{"id":28504,"date":"2022-06-04T17:00:28","date_gmt":"2022-06-04T11:30:28","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=28504"},"modified":"2022-05-23T15:18:17","modified_gmt":"2022-05-23T09:48:17","slug":"ncert-solutions-for-class-10-science-chapter-11","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-10-science-chapter-11\/","title":{"rendered":"NCERT Solutions for Class 10 Science Chapter 11 The Human Eye and the Colourful World"},"content":{"rendered":"

These NCERT Solutions for Class 10 Science<\/a> Chapter 11 The Human Eye and the Colourful World Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.<\/p>\n

The Human Eye and the Colourful World NCERT Solutions for Class 10 Science Chapter 11<\/h2>\n

Class 10 Science Chapter 11 The Human Eye and the Colourful World InText Questions and Answers<\/h3>\n

In-text Questions (Page 190)<\/span><\/p>\n

Question 1.
\nWhat is meant by power of accommodation of the eye?
\nAnswer:
\nThe property of the eye lens to change its focal length is called its power of accomodation.<\/p>\n

Question 2.
\nA person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the nature of the corrective lens used to restore proper vision ?
\nAnswer:
\nGiven v = -1.2 m
\nu = \u221e
\n\"NCERT
\nSo, if the far point of myopic eye is 1.2 m, focal length of corrective lens is -1.2 m. And it will be concave.<\/p>\n

Question 3.
\nWhat is the far point and near point of the human eye with normal vision?
\nAnswer:
\nThe near point is 25 cm far point is infinity of the human eye with normal vision.<\/p>\n

Question 4.
\nA student has difficulty in 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:
\nMyopia : It can be corrected by using concave lens. A concave lens of suitable power will bring the image back on to the retina and thus defect is corrected.<\/p>\n

Class 10 Science Chapter 11 The Human Eye and the Colourful World Textbook Questions and Answers<\/h3>\n

Page no. 197 & 198<\/span><\/p>\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) accommodation.
\n(c) near-sightedness.
\n(d) far-sightedness.
\nAnswer:
\n(b) accommodation<\/p>\n

\"NCERT<\/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 near vision he needs a lens of power +1.5 dioptre. What is the focal length of the lens required for correcting (i) distant vision, and (ii) near vision?
\nAnswer:
\nGiven p = -5.5 dioptres
\n\"NCERT<\/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:
\nWhen object is at 80 cm, focal length of eye lens of f and image distance i.e., distance between eye lens and retina is v., By using lens formula :
\n\\(\\frac{1}{f}=\\frac{1}{v}-\\frac{1}{u}\\)
\n\\(\\frac{1}{f}=\\frac{1}{v}-\\frac{1}{-80 \\mathrm{~cm}}\\) ………(i)
\nWhen the object is at infinity, the object is not distinctly visible to short sighted person so he will use corrective lens of focal length f’ in front of eye lens. By applying lens formula, we have
\n\"NCERT
\nSo the lens will be concave because its focal length is negative.<\/p>\n

\"NCERT<\/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:
\nLet the focal length of eye lens is ‘f’
\nGiven u = – 1m
\nBy lens formula, we get
\n\"NCERT
\nWhen the object is placed at normal eye point, then the focal length of corrective lens be f’. By using lens formula we get
\n\"NCERT
\nBy substituting (i) and (ii) we get
\n\"NCERT
\n\"NCERT
\n\"NCERT<\/p>\n

Question 8.
\nWhy is a normal eye not able to see clearly the objects placed closer than 25 cm?
\nAnswer:
\nThe eye lens is composed of a fibrous, jelly-like material. Its curvature can be modified to some extent by the ciliary muscles. The change in the curvature of the eye lens can thus chage its focal length. When the muscles are released the lens becomes thin. Thus, its focal length increases This enables us to see distant object clearly. When we are looking at objects closer to the eye lens decreases. However, the focal length of eye lens cannot be decreased below a cetain limit, minimum limit i.e., below 25 cm. In this situation image is formed behind the retina so a normal eye is not able to see clearly the objects placed closer than 25 cm.<\/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:
\nImage distance in the eye also increases when we increase The distance of an object from the eye.<\/p>\n

Question 10.
\nWhy do stars twinkle ?
\nAnswer:
\nLight emitted by star passes through the atmosphere of the earth before reaching our eyes. The atmosphere of the earth is not uniform but consists of many layers of different densities.
\n\"NCERT
\nThe layers close to the surface of the earth are denser. As we go higher the density of layers and refractive index decreases progressively. As the light from a star enters the upper-most layer of the atmosphere, it bends towards the normal as it enters the next layer. This process continues till the light enters our eyes. So due to refraction of light, the apparent position of the star is different from the actual position of the star.<\/p>\n

Moreover, the different layers of the atmosphere are mobile and the temperature and the density of layers of atmosphere changes continuously. Hence the apparent position of the star changes continuously. This change in the apparent position of the star continuously leads to the twinkling of a star.<\/p>\n

Question 11.
\nExplain why the planets do not twinkle.
\nAnswer:
\nPlanets are very dose to the earth as compared to the stars. So they appear bigger due to their comparatively smaller distances and stars appear smaller due to their very large distances. Because of this difference planet is considered as a collection of many point sources of light while a star is considered as a point source of light. So, a planet is made up of a number of point sources of light, each twinkling in a random fashion.<\/p>\n

Random twinkling of each individual point light source, nullify the twinkling effect of other source and as a result, there is null twinkling effect as a whole. Due to this, the variations in the atmospheric conditions are unable to produce twinkling effect of the planet in the eye of the observer.<\/p>\n

Question 12.
\nWhy does the sun appear reddish early is the morning?
\nAnswer:
\nLight of lower frequencies such as yellow, orange, red is scattered the least by oxygen and nitrogen molecules of the atmosphere. Thus, the red, orange and yellow lights are transmitted through the atmosphere much more than violet and blue. Red which is scattered the least due to long wavelength, passes through more atmosphere than other colours. Therefore, most of the light with high frequencies, violet, indigo, blue, green are scattered away in the atmosphere at sunrise. Only red and a little orange light which are least scattered enter our eyes, and appear reddish early in the morning.<\/p>\n

\"NCERT<\/p>\n

Question 13.
\nWhy does the sky appear dark instead of blue to an astronaut ?
\nAnswer:
\nAt very high altitude earth has no atmosphere, so the astronaut would not have been any scattering of light. Then the sky would look dark. The sky appear dark to astronaut flying at very high altitudes, as scattering is not prominent at such heights.<\/p>\n

Class 10 Science Chapter 11 The Human Eye and the Colourful World Textbook Activities<\/h3>\n

Activity 11.1 (Page 192)<\/span><\/p>\n