NCERT Solutions Class 10th Science Chapter – 10 The Human Eye and Colourful World
| Textbook | NCERT |
| Class | 10th |
| Subject | Science |
| Chapter | 10th |
| Chapter Name | The Human Eye and Colourful World |
| Category | Class 10th Science |
| Medium | English |
| Source | Last Doubt |
| NCERT Solutions Class 10th Science Chapter – 10 The Human Eye and Colourful World Question & Answer In This Chapter we will learn about Can rods detect color, Do animals see colour, Can dogs see color, Do rods see shapes, Where are rods found, Can rods see red, Do rods detect light, Can dogs see TV, Can dogs see pink, Can dogs dream, What are rods also called, What animal only has rods, Which animal has more rods and more such things for more knowledge about this Chapters read their. |
NCERT Solutions Class 10th Science Chapter – 10 The Human Eye and Colourful World
Chapter – 10
The Human Eye and Colourful World
Question & Answer
| Exercise |
| 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 Answer – (b) Accommodation |
| 2. The human eye forms an image of an object at its. (a) Cornea (b) Iris (c) Pupil (d) Retina Answer – (d) Retina |
| 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 Answer – (c) 25 cm |
| 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 Answer – (c) Ciliary Muscles |
| 5. 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? Answer – The power (P) of a lens of focal length f is given by the relation Power (P) = 1/f. (i) Power of the lens (used for correcting distant vision) = – 5.5 D Focal length of the lens (f) = 1/P f = 1/-5.5 f = -0.181 m The focal length of the lens (for correcting distant vision) is – 0.181 m. (ii) Power of the lens (used for correcting near vision) = + 1.5 D Focal length of the required lens (f) = 1/P f = 1/1.5 = + 0.667 m The focal length of the lens (for correcting near vision) is 0.667 m. |
| 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 correct the problem? Answer – The individual is suffering from myopia. In this defect, the image is formed in front of the retina. Therefore, a concave lens is used to correct this defect of vision.Image distance (v) = – 80 cm Focal length = f Object distance u = infinity = ∞ According to the lens formula, The far point of the person suffering from myopia should be at infinity. Image distance v = – 80 cm Focal length = f , according to the lens formula = 1/v – 1/u = 1/f -1/80 – 1/∞ = 1/f 1/ f = -1/80 cm or f = -0.8 m We know – P = 1/f P = 1/-0.8 P = -1.25 = – 1.25D A concave lens of power – 1.25 D is required by the individual to correct his defect. |
| 7. 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. Answer – An individual suffering from hypermetropia can see distinct objects clearly but he or she will face difficulty in clearly seeing objects nearby. This happens because the eye lens focuses the incoming divergent rays beyond the retina. This is corrected by using a convex lens. A convex lens of a suitable power converges the incoming light in such a way that the image is formed on the retina, as shown in the following figure.  Correction For hypermetropic eyeu = – 25 cm, v = – 1 m = – 100 cm = 1/v – 1/u = 1/f .(by lens formula) 1/-100 – 1/-25 cm = 1f 1f = – 1/ 100 + 1/25 = -1 + 4/100 = 3/100 cm f = 100/3 = (1/3) mHence, power of the modified lens P = 1/f = 1/1/3 = + 3 D ….. (convex lens) A convex lens of power +3.0 D is required to correct the defect. |
| 8. Why is a normal eye not able to see clearly the objects placed closer than 25 cm? Answer – A normal eye is not able to see the objects placed closer than 25 cm clearly because the ciliary muscles of the eyes are unable to contract beyond a certain limit. |
| 9. What happens to the image distance in the eye when we increase the distance of an object from the eye? Answer – The image is formed on the retina even on increasing the distance of an object from the eye. The eye lens becomes thinner and its focal length increases as the object is moved away from the eye. |
| 10. Why do stars twinkle? Answer – The twinkling of a star is due to atmospheric refraction of starlight. The starlight, on entering the earth’s atmosphere, undergoes refraction continuously before it reaches the earth. The atmospheric refraction occurs in a medium of gradually changing refractive index. |
| 11. Explain why the planets do not twinkle? Answer – Unlike stars, planets don’t twinkle. Stars are so distant that they appear as pinpoints of light in the night sky, even when viewed through a telescope. Since all the light is coming from a single point, its path is highly susceptible to atmospheric interference (i.e. their light is easily diffracted). |
| 12. Why does the sky appear dark instead of blue to an astronaut? Answer – The sky appears dark instead of blue to an astronaut, as scattering of light does not take place outside the earth’s atmosphere. |
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