Class 10 Science Chapter 11 The Human Eye and the Colourful World

Class 10 Science Chapter 11 The Human Eye and the Colourful World Solutions to each chapter is provided in the list so that you can easily browse throughout different chapters NCERT Class 10 Science Chapter 11 The Human Eye and the Colourful World and select need one.

Class 10 Science Chapter 11 The Human Eye and the Colourful World

Also, you can read the CBSE book online in these sections Solutions by Expert Teachers as per NCERT (CBSE) Book guidelines. These solutions are part of SEBA All Subject Solutions. Here we have given Assam BoardClass 10 Science Chapter 11 The Human Eye and the Colourful World Solutions for All Subjects, You can practice these here.

The Human Eye and the Colourful World

Chapter – 11

GENERAL SCIENCE

Textual Questions and Answers:

Page – 190 

Q.1. What is the meant by power of accommodation of the eye? 

Ans: The power of accommodation is the ability of an eye to focus near and far objects clearly and making image on the retina by adjusting its focal length. 

Q.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? 

Ans: Concave Lens. 

Q.3. What is the far point and near point of the human eye with normal vision?

Ans: The far point of human eye with normal vision is at infinity and the near point is 25 cm distance from the eye. 

Q.4. A student has difficulty 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? 

Ans: Since the child cannot see the distant objects (like blackboard writing ) clearly, he is suffering from the defect of vision (or defect of eye ) called ‘myopia’ or ‘short – sightedness’. Myopia can be corrected by using spectacles containing concave lenses of suitable power.

EXERCISES 

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

(a) Presbyopia.

(b) Accommodation.

(c) Near-sightedness.

(d) Far-sightedness.

Ans: (b) Accommodation. 

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

(a) Cornea.

(b) Iris.

(d) Retina.

Ans: (d) Retina.

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

(a) 25 m 

(b) 2.5 cm 

(c) 25 cm 

(d) 2.5 m 

Ans: (c) 25 cm.

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

(a) Pupil.

(b) Retina.

(c) Ciliary muscles.

(d) Iris.

Ans: (c) Ciliary muscles. 

Q.5. A person needs a lens of power -5.5 diopters for correcting his distant vision. For correcting his near di vision he needs a lens What is od ji the focal length of he lens required for correcting 

(i) Distant vision and 

(ii) Near vision? 

Ans: (i) Given that,

                              p = – 5.5 D 

                              f = ?

  We have,

                              p = 1/f

                           ⇒ f = 1/p

                                 = 1/-5.5D

                                 = 1/-5.5m

                                 = -0.18m

(ii) Given that,    

                              p  =  +1.5D

                              f   = ?

       We have, 

                             p = 1/f

                          ⇒ f = 1/p

                                = 1/+1.5D

                                = 1/+1.5m

                                = +0.666m

                                = +0.67m.

Q.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 corrects the problem? 

Ans: Given that, 

           Object distance  u   =  ∞ (infinity) 

           Image distance   v  = – 80 cm. 

           focal length          f  = ?

We have,    

                      1/v – 1/u = 1/f

                ⇒ – 1/80 – 1/∞ = 1/f

                ⇒ – 1/80 – 1/1/0 = 1/f

                ⇒ – 1/80 – 0 = 1/f

                ⇒ 1/f = – 1/80

                ⇒  f  = – 80 

              ∴     f  = – 80cm.

Thus, the focal length of the required concave lens is 80cm.

Now, Power (P) = 1/f

                          = 1/-80cm

                          = – 1/0.8cm

                          = – 1/0.8D

                          = – 1.25 D

Q.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 25 cm. 

Ans: Diagram: 

2nd part:

Given that, 

              Object distance 

                                  u = – 25 cm / For convex lens.

     Image distance      

                                  v  = – 1m 

                                      = – 100 cm 

     Focal length     

                                    f = ?

      We have,

                             1/v – 1/u = 1/f 

                         ⇒ 1/f = 1/-100 = 1/-25

                         ⇒ 1/f = 1/-100 + 1/25

                         ⇒ 1/f = – 1+4/100

                         ⇒ 1/f = 3/100

                     ∴         f = 100/3

                                  = 33.3cm.

Now , Power  

                         P = 1/f

                            = 1/+33.3cm.

                            = 1/0.33m

                            = + 100/33D

                            = + 3.0D

Q.8. Why is a normal eye not able to see clearly the objects explain placed closer than 25 cm?

Ans: Because the focal length of Normal eye is 25cm.

When any object placed closer than 25 cm then the image is FORMED behind the retina. Then object seem blurred.

Q.9. What happens to the image distance in the eye when we increase the distance of an object from the eye? 

Ans: Since the size of eyes cannot increase or decrease, the image distance remains constant. When we increase the distance of an object from the eye, the image distance in the eye does not change. The increase in the object distance is compensated by the change in the focal length of the eye lens. The focal length of the eyes changes in such a way that the image is always formed at the retina of the eye.

Q.10. Why do stars twinkle? 

Ans: The stars seem to twinkle in the night sky due to the effects of the Earth’s atmosphere. When starlight enters the atmosphere, it is affected by winds in the atmosphere and areas with different temperatures and densities. This causes the light from the star to twinkle when seen from the ground.

Explanation:

• The twinkling of stars is due to atmospheric refraction of star-light. The refraction of light caused by the earth’s atmosphere having air layers of varying optical density is called atmospheric refraction.
• The physical conditions of the atmosphere keep on changing continuously due to which density of air in different layers of atmosphere also keeps on changing.
• As a result of this, the refractive index of the various layers of atmosphere also keeps on changing continuously. So, light coming from stars suffers multiple refractions and the amount of starlight reaching the eye also keeps changing and so, due to fluctuation of perceived brightness of the star, they appear like they are twinkling.

Q.11. Explain why the planets do not twinkle. 

Ans: The planets are much closer to the earth. A planet can be considered as a collection of large number of point-sized sources of light. So 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. On the other hand stars twinkle because stars are point-sized sources of energy therefore the continuously changing atmosphere cause atmospheric refraction which cause variation in light.

Q.12. Why does the Sun appear reddish early in the morning? 

Ans: At sunrise, the rays of light coming from the Sun need to travel long distances in the Earth’s atmosphere before they reach our eyes. On this journey, shorter wavelengths of light are scattered and only longer wavelengths reach our eyes. Blue has a short wavelength and red has a long wavelength, so red reaches the eye after atmospheric scattering. Therefore, the sun looks reddish in the early morning.

Q.13. Why does the sky appear dark instead of blue to an astronaut? 

Ans: To an astronaut, the sky looks dark and black instead of blue because there is no atmosphere containing air in the outer space to scatter sunlight. So, there is no scattered light to reach our eyes in outer space, therefore the sky looks dark and black there. Since there is no scattering of blue component of white sunlight which ca reach the eyes of an astronaut in outer space, therefore the sky appears dark to the astronaut instead of blue.