Class 10 Science Chapter 11 Question Answer

CBSE Class 10 Science Chapter 11 Question Answer

Science is one of the most important subjects students study in school. In every aspect of our life, we need science. It provides a logical explanation of everything. In this chapter, students will learn about an important organ of the human body and the significant property of light.

Chapter 11 is about the eye and the refraction of light. The eye is one of the major sense  organs of human life, and it helps us to visualise the world. Students will study different parts of the human eye, how it works, and common diseases related to eyesight. On the other hand, students will also study about refraction of light through a prism. Both topics are related as they are associated with light.

Extramarks is one the  leading educational platforms  that provides important study materials related to CBSE exam pattern. . Our subject experts made the Important Questions Class 10 Science Chapter 11, collected and compiled the questions from different sources, and provided the answers. Thus, it will help students to prepare well and  score better marks  in exams.

Students can download the study materials after registering on the official website of Extramarks. You will find the CBSE syllabus, CBSE sample papers, CBSE extra questions, CBSE revision notes, NCERT books, NCERT exemplars, NCERT solutions, NCERT important questions, CBSE past years’ question papers, vital formulas and much  more.Students need to sign up at Extramarks to enjoy the maximum benefits of these resources and stay ahead of the competition.

CBSE Class 10 Science Chapter 11 Question Answer

CBSE Class 10 Science Important Questions are also available for the following chapters:

Important Questions Class 10 Science Chapter 11 with Solutions

The subject experts of Extramarks have collected the questions from different sources and provided the answers. They have taken help from the textbook exercises, CBSE past years’ question papers, CBSE sample papers and important reference books. They have further solved the questions, and experienced professionals have further checked the answers to ensure the best quality  content is available for the students. Thus, the Important Questions Class 10 Science Chapter 11 will help students to understand the subject matter and boost their confidence considerably.  

Some of the  important questions are given below-

Question 1. Why does the sky appear dark instead of blue to an astronaut?

Answer 1: The sky appears  dark as no light scattering occurs due to the absence of atmosphere at such great  heights.

Question 2. There is no twinkling of  planets. Explain 

Answer 2: The planets are closer to the earth than the stars. They can be treated like a collection of a large number of point-sized light sources. There are varying conditions of the atmosphere due to which the darkest part of the twinkling effect from one light source is overlapped by the focussed light from the point source of the planet’s other region. There is a constant amount of light entering the eye. Because of this reason, the planets do not appear to twinkle, and they appear steady.

Question 3. Why is a normal eye unable to see objects when placed closer than 25 cm?

Answer 3: This is due to the focal length, which cannot be reduced below a certain limit.

Question 4: Multiple choice questions:

• Nothing can be seen through the fog. Why is it so? .

• The fog has a high refractive index.

• Total reflection is suffered by light at droplets.

• The fog absorbs light.

• Droplets scatter the light.

Answer: (d) droplets scatter the light

Explanation:

 Because the  light is scattered by the droplets,  nothing can be seen through the fog.

• ________ is called the light’s deflection by minute particles and the atmosphere’s molecules in all the directions

• Tyndall effect

• Scattering

• Interface

• Dispersion

Answer : (b) scattering

Explanation:

The above-mentioned phenomenon is called scattering 

• What is the nature of the image formed by the retina?

• Virtual and erect

• Real and inverted

• Virtual and inverted

• Real and erect

Answer: (b) real and inverted

Explanation:

The nature of the eye lens is convex. Therefore, the image’s character formed will be real and inverted.

• When a person is not able to see the object beyond 2 m distinctly, then this defect is corrected by the lens of power:

• + 0.5 D

• – 0.5 D

• + 0.2 D

• – 0.2 D

Answer: (b) – 0.5 D

Explanation:

P = 1/f = 1/-2 = -0.5 D

As the condition mentioned, the person has myopia, and it is corrected by using the concave lens of power -0.5 D.

• Bifocal lenses are used to correct which condition is mentioned below.

• Astigmatism

• Coma

• Myopia

• Presbyopia

Answer: (d)

Explanation:

Presbyopia is corrected by the use of the bifocal lens. Bifocal lenses have an upper and lower point. The upper point has a concave lens used for distant vision, while the lower point contains a convex lens responsible for near vision.

• There are danger signals of red colour installed at the top of the tall buildings. These can be easily seen among all the colours, even from a distance. The reason is 

• The red light is scattered by fog or smoke

• The red light scattering is least by smoke or fog

• There is red light absorbed most by the smoke or the fog

• The red light moves the fastest in the air.

Answer: (b) the red light’s scattering is least by smoke or fog.

Explanation:

The longer wavelength of red light is responsible for this phenomenon and is least scattered by smoke or fog.

• Why is the deep sea bluish in colour?

• Algae and plants present in the sea are responsible for this.

• There is a  reflection of sky  in water

• Light scattering

• Sea absorbs the light.

Answer: ( c) light scattering

Explanation:

The water molecules in the sea are finer, so they cause scattering of light. Also, blue light has a shorter wavelength.

• The clear sky is blue because

• The atmosphere absorbs the blue light.

• There is the absorption of ultraviolet radiation in the atmosphere.

• There is a scattering of violet and blue light more by the atmosphere when compared to the other colours.

• Light of all other colours is scattered more than the violet and blue light by the atmosphere.

Answer : ( c) there is a scattering of violet and blue light more by the atmosphere when compared to the other colours.

Explanation: 

The short wavelength of blue and violet colours is responsible for their scattering more than the light of other colours by the molecules in the atmosphere.

• When a ray of light enters our eyes, then most of the refraction occurs in which part:

• Crystalline lens

• Outer surface of cornea

• Iris

• Pupil

Answer: (b)

Explanation:

The most refraction occurs at the outer surface of the cornea so this region acts as a primary lens converging in nature.

• There is an increase in the focal length of the eye lens when eye muscles

• are relaxed and the lens becomes thinner.

• contract and the lens becomes thicker.

• are relaxed and the lens becomes thicker.

• contract and the lens becomes thinner.

Answer: (a)

Explanation:

The curvature of the eye lens is modified by the ciliary muscles. Eye lenses become thinner when the eye muscles are relaxed and when there is an increase in the focal length of the eye lens.

Question 5. Explain the structure and function of the eye. Why are we able to see nearby as well as distant objects?

Answer 5: 

The parts of the human eye are:

• Cornea is the thin membrane through which the light enters the eye. It forms a transparent bulge on the eyeball’s front surface, and most of the refraction occurs on the outer surface of the cornea.
• Aqueous humour is a viscous liquid which is present between the eye lens and cornea.
• Iris is present behind the cornea and it controls the size of the pupil.
• Pupil is that part of our eye which controls the amount of light entering it.
• Ciliary muscles are responsible for holding the eye lens and also help in the adjustment of the focal length.
• Eye lenses are convex and made up of transparent, soft and flexible tissue. It forms a real and inverted image on the retina.
• Vitreous humour supports the back of the eye and it is a viscous fluid present between the eye lens and retina.
• Retina is responsible for capturing light and converting it into electric signals which are further translated by the brain into images. In other words, we can say that it is the light-sensitive screen on which the image is formed.
• Rods and cones are the light-sensitive cells which are present in the retina. Rods respond to the light’s intensity and cones respond to the colours.

We are able to see both the nearby and the distant objects because: :

• The eye lens becomes thin when the ciliary muscles are relaxed. The focal length of the lens increases and has a maximum value which is equal to the distance from the retina. The parallel rays which are coming from the distant object while entering the eyes are focussed on the retina. Because of this reason, the vision to see distant objects is clear.
• The ciliary muscles are strained or they get contracted when the eyes see the nearby object. This increases the eye lens curvature and it becomes thicker. There is a decrease in focal length with an increase in converging power. The sharp image of the nearby object is formed on the retina. This makes the vision of the nearby object clear.

This phenomenon is called the power of accommodation of the eyes.

Question 6. What is the Tyndall effect?

Answer 6: Tyndall effect is the phenomenon of light scattering by the colloidal particle.

Question 7. Why is the sun’s colour different during sunrise/ sunset and noon? Explain your answer with a reason.

Answer 7: The reason for this colour change is that the sun’s rays pass through a maximum length of the atmosphere during sunrise or sunset. There is a scattering of blue light and a shorter wavelength. The only colour which reaches the observer’s eyes is red. This is the only reason the sun appears red at sunrise and sunset. The distance to be travelled is the least at  noon. . The sun appears  white as all the wavelengths are equally scattered.

Question 8. Explain the range of vision of the normal eye.

Answer 8: Accommodation means the ability of one’s eye lens to adjust its focal length or the converging power to get a clear view of the object. This is the accommodation.

Power of accommodation is the maximum variation in the power of the lens so that the far-off and the nearby objects are viewed clearly.

The nearest point of the eye in which the object can be clearly seen without any strain to the eye is called the near point and is 25 cm for a normal eye.

A far point is defined as the farthest point up to which the object can be seen clearly which is infinity for a normal eye.

Question 9. Explain myopia with causes and correction.

Answer 9: Myopia is also called near-sightedness. It is defined as the inability of our eyes in viewing long  distant objects. The far point of the eye is less than infinity and the resulting image is formed in front of the retina.

It is caused due to the excessive curvature of the eye lens and the elongation of the eyeball.

Myopia can be easily corrected by use of a concave lens which diverges the image and shifts the image to the retina.

Question 10. Explain hypermetropia. What are the causes and how can they be corrected?

Answer 10: Hypermetropia is also known as far-sightedness. It is the inability of the eye in viewing nearby objects. The near point of our eye is more than 25 cm and the image is formed behind the retina.

Hypermetropia occurs due to the low converging power of our eye lens. The size of the eyeball is smaller.

It is corrected by use of a convex lens. This convex lens converges and also shifts the image to the retina from beyond.

Question 11. Why does a myopic person remove spectacles while reading and a hypermetropic person remove spectacles while looking at the sky?

Answer 11: Spectacles are not required by a myopic person while reading the book, as the near point is 25cm. If the person reads the book with the concave lens, the book needs to be kept at a distance greater than 25 cm. so the concave lens forms the image at 25 cm, and the size of the book also appears smaller than the actual size. So the person will prefer to read the book without spectacles.

The hypermetropic person does not need spectacles to see the distant object as the far point is at infinity. If the person uses a convex lens to see the object at a distance, the image is formed before the retina due to the increase in the converging power, and the person will not be able to see the distant object. So the person prefers to remove the spectacles to look at the sky.

Question 12. A corrective lens of power “-2D” is used by a person suffering from a vision defect, given the nature and focal length of the corrective lens. Find the nature and focal length of the corrective lens.

Answer 12: 

Focal length = 100/P = 100/-2 = -50 cm

Question 13. Why does the power to see the object nearby, as well as the far object, diminishes with age?

Name the defect of the eyes which arises from this condition. Elaborate on the cause and mention the correction of this eye defect.

Answer 13: With advancing age, there is a gradual weakening of ciliary muscles which results in decreasing flexibility of the eye lens. So the power of the eye reduces to see nearby and far-off objects.

The defect of the eye in this condition is called presbyopia. In this condition, the lens becomes more myopic and hypermetropic. The causes are due to the deterioration of the ciliary muscles and the slow diminishing flexibility of the eye lens. The condition can be easily corrected by using the bi-focal lens.

Question 14. In the figure below, a beam of light passes through a glass prism. Trace the path of the beam and name the phenomena with an explanation. What is the conclusion drawn from the white light constituents?

Answer 14:

The phenomenon in which the white light splits into constituents is known as the dispersion of light. It is caused by the different constituent’s colour of light which gets formed due to the different refractive indices of the prism material.

The rainbow which is formed is an example of this phenomenon, as the dispersion of the white sunlight causes it into the constituent colours. 

The conclusion which can be drawn from the white light dispersion into the constituent colour is

• There are seven colours present in the white light.
• The maximum deviation is shown by the violet light
• The minimum deviation is shown by the red light.

Question 15. Explain the spectrum in detail. Support the answer by demonstrating an activity with the help of the diagram.

Answer 15: 

The band of different coloured components of any light beam is known as the spectrum. When the components of white light are combined by placing another identical prism in an inverted position as compared to the first prism, then the white light on passing through the first spectrum splits. The splitting is into the colours of the spectrum. Then this spectrum passes through the second identical prism which is in an inverted position as compared to the first prism, and then a beam of white light is obtained.

Question 16. What is dispersion? Explain the dispersion of white light by the glass prism.

Answer 16: 

The splitting of the white light into the seven constituent colours by refraction is known as the dispersion of the white light.

When a beam of white light enters a prism, it gets refracted and splits into the seven constituent colours. This occurs to the different bending angles for each colour on the splitting of light. Each ray of colour when passing through the prism bends at different angles with respect to the incident beam; it gives rise to the spectrum.

Question 17. The star appears higher than the actual position. Explain with the diagram.

Answer 17: This is because the starlight, when it enters the earth’s atmosphere, it undergoes refraction continuously in a medium where the refractive index gradually keeps changing  before it reaches the earth. 

Since the atmosphere bends straight towards the normal, the star seems slightly higher as compared to its actual position.

Question 18. A glass prism produces a spectrum when the white light passes through it but the glass slab cannot produce it. Why?

Answer 18: This is because in a glass prism when white light passes through the prism, each of the constituent wavelengths of light undergoes a different extension of deviation. This results in the dispersion of the white light. In a similar way, when the light enters through parallel sides of the glass slab, all the constituent wavelengths of white light will not undergo any deviation. So , the white light does not split into the constituent spectrum.

Question 19. Explain the concept of advanced sunrise and delayed sunset.

Answer 19: Two minutes before the actual sunrise, the sun is visible to us and 2 minutes after the actual sunset. This is because of atmospheric refraction. The actual sunrise means the actual crossing of the horizon by the sun. The actual position and the sun’s apparent position with respect to the horizon are given in the figure below. The time difference between the actual and apparent sunset is 2 minutes. The apparent flattening of the sun’s disc at sunrise and sunset is also due to the same phenomena.

Question 20. Explain the formation of the rainbow with diagrams.

Answer 20:

A rainbow is a natural spectrum caused by the dispersion of sunlight by tiny water droplets which are present in the atmosphere.

After a rain shower, the sunlight is dispersed. The small water droplet here plays the role of small prisms. They disperse and refract the incident sunlight. Then they refract it internally and it is finally retracted again when it comes out of the raindrop. Different colours reach the observer’s light due to the dispersion of light and the internal reflection.

Question 21. Why the extent of deviation of a light ray through a prism depends upon the colour.

Answer 21: This is because the refractive index of glass for different colours is different. It depends upon the wavelength of a particular light.

Question 22. Why is the red light used in the danger signals?

Question 22. The red light is used in the danger signals because it has the maximum value in the entire spectrum. Red light’s penetration power in the air is maximum of all and due to this reason, it can be seen from a further distance. So this colour is used to depict the danger.

Question 23. A student is not able to clearly see the letters written on the blackboard and he is sitting at the back of the classroom. What advice would a doctor give him?

Answer 23: The student is suffering from short-sightedness. He has to wear a concave lens to correct the vision as advised by the doctor. 

Question 24. A lens of power -4.5 D is required by a person to correct his vision. State the defect of vision he is suffering from, calculate the focal length and name the corrective lens.

Answer 24: The vision defect is myopia or short-sightedness. 

Focal length = 1/power = -100/4.5 = -22.2 cm.

The person has to use a concave lens to correct the vision.

Question 25. What is the bifocal lens?

Answer 25: Bifocal lenses are used to see clearly both nearby and distant objects. The upper part of the lens is concave and used for distant vision and the lower part is a convex lens to facilitate near vision. This lens can be used to correct both myopia and hypermetropia.

Question 26. Use the diagram to explain the angle of deviation.

Answer 26: The incident ray and the emergent ray are not parallel to each other. The angle between the emergent ray and the incident ray is called the angle of deviation.

Question 27. What is a recombination of the spectrum?

Answer 27:Recombination of the spectrum means the seven coloured lights of the spectrum can be recombined to give back white light by placing two prisms, one upside down.

Question 28. Explain the concept of scattering of light.

Answer 28: The phenomenon of change in the direction of light propagation is known as a scattering of light. The concept here is that very fine particles scatter the blue colour. Large-sized particles scatter light of longer wavelengths. The shorter the wavelength, the greater the scattering will be. The effects of scattering are the Tyndall effect, the blue colour of the sky, the white colour of clouds and the reddening of the sun at sunrise and sunset.

Question 29. What is atmospheric refraction?

Answer 29. Atmospheric refraction is defined as the refraction of light caused by the earth’s atmosphere due to the variation in the optical densities of air layers. The effect of atmospheric refraction is the twinkling of stars, the position of the stars that seem higher than their actual position, advanced sunrise and the delayed sunset.

Question 30. What is the cure for corneal impairment?

Answer 30.   The cure of corneal impairment is possible with the replacement of the objective cornea with the cornea of the donated eye. We should explain the importance of cornea donation to the community. The eyes can live even after death by helping the needy person. The human eye is a sensitive organ and can be donated to make another person’s life beautiful.

Question 31. What is the importance of eye donation advertisements?

Answer 31. The advertisement for eye donation is important on television. In newspapers, this is because very few people know that eye transplantation is possible by which blind people may see the colourful world. We should encourage the people to donate eyes and spread awareness so that we can help in improving the quality of life of those people who need it. 

Question 32. Answer the following questions:

1. Which eye part has the delicate membrane, and what type of light-sensitive cells are present?
2. Why is a person advised to wear a convex lens?
3. What happens to the pupil in dim light and bright light?
4. Name the defect of vision when the power of the eye is too long.
5. Name the defect of the vision when the focal length of the eye lens is too long.

Answer 32.

1. Retina is a delicate membrane, and it contains a large number of photosensitive cells, which are called the rods and the cones.
2. A person is suffering from hypermetropia.
3. The size of the pupil increases in dim light and decreases in bright light.
4. The vision defect is short-sightedness.
5. The vision defect is long-sightedness.

Question 33. Multiple choice questions:

Question 1: The human eye focuses at different distances and also adjusts the focal length. This is due to

• Presbyopia

• Accommodation

• Near-sightedness
• Far-sightedness

Answer 1: (b) accommodation

 Question 2.  The human eye form image at

• Retina

• Cornea

• Iris

• Pupil

Answer 2: (a) retina

Question 3: The change in focal length of the eye lens is due to 

• Pupil

• Retina

• Ciliary muscles

• Iris

Answer 3: (c ) ciliary muscles

Question 4: The muscular diaphragm of the eyes is

• Cornea

• Ciliary muscles

• Iris

• Retina

Answer 4: c) iris

Explanation:

Iris controls the size of the pupil.

Question 5: __________ is the black opening between the aqueous humour and the lens.

• Retina

• Iris

• Cornea

• Pupil

Answer 5 : d) Pupil

Question 6:The defective eye has a near point of 0.5 m and point of 3m. for the purpose of reading and seeing, the power of  corrective lens are required is

• 0.5 D and +3D

• +2D and -1/3 D

• -2D and +1/3 D

• 0.5 D and -0.3 D

Answer 6: (b) +2D and -1/3 D

Explanation:

For the purpose of reading,

U = -25 cm, v = 0.5 m = -50cm, f = ? , P = ?

1/f = 1/v – 1/u = 1-50 – 1-25 = 150

P = 100/f = 100 × 1/50 = +2D

For distant object,

u= infinity, v = -3, f = ?, P = ?

1/f = 1-3 – 1 = -1/3

P = 1/f = -1/3 D

Question 7: When the white light enters the prism, it splits into colours. This is due to 

• Different wavelength is related to the different refractive index for each colour

• The same velocity is exhibited by the same colour in the prism

• High density is exhibited by the prism material

• Light scattering

Answer 7: (a)

Explanation:

Dispersion takes place because the material’s refractive index for prism varies with varying wavelength.

Question 8: The air layer of the atmosphere whose temperature is less than the hot layer behaves optically as

• Denser medium

• Rare medium

• Inactive medium

• Either dense or rare medium

Answer 8: (a)

Explanation:

The atmosphere’s cold layer air acts as an optically denser medium than the hot air because the molecules are packed together closely.

Question 34. State true or false:

1. There are no rods and cones present at the junction of the optic nerve and the retina in the eye.

2.No light is reflected in the pupil.

3. The rainbow is formed when the sun is shining and it is raining at the same time.
4. A rainbow is a man-made spectrum of the sunlight in the sky.
5. Hypermetropia is corrected using a concave lens.
6. Lens of higher focal length have less power.
7. 28 frames are projected per second in motion picture
8. Myopia is due to the eyeball being short.
9. The focal length of the eye lens can be reduced to a certain limit.
10. Red light is most scattered in fog and smoke.

Answer 34:

1. True
2. True
3. True
4. False
5. False. Myopia is corrected by the concave lens and the convex lens is used to correct hypermetropia.
6. True 
7. False 
8. False 
9. False. A normal eye cannot see objects clearly which are placed closer than 25 cm because the focal length of the eye lens cannot be reduced to a certain limit.
10. False. The red colour is that component of white light which is scattered least by the fog.

Question 35. Fill in the blanks:

• ________ is the small area of the retina which is insensitive to the light where the optic nerve leaves the eye.

• The pupil is _______ in colour.

• Excessive curvature in cornea cause _______

• Power of accommodation of normal eye is _______

• The ________carries signals to the brain.

• Very fine particles scatter more of _______colour

• Red light is scattered _______, so it is used for signals

• Bi-focal lens are used to correct _______

• Hypermetropic eye is corrected using _______ lens.

• When light falls at a critical angle on the surface of a rarer medium while coming from the denser medium, the refracting angle is ________

• The white light dispersion happens because white light colour at different _______through the light prism.

• ________are sensitive to the bright light.

1. 1. _______are sensitive to the dim light.

• _________makes the path of the light visible.

• ___________ is the phenomenon due to which we get light from the sun before sunrise.

• _________ shows that the sunlight can be of different colours.

• ________ change the focal length of the eye lens.

• _____________ is the thin membrane which allows light to enter the eye.

• ________ is the other name for old age hypermetropia

• The far point of the person suffering from myopia is _______

• Near and far point of a young person’s normal eye are respectively ______and ______

Answer 35:

1. Blind spot is the small area of the retina which is insensitive to the light where the optic nerve leaves the eye.
2. The pupil is black in colour.
3. Excessive curvature in the cornea causes myopia.
4. The power of accommodation of the normal eye is 4 Dioptre.
5. The optical nerve carries signals to the brain.
6. Very fine particles scatter more of blue colour
7. Red light is scattered less so it is used for signals
8. Bi-focal lenses are used to correct presbyopia
9. The hypermetropic eye is corrected using a convex lens.
10. When light falls at a critical angle on the surface of a rarer medium while coming from the denser medium, the refracting angle is 90 degrees.
11. white light dispersion happens because white light colours at different speeds through the light prism.
12. Cone cells are sensitive to bright light.
13. Rod cells are sensitive to dim light.
14. Tyndall effect makes the path of the light visible.
15. Atmospheric refraction is the phenomenon in which we get light from the sun before sunrise.
16. Rainbow shows that the sunlight can be of different colours.
17. Ciliary muscles change the focal length of the eye lens.
18. Cornea is the thin membrane which allows light to enter the eye.
19. Presbyopia is the other name for old age hypermetropia
20. The far point of the person suffering from myopia is 150 cm.
21. Near and far points of a young person’s normal eye are respectively 25 cm and infinity

Question 36. The near point of the normal eye is 25 cm. The near point of the hypermetropic eye is 1 m. What is the power required to correct the defect?

Answer 36:

In order to correct the defect, the image of an object at 25 cm should be brought at 100 cm.

1/f = 1/v – 1/u = 1-100 – 1-25 = -1/100 + 1/25 = 3/100

F = +100/3 = +33.3 cm

P = 100/33.3 = 3.0D

Question 37. A person needs the power of the lens is  -5.5 dioptres for the correction of distant vision. For correcting his near vision, he needs a lens of power +1.5 dioptre. Calculate the focal length for the distant and the near vision.

Answer 37:

For the distant vision, focal length can be calculated as,

Focal length = 1/power = 100/-5.5 = -18 cm approximately

For the near vision, to calculate the focal length,

= 100/1.5 = 66.66 cm

Question 38. Answer the following questions:

• Give the nature of the eye lens of humans.

• What is the role of the optic nerve in the human eye?.

• Is light dispersed by all the transparent bodies?

• Give one phenomenon of the Tyndall effect.

• When we come out of the darkroom, why can we not see things?

• What is a cataract?

• Mention the relation between the focal length of the eye lens and the ciliary muscles.

Answer 38:

1. The nature of the eye lens is convex. The image formed on the retina is inverted, real and diminished.
2. The optic nerve transmits visual information through the electrical signals generated at the retina to the brain.
3. No, this is not possible for all the transparent bodies because bodies with parallel surfaces do not disperse  light.
4. The scattering of a light beam by a medium containing microscopic suspended particles—for example, smoke or dust in a room—making a light beam entering a window visible.
5. In dim light, our iris expands the pupil for it to allow more light to enter our eyes. So when one comes out of a dark and unlit room into the bright sunlight, a large amount of light enters the eyes, and things are not seen clearly due to the glare. Then gradually, there is a pupil contraction by the iris to allow less light to enter the eye to see the object clearly. It will take some time for the pupil, so because of this reason, in a shorter time interval, the person is unable to see things.
6. When the crystalline lens of the eye becomes opaque or hazy due to the formation of a thin membrane over it, there are chances that there may be a partial or complete loss of vision. This eye defect is known as a cataract. Cataract surgery is performed to restore the functioning of the eye.
7. The focal length of the eye lens will be more when the ciliary muscles of the normal eye are the most relaxed. When the ciliary muscles are normally relaxed, the eye lens will become thin, so there is an increase in the focal length. 

When the ciliary muscles are in the most contracted state, the radius of curvature of the eye lens increases. The lens becomes thicker, so there is a decrease in the focal length of the eye lens.

Question 39. A person is able to see objects clearly when they are lying at distances between 50 cm and 300 cm from the eyes. Which type of vision defect is the person suffering from?

Answer 39: In the given question, it is mentioned that a person is suffering from both myopia and hypermetropia so it can be said that the person is suffering from presbyopia.

Benefits of Solving Important Questions Class 10 Science Chapter 11

Practice is very important for students. It will help students to retain the information through frequent revision  and help them score excellent grades  in exams. The textbook exercises have limited questions, and students often need to go beyond  the textbooks. So, they must take help from other sources to enhance and strengthen their ideas. The subject experts of Extramarks  are aware of this  and have made the Important Questions Class 10 Science Chapter 11 to help students to practise in less time and be thorough with their syllabus. . The benefits of solving the questions series are manifold-

• The experts of Extramarks have collected important questions from different sources. They have taken help from the textbook exercises, CBSE sample papers, NCERT Exemplars and important reference books. Apart from this, they have included questions from CBSE past years’ question papers so that students may have an idea regarding possible types of questions in exams. Thus, the Class 10 Science Chapter 11 Important Questions will help them practise each and every question regularly and thus help them to excel  in exams. 
• The experts have not only collected the questions, but they have provided the answers too. They have given detailed answers to the questions and provided diagrams when needed. Thus, they have explained the subject matter beautifully to students so they can understand it quickly. Experienced professionals have further checked the answers to ensure the best quality  content is available to students . Thus, Important Questions Class 10 Science Chapter 11 will guide the students to clear their doubts and build concepts more easily. It will generate their interest in the subject matter and boost their confidence during  the exams.
• Practice is the  key to success. It helps students in numerous ways. The more one practices, the better one gets, and it implies greatly to  studies as well. . Students often appear apprehensive about  the subject simply because they need help in understanding it or maybe they don’t have the right study materials. Practising with Extramarks study material  will help them to clarify their doubts and understand the subject. The textbook exercises have limited questions, and the subject matter experts of Extramarks have made the Science Class 10 Chapter 11 Important Questions provide students with a sufficient number of questions. If they solve the questions regularly, they will better understand the subject and be confident.

Extramarks is one of the  leading educational platforms which   provides all the important study materials related to CBSE guidelines and follows the NCERT textbooks.  Students may  register on our official website and download the study materials. We provide CBSE syllabus, CBSE sample papers, CBSE extra questions, CBSE revision notes, CBSE past years’ question papers, NCERT books, NCERT exemplars, NCERT solutions, NCERT important questions, vital formulas and  more. Like the Chapter 11 Class 10 Science Important Questions, you will also find important questions for other chapters. Links to the study materials are given below-