CHAPTER 10 The Human Eye and the Colourful World KSEEB SSLC CLASS 10 SCIENCE SOLUTIONS

CHAPTER 10 The Human Eye and the Colourful World KSEEB SSLC CLASS 10 SCIENCE SOLUTIONS English medium Karnataka state board,the Answers Are Prepared By Our Teachers Which Are Simple ,Pointwise,Easy To Read And Remember

CHAPTER 10 The Human Eye and the Colourful World KSEEB SSLC CLASS 10 SCIENCE SOLUTIONS

What is meant by power of accommodation of the eye?

Power of Accommodation of the Eye:

The eye’s ability to adjust its lens to focus on objects at different distances.
Controlled by ciliary muscles that change the lens’s curvature.
Lens becomes thinner for distant objects and thicker for nearby objects.

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?

Corrective Lens for Myopic Eye:

Myopic eye sees nearby objects clearly but distant objects appear blurry.
Corrected using a concave lens.
Concave lens helps to diverge light rays, allowing the image to focus correctly on the retina.

What is the far point and near point of the human eye with normal vision?

Far Point and Near Point of the Human Eye:

Far point: The farthest distance for clear vision, usually infinity.
Near point: Closest distance for clear vision without strain, typically around 25 cm for normal vision.

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?

Defect and Correction for Difficulty in Reading the Blackboard:

Student’s problem suggests possible myopia, where distant objects are blurry.
Corrected with a concave lens to compensate for eye’s excessive curvature.
Concave lens helps focus distant objects clearly on the retina.

The human eye can focus on objects at different distances by adjusting the focal length of the eye lens. This is due to

(a) presbyopia.

(b) accommodation.

(d) far-sightedness.

ANSWER;-

(b) accommodation.

The human eye forms the image of an object at its

(a) cornea.

(b) iris.

(d) retina.

ANSWER;-

(d) retina.

The least distance of distinct vision for a young adult with normal vision is about

(a) 25 m.

(b) 2.5 cm.

(d) 2.5 m.

ANSWER;-

The change in focal length of an eye lens is caused by the action of the

(a) pupil.

(b) retina.

(d) iris.

ANSWER;-

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?

To find the focal length of the lenses required for correcting distant and near vision, we use the formula:

$\text{Power (P)} = \frac{1}{\text{Focal length (f)}}$

(i) For distant vision:
Given: Power (P) = -5.5 dioptres
Formula:

$$ \text{Focal length (f)} = \frac{1}{\text{Power (P)}} $ \[ \text{Focal length (f)} = \frac{1}{-5.5} \] \[ \text{Focal length (f)} = -0.182 \, \text{m} \] \[ \text{Focal length (f)} = -18.2 \, \text{cm} \]$

(ii) For near vision:
– Given: Power (P) = +1.5 dioptres
– Formula:

$$ \text{Focal length (f)} = \frac{1}{\text{Power (P)}} $ \[ \text{Focal length (f)} = \frac{1}{1.5} \] \[ \text{Focal length (f)} = 0.67 \, \text{m} \] \[ \text{Focal length (f)} = 67 \, \text{cm} \]$

So, the focal lengths required are:

For distant vision: -18.2 cm
For near vision: 67 cm

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?

Given:
– Far point of a myopic person: 80 cm in front of the eye (since it’s in front of the eye, it is taken as negative in sign convention, i.e., $$ -80 $ cm or $ -0.8 $ m)$

Nature of the lens:
– Concave lens (used to correct myopia by diverging light rays)

Power of the lens:
– Formula: $\text{Power (P)} = \frac{1}{\text{Focal length (f)}}$
– Focal length (f) for a concave lens is the same as the far point of the myopic person:

$$ -0.8 $ m \[ \text{Power (P)} = \frac{1}{\text{Focal length (f)}} \] \[ \text{Power (P)} = \frac{1}{-0.8} \] \[ \text{Power (P)} = -1.25 \, \text{dioptres} \]$

Summary:
– Nature of the lens:Concave lens
– Power of the lens: $-1.25 \, \text{dioptres}$

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.

In hypermetropia (farsightedness), the image of a nearby object is formed behind the retina. A convex lens is used to correct this defect by converging light rays before they enter the eye, thus allowing the image to form on the retina.

Calculation of the Power of the Lens:

Given:
– Near point of the hypermetropic eye = 1 m
– Near point of a normal eye = 25 cm = 0.25 m

The lens must create a virtual image of an object placed at 25 cm (the normal near point) at the near point of the hypermetropic eye (1 m).

Using the lens formula:

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

Where:
– $$ u = -0.25 $ m (object distance) - $ v = -1 $ m$ (image distance, virtual and on the same side as the object)

$\frac{1}{f} = \frac{1}{-1} - \frac{1}{-0.25} \] \[ \frac{1}{f} = -1 + 4 \] \[ \frac{1}{f} = 3 \] \[ f = \frac{1}{3} \, \text{m}$

The power of the lens

$$ P $ is given by: \[ P = \frac{1}{f} \] \[ P = \frac{1}{1/3} \] \[ P = 3 \, \text{dioptres} \]$

Summary:
– Nature of the lens:Convex lens
– Power of the lens: $+3 \, \text{dioptres}$

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

A normal eye cannot see objects clearly placed closer than 25 cm because the ciliary muscles cannot adjust the focal length of the lens to focus the image on the retina for objects closer than the near point.

What happens to the image distance in the eye when we increase the distance of an object from the eye?

When we increase the distance of an object from the eye, the image distance in the eye also increases. This is because the eye lens adjusts its focal length to ensure that the image is formed on the retina, and as the object moves away, the lens becomes thinner, increasing its focal length to maintain the focus on the retina.

Why do stars twinkle?

Stars twinkle due to atmospheric refraction. As the light from stars enters the Earth’s atmosphere, it passes through layers of air with different densities, causing the starlight to bend and flicker slightly, which appears as twinkling to observers on the ground.

Explain why the planets do not twinkle.

Planets do not twinkle because they are relatively closer to the Earth and appear as extended sources of light, not point sources like stars. The combined effect of light from numerous points on the planet’s surface averages out the twinkling effect.

Why does the Sun appear reddish early in the morning?

The Sun appears reddish early in the morning due to atmospheric scattering. When the Sun is near the horizon, its light passes through a thicker layer of the Earth’s atmosphere. The shorter blue and violet wavelengths scatter away, leaving the longer red wavelengths to dominate, giving the Sun a reddish hue.

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

The sky appears dark instead of blue to an astronaut because there is no atmosphere in space to scatter sunlight. On Earth, the atmosphere scatters shorter blue wavelengths more effectively than longer red wavelengths, resulting in the blue color of the sky. Without an atmosphere, there is no scattering, and the sky appears dark.