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

 Page 1


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. 
 
Answer: Ciliary muscles. 
(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. 
 
Answer: (b) accommodation. 
 
Q6: The human eye forms the image of an object at its 
(a) cornea. 
(b) iris. 
(c) pupil. 
(d) retina. 
 
Answer: (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. 
 
Answer: 25cm  
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. 
 
Answer: Ciliary muscles. 
(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. 
 
Answer: (b) accommodation. 
 
Q6: The human eye forms the image of an object at its 
(a) cornea. 
(b) iris. 
(c) pupil. 
(d) retina. 
 
Answer: (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. 
 
Answer: 25cm  
 
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.  
 
Answer: (c) ciliary muscles. 
 
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. 
 
Answer: Ciliary muscles. 
(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. 
 
Answer: (b) accommodation. 
 
Q6: The human eye forms the image of an object at its 
(a) cornea. 
(b) iris. 
(c) pupil. 
(d) retina. 
 
Answer: (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. 
 
Answer: 25cm  
 
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.  
 
Answer: (c) ciliary muscles. 
 
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? 
 
Answer: Having two eyes instead of one is advantageous for the following: 
? 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? 
 
Answer:  
(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. 
 
Answer: Ciliary muscles. 
(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. 
 
Answer: (b) accommodation. 
 
Q6: The human eye forms the image of an object at its 
(a) cornea. 
(b) iris. 
(c) pupil. 
(d) retina. 
 
Answer: (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. 
 
Answer: 25cm  
 
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.  
 
Answer: (c) ciliary muscles. 
 
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? 
 
Answer: Having two eyes instead of one is advantageous for the following: 
? 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? 
 
Answer:  
(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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