The Human Eye
Q1. What is meant by power of accommodation of the eye?
Ans: When the ciliary muscles are relaxed, the eye lens becomes
thin, the focal length increases, and the distant objects are clearly
visible to the eyes. To see the nearby objects clearly, the ciliary
muscles contract making the eye lens thicker. Thus, the focal length of
the eye lens decreases and the nearby objects become visible to the
eyes. Hence, the human eye lens is able to adjust its focal length to
view both distant and nearby objects on the retina. This ability is called
the power of accommodation of the eyes.
Q2. 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?
Ans: The person is able to see nearby objects clearly, but he is
unable to see objects beyond 1.2 m. This happens because the image
of an object beyond 1.2 m is formed in front of the retina and not on
the retina, as shown in the given figure.
Myopic Eye
To correct this defect of vision, he must use a concave lens. The
concave lens will bring the image back to the retina as shown in the
given figure.
Correction of Myopic Eye
Q3. What is the far point and near point of the human eye with
normal vision?
Ans: The near point of the eye is the minimum distance of the object
from the eye, which can be seen distinctly without strain. For a normal
human eye, this distance is 25 cm.
The far point of the eye is the maximum distance to which the eye can
see the objects clearly. The far point of the normal human eye is
infinity.
Q4. 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?
Ans: A student has difficulty reading the blackboard while sitting in the
last row. It shows that he is unable to see distant objects clearly. He is
suffering from myopia. This defect can be corrected by using a
concave lens.
Exercises
Q1. The human eye can focus 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
Ans: (b) accommodation
Human eye can change the focal length of the eye lens to see the
objects situated at various distances from the eye. This is possible
due to the power of accommodation of the eye lens.
Q2. The human eye forms the image of an object at its
(a) cornea
(b) iris
(c) pupil
(d) retina
Ans: (d) retina
The retina is the layer of nerve cells lining the back wall inside the eye.
This layer senses light and sends signals to the brain so you can see.
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
Ans: (c) 25 cm
The least distance of distinct vision is the minimum distance of an
object to see a clear and distinct image. It is 25 cm for a young adult
with normal vision.
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
Ans: (c) ciliary muscles
The relaxation or contraction of ciliary muscles changes the curvature
of the eye lens. The change in curvature of the eye lens changes the
focal length of the eyes. Hence, the change in focal length of an eye
lens is caused by the action of ciliary muscles.
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?
Ans: For distant vision = −0.181 m, for near vision = 0.667
m
The power P of a lens of focal length f is given by the relation
Power (P) = 1 / f(in metres)
(i) Power of the lens used for correcting distant vision = −5.5
D
Focal length of the required lens, f = 1 / p
f = 1 / -5.5 => f= -0.181 m
The focal length of the lens for correcting distant vision is
−0.181 m.
(ii) Power of the lens used for correcting near vision = +1.5 D
Focal length of the required lens, f = 1 / p
f = 1 / 1.5 = +0.667 m
The focal length of the lens for correcting near vision is 0.667 m.
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?
Ans: The person is suffering from an eye defect called
myopia. In this defect, the image is formed in front of the
retina. Hence, a concave lens is used to correct this defect of
vision.
Object distance, u = infinity = ∞
Image distance, v = −80 cm
Focal length = f
According to the lens formula,
We know,
A concave lens of power −1.25 D is required by the person to
correct his defect.
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.
Ans: A person suffering from hypermetropia can see distinct objects
clearly but faces difficulty in seeing nearby objects clearly. It happens
because the eye lens focuses the incoming divergent rays beyond the
retina. This defect of vision is corrected by using a convex lens. A
convex lens of suitable power converges the incoming light in such a
way that the image is formed on the retina, as shown in the following
figure.
Correction of Hypermetropic Eye
The convex lens actually creates a virtual image of a nearby
object (N’ in the figure) at the near point of vision (N) of the
person suffering from hypermetropia.
The person will be able to clearly see the object kept at 25
cm (near point of the normal eye) if the image of the object is
formed at his near point, which is given as 1 m.
Object distance, u = −25 cm
Image distance, v = −1 m = −100 m
Focal length, f
Using the lens formula,
A convex lens of power +3.0 D is required to correct the defect.
Q8. Why is a normal eye not able to see clearly the objects
placed closer than 25 cm?
Ans: A normal eye is unable to clearly see the objects placed closer
than 25 cm because the ciliary muscles of eyes are unable to contract
beyond a certain limit.
If the object is placed at a distance less than 25 cm from the eye, then
the object appears blurred and produces strain in the eyes.
Q9. What happens to the image distance in the eye when we
increase the distance of an object from the eye?
Ans: Since the size of the eyes cannot increase or decrease, the
image distance remains constant. When we increase the distance of
an object from the eye, the image distance in the eye does not
change. The increase in the object distance is compensated by the
change in the focal length of the eye lens. The focal length of the eyes
changes in such a way that the image is always formed on the retina
of the eye.
Q10. Why do stars twinkle?
Ans: Stars emit their own light and they twinkle due to the
atmospheric refraction of light. Stars are very far away from the earth.
Hence, they are considered as point sources of light. When the light
coming from stars enters the earth’s atmosphere, it gets refracted at
different levels because of the variation in the air density at different
levels of the atmosphere. When the starlight refracted by the
atmosphere comes more towards us, it appears brighter than when it
comes less towards us. Therefore, it appears as if the stars are
twinkling at night.
Q11. Explain why the planets do not twinkle?
Ans: Planets do not twinkle because they appear larger in size than
the stars as they are relatively closer to earth. Planets can be
considered as a collection of a large number of point-size sources of
light. The different parts of these planets produce either brighter or
dimmer effect in such a way that the average of brighter and dimmer
effect is zero. Hence, the twinkling effects of the planets are nullified
and they do not twinkle.
Q12. Why does the sky appear dark instead of blue to an
astronaut?
Ans: The sky appears dark instead of blue to an astronaut because
there is no atmosphere in the outer space that can scatter the
sunlight. As the sunlight is not scattered, no scattered light reach the
eyes of the astronauts and the sky appears black to them.
