{"id":436,"date":"2020-09-15T05:02:55","date_gmt":"2020-09-14T23:32:55","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=436"},"modified":"2021-07-02T15:01:52","modified_gmt":"2021-07-02T09:31:52","slug":"kseeb-sslc-class-10-science-solutions-chapter-11","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-science-solutions-chapter-11\/","title":{"rendered":"KSEEB SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World"},"content":{"rendered":"

KSEEB SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World<\/strong>\u00a0are part of KSEEB SSLC Class 10 Science Solutions<\/a>. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World.<\/p>\n

Karnataka SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World<\/h2>\n

KSEEB SSLC Class 10 Science Chapter 11 Intext Questions<\/h3>\n

Text Book Part II Page No. 100<\/p>\n

Question 1.
\nWhat is meant by power of accommodation of the eye?
\nAnswer:
\nThe ability of the eye to focus the distant objects as well as the nearby objects on the retina by changing focal length or converging power of its lens is called accommodation. The normal eye has a power of accommodation which enables the object as close as 25cm & as far as infinity to be focused on its retina.<\/p>\n

Question 2.
\nA 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?
\nAnswer:
\nA person with a myopic eye cannot see objects beyond 1.2 m distinctly because, In a myopic eye, the image of a distant object is formed in front of the retina. This defect can be corrected by using a concave lens.<\/p>\n

Question 3.
\nWhat is the far point and near point of the human eye with normal vision?
\nAnswer:
\nFar point of the human eye with normal vision is near than infinity and near point is 1.2 m.<\/p>\n

Question 4.
\nA 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?
\nAnswer:
\nA student has difficulty reading the black board while sitting in the last row means he is suffering from myopia. This defect can be corrected using concave lens of suitable power.<\/p>\n

KSEEB SSLC Class 10 Science Chapter 11 Textbook Exercises<\/h3>\n

Question 1.
\nThe human eye can focus on objects at different distances by adjusting the focal length of the eye lens. This is due to
\n(a) presbyopia
\n(b) accomodation
\n(c) near-sightedness
\n(d) far-sightedness
\nAnswer:
\n(b) accomodation<\/p>\n

Question 2.
\nThe human eye forms the image of an object at its
\n(a) cornea
\n(b) Iris
\n(c) pupil
\n(d) retina
\nAnswer:
\n(d) retina<\/p>\n

Question 3.
\nThe least distance of distinct vision for a young adult with normal vision is about
\n(a) 25 m.
\n(b) 2.5 cm
\n(c) 25 cm
\n(d) 2.5 m
\nAnswer:
\n(c) 25 cm<\/p>\n

Question 4.
\nThe change in focal length of an eye lens is caused by the action of the
\n(a) pupil.
\n(b) retina
\n(c) ciliary muscles
\n(d) iris
\nAnswer:
\n(c) ciliary muscles<\/p>\n

Question 5.
\nA person needs a lens of power – 5.5 dioptres for correcting his distant vision. 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) distinct vision, and (ii) near vision?
\nAnswer:
\ni) Lens required for correcting
\ndistant vision = – 5.5
\nFocal length of lens F \\(=\\frac{1}{P}\\)
\n\\(\\mathrm{F}=\\frac{1}{-5.5}=0.181 \\mathrm{m}\\)
\nLens required for correcting this defect =-0.181 M
\nii) Lens required for correcting near vision = + 1.5 D
\nFocal length of lens F \\(=\\frac{1}{P}\\)
\n\\(F=\\frac{1}{1.5}=0.667 \\mathrm{m}\\)<\/p>\n

Question 6.
\nThe 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?
\nAnswer:
\nA person is suffering from near vision. So image is formed in front of retina.
\nDistance of Image, V = – 80 cm
\nfocal length = f
\nAs per lens formula.
\n\"KSEEB
\nf = 80 cm = – 0.8 m,
\n\"KSEEB
\n\u2234 Power of -125 D is required to correct this problem.<\/p>\n

Question 7.
\nMake 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.
\nAnswer:
\n\"KSEEB
\nObject distance, u = 25 cm
\nImage distance, v = 1m = -100 m
\nAs per lens formula,
\n\"KSEEB
\nThis defect should be corrected by +3.0 power.<\/p>\n

Question 8.
\nWhy is a normal eye not able to see clearly the objects placed closer than 25 cm?
\nAnswer:
\nThe maximum accommodation of a normal eye is reached when the object is at a distance of 25 cm from the eye. The focal length of the eye lens cannot be decreased below this minimum limit. Thus an object placed closer than 25cm [or very close to eye] cannot be seen clearly by a normal eye.<\/p>\n

Question 9.
\nWhat happens to the image distance in the eye when we increase the distance of an object from the eye?
\nAnswer:
\nThe distance eye lens and retina is the image distance inside the eye. The image distance is fixed. It cannot be changed at all. Therefore, when we increase the distance of an object from the eye, there is no change in the image distance inside the eye.<\/p>\n

Question 10.
\nWhy do stars twinkle?
\nAnswer:
\nThe twinkling of a star is due to atmospheric refraction of starlight. The atmospheric refraction occurs in a medium of gradually changing refractive index. Since the atmosphere bends starlight towards the normal, the apparent position of the star is slightly different from its actual position. This apparent position of the star is not stationary, but keeps on changing slightly, as the physical conditions of the earths 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 some other time, fainter, which is the twinkling effect.<\/p>\n

Question 11.
\nExplain why the planets do not twinkle.
\nAnswer:
\nThe planets are much closer to the earth and are thus seen as extended sources. If we consider a plant 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.<\/p>\n

Question 12.
\nWhy does the Sun appear reddish early in the morning?
\nAnswer:
\nLight from the Sun near the horizon passes through thicker layers of air and larger distance in the earth’s atmosphere before reaching our eyes. However, light from the Sun overhead would travel relatively shorter distance. At noon, the Sun appears white as only a little of the blue and violet colours are scattered. Near the horizon, most of the blue light and shorter wavelengths are scattered away by the particles. Therefore, the light that reaches our eyes is of longer wavelengths. This gives rise to the reddish appearance of the Sun.<\/p>\n

Question 13.
\nWhy does the sky appear dark instead of blue to an astronaut?
\nAnswer:
\nWhen sunlight passes through the atmosphere, the fine particles in air scatter the blue colour (shorter wavelength more strongly than red. The scattered blue light enters our eyes. If the earth had no atmosphere, there would not have been any scattering. Then the sky would have looked dark. The sky appears dark to astronaut flying at very high attitude, as scattering is not prominent at such heights.<\/p>\n

KSEEB SSLC Class 10 Science Chapter 11 Additional Questions and Answers<\/h3>\n

Question 1.
\nDraw a neat diagram showing
\n(a) neat point of a Hypermetropic eye
\n(b) Hypermetropic eye
\n(c) correction for Hypermetropic eye.
\nAnswer:
\n\"KSEEB<\/p>\n

Question 2.
\nWhich causes presbyopia? Name the lens to correct this defect.
\nAnswer:
\nThis causes due to the gradual weakening of the ciliary muscles and diminishing flexibility of the eye lens.
\nA common type of bi-focal lense is necessary to correct this defect.<\/p>\n

Question 3.
\nWho discovered spectrum of sunlight?
\nAnswer:
\nIsaac Newton.<\/p>\n

Question 4.
\nWhat is the reason for Advance sunrise and delayed sunset?
\nAnswer:
\nThe sun is visible to us about 2 minutes before the actual sunrise, and about 2 minutes after the actual sunset because of atmospheric refraction.<\/p>\n

Question 5.
\nAt noon, the sun appears white. Give reason.
\nAnswer:
\nBecause at noon, the sun appears white as only a little of the blue and violet colours are scattered.<\/p>\n

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World\u00a0are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World. Karnataka SSLC Class 10 Science Solutions Chapter 11 Human Eye and Colourful World KSEEB SSLC Class 10 …<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/436"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=436"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/436\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=436"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=436"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=436"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}