The Human Eye and the Colourful WorldClass 10 Physics NCERT Solutions

Answer: (b) accommodation.

Explanation: The ability of the eye lens to adjust its focal length to see both nearby and distant objects clearly is called the power of accommodation. This is achieved by the action of the ciliary muscles, which change the curvature of the lens.

Answer: (d) retina.

Explanation: The human eye works like a camera. Its lens system forms a real, inverted image of an object on a light-sensitive screen called the retina. The light-sensitive cells on the retina then send electrical signals to the brain via the optic nerve.

Answer: (c) 25 cm .

Explanation: The least distance of distinct vision, also known as the near point, is the minimum distance at which an object can be seen clearly without any strain on the eye. For a young adult with normal vision, this distance is approximately 25 cm.

Answer: (c) ciliary muscles.

Explanation: The ciliary muscles are responsible for changing the shape and curvature of the eye lens. When they contract, the lens becomes thicker, decreasing the focal length to focus on nearby objects. When they relax, the lens becomes thinner, increasing the focal length to focus on distant objects.

Given: Power of the lens, $P_1 = -5.5 \text{ D}$

To Find: Focal length of the lens, $f_1$

Formula: Power $(P)$ is the reciprocal of focal length $(f)$ in metres. $P = \frac{1}{f(\text{in m})}$

Calculation: $f_1 = \frac{1}{P_1} = \frac{1}{-5.5} = -\frac{10}{55} = -\frac{2}{11} \text{ m}$ $f_1 = -\frac{2}{11} \times 100 \text{ cm} \approx -18.18 \text{ cm}$

Final Answer: The focal length of the lens for correcting distant vision is $-0.1818 \text{ m}$ or $-18.18 \text{ cm}$. The negative sign indicates it is a concave lens.

** (ii) For correcting near vision:**

Given: Power of the lens, $P_2 = +1.5 \text{ D}$

To Find: Focal length of the lens, $f_2$

Formula: $P = \frac{1}{f(\text{in m})}$

Calculation: $f_2 = \frac{1}{P_2} = \frac{1}{+1.5} = +\frac{10}{15} = +\frac{2}{3} \text{ m}$ $f_2 = +\frac{2}{3} \times 100 \text{ cm} \approx +66.67 \text{ cm}$

Final Answer: The focal length of the lens for correcting near vision is $+0.667 \text{ m}$ or $+66.67 \text{ cm}$. The positive sign indicates it is a convex lens.

Given: The person suffers from myopia (near-sightedness). The far point of the myopic eye is 80 cm. This means the person can see objects clearly only up to a distance of 80 cm. Object distance, $u = -\infty$ (for viewing distant objects) Image distance, $v = -80 \text{ cm} = -0.8 \text{ m}$ (The corrective lens must form the image of a distant object at the person's far point).

To Find: The nature and power of the corrective lens.

Formula: Lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ Power of lens: $P = \frac{1}{f(\text{in m})}$

Calculation: Using the lens formula to find the focal length $(f)$: $\frac{1}{f} = \frac{1}{-80} - \frac{1}{-\infty}$ $\frac{1}{f} = -\frac{1}{80} - 0$ $f = -80 \text{ cm} = -0.8 \text{ m}$ Now, calculating the power $(P)$: $P = \frac{1}{-0.8} = -\frac{10}{8} = -1.25 \text{ D}$

Final Answer: Since the focal length and power are negative, the lens required is a concave lens. The power of the lens is -1.25 D.

Diagram: The correction for hypermetropia is shown in Figure 10.3 (c) of the textbook. The diagram shows:

Given: The person suffers from hypermetropia (far-sightedness). The near point of the hypermetropic eye is 1 m. The corrective lens should enable the person to read an object placed at the normal near point (25 cm). Object distance, $u = -25 \text{ cm} = -0.25 \text{ m}$ (Normal near point) Image distance, $v = -1 \text{ m} = -100 \text{ cm}$ (Near point of the defective eye)

To Find: The power of the corrective lens.

Formula: Lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ Power of lens: $P = \frac{1}{f(\text{in m})}$

Calculation: Using the lens formula to find the focal length $(f)$: $\frac{1}{f} = \frac{1}{-100} - \frac{1}{-25}$ $\frac{1}{f} = -\frac{1}{100} + \frac{1}{25} = \frac{-1 + 4}{100} = \frac{3}{100}$ $f = +\frac{100}{3} \text{ cm} = +\frac{1}{3} \text{ m}$ Now, calculating the power $(P)$: $P = \frac{1}{f(\text{in m})} = \frac{1}{1/3} = +3.0 \text{ D}$

Final Answer: The power of the convex lens required to correct the defect is +3.0 D.

A normal eye is not able to see objects placed closer than 25 cm clearly because the eye lens has a limit to how much it can increase its curvature and decrease its focal length. To focus on a very close object, the ciliary muscles must contract significantly to make the eye lens thicker and more curved. However, the ciliary muscles cannot contract beyond a certain limit. Therefore, the focal length of the eye lens cannot be decreased below a certain minimum value. If an object is placed closer than this minimum distance (the near point, about 25 cm), the lens cannot converge the light rays sufficiently to form a sharp image on the retina, resulting in a blurred image and strain on the eye.

When we increase the distance of an object from the eye, the image distance in the eye remains almost constant. For an object to be seen clearly, its image must be formed on the retina. The distance between the eye lens and the retina is fixed. To accommodate for the change in object distance, the eye changes the focal length of its lens by the action of the ciliary muscles. As the object moves farther away, the ciliary muscles relax, making the lens thinner and increasing its focal length to ensure the image is still formed on the retina.

Stars twinkle due to the atmospheric refraction of starlight. Stars are very distant and act as point-sized sources of light. As starlight enters the Earth's atmosphere, it passes through layers of air with continuously changing temperatures and densities. This causes the refractive index of the atmosphere to fluctuate at different points. Consequently, the light from the star is refracted multiple times and follows a wavering path before it reaches the observer's eye. This causes the apparent position of the star to change slightly and the amount of starlight entering the eye to flicker. This fluctuation in brightness is perceived as the twinkling of stars.

Planets do not twinkle because they are much closer to the Earth than stars and are therefore seen as extended sources of light, not point-sized sources. A planet can be considered as a collection of a very large number of point-sized light sources. While the light from each individual point source flickers due to atmospheric refraction, the twinkling effects from all these points average out. The total amount of light entering the eye from the planet remains relatively constant, thereby nullifying the twinkling effect.

The blue colour of the sky is a result of the scattering of sunlight by the molecules of air and other fine particles in the Earth's atmosphere. This phenomenon is called Rayleigh scattering, which scatters shorter wavelengths (like blue and violet light) more effectively than longer wavelengths (like red and orange light). When we look at the sky from Earth, this scattered blue light enters our eyes from all directions, making the sky appear blue.

An astronaut in outer space is above the Earth's atmosphere. In space, there are no particles to scatter the sunlight. Without scattering, light travels in straight lines from its source (the Sun) to the observer's eye. Since there is no scattered light reaching the astronaut's eyes from the surrounding space, the sky appears dark or black.