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# Human Eye Colourful World - Class 10 NCERT Solutions Class 10 Notes | EduRev

## Class 10 : Human Eye Colourful World - Class 10 NCERT Solutions Class 10 Notes | EduRev

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CH11: The Human Eye and the Colourful World
(NCERT Q & A)

Dispersion of Light
Which light bends the most?

Q1: What is meant by power of accommodation of the eye?

Answer: Power of accommodation is the ability of the eye lens to focus near and far objects clearly on
the retina by adjusting its focal length. Power of accommodation of the eye is limited. It implies the
focal length of the eye lens cannot be reduced beyond certain minimum limit.

Q2: A person cannot see the objects distinctly, when placed at a distance less than 50 cm.
(a) Identify the defect of vision.
(b) Give two reasons for this defect.
(c) Calculate the power and nature of the lens he should be using to see clearly the object
placed at a distance of 25 cm from his eyes.
(d) Draw the ray diagrams for the defective and the corrected eye.

Answer: (a) Defect of vision = Hypermetropia or Long-sightedness

(b) Reasons for the defect are:
? Curvature of lens or retina becomes less than normal increases focal length.
? Ciliary muscles become stiff
? Shortening of eye ball due to which distance between retina and and lens reduces and image
is formed beyond retina.
(c) Given, u = -25 cm, v = -50cm, f = ?
Applying lens formula, 1/f = 1/v - 1/u
?1/f = -1/50 - (-1/25) = -1/50 + 1/25 = (-1+2) /50 = 1/50
? f = 50cm = 0.5m
? Power of the lens = 1/f = 1/0.5m = +2D
? A convex lens having power of 2D can be used to correct the vision.

(d) ray diagrams for the defective and the corrected eye.
Page 2

CH11: The Human Eye and the Colourful World
(NCERT Q & A)

Dispersion of Light
Which light bends the most?

Q1: What is meant by power of accommodation of the eye?

Answer: Power of accommodation is the ability of the eye lens to focus near and far objects clearly on
the retina by adjusting its focal length. Power of accommodation of the eye is limited. It implies the
focal length of the eye lens cannot be reduced beyond certain minimum limit.

Q2: A person cannot see the objects distinctly, when placed at a distance less than 50 cm.
(a) Identify the defect of vision.
(b) Give two reasons for this defect.
(c) Calculate the power and nature of the lens he should be using to see clearly the object
placed at a distance of 25 cm from his eyes.
(d) Draw the ray diagrams for the defective and the corrected eye.

Answer: (a) Defect of vision = Hypermetropia or Long-sightedness

(b) Reasons for the defect are:
? Curvature of lens or retina becomes less than normal increases focal length.
? Ciliary muscles become stiff
? Shortening of eye ball due to which distance between retina and and lens reduces and image
is formed beyond retina.
(c) Given, u = -25 cm, v = -50cm, f = ?
Applying lens formula, 1/f = 1/v - 1/u
?1/f = -1/50 - (-1/25) = -1/50 + 1/25 = (-1+2) /50 = 1/50
? f = 50cm = 0.5m
? Power of the lens = 1/f = 1/0.5m = +2D
? A convex lens having power of 2D can be used to correct the vision.

(d) ray diagrams for the defective and the corrected eye.

Q3: 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?

Answer: The distance of far point x = 1.2m
To view distant objects, concave lens of focal length -1.2 should be used.
i.e. f = -x = -1.2m
Power of lens = 1/f = 1/(-1.2m) = -0.83D

Q4: Name the part of the human eye that helps in changing the focal length of the eye lens.

(Note: Pupil regulates and controls the amount of light entering the eye.)

Q5: 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.

Q6: The human eye forms the image of an object at its
(a) cornea.
(b) iris.
(c) pupil.
(d) retina.

Q7: 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.

Page 3

CH11: The Human Eye and the Colourful World
(NCERT Q & A)

Dispersion of Light
Which light bends the most?

Q1: What is meant by power of accommodation of the eye?

Answer: Power of accommodation is the ability of the eye lens to focus near and far objects clearly on
the retina by adjusting its focal length. Power of accommodation of the eye is limited. It implies the
focal length of the eye lens cannot be reduced beyond certain minimum limit.

Q2: A person cannot see the objects distinctly, when placed at a distance less than 50 cm.
(a) Identify the defect of vision.
(b) Give two reasons for this defect.
(c) Calculate the power and nature of the lens he should be using to see clearly the object
placed at a distance of 25 cm from his eyes.
(d) Draw the ray diagrams for the defective and the corrected eye.

Answer: (a) Defect of vision = Hypermetropia or Long-sightedness

(b) Reasons for the defect are:
? Curvature of lens or retina becomes less than normal increases focal length.
? Ciliary muscles become stiff
? Shortening of eye ball due to which distance between retina and and lens reduces and image
is formed beyond retina.
(c) Given, u = -25 cm, v = -50cm, f = ?
Applying lens formula, 1/f = 1/v - 1/u
?1/f = -1/50 - (-1/25) = -1/50 + 1/25 = (-1+2) /50 = 1/50
? f = 50cm = 0.5m
? Power of the lens = 1/f = 1/0.5m = +2D
? A convex lens having power of 2D can be used to correct the vision.

(d) ray diagrams for the defective and the corrected eye.

Q3: 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?

Answer: The distance of far point x = 1.2m
To view distant objects, concave lens of focal length -1.2 should be used.
i.e. f = -x = -1.2m
Power of lens = 1/f = 1/(-1.2m) = -0.83D

Q4: Name the part of the human eye that helps in changing the focal length of the eye lens.

(Note: Pupil regulates and controls the amount of light entering the eye.)

Q5: 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.

Q6: The human eye forms the image of an object at its
(a) cornea.
(b) iris.
(c) pupil.
(d) retina.

Q7: 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.

Q8: 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.

Q9: What is the far point and near point of the human eye with normal vision?

Answer: The near point of the eye is the minimum distance of the object from the eye, which
can be seen clearly. For a normal human eye, this distance is 25cm.

The far point of the eye is the maximum distance upto which the eye can see the objects clearly
without strain. The far point of the normal human eye is infinity.

Q10: A boy uses spectacles of focal length – 60 cm. Name the defect of vision he is suffering
from. Which lens is used for the correction of this defect? Compute the power of this lens.

Answer: Given f = -60 cm =-0.60m
The -ve sign of focal length indicates that the lens is concave. ? the boy suffers from myopic (short
sightedness) vision.
Power of the lens = 1/f(in m) = 1/(-0.60) = -1.67 D

Q11: 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?

Answer: The student has difficulty in seeing distant objects. He might be suffering from myopia. The
defect can be corrected using a concave lens of appropriate focal length.

Myopic Eye Correction using Concave lens

Page 4

CH11: The Human Eye and the Colourful World
(NCERT Q & A)

Dispersion of Light
Which light bends the most?

Q1: What is meant by power of accommodation of the eye?

Answer: Power of accommodation is the ability of the eye lens to focus near and far objects clearly on
the retina by adjusting its focal length. Power of accommodation of the eye is limited. It implies the
focal length of the eye lens cannot be reduced beyond certain minimum limit.

Q2: A person cannot see the objects distinctly, when placed at a distance less than 50 cm.
(a) Identify the defect of vision.
(b) Give two reasons for this defect.
(c) Calculate the power and nature of the lens he should be using to see clearly the object
placed at a distance of 25 cm from his eyes.
(d) Draw the ray diagrams for the defective and the corrected eye.

Answer: (a) Defect of vision = Hypermetropia or Long-sightedness

(b) Reasons for the defect are:
? Curvature of lens or retina becomes less than normal increases focal length.
? Ciliary muscles become stiff
? Shortening of eye ball due to which distance between retina and and lens reduces and image
is formed beyond retina.
(c) Given, u = -25 cm, v = -50cm, f = ?
Applying lens formula, 1/f = 1/v - 1/u
?1/f = -1/50 - (-1/25) = -1/50 + 1/25 = (-1+2) /50 = 1/50
? f = 50cm = 0.5m
? Power of the lens = 1/f = 1/0.5m = +2D
? A convex lens having power of 2D can be used to correct the vision.

(d) ray diagrams for the defective and the corrected eye.

Q3: 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?

Answer: The distance of far point x = 1.2m
To view distant objects, concave lens of focal length -1.2 should be used.
i.e. f = -x = -1.2m
Power of lens = 1/f = 1/(-1.2m) = -0.83D

Q4: Name the part of the human eye that helps in changing the focal length of the eye lens.

(Note: Pupil regulates and controls the amount of light entering the eye.)

Q5: 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.

Q6: The human eye forms the image of an object at its
(a) cornea.
(b) iris.
(c) pupil.
(d) retina.

Q7: 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.

Q8: 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.

Q9: What is the far point and near point of the human eye with normal vision?

Answer: The near point of the eye is the minimum distance of the object from the eye, which
can be seen clearly. For a normal human eye, this distance is 25cm.

The far point of the eye is the maximum distance upto which the eye can see the objects clearly
without strain. The far point of the normal human eye is infinity.

Q10: A boy uses spectacles of focal length – 60 cm. Name the defect of vision he is suffering
from. Which lens is used for the correction of this defect? Compute the power of this lens.

Answer: Given f = -60 cm =-0.60m
The -ve sign of focal length indicates that the lens is concave. ? the boy suffers from myopic (short
sightedness) vision.
Power of the lens = 1/f(in m) = 1/(-0.60) = -1.67 D

Q11: 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?

Answer: The student has difficulty in seeing distant objects. He might be suffering from myopia. The
defect can be corrected using a concave lens of appropriate focal length.

Myopic Eye Correction using Concave lens

Q12: Why do we have two eyes instead of one?

? field of view is more with two eyes than one eye.
? two eyes gives us three dimensional vision (stereo vision) of an object.

Q13: 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?

(i) Distant vision: Given focal length (f) = ?, Power (P) = -5.5D
P = 1/f ? f = 1/P = 1/(-5.5)m = -100/55cm = -18.2 cm

(ii) For near vision, P = +1.5 D
? f = 1/P = 1/1.5m = 1000/15cm = 66.7 cm

Q14: Define dispersion of white light and name the colours of white light in order?

Answer: The splitting of light into its component colours is called dispersion. White light splits into
seven colours (VIBGYOR) i.e. violet, indigo, blue, green, yellow, orange and red. Isaac Newton was
the first to use a glass prism to obtain the spectrum of sunlight.

Q15: 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: Given, distance of far point (x) = 80 cm, P = ?

To view distant objects correctly, focal length of the corrective lens = f = -x = -80cm

Power of the lens (P = 1/(f in metres) = 1/(-0.80) = 100/-80 = -1.25D
The lens is concave of powed -1.25D

Q16: 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: For diagram see question 2(b) above.
Given v = -1m = -100cm
u = -25cm
Using lens formula, 1/v -/1u = 1/f
? -1/100 + 1/25 = 1/f
? f = 33.3 cm - 0.333m

Power of the lens = 1/f = 100/33.3 = 3D

Q17: Why is a normal eye not able to see clearly the objects placed closer than 25 cm?

Answer: Since the focal length of eye lens cannot be decreased beyond certain minimum length.

Q18: List four common defects which can be corrected with the use of spectacles.

Answer: Four common defects that can be corrected by using spectacles:
Page 5

CH11: The Human Eye and the Colourful World
(NCERT Q & A)

Dispersion of Light
Which light bends the most?

Q1: What is meant by power of accommodation of the eye?

Answer: Power of accommodation is the ability of the eye lens to focus near and far objects clearly on
the retina by adjusting its focal length. Power of accommodation of the eye is limited. It implies the
focal length of the eye lens cannot be reduced beyond certain minimum limit.

Q2: A person cannot see the objects distinctly, when placed at a distance less than 50 cm.
(a) Identify the defect of vision.
(b) Give two reasons for this defect.
(c) Calculate the power and nature of the lens he should be using to see clearly the object
placed at a distance of 25 cm from his eyes.
(d) Draw the ray diagrams for the defective and the corrected eye.

Answer: (a) Defect of vision = Hypermetropia or Long-sightedness

(b) Reasons for the defect are:
? Curvature of lens or retina becomes less than normal increases focal length.
? Ciliary muscles become stiff
? Shortening of eye ball due to which distance between retina and and lens reduces and image
is formed beyond retina.
(c) Given, u = -25 cm, v = -50cm, f = ?
Applying lens formula, 1/f = 1/v - 1/u
?1/f = -1/50 - (-1/25) = -1/50 + 1/25 = (-1+2) /50 = 1/50
? f = 50cm = 0.5m
? Power of the lens = 1/f = 1/0.5m = +2D
? A convex lens having power of 2D can be used to correct the vision.

(d) ray diagrams for the defective and the corrected eye.

Q3: 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?

Answer: The distance of far point x = 1.2m
To view distant objects, concave lens of focal length -1.2 should be used.
i.e. f = -x = -1.2m
Power of lens = 1/f = 1/(-1.2m) = -0.83D

Q4: Name the part of the human eye that helps in changing the focal length of the eye lens.

(Note: Pupil regulates and controls the amount of light entering the eye.)

Q5: 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.

Q6: The human eye forms the image of an object at its
(a) cornea.
(b) iris.
(c) pupil.
(d) retina.

Q7: 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.

Q8: 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.

Q9: What is the far point and near point of the human eye with normal vision?

Answer: The near point of the eye is the minimum distance of the object from the eye, which
can be seen clearly. For a normal human eye, this distance is 25cm.

The far point of the eye is the maximum distance upto which the eye can see the objects clearly
without strain. The far point of the normal human eye is infinity.

Q10: A boy uses spectacles of focal length – 60 cm. Name the defect of vision he is suffering
from. Which lens is used for the correction of this defect? Compute the power of this lens.

Answer: Given f = -60 cm =-0.60m
The -ve sign of focal length indicates that the lens is concave. ? the boy suffers from myopic (short
sightedness) vision.
Power of the lens = 1/f(in m) = 1/(-0.60) = -1.67 D

Q11: 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?

Answer: The student has difficulty in seeing distant objects. He might be suffering from myopia. The
defect can be corrected using a concave lens of appropriate focal length.

Myopic Eye Correction using Concave lens

Q12: Why do we have two eyes instead of one?

? field of view is more with two eyes than one eye.
? two eyes gives us three dimensional vision (stereo vision) of an object.

Q13: 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?

(i) Distant vision: Given focal length (f) = ?, Power (P) = -5.5D
P = 1/f ? f = 1/P = 1/(-5.5)m = -100/55cm = -18.2 cm

(ii) For near vision, P = +1.5 D
? f = 1/P = 1/1.5m = 1000/15cm = 66.7 cm

Q14: Define dispersion of white light and name the colours of white light in order?

Answer: The splitting of light into its component colours is called dispersion. White light splits into
seven colours (VIBGYOR) i.e. violet, indigo, blue, green, yellow, orange and red. Isaac Newton was
the first to use a glass prism to obtain the spectrum of sunlight.

Q15: 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: Given, distance of far point (x) = 80 cm, P = ?

To view distant objects correctly, focal length of the corrective lens = f = -x = -80cm

Power of the lens (P = 1/(f in metres) = 1/(-0.80) = 100/-80 = -1.25D
The lens is concave of powed -1.25D

Q16: 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: For diagram see question 2(b) above.
Given v = -1m = -100cm
u = -25cm
Using lens formula, 1/v -/1u = 1/f
? -1/100 + 1/25 = 1/f
? f = 33.3 cm - 0.333m

Power of the lens = 1/f = 100/33.3 = 3D

Q17: Why is a normal eye not able to see clearly the objects placed closer than 25 cm?

Answer: Since the focal length of eye lens cannot be decreased beyond certain minimum length.

Q18: List four common defects which can be corrected with the use of spectacles.

Answer: Four common defects that can be corrected by using spectacles:
1. Hypermetropia
2. Myopia
3. Astigmatism
4. Presbyopia

Q19: What happens to the image distance in the eye when we increase the distance of an
object from the eye?

Answer: The size of the eye can change, so the image distance is fixed. When we increase the
distance of an object from the eye, the image distance in the eye does not change. Due to power of
accommodation of the eye, focal length of the eye lens is changed which compensates the increase
in object distance. Hence image distance remains fixed and image is formed on the retina of the eye.

Q20: Why do stars twinkle?
OR
Q: Explain why do stars twinkle and planets do not?

Answer: The twinkling of a star is due to the earth's atmospheric refraction. The earth's atmosphere is
moving and it consists of pockets of warm and cold air. Thus the atmosphere has variations in
refractive indices of air. Stars are very far away from the earth and emit their own light. Being very far
from the earth, these are considered point sources. When a star light enters the earth's atmosphere, it
undergoes multiple refractions and bends continuously towards the normal till it enters our eyes. Its
apparent position appears higher than the normal one. Due to mobility of air and variation is
temperature, this apparent position of star is not steady and moves continuously giving a twinkling
effect.
When in space (outside the earth's atmosphere) stars do not appear twinkle. Planets as compared to
stars are closer to the earth and appear bigger (we cannot consider them point sized like stars). It
nullifies the twinkling effect.

Q21: Explain why the planets do not twinkle.

Answer: The distanced between planets and the earth is less as compared to stars. Planets cannot
be considered as point sources. The apparent shift in their position due to their position cannot be
observed because they subtend greater angle at the eye. Comparatively larger size, their brightness
do not change and do not give twinkle effect.

Q22: What is Mirage? How does it occur?
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