The Human Eye and the Colourful World-Exercise
Physics - Exercise
Q1. Defect of Human Eye
The human eye can focus on objects at different distances by adjusting the focal length of the eye lens. This is due to
- presbyopia
- accommodation
- near-sightedness
- far-sightedness
Answer: b. accommodation
The human eye can focus on objects at different distances by adjusting the focal length of the eye lens.
This is due to accommodation.
Accommodation is the ability of the eye lens, controlled by ciliary muscles, to change its shape and thus
its focal length—allowing the eye to focus on both near and distant objects.
Q2. Image Formation
The human eye forms the image of an object at its
- cornea
- iris
- pupil
- retina
Answer: d. Retina
The human eye forms the image of an object at its retina.
The retina is the light-sensitive layer at the back of the eye where images are focused to allow vision.
Q3. Least Distance of Vision
The least distance of distinct vision for a young adult with normal vision is about
- 25 m.
- 2.5 cm.
- 25 cm.
- 2.5 m
Answer: c. 25 cm
The least distance of distinct vision for a young adult with normal vision is about 25 cm.
This is the minimum distance at which the eye can see objects clearly without strain.
Q4. Change in Focal length
The change in focal length of an eye lens is caused by the action of the
- pupil
- retina
- ciliary muscles.
- iris
Answer: c. ciliary muscles
The change in focal length of an eye lens is caused by the action of the (c) ciliary
muscles.
Ciliary muscles adjust the lens curvature for focusing on objects at different distances.
Q5. Focal Length of Lens
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
- distant vision
- near vision
Answer:
Given that:
Power required for distant vision, \[ P_1 = -5.5\, \text{D} \]
Power required for near vision, \[ P_2 = +1.5\, \text{D} \]
Formula of Focal Length:
\[
\boxed{
\strut \quad P = \frac{1}{f} \quad
}
\]
where \( P \) is power in dioptres and \( f \) is focal length in meters.
Calculations:
- For distant vision: \[\begin{aligned} f_1 &= \frac{1}{P_1} \\\\&= \frac{1}{-5.5} \\\\&= -0.182\, \text{m} \\\\&=\boxed{ -18.2\, \text{cm}} \end{aligned}\]
- For near vision: \[\begin{aligned} f_2 &= \frac{1}{P_2} \\\\&= \frac{1}{1.5} \\\\&= 0.667\, \text{m} \\\\&= \boxed{66.7\, \text{cm}} \end{aligned}\]
- The negative focal length (-18.2 cm) indicates a concave lens for myopia correction
- The positive focal length (66.7 cm) indicates a convex lens for hypermetropia correction
- This person likely needs bifocal lenses to correct both vision defects
Q6. Nature & Power of Lens
Answer:
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:
The nature of the lens required to correct a myopic person is a concave (diverging) lens.
Power of the lens:
$$\boxed{\strut\quad
P = \frac{1}{f}\quad}
$$
where far point $$ f = -0.8\,m $$ (negative for concave lens).
$$\begin{aligned}
P &= \frac{1}{-0.8} \\\\&= -1.25\, \text{dioptres}
\end{aligned}$$
A concave lens of power \(-1.25\) dioptres is required to correct the problem.
Q7. Hypermetropia
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:
- Defective (hypermetropic) eye has a near point $$ v = -1\,\text{m} $$
- Normal near point of the eye $$ u = -0.25\,\text{m} $$
We use the lens formula:
$$\boxed{\quad
\frac{1}{f} = \frac{1}{v} - \frac{1}{u}}\quad
$$
Substituting values:
$$\begin{aligned}
\frac{1}{f} &= \frac{1}{-1} - \frac{1}{-0.25} \\\\&= -1 + 4 = 3\\\\
\implies f &= \frac{1}{3} \\\\&= 0.333\,\text{m}
\end{aligned}$$
Power of the lens:
$$
P = \frac{1}{f} = +3.0\,\text{D}
$$
A convex lens (converging lens) of power +3.0 dioptres is required to correct the
hypermetropia.
Explanation:
For a hypermetropic (far-sighted) eye, the image of a nearby object forms behind the retina. A
convex lens converges the incoming light rays before they enter the eye so that the image forms
exactly on the retina, allowing the person to see nearby objects clearly.
Q8. Minimum distance of Vision
Why is a normal eye not able to see clearly the objects placed closer than 25 cm?
Answer:
A normal eye cannot see objects clearly placed closer than 25 cm because the ciliary muscles of the eye cannot increase the curvature of the eye lens beyond a certain limit. When an object is too close, the lens would need to become very thick to focus its image onto the retina. The muscles are unable to further contract to provide this much curvature and hence, the image is not formed on the retina, resulting in a blurred vision. This minimum distance, called the least distance of distinct vision, is about 25 cm for a normal adult eye.
Q9. Image Position
What happens to the image distance in the eye when we increase the distance of an object from the eye?
Answer:
When we increase the distance of an object from the eye, the image distance inside the eye (distance
between the lens and the retina) does not change. The image is always formed on the retina for clear
vision.
Instead, the focal length of the eye lens automatically adjusts (through
accommodation) so
that objects at different distances can be focused onto the retina. The physical position of the
retina remains fixed, so the eye lens changes its curvature to keep the image at a constant image
distance.
Q10. Why do stars twinkle?
Answer:
Stars twinkle because their light passes through various layers of Earth's atmosphere, which have
different temperatures and densities. These atmospheric layers constantly change and cause the light
to bend (refract) unevenly.
As a result, the light from a star sometimes reaches our eyes
and
sometimes gets scattered. This rapid variation in brightness and position makes the star appear to
twinkle. The effect is stronger for stars because they are very far away and act as point sources of
light.
Q11. Explain why the planets do not twinkle.
Answer:
Planets do not twinkle because they are much closer to Earth in comparison to stars and appear as
tiny discs rather than point sources of light. Their light passes through the atmosphere in
multiple, closely spaced paths.
As a result, the small changes in brightness caused by
atmospheric refraction tend to get averaged out over the disc.
Therefore, any fluctuation in light is minimised,
and planets shine with a steady light instead of twinkling.
Q12. Why does the sky appear dark instead of blue to an astronaut?
Answer:
The sky appears dark instead of blue to an astronaut because there is no atmosphere in outer space to scatter sunlight.On Earth, the blue color of the sky is due to the scattering of shorter wavelength blue light by air molecules in the atmosphere (Rayleigh scattering).
In space, the absence of an atmosphere means no scattering occurs, so astronauts see a black sky even when the Sun is shining.
