NCERT Class 10 Science – Wide View Edition
NCERT CLASS 10 SCIENCE • CHAPTER 10 • HUMAN EYE AND COLORFUL WORLD • WIDE VIEW EDITION • SWIPE EQUATIONS TO READ

The Human Eye

Detailed Solutions Q1 – Q13

Q1
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
(c) Near-sightedness    (d) Far-sightedness
Answer: (b) Accommodation
Explanation: The ability of the eye lens to adjust its focal length using ciliary muscles to focus on both near and distant objects is called the power of accommodation.
[Image of human eye anatomy ciliary muscles]
Q2
The human eye forms the image of an object at its:
(a) Cornea    (b) Iris
(c) Pupil    (d) Retina
Answer: (d) Retina
The retina acts as a light-sensitive screen at the back of the eye where the image is formed. It contains rods and cones that convert light into electrical signals.
Q3
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
This is also known as the Near Point of the eye. It is the minimum distance at which objects can be seen most distinctly without strain.
Q4
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
The ciliary muscles modify the curvature of the eye lens. Relaxing makes the lens thin (increases focal length), and contracting makes it thick (decreases focal length).
Q5
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?
(i) For Distant Vision:
Given Power ($P$) = -5.5 D
↔ Swipe Calculation
$$f = \frac{1}{P} = \frac{1}{-5.5} = -0.181 \text{ m} = -18.1 \text{ cm}$$
Since $f$ is negative, it is a concave lens (Myopia correction).
(ii) For Near Vision:
Given Power ($P$) = +1.5 D
↔ Swipe Calculation
$$f = \frac{1}{P} = \frac{1}{+1.5} = \frac{10}{15} = +0.667 \text{ m} = +66.7 \text{ cm}$$
Since $f$ is positive, it is a convex lens (Hypermetropia correction).
Q6
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?
Problem Analysis: A myopic person cannot see distant objects (at infinity). To correct this, the lens must take an object at infinity ($u = \infty$) and form a virtual image at the person’s far point ($v = -80$ cm).
↔ Swipe Calculation
$$\frac{1}{f} = \frac{1}{v} – \frac{1}{u} \Rightarrow \frac{1}{f} = \frac{1}{-80} – \frac{1}{\infty}$$
$$\frac{1}{f} = -\frac{1}{80} \Rightarrow f = -80 \text{ cm} = -0.8 \text{ m}$$
Power Calculation:
$$P = \frac{1}{f(\text{in meters})} = \frac{1}{-0.8} = -1.25 \text{ D}$$
Result: A Concave Lens of power -1.25 D is required.
[Image of myopia correction ray diagram]
Q7
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.
Problem Analysis: The person wants to read an object placed at the normal near point ($u = -25$ cm), but their eye can only focus if the image is formed at their own near point ($v = -1$ m = -100 cm).
↔ Swipe Calculation
$$\frac{1}{f} = \frac{1}{v} – \frac{1}{u} \Rightarrow \frac{1}{f} = \frac{1}{-100} – \frac{1}{-25}$$
$$\frac{1}{f} = -\frac{1}{100} + \frac{4}{100} = \frac{3}{100}$$
$$f = \frac{100}{3} \text{ cm} = +33.3 \text{ cm} = +0.33 \text{ m}$$
Power Calculation:
$$P = \frac{1}{f} = \frac{1}{0.33} = +3.0 \text{ D}$$
Result: A Convex Lens of power +3.0 D is required.
Q8
Why is a normal eye not able to see clearly the objects placed closer than 25 cm?
The ciliary muscles can contract to increase the curvature of the eye lens only up to a certain limit. If an object is placed closer than 25 cm, the lens cannot become thick enough to focus the image on the retina. This causes strain and blurred vision.
Q9
What happens to the image distance in the eye when we increase the distance of an object from the eye?
The image distance in the eye remains constant.
The image is always formed on the retina. To maintain this fixed image distance, the ciliary muscles change the focal length of the eye lens.
Far object: Muscles relax, focal length increases.
Near object: Muscles contract, focal length decreases.
Q10
Why do stars twinkle?
Stars twinkle due to Atmospheric Refraction.
1. Starlight travels through different layers of the atmosphere with varying densities (temperatures).
2. These densities keep changing continuously due to wind.
3. This causes the path of light rays to fluctuate, making the apparent position of the star shake and the amount of light entering our eye flicker. This effect is seen as twinkling.
Q11
Explain why the planets do not twinkle.
1. Distance: Planets are much closer to Earth than stars. They appear as extended sources of light (a collection of points) rather than point sources.
2. Averaging Effect: The atmospheric fluctuations cancel out. If light from one part of the planet is dimmed, light from another part brightens. The total amount of light entering the eye remains constant, so they do not twinkle.
Q12
Why does the sun appear reddish early in the morning?
This is due to the Scattering of Light.
1. At sunrise/sunset, sunlight travels a longer distance through the thick atmosphere to reach us.
2. Blue light (shorter wavelength) is scattered away by the particles in the atmosphere.
3. Red light (longer wavelength) scatters the least and reaches our eyes. Hence, the sun appears red.
Q13
Why does the sky appear dark instead of blue to an astronaut?
The blue color of the sky is caused by the scattering of sunlight by air molecules. In space, there is no atmosphere. Without air particles, there is no scattering of light. Therefore, no light reaches the astronaut’s eyes from the surroundings, making the sky appear dark.
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