The Human Eye and the Colourful World - Questions and Answers

Answer: (b) accommodation

Accommodation is the ability of the eye to adjust its focal length to see objects at different distances clearly, achieved through the action of ciliary muscles that change the curvature of the eye lens.

Answer: (d) retina

The retina is the light-sensitive screen at the back of the eye where the image is formed. It contains photoreceptor cells (rods and cones) that convert light into electrical signals sent to the brain.

Answer: (c) 25 cm

The least distance of distinct vision, also called the near point, is the minimum distance at which an object can be seen most distinctly without strain. For a young adult with normal vision, this is approximately 25 cm.

Answer: (c) ciliary muscles

The ciliary muscles control the curvature of the eye lens. When these muscles contract or relax, they change the shape of the lens, thereby adjusting its focal length to focus on objects at different distances.

Solution:

The power of a lens (P) is related to its focal length (f) by the formula:

(i) For distant vision:

Power, P = -5.5 D (negative sign indicates concave lens)

Focal length, f = 1/P = 1/(-5.5) = -0.1818 m = -18.18 cm

(ii) For near vision:

Power, P = +1.5 D (positive sign indicates convex lens)

Focal length, f = 1/P = 1/(1.5) = 0.6667 m = 66.67 cm

Answer: The focal length for distant vision correction is -18.18 cm (concave lens) and for near vision correction is 66.67 cm (convex lens).

Solution:

For a myopic person, the far point is closer than infinity. To correct myopia, we need a concave lens that will make parallel rays appear to come from the person's far point.

Given: Far point = 80 cm = 0.8 m

For the corrective lens, the object at infinity should form an image at the far point of the defective eye.

So, for the lens:

• Object distance, u = ∞
• Image distance, v = -0.8 m (negative as it's on the same side as the object for a concave lens)

Using lens formula: 1/f = 1/v - 1/u

1/f = 1/(-0.8) - 1/∞ = -1/0.8 = -1.25

So, f = -0.8 m = -80 cm

Power, P = 1/f (in meters) = 1/(-0.8) = -1.25 D

Answer: The person requires a concave lens of power -1.25 dioptres.

Solution:

Hypermetropia (far-sightedness) is corrected using a convex lens. The lens should form an image of an object at the normal near point (25 cm) at the hypermetropic eye's near point (1 m).

Given:

• Object distance, u = -25 cm = -0.25 m (negative as per sign convention)
• Image distance, v = -1 m (negative as it's on the same side as the object)

Using lens formula: 1/f = 1/v - 1/u

1/f = 1/(-1) - 1/(-0.25) = -1 + 4 = 3

So, f = 1/3 m = 0.333 m

Power, P = 1/f = 1/(1/3) = +3 D

Answer: The person requires a convex lens of power +3 dioptres.

Diagram explanation: A convex lens converges light rays from nearby objects so that they focus correctly on the retina instead of behind it.

A normal eye cannot see objects clearly when placed closer than 25 cm because:

• The ciliary muscles can only contract to a certain limit to increase the curvature of the eye lens.
• There is a minimum focal length that the eye lens can achieve through accommodation.
• When objects are closer than 25 cm, the eye lens cannot adjust enough to focus the image sharply on the retina, resulting in a blurred image.
• This minimum distance of 25 cm is called the least distance of distinct vision or near point.

When we increase the distance of an object from the eye:

• The image distance in the eye remains approximately constant.
• The image is always formed on the retina, which is at a fixed distance from the eye lens.
• To maintain focus on the retina, the eye lens adjusts its focal length through accommodation - it becomes thinner (less curved) to increase its focal length when viewing distant objects.
• This adjustment is achieved by the relaxation of ciliary muscles, which reduces the curvature of the eye lens.

Stars twinkle due to atmospheric refraction of starlight:

• Starlight enters Earth's atmosphere and undergoes continuous refraction due to varying refractive indices of different air layers.
• The atmosphere has turbulent air currents of varying temperatures and densities, causing the path of starlight to fluctuate.
• This causes the apparent position of the star to change slightly and rapidly.
• The amount of starlight entering our eye flickers, making the star appear to twinkle - sometimes brighter, sometimes fainter.
• Stars appear as point sources of light, which makes this effect more noticeable.

Planets do not twinkle because:

• Planets are much closer to Earth than stars, so they appear as extended sources of light rather than point sources.
• An extended source can be thought of as a collection of many point sources.
• When light from different points of a planet undergoes atmospheric refraction, the variations average out.
• The total amount of light entering our eye from all points of the planet remains nearly constant, eliminating the twinkling effect.
• Thus, planets appear to shine with a steady light without twinkling.

The sky appears dark to an astronaut because:

• The blue color of the sky is due to scattering of sunlight by air molecules and fine particles in the atmosphere.
• This scattering is more effective for shorter wavelengths (blue light) than longer wavelengths (red light).
• In space, there is no atmosphere to scatter sunlight.
• Without atmospheric scattering, there is no diffused blue light reaching the astronaut's eyes.
• The astronaut sees direct sunlight from the Sun and darkness in other directions, making the sky appear dark or black.