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Chapter 11 The Human Eye and the Colourful World

Eye is the most delicate external organ. It is the most valuable and sensitive sense organ which enables us to see the beauty of a colourful world. In the previous chapter we have studied about the phenomenon of reflection and refraction of light. Human eye has a convex lens that refracts the light and image is formed on the retina. In this chapter we shall study the structure and function of the human eye. Defect of vision and their correction using spectacles. We will also study natural phenomenon, splitting of white light, scattering of light, formation of rainbow, why the sky appears blue and many more.

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Exercise 1

  • Q1 What is meant by power of accommodation of the eye?
    Ans:

    Accommodation is a reflex mechanism in by which eyes changes the focal power to make the farthest and the nearest object clearly visible on the retina. Human beings have the good power of accommodation. Accommodation requires the refraction of light rays when they passed from one medium to another for getting focus on retina. Refraction occurs at the lens  and it depends on the angle of the light or distance of the object from cornea. Thus, the degree of refraction is changed by changing the convexity of the lens. And, all of this mechanism is done with the help of ciliary muscles or suspensory ligament also known as accommodation apparatus.


    Q2 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:

    Myopia is also known as near sightedness. A person who is suffering from this eye defect cannot see the distant objects clearly. The normal vision of the eye is 25 cm. The image is formed in front of the retina to the person who is having this defect. So, the person is advised or recommended to use concave lens of the correct power. So, the image will form on retina. Thus, this eye defect can be rectified.


    Q3 What is the far point and near point of the human eye with normal vision?
    Ans:

    The farthest point of the human eye with normal vision is infinity.

    The nearest point of the human eye with normal vision is 25cm.

    At the farthest point, when we focussed at any distant object, beyond the 6m. eyes are said to be at rest because ciliary muscles are relaxed, lens get thin and refraction got decreased. And, at the nearest point, we can see the objects very clearly. Light rays from near objects in the range of 6m. are diverging when they strike the eye. High refraction value is needed for the nearest objects. To see the nearest object, eyes get contracted, lens gets thick and refraction got increased.

     

    Q4 A 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?
    Ans:

    When a student is having difficulty to read the blackboard while sitting in the last row, then this condition showing the myopia defect in the child also known as near- sightedness. In this defect of eye, the image of farthest object does not formed on retina, only forms in the front of the retina in which  blurred image can be seen. This condition can be corrected by using the appropriate concave lens of suitable power. So, the eye will be able to see the correct image instead of blurred image, again on the retina.


Exercise 2

  • Q1 The human eye can focus on 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

    The human eye can focus on objects at different distances by adjusting the focal length of the eye lens due to the accommodation power of the eye, in which eye changes the focus power to make the clear image of the farthest or the nearest object.

     

    Q2 The human eye forms the image of an object at its (a) cornea.          (b) iris.          (c) pupil.          (d) retina.
    Ans:

    (d) Retina

    The human eye forms the image of an object at retina. Retina is a very delicate coat and lines the whole part, vascular coat. It also shows three parts- optic, ciliary and iridial. And, also contains two layers nervous and pigmented layer.

     

    Q3 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:

    (a) 25 cm.

    Least distance of distinct vision is the minimum distance of an object to see clear object. The least distance of distinct vision for a young adult with normal vision is about 25cm.

     

    Q4 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

    The change in the focal length of an eye lens is caused by the action of ciliary muscles. The degree of refraction is changed by the changing of the convexity of the lens and this is done by the ciliary muscles.

     

    Q5 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 (i) distant vision, and (ii) near vision?
    Ans:

    Power of the lens is given (P) for distant vision = -5.5 D

    Power of the lens (P2) is given for near vision = + 1.5 D

    1. Focal length of the lens required for correcting lens of distant vision

    Formula used, P = 1/ f where, P is the power of the lens 

    And, f is the focal length

    To correct f

    f = 1/ P

    f = 1/ -5.5

    f = - 0.1818 m.

    2. Focal length of the lens required for correcting lens of near vision

    Formula used, P = 1/ f where, P is the power of the lens 

    And, f is the focal length

    To correct f

    f = 1/ P

    f = 1/ 1.5

    f = - 0.667 m.


    Q6 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?
    Ans:

     Focal length of the myopic person is given, 80 cm. The focal length should be in meters and the person has the eye defect of myopia so the focal length would be in minus (-).

           So,                                Focal length (f) = 80 cm.   = -0.8 m.

       Formula used,                Power of the lens (P) = 1/ f

                                                                                  P = 1/ -0.8

                                                                                  P = - 1.25 D          , where D is the dioptre 

    The image will not formed on retina in this defect it will form in the front of retina. And, this defect can be corrected by the concave lens of the suitable power.

     

    Q7 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.
    Ans:

    Diagram:

     
    Diagram A


    Diagram B

    In the above diagram, image is formed on the left side so the both image and the object should be taken in the minus (-).

    Image distance (v), given = -1m = -100cm

    Object distance (u), given = -25 cm

    Formula of focal length on the basis of image and the object distance,

                                                           1/v – 1/ u = 1/ f

                                                        - 1/ 100 – 1/ -25 = 1/ f

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

                                                           - 1 + 4 / 100 = 1/ f

                                                                    3/ 100 = 1/f 

                                                               F = 100/ 3   = 33.3 cm. = 0.33m.

       And, power of the lens (P) = 1/f

    Put the value of f in this formula,

                                                                P = 1/ 0.33                           

                                                                    = + 3D

    Hypermetropia is the eye defect also called far sightedness. A person who is having this defect can see the distant objects clearly but cannot see the near objects very clearly. This is because, light rays from the object focussed at the point behind the retina.

     

    Q8 Why is a normal eye not able to see clearly the objects placed closer than 25 cm?
    Ans:

    The normal eye not able to see the objects very clearly that is placed closer than 25 cm because to see the nearest objects within the range before 25 cm, ciliary muscles should be contracted, suspensory ligament should be in the relaxed state, lens should be thick or more convex (increase in the thickness of lens, focal length get short that is adjusted for focusing on the near objects very clearly) and refraction should be increased to see clear image. And, the too much contraction of the ciliary muscles cause stress on the eyes. Thus, the normal eye is unable to see the objects placed closer than 25cm.


    Q9 What happens to the image distance in the eye when we increase the distance of an object from the eye?
    Ans:

    The image formed on retina even when we move the object with increasing the distance from the eye. In the distant vision ciliary muscles get relaxed and the lens is thin with the decreased refraction.

     

    Q10 Why do stars twinkle?
    Ans:

    Stars twinkling appears due to the atmospheric refraction of the starlight. When the starlight enters on the earth surface, refraction undergoes continuously with gradually changes in refractive index before it reaches the earth surface. The atmospheric physical condition changes, i.e, do not remain stationary, the apparent position (when star is slightly different from its original position) of the stars are also not stationary. When the refractive index changes continuously due to which path of the light rays from the stars keeps on changing the path continuously. As the path of rays of light that comes from the stars, goes on varying slightly. The apparent position of the stars fluctuates and the starlight enters the eye- sometimes the star may appear brighter, and sometime this effect is known as the twinkling effect.   


    Q11 Explain why the planets do not twinkle.
    Ans:

    As in the stars, they twinkles due the apparent position (which is not a fixed position) cause the light travels from the stars is refracted through the atmosphere. Planets do not twinkle because they are very much closer to the earth in the comparison of stars and appear large to the human eyes. Thus, the planet can be seen as extended sources. When the light ray come to our eyes from all the individual extended sources are constant. And there is no shifting occurs, so planets do not seem to be twinkle.


    Q12 Why does the Sun appear reddish early in the morning?
    Ans:

    In the early morning sun is at the farthest distance from the earth.

    When the white light beam passed through the transparent glass tank having clear water, a converging lens is placed between them. After the glass tank a cardboard is placed having a circular hole in it, in which light ray passed on the screen by using another converging lens. Add 200g of sodium thiosulphate and 1- 2ml of concentrated sulphuric acid to 2L of clean water in the tank, then the sulphur particles begins to form in 2- 3 minutes. We can observe the light with the smaller wavelength, i.e, blue from the three sides of the glass tank will scatter and from the fourth side we will see the red light that pass into the circular hole, reaches to the human eyes. 

    As in the above paragraph mentioned, the sun colour appearance is described in the same way. The sun appears red in the early morning because when the light from the sun near the horizon passes through the atmospheric thick layers, blue light with shorter wavelength will scatter and the red light will reach to the eyes having longer wavelength.

     

    Q13 Why does the sky appear dark instead of blue to an astronaut?
    Ans:

    When the sunlight passes through the atmosphere, the fine particles scatters in the atmospheric layers blue light will appear having shorter wavelength and the red light having the larger wavelength. The scattered blue light enters to our eyes if we see the sky from the earth.

    The sky does not appear blue to an astronaut because there is no atmosphere in the space and hence the light does not scatter into the various colours as like when we see the colours of sun or sky from the earth. If there is no scattering occurs, the sky will look dark to the astronaut from the space.