Q 5). 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 ?
NCERT Solutions for Class 10 Science Chapter 11 Human Eye And Colorful World
Answer :
The power P of a lens of focal length f is given by the relationP= 1/f
(i) Power of the lens used for correcting distant vision = - 5.5 D
Focal length of the required lens, f= 1/Pf= 1/-5.5 = -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.
6. 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?
NCERT Solutions for Class 10 Science Chapter 11 Human Eye And Colorful World
Answer :
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,
Using (1/f) = (1/v) – (1/u)
(1/f) = 1/ (-80) + (1/-infinity)
(1/f) = - (1/80) + (-0)
f = -80 cm
f = -0.8 m
Power = (1/f)
= 1/ (-0.8)
= -1.25 D
The nature of lens required to see the distant objects clearly is convex lens is power -1.25 D.
7.Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropia 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.
NCERT Solutions for Class 10 Science Chapter 11 Human Eye And Colorful World
Answer :
It is corrected by using convex lens of suitable focal length.
Near point hypermetropia eye v = -1m = -100cm
Object distance u = -25cm
According to formula, (1/v) – (1/u) = (1/f)
Near point of hypermetropia eye:-
(1/f) = - (1/100) – (1/-25) = (-1/100) + (1/25)
(1/f) = (-1+4)/ (100) = (3/100)
f = (100/3)
Near point of a hypermetropic eye
Hypermetropia eye:-
f = (100/3)
P = (1/f) = (3 x 100)/ (100) = 3D
A convex lens of power +3.0 D is required to correct the defect.
Correction of a hypermetropia eye
NCERT Solutions for Class 10 Science Chapter 11 Human Eye And Colorful World
Answer :
The image distance in the eye, is fixed and cannot be changed. The ability of the eye lens to adjust its focal length is called accommodation.
Due to this eye can increase or decrease focal length of the lens in order to see either close or distant objects.
In order to see nearby objects the ciliary muscles contract and lens become thick and focal length is decreased which helps to see nearby objects.
In order to see distant objects the ciliary muscles expand and the lens become thin and focal length increases which help to see distant objects.
NCERT Solutions for Class 10 Science Chapter 11 Human Eye And Colorful World
Answer :
Stars twinkle due to atmospheric refraction of starlight. The light from the star on entering earth’s atmosphere undergoes refraction continuously before it reaches the earth.
Refraction occurs and refractive index changes.
As 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.
Therefore sometimes star appears brighter and sometimes twinkle.
NCERT Solutions for Class 10 Science Chapter 11 Human Eye And Colorful World
Answer :
Planets, do not twinkle as they are closer to the Earth than those distant stars, so planets appear larger in comparison. Due to the planets’ closeness to Earth, the light coming from them does not bend much due to Earth’s atmosphere. Therefore, the light coming from our solar system’s planets does not appear to twinkle.
NCERT Solutions for Class 10 Science Chapter 11 Human Eye And Colorful World
Answer :
In the early morning, as the Sun rises above horizon, light from the sun has to cover a larger distance in the denser medium of the
earth’s atmosphere before reaching our eyes. The blue light due to shorter wavelength gets scattered in the half way
and not visible in the sky. Only the red light which has larger wavelength is able to reach our eyes due to less scattering.
NCERT Solutions for Class 10 Science Chapter 11 Human Eye And Colorful World
Answer :
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
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 ?
NCERT Solutions for Class 10 Science Chapter 11 Human Eye And Colorful World
Answer :
The child is suffering from myopia and the child should use concave lens of suitable focal length.
Q.9 What happens to the image distance in the eye when we increase the distance of an object from the eye?
NCERT Solutions for Class 10 Science Chapter 11 Human Eye And Colorful World
Answer :
Since the size of 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 at the retina of the eye.