Q 1. What is meant by power of accommodation of the eye?
Ans. The ability of the eye lens to adjust its focal length, so as to clearly focus rays coming from distant as well a near objects on the retina, is called the power of accommodation of the eye.
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. A person with a myopic eye should use a concave lens of focal length 1.2 m so as to restore proper vision.
Q 3. What is the far point and near point of the human eye with normal vision?
Ans. For a human eye with normal vision the far point is at infinity and the 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?
Ans. The student is suffering from myopia or short-sightedness. The defect can be corrected by the use of concave (diverging ) lens of an appropriate power.
Q 5. The human eye can focus objects at different distance 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.
[ Hint: While seeing distant object eye lens is thinner and its focal length is more (about 2.5 cm). While looking near objects eye lens assumes a more round shape, its focal length decreases and nearly objects are clearly focussed at the retina.]
Q 6. The human eye forms the image of an object at its
(a) cornea (b) iris (c) pupil (d) retina Ans. (d) retina.
[ Hint: Retina is photosensitive and human eye forms the image of an object at its retina.] Q 7. 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.
[ Hint: For a young adult having normal vision the least distance of distinct vision (distance of near point) is about 25 cm.] Q 8. 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.
[ Hint: While seeing distant object the ciliary muscles are relaxed and focal length of eye lens is maximum (2.5 cm). At the time of looking at nearby objects the ciliary muscles contract and focal length of eye lens decreases.] Q 9. 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. (i) Power of lens needed for correction distant vision of the person P1 = ( 5.5 D (Focal length of lens required for correcting distant vision [pic] or [pic] (ii) For correcting near vision power required P2 = + 1.5 D (Focal length of lens required for correcting near vision [pic] or 66.7 cm. Q10. The far point of a myopic the person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?
Ans. To correct the myopia the person concerned should use concave lens (diverging lens) of focal length ( = ( 80 cm, so that for an object at infinity (u = ( ), the virtual image is formed at the far point of myopic person (v = ( 80 cm). ( From lens formula [pic] we have
[pic] or [pic] 1 100 100 ( Power of the lens P = ( in m = ( in cm = (80 = (1.25 D.
Q11. 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. The relevant diagram has been shown in adjoining fig. Here the convex lens used forms virtual image of object placed at N( (near point of normal eye i.e., 25 cm from eye) at N, the near point of defective eye (at a distance x). Now, the defective eye can focus the rays from N at the retina. In the problem it is given that the near point of the normal eye is 25 cm, hence u = ( 25 cm. The lens used forms its virtual image at near point of hypermetropic eye i.e., v = ( 1m = ( 100 cm. Using lens formula, we have
1 1 ( Power of correcting lens P = ( in m = (1/3) = + 3 D. Q12. Why is a normal eye not able to see clearly the objects placed closer than 25 cm ? Ans. Due to a power of accommodation, the focal length of the eye lens cannot be decreased below a certain minimum limit. Consequently, a normal eye cannot see clearly the objects placed closer than a minimum distance, called near point of eye. For a young adult value of near point is 25 cm. So, we cannot see clearly objects placed closer than 25 cm. Q13. What happens to the image distance in the eye when we increase the distance of an object from the eye? Ans. The image is formed on the retina even on increasing the distance of an object from the eye. For this eye lens becomes thinner and its focal length increases as the object is moved away from the eye.
Q14. Why do stars twinkle?
Ans. Stars twinkle due to atmospheric refraction of starlight. As the stars are very far away, they behave as almost point sources of light. A son account of atmospheric refraction, 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. So, sometimes, the star appears brighter and at some other time, fainter. Thus, the stars twinkle.
Q15. Explain why the planets do not twinkle.
Ans. Planets are much closer to the earth and are seen as extended sources. So, a planet may be considered as a collection of a large number of point-sized light sources. Although light coming from individual point-sized sources flickers but the total amount of light entering our eye from all the individual point-sized sources average out to be constant. Thereby, planets appear equally bright and there is no twinkling of planets. Q16. Why does the Sun appear reddish early in the morning?
Ans. In the early morning, the sun is situated near the horizon. Light from the Sun passes through thicker layers of air and larger distance in the earth’s atmosphere before reaching our eyes. While passing through atmosphere blue light is most scattered away and the Sun appear reddish as shown in fig. [pic]
Q17. Why does the sky appear dark instead of blue to an astronaut? Ans. Colour of the sky is on account of scattering of light of shorter wavelengths by particles in the atmosphere of earth. If the earth had no atmosphere, there would not have been any scattering and the sky would have looked dark. When an astronaut in his spacecraft goes above the atmosphere of earth, sky appears dark to him because there is no scattering of light.