Important Questions for Class 10 Science Chapter 11 – Human Eye And Colourful World

Below are some important questions for class 10 science chapter 11 – Human Eye And Colourful World students must study thoroughly.Human Eye and Colourful World is Chapter 11 of CBSE Class 10 Science. The chapter covers points like the ability of the eye to focus on objects both close and far by adjusting their focal length. Students learn about myopia, hypermetropia, and presbyopia, the three common refractive defects. The chapter also talks about how with age, our eyes lose their ability to accommodate and dispersion, the method of splitting white light into its colours.

Important Questions topics for Class 10 Science Chapter 11 – Human Eye And Colourful World are:

• Corrective measures for vision defects
• A glass prism disperses white light
• Light scattering
• What causes the clear sky to be blue in colour?
• Refraction in the atmosphere

below are some of the important questions for class 10 science chapter 11 – Human Eye And Colourful World from an examination point of view. 

Class 10 Science Important Questions Chapter 11 – Human Eye and Colourful World

Q1. Define the term power of accommodation. Write the modification in the curvature of the eye lens which enables us to see the nearby objects clearly?

Solution:

The ability of the eye lens to adjust its focal length is called power of accommodation. The ciliary muscles modifies the curvature to some extent. The change in the curvature of the eye lens can thus change its focal length. When the ciliary muscles contract, the lens becomes thick and its focal length decreases, thus enables us to see nearby objects clearly.

Q2. Trace the sequence of events which occur when a bright light is focused on your eyes. 

Solution:

When a bright light enters the eye then most of the refraction for the light rays entering the eye occurs at the outer surface of the cornea. Then, the crystalline lens merely provides the finer adjustment of focal length required to focus object at different distances on the retina. The pupil regulates and controls the amount of light entering the eye. At retina, the light-sensitive cells get activated upon illumination and generate electric signals. These signals are sent to the brain via the optic nerves. The brain interprets these signals and finally, processes the information so that we perceive objects as they are.

Q3. Person suffering from cataract has

(a) elongated eyeball

(b) excessive curvature of eye lens

(c) weakened ciliary muscles

(d) opaque eye lens

Solution:

(d) A person suffering from cataract has cloudy opaque eye lens.

Q4. Name the three common defects of vision. What are their causes? Name the type of lens used to correct each of them.

Solution:

Three common defects of vision are

• Myopia
• Hypermetropia
• Presbyopia

Myopia can be caused due to following reasons.

• Elongation of eyeball.
• Excessive curvature of eye lens.

Hypermetropia can be caused due to following reasons.

• Shortening of eyeball.
• Focal length of eye lens becomes too long.

Presbyopia is caused due to gradual weakening of ciliary muscles and diminishing flexibility of eye lens due to ageing.

Correction of these defects:

• Myopia can be corrected by using concave lens of appropriate focal length.
• Hypermetropia can he corrected by using convex lens of appropriate local length.
• Presbyopia can be corrected by using bifocal lens.

Q5. Millions of people of the developing countries of world are suffering from corneal blindness. These persons can be cured by replacing the defective cornea with the cornea of a donated eye. A charitable society of your city has organised a campaign in your neighbourhood in order to create awareness about this fact. If you are asked to participate in this mission how would you contribute in this noble cause?

(a) State the objective of organising such campaigns.

(b) List two arguments which you would give to motivate the people to donate their eyes after death.

(c) List two values which are developed in the persons who actively participate and contribute in such programmes.

Solution:

We can encourage people to participate in the camp and also register ourselves as a donator.

(a) The objective of organising such campaign is to make people aware and realize their duties towards society.

(b) (i) By donating our eyes after we die, we can light the life of a blind person.

(ii) One pair of eyes gives vision to two corneal blind people.

(c) (i) It shows the concern for others.

(ii) It also shows responsible behavior towards the society.

Q6. Differentiate between a glass slab and a glass prism. What happens when a narrow beam of

(i) a monochromatic light and (ii) white light passes through (a) glass slab and (b) glass prism? 

Solution:

Glass slab:

• It is a substance made of glass having three dimensions and has cuboidal structure.
• It does not deviate the path of light falling on it but produces a lateral displacement of the light ray after refraction. The incident and emergent ray are parallel to each other.

Glass prism:

• A prism is a structure made of glass with two triangle bases and three rectangular lateral surfaces. These surfaces are inclined to each other.
• A prism deviates the path of light ray falling on it. Here the incident ray and emergent ray are not parallel to each other.

(i) When a narrow beam of monochromatic light falls on a

(a) glass slab, it gets refracted at its surface and the emergent ray is laterally displaced from the incident ray.

(b) prism, it gets refracted at the surface and the light gets deviated from its initial path. The angle between the incident ray and emergent ray is known as angle of deviation.

(ii) When a white light passes through a

(a) glass slab, the light does not undergo dispersion as its two refracting surfaces are parallel to each other. The white light is laterally displaced from its initial path.

(b) prism, the white light undergoes dispersion and splits into its constituent colours along with deviation from its initial path.

Q7. Explain why the planets do not twinkle.

Solution:

Planets do not emit light. However, they become visible due to reflection of light falling on them. The planets are much closer to the earth and thus can be considered as the extended source of light. The fluctuations in the light coming from various points of the planet due to atmospheric refraction get averaged out. As a result, no twinkling of planets is seen.

Q8. Consider the following reasons for the reddish appearance of the sun at the sunrise or the sunset:

(A) Light from the sun near the horizon passes through thinner layers of air.

(B) Light from the sun covers larger distance of the earths atmosphere before reaching our eyes.

(C) Near the horizon, most of the blue light and shorter wavelengths are scattered away by the particles.

(D) Light from the sun near the horizon passes through thicker layers of air.

The correct reasons are

(a) A and C only

(b) B, C and D

(c) A and B only

(d) C and D only

Solution:

(b) Near the horizon, the light rays from the sun has to travel a larger distance through the Earth’s atmosphere as compared to when it is away from the horizon. Thus, when this light travels through the atmosphere, most of short wavelength lights are scattered away causing the reddish appearance of the sun.

Q9. Why there is no dispersion of light refracted through a rectangular glass slab?

Solution:

After refraction at two parallel faces of a glass slab, a ray of light emerges in a direction parallel to the direction of incidence of white light. As rays of all colours emerge in same direction, i.e., the direction of the incidence of white light, there is no dispersion. However, there is a lateral displacement.

Q10. 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?

Solution:

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.