The Human Eye and the Colourful World

Intext Questions 1

Question 1

What is meant by power of accommodation of the eye?

Answer

Power of accommodation of the eye is the ability of the eye lens to focus near and far object clearly on the retina by adjusting the thickness of lens (and hence focal length). The ciliary muscles attached to the lens help the thickening or thinning of lens hence changing its curvature.

Question 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 myopic eye can be corrected by using a concave lens of suitable power. This will restore proper vision.

Question 3

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

Answer

The minimum distance, at which objects can be seen most distinctly without strain, is called the least distance of distinct vision or the near point of the eye. For human eye with normal vision, the near point is about 25 cm. The farthest point upto which the eye can see objects clearly is called the far point of the eye. It is infinity for a normal eye.

Question 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 has difficulty reading the blackboard while sitting in the last row, this means he cannot see distant objects clearly. Therefore, child could be suffering from myopia. Myopia can be corrected by using a concave lens of suitable power.

Exercises

Question 1

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

1. presbyopia
2. accommodation
3. near-sightedness
4. far-sightedness

Answer

accommodation

Reason — The ability of the eye lens to adjust its focal length is called accommodation.

Question 2

The human eye forms the image of an object at its

1. cornea
2. iris
3. pupil
4. retina

Answer

retina

Reason — The eye lens forms an inverted real image of the object on the retina.

Question 3

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

1. 25 m
2. 2.5 cm
3. 25 cm
4. 2.5 m

Answer

25 cm

Reason — For human eye with normal vision, the near point or least distance of distinct vision is about 25 cm.

Question 4

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

1. pupil
2. retina
3. ciliary muscles
4. iris

Answer

ciliary muscles

Reason— The ciliary muscles contract to make the lens thinner and relax to make it thicker, thus changing its curvature and hence focal length.

Question 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

Power is reciprocal of focal length.

(i) Focal length of the lens required for correcting distant vision = $-\dfrac{1}{5.5}$ = -0.181 m = -18.1 cm

(ii) Focal length of the lens required for correcting near vision = $\dfrac{1}{1.5}$ = 0.667m = 66.7cm

Question 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

Concave lens is used to correct the problem of myopia.

The image must be formed at far point.

Given,

Far point = 80 cm = 0.8 m

∴ Power of lens required = $-\dfrac{1}{0.8}$ = -1.25 dioptres.

Question 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

Hypermetropia can be corrected by using a convex lens as shown in the diagram below:

An object at 25 cm forms an image at the near point of hypermetropic eye.

Given,

Near point of hypermetropic eye = 1 m =100 cm

Object distance, u = -25 cm

Image distance, v = -100 cm

According to the formula,

$\phantom{\Rightarrow} \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} \\[1 em] \Rightarrow \dfrac{1}{-100} - \dfrac{1}{-25} = \dfrac{1}{f} \\[1 em] \Rightarrow \dfrac{-1 + 4}{100} = \dfrac{1}{f} \\[1 em] \Rightarrow \dfrac{3}{100} = \dfrac{1}{f} \\[1 em] \Rightarrow f = \dfrac{100}{3} \text{ cm} \\[1 em] \Rightarrow f = \dfrac{1}{3} \text{ m} \\[1 em] \text{Power} = \dfrac{1}{f} = \dfrac{1}{\dfrac{1}{3}} \\[1 em] \therefore \text{Power} = 3 \text{ D}$

Hence, the power of the lens required to correct this defect is 3 dioptre.

Question 8

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

Answer

When focusing on a nearby object, the ciliary muscles strain to adjust the eye lens, reducing its focal length. However, there's a limit to this adjustment. Beyond approximately 25 cm, the eye struggles to bring the image into clear focus, resulting in blurred vision for objects placed closer. This limitation is due to the eye's finite ability to accommodate and maintain sharp focus.

Question 9

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

Answer

The image distance in the eye remains the same because the image is always formed on retina. When we increase the distance of an object from the eye, only the lens changes its thickness to form the image of the object on retina.

Question 10

Why do stars twinkle?

Answer

Stars twinkle due to atmospheric refraction of starlight. As starlight passes through Earth's atmosphere, it undergoes continuous refraction in a medium of gradually changing refractive index, bending towards the normal. This bending causes the apparent position of the star to fluctuate slightly, leading to the twinkling effect. The dynamic and non-uniform conditions of the Earth's atmosphere, especially near the horizon, contribute to this phenomenon.

Question 11

Explain why the planets do not twinkle.

Answer

Planets do not twinkle because, unlike stars, they are seen as extended sources of light. When observing a planet, which appears as a disk rather than a point of light, the combined effect of a large number of point-sized sources averages out the total variation in the amount of light entering the observer's eye. This averaging effect nullifies the twinkling phenomenon, making planets appear steady in brightness.

Question 12

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

Answer

The molecules of air and other fine particles in the atmosphere have size smaller than the wavelength of visible light. These are more effective in scattering light of shorter wavelengths i.e blue. When sunlight passes through the atmosphere, the fine particles in air scatter the blue colour (shorter wavelengths) more strongly than red. The scattered blue light enters our eyes and the sky appears blue. But there are no particles, hence no scattering of light in space. Therefore, the sky appear dark to an astronaut.