","post","type-post","status-publish","format-standard","hentry","category-class-10-science","tag-class-10-ncert-science-solutions","tag-class-10-ncert-solutions","tag-ncert-class-10-science-chapter-11-human-eye-and-colourful-world","tag-ncert-class-10-science-chapter-11-solutions","tag-ncert-class-10-science-solutions","tag-ncert-solutions-in-english"],"yoast_head":"\nNCERT Solutions Class 10 (Science) Chapter 11 (Human Eye and Colourful World) | TheExamPillar NCERT<\/title>\n<meta name=\"description\" content=\"NCERT Solutions Class 10 Science\u00a0The NCERT Solutions in English Language for Class 10 Science Chapter - 11 (Human Eye and Colourful World) has been provided here to help the students in solving the questions from this exercise.\u00a0Chapter - 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