X Solved:The Human Eye and the Colorful World :Chapter – 11
Chapter – 11 The
Human Eye and the Colorful World
Useful study link from: http://jsuniltutorial.weebly.com/the-human-eye-and-the-colourful-world.html
10th Human eye and Atmospheric refraction Notes
Human Eye & Atmospheric Refraction Solved paper -2
Ans.
The ability of the eye lens to adjust its focal
length, so as to clearly focus rays coming from distant as well a near objects
on the retina, is called the power of accommodation of the eye.
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?
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.
A person with a myopic eye should use a concave lens
of focal length 1.2 m so as to restore proper vision.
Q 3. What is the far point and near point of the
human eye with normal vision?
Ans.
For a human eye with normal vision the far point is at
infinity and the near point is at 25 cm from the eye.
Q
4. 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.
The student is suffering from myopia or
short-sightedness. The defect can be corrected by the use of concave (diverging
) lens of an appropriate power.
Q
5. The human eye can focus objects at different
distance by adjusting the focal length of the eye lens. This is due to
(a) presbyopia (b) accommodation
(c) near-sightedness (d) far-sightedness
(a) presbyopia (b) accommodation
(c) near-sightedness (d) far-sightedness
Ans.
(b) accommodation.
Q
6. The human eye forms the image of an object at its (a) cornea (b) iris
(c)
pupil (d) retina
Ans.
(d) retina.
Q
7. 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.
Q
8. 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.
Q
9. 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.
(i) Power of lens needed for correction distant vision of
the person P1 = - 5.5 D
\Focal length of lens required for correcting distant
vision f1 = 1/p1 =
1/-5.5 = -18cm
(ii) For correcting near vision power
required P2 = + 1.5 D
\Focal length of lens required for correcting near
vision f2 = 1/p2 = 1/-1.5 = 0.67 m=66.7 cm
Q10.
The far point of a myopic the person is 80 cm in
front of the eye. What is the nature and power of the lens required to correct
the problem?
Ans.
To correct the myopia the person concerned should use
concave lens (diverging lens) of focal length ¦ = - 80 cm, so that for an
object at infinity (u = ¥ ), the virtual image
is formed at the far point of myopic person (v = - 80 cm).
Q11.
Make a diagram to show how hypermetropia is
corrected. The near point of a hypermetropic eye is 1m. 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. The
relevant diagram has been shown in adjoining fig. Here the convex lens used
forms virtual image of object placed at N¢ (near point of normal eye i.e., 25 cm from eye) at N, the near
point of defective eye (at a distance x).
In the problem it is given that the near point of the
normal eye is 25 cm, hence u = - 25 cm. The lens used forms its
virtual image at near point of hypermetropic eye i.e., v = - 1m = - 100 cm.
Q12.
Why is a normal eye not able to see clearly the
objects placed closer than 25 cm ?
Ans.
Due to a power of accommodation, the focal length of
the eye lens cannot be decreased below a certain minimum limit.Consequently, a
normal eye cannot see clearly the objects placed closer than a minimum
distance, called near point of eye.
For a young
adult value of near point is 25 cm. So, we cannot see clearly objects placed
closer than 25 cm.
Q13.
What happens to the image distance in the eye when
we increase the distance of an object from the eye?
Ans.
The image is formed on the retina even on increasing
the distance of an object from the eye. For this eye lens becomes thinner and its focal length increases as the
object is moved away from the eye.
Q14.
Why do stars twinkle?
Ans.
Stars twinkle due to atmospheric refraction of
starlight. As the stars are very far away, they behave as almost point sources
of light. A son account of atmospheric refraction, the path of rays of light
coming from the star goes on varying slightly, the apparent position of the
star fluctuates and the amount of starlight entering the eye flickers. So,
sometimes, the star appears brighter and at some other time, fainter. Thus, the
stars twinkle.
Q15.
Explain why the planets do not twinkle.
Ans.
Planets are
much closer to the earth and are seen as extended sources. So, a planet may be
considered as a collection of a large number of point-sized light sources.
Although light coming from individual point-sized sources flickers but the
total amount of light entering our eye
from all the individual point-sized sources average out to be constant.
Thereby, planets appear equally bright
and there is no twinkling of planets.
Q16.
Why does the Sun appear reddish early in the
morning?
Ans.
In the early morning, the sun is situated near the
horizon. Light from the Sun passes through thicker layers of air and larger
distance in the earth’s atmosphere before reaching our eyes. While passing
through atmosphere blue light is most scattered away and the Sun appear reddish
.
Q17.
Why does the
sky appear dark instead of blue to an astronaut?
Ans.
Colour of the sky is on account of scattering of light
of shorter wavelengths by particles in the atmosphere of earth. If the earth had no atmosphere, there would not have
been any scattering and the sky would have looked dark. When an astronaut in his spacecraft goes above the
atmosphere of earth, sky appears dark to him because there is no scattering
of light.
10th Human eye and Atmospheric refraction Notes
Human Eye & Atmospheric Refraction Solved paper -2
Amazing Site
ReplyDelete