Chapter 11. The Human Eye and the Colourful World

Class 10 Chapter 11. The Human Eye and the Colourful World Intaernal Questions and Answers :

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

Answer: The power of accommodation of the eye is the ability to adjust the focus and maintain clear vision when viewing objects at different distances.
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 ?

Answer: A person with myopia (nearsightedness) who cannot see objects beyond 1.2 meters distinctly would require a corrective lens that diverges light rays before they enter the eye. This type of lens is concave or negative in shape. Therefore, a concave lens would be used to restore proper vision for someone with myopia.
3. What is the far point and near point of the human eye with normal vision ?

Answer:  The far point is the maximum distance at which the eye can focus on an object with clear vision.

For a person with normal vision, the far point is considered to be at infinity, meaning they can see objects at a very far distance without any blurring.

The near point is the closest distance at which the eye can focus on an object with clear vision.

 In individuals with normal vision, the near point is typically around 25 centimeters (10 inches) or closer, as the eye loses its ability to focus on objects placed at a closer distance.    

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 ?

Answer: The student may be suffering from myopia (nearsightedness), where distant objects appear blurry. It can be corrected with glasses or contact lenses with concave lenses to diverge light rays. Refractive surgery like LASIK is also an option.

Class 10 Chapter 11. The Human Eye and the Colourful World Exercises  :

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 the 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:  (i) For distant vision correction : Here, and  

We have,  

The focal length for correcting distant vision is approximately  – 0.18 meters .

(ii) For near vision correction : Here,  and

We have,   

The focal length for correcting near vision is + 0.67 meters.

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:  To correct myopic (near-sightedness), a concave lens is used .

So, 

We have, 

The nature and power of the lens required to correct the problem are a diverging (concave) lens with a power of approximately  – 1.25 D .
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: The correction Diagram of hypermetropia :

To correct hypermetropia (farsightedness), a converging lens (convex lens) is used.

The near point of a hypermetropic eye is the closest distance at which the person can see objects clearly. In this case, the near point is given as 1 meter.

The near point of a normal eye is 25 cm

Here, ,

Using lens formula , 

Since the focal length is negative, it indicates a converging lens (convex lens).

We have,  

The power of the lens required to correct the hypermetropic eye is + 3D .

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

Answer: A normal eye has a limited ability to focus on objects closer than 25 cm due to the accommodation mechanism. The ciliary muscles in the eye adjust the shape of the lens to bring nearby objects into focus.
9. What happens to the image distance in the eye when we increase the distance of an object from the eye?

Answer: When we increase the distance of an object from the eye, the image distance in the eye increases as well. This means that the image formed on the retina moves farther away from the eye's lens.
10. Why do stars twinkle?

Answer: Stars twinkle due to the phenomenon called atmospheric scintillation. As starlight passes through the Earth's atmosphere, it gets refracted and distorted by the varying density of air, causing the apparent brightness of stars to fluctuate or twinkle.
11. Explain why the planets do not twinkle.

Answer: Planets do not twinkle as prominently as stars because they appear larger and act as extended sources of light. Their apparent size averages out the atmospheric disturbances, resulting in a steadier and less twinkling appearance compared to distant point-like stars.
12. Why does the Sun appear reddish early in the morning?

Answer: The Sun appears reddish early in the morning due to a phenomenon called atmospheric scattering. When the Sun is near the horizon, its light has to pass through a thicker layer of the Earth's atmosphere, causing shorter blue and green wavelengths to scatter more, while longer red wavelengths dominate the visible spectrum.

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

Answer: The sky appears dark instead of blue to an astronaut in space because there is no atmosphere to scatter sunlight. On Earth, the atmosphere scatters short-wavelength blue light, giving the sky its blue color. Without this scattering, the sky appears dark in space.