Title CHAPTER 11 THE HUMAN EYE AND THE COLOURFUL WORLD.doc ✅

SELF STUDY CBSE X CHAPTER 11 THE HUMAN EYE AND THE COLOURFUL WORLD11 CHAPTER 11: THE HUMAN EYE AND THE COLOURFUL WORLD 1. PROBLEMS FROM NCERT TEXTBOOK Q. 1. What is meant by power of accommodation of the eye? (Page190) Ans. The ability of the eye lens to adjust its focal length is called power of accommodation. Q. 2. A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the nature of corrective lens to restore proper vision? (Page190) Ans. Here, distance of far point, x = 1 2 m For viewing distant objects, focal length of corrective lens, f = -x = - 1.2 m Therefore power of lens 11 P0.83D f1.2  . Since the power is negative therefore the lens must be concave. Q. 3. What is the far point and near point of the human eye with normal vision? (Page190) Ans. For human eye with normal vision, far point is at infinity and near point is at 25 cm from the eye. Q. 4. 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? (Page190) Ans. As the child has difficulty in reading the blackboard, he is suffering from myopia or short sightedness. To correct this defect, he has to use spectacles with concave lens of suitable focal length. Q5. The human eye can focus objects at different distances by adjusting the focal length of the eye lens. This is due to (Page197) (i) persistence of vision (ii) near sightedness (iii) accommodation (iv) far-sightedness Ans. The property of the eye to adjust the focal length of eye lens is called accommodation. Choice (iii) is correct. Q6. The human eye forms the image of an object at its (Page 198) (i) cornea (ii) pupil (iii) iris (iv) retina Ans. The human eye forms the image of an object at its retina. Choice (iv) is correct. Q7. The least distance of distinct vision for a young adult with normal vision is (Page 198) (i) 25 m (ii) 25 cm (iii) 25 cm (iv) 25 m Ans. The least distance of distinct vision for a young adult with normal vision is 25 cm. Choice (iii) is correct. Q8. The change in focal length of an eye lens is caused by the action of the (Page 198) (i) pupil (ii) retina (iii) blind spot (iv) ciliary muscles Ans The action of ciliary muscles holding the eye lens changes the focal length of eye lens enabling the eye to focus the image of objects at varying distances. Choice (iv) is correct. Q9. A person needs a lens of power - 5.5 D for correcting his distant vision. For correcting his near vision, he needs a lens of power + 1.5 D. What is the focal length of the lens required for correcting (i) distant vision (ii) near vision? (Page 198) Ans. (i) For distant viewing, given f = ? P = - 5.5 D Using the relation 11 Porf fP or 100 f18.2cm 5.5  (ii) For near vision, given f =?, P = + 1.5 D Using the relation 11 Porf fP or 100 f66.7cm 1.5 Q10. 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?( Page 198) Ans. Distance of far point, x = 80 cm, P =? For viewing distant objects, focal length of corrective lens, f= - x = -80 cm Using the relation 1100100 P1.25D ff80  The lens is concave. Q11. Make a diagram to show how Hypermetropia is corrected. The near point of a Hypermetropia eye is 1 metre. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm. (Page 198) Ans. The diagram is as shown below. Given u = 1 m = 100 cm, d = 25 cm, f = ? Using the relation ud1005 f33.3cm ud10025    Hence power of lens required 1100 P3D f33.3 Q12. Why is a normal eye not able to see clearly the objects placed closer than 25 cm? (Page 198) Ans. This is because the focal length of eye lens cannot be decreased below a certain minimum limit. Q13. What happens to the image distance in the eye when we increase the distance of an object from the eye? (Page 198) Ans. For a normal eye, image distance in the eye is fixed. This is equal to the distance of retina from the eye lens. When we increase the distance of the object from the eye, focal length of eye lens is changed on account of power of accommodation of the eye so as to keep image distance constant. Q14. Why does the sun appear reddish early in the morning? (Page 198)
SELF STUDY CBSE X CHAPTER 11 THE HUMAN EYE AND THE COLOURFUL WORLD 11 Ans. Light from the Sun, near the horizon, passes through thicker layers of air and covers a large 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 the Sun. Q15. Why does the sky appear dark instead of blue to an astronaut? (Page 198) Ans. This is because at this height there is no atmosphere which can scatter the white light. Therefore, the sky appears dark. NCERT ACTIVITIES ACTIVITY 1  Fix a sheet of white paper on a drawing board using drawing pins.  Place a glass prism on it in such a way that it rests on its triangular base. Trace the outline of the prism using a pencil.  Draw a straight line PE inclined to one of the refracting surfaces, say AB, of the prism.  Fix two pins, say at points P and Q, on the line PE as shown in Fig.  Look for the images of the pins, fixed at P and Q, through the other face AC.  Fix two more pins, at points R and S, such that the pins at R and S and the images of the pins at P and Q lie on the same straight line.  Remove the pins and the glass prism.
SELF STUDY CBSE X CHAPTER 11 THE HUMAN EYE AND THE COLOURFUL WORLD11  The line PE meets the boundary of the prism at point E see figure. Similarly, join and produce the points R and S. Let these lines meet the boundary of the prism at E and F, respectively. Join E and F.  Draw perpendiculars to the refracting surfaces AB and AC of the prism at points E and F, respectively.  Mark the angle of incidence ( i ), the angle of refraction ( r ) and the angle of emergence ( e ) as shown in Fig. PE — Incident ray, i — Angle of incidence EF — Refracted ray, r — Angle of refraction FS — Emergent ray, e — Angle of emergence A - Angle of the prism, D - Angle of deviation EXPLANATION: Let A be the angle of prism. A ray PE of monochromatic light is incident on face AB of the prism at an angle i . The ray is called the incident ray and the angle is called the angle of incidence. This ray is refracted towards the normal NEN / and travels in the prism along EF. This ray is called the refracted ray at the face AB. Let r be the angle of refraction at this surface. The refracted ray EF is incident at an angle r on the surface AC. The ray FS again suffers refraction and emerges out of face AC at an angle e along FS. The ray is called the emergent ray and the angle is called the angle of emergence. When the ray FS is extended backwards it meets the extended ray PE at point G such that D is the angle of deviation of the ray. This kind of bending also occurs in a glass slab. But the net deviation in a rectangular glass slab is zero, whereas it is not so in case of a prism, due to its peculiar shape. The deviation suffered by the ray D. From this activity, we learn that in passing through a prism, a ray of light undergoes two refractions Due to its peculiar shape, the net deviation while passing through a prism is never zero. ACTIVITY 2:  Take a thick sheet of cardboard and make a small hole or narrow slit in its middle.  Allow sunlight to fail on the narrow slit. This gives a narrow beam of white light.  Now, take a glass prism and allow the light from the slit to fall on one of its faces as shown in Fig.  Turn the prism slowly until the light that comes out of it appears on a nearby screen.  What do you observe? You will find a beautiful band of colours. Why does this happen? EXPLANATION: Fig. shows sunlight falling on a narrow slit in a thick sheet of cardboard. A narrow beam of white light obtained from the slit is made to fall on face AB of a glass prism ABC. As we turn the prism slowly, we observe a beautiful band of seven colours on a screen. This is called the spectrum of visible light. The sequence of colours seen from the lower end of the screen is Violet (V), Indigo (I), Blue (B), Green (G), Yellow (Y), Orange (O) and Red (R). This is due to the fact that the prism offers different refractive indices to different colours of light. As s result they deviate through different angle thereby producing the spectrum. ACTIVITY 3:  Place a strong source (S) of white light at the focus of a converging lens (L 1 ). This lens provides a parallel beam of light.  Allow the light beam to pass through a transparent glass tank (T) containing clear water. Allow the beam of light to pass through a circular hole (C) made in a cardboard. Obtain a sharp image of the circular hole on a screen (MN) using a second converging lens (L 2 ), as shown in Fig.
SELF STUDY CBSE X CHAPTER 11 THE HUMAN EYE AND THE COLOURFUL WORLD11  Dissolve about 200 g of sodium thiosuiphate (hypo) in about 2 L of clean water taken in the tank. Add about 1 to 2 mL of concentrated sulphuric acid to the water. What do you observe? EXPLANATION: The arrangement is as shown in Fig we find that fine microscopic sulphur particles precipitate in water in a couple of minutes. Blue light is scattered by the minute colloidal sulphur particles. This blue colour is observed from the three sides of the tank. From the fourth side of the tank facing the hole in a cardboard, we observe transmitted light. As blue colour has been scattered, therefore we observe the remaining portion of visible spectrum i.e, the orange red colour and then bright crimson red colour on the screen. Hence, we conclude that very fine particles scatter mainly blue light of smaller wavelength, while the transmitted light contains longer wavelengths only. HIGHER ORDER THINKING SKILL (HOTS) QUESTIONS Q1 A child sitting in a class room is not able to read clearly the writing on the blackboard. (a) Name the type of defect from which his eye is suffering (b) With the help of a ray diagram show how this defect can be remedied? Ans (a) The child’s eye is suffering from the defect of myopia. (b) The defect can be removed by using a concave lens as shown in the figure below. Q2. Why we have two eyes instead of one? Ans. Two eyes are better than one eye because of the following reasons (i) the field of view with two eyes is more than with one eye. (ii) two eyes give three dimensional picture of an object whereas one eye gives only two dimensional picture of an object. Q3. When one enters a dim-lit room from bright light, one is unable to see the object in the room for some time. Explain, why? Ans. In bright light, the iris contracts the pupil of an eye to allow less light to enter the eye. When, we enter a dim-lit room, the iris takes time to expand the pupil of the eye to allow more light to enter it. Thus for a short time one is unable to see in a dim-lit room. Q4. A person is able to see objects clearly only when these are lying at distances between 50 cm and 300 cm from his eye. (i) What kind of defect of vision is he suffering from? (ii) What kind of lenses will he required to increase his range of vision from 25 cm to infinity? Explain. Ans. (a) For a normal eye, the near point is at 25 cm and the far point is at infinity. The given person cannot see object clearly either closer to the eye or far away from the eye. So, he is suffering from both myopia and hypermetropia. (b) A bi-focal lens consisting of a concave lens and convex lens of suitable focal lengths will be required to correct the defects and to increase his range or vision from 25 cm to infinity. In a bi-focal lens the upper half of the lens is concave which corrects distant vision and the lower half is convex which corrects near vision. Q5. A student cannot read properly from the blackboard while sitting on the front desk in a classroom of a big size. He however, can read clearly while sitting on the last desk of the classroom. (a) Draw ray diagrams to illustrate the formation of image of the black board writing by his eye-lens when he is seated at (i) the front desk, (ii) last desk. (b) Name the defect of the eye from which the student is suffering. (c) Name the type of lens that would enable him to see the black board writing clearly, when seated on the front desk. Draw a ray diagram to illustrate how this lens helps him to see clearly. Ans. (a) (i) Formation of image of black board writing by the eye-lens of the student sitting on the front desk is as shown below.