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Light Reflection and Refraction Class 10 Notes Science Chapter 10

January 22, 2024 by Sastry CBSE

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CBSE Class 10 Science Notes Chapter 10 Light Reflection and Refraction Pdf free download is part of Class 10 Science Notes for Quick Revision. Here we have given NCERT Class 10 Science Notes Chapter 10 Light Reflection and Refraction.

According to new CBSE Exam Pattern,  MCQ Questions for Class 10 Science pdf  Carries 20 Marks.

CBSE Class 10 Science Notes Chapter 10 Light Reflection and Refraction

Reflection of Light: The phenomenon of bouncing back of light into the same medium by the smooth surface is called reflection.

Incident light: Light which falls on the surface is called incident light.

Reflected light: Light which goes back after reflection is called reflected light.

The angle of incidence: The angle between the incident ray and the normal.

An angle of reflection: The angle between the reflected ray and the normal.

Mirror: The surface which can reflect the light is a mirror.

Plane Mirror: If the reflecting surface is a plane then the mirror is plane.

Spherical Mirror: If the reflecting surface is part of the hollow sphere then the mirror is a spherical mirror. The spherical mirror is of two types:

• Convex mirror: In this mirror reflecting surface is convex. It diverges the light so it is also called a diverging mirror.
• Concave mirror: In this mirror reflecting surface is concave. It converges the light so it is also called converging mirror.

Parameters of Mirror:

• Center of Curvature: The centre of hollow sphere of which mirror is a part.
• The radius of curvature: The radius of hollow sphere of which mirror is a part.
• Pole: The centre of mirror (middle point) is pole.
• Principal axis: The line joining the pole and center of curvature is called principal axis.
• Aperture: Size of mirror is called aperture of mirror.
• Principal Focus: The point on the principal axis, where all the incident rays parallel to principal axis converge or diverge after reflection through mirror.
• Focal Length: The distance between pole and focus point is focal length.

Special Rays for Formation of Image:

• A ray of light which is parallel to the principal axis of a spherical mirror, after reflection converges or diverges from focus.
• A ray of light passing through or appearing from the center of curvature of spherical mirror is reflected back along the same path.
• A ray of light passing through or appearing from the focus of spherical mirror becomes parallel to the principal axis.
• A ray of light which is incident at the pole of a spherical mirror is reflected back making same angle with principal axis.

Use of Concave Mirror: It is used as a makeup mirror, the reflector in torches, in headlights of cars and searchlights, doctor’s head-mirrors, solar furnace, etc.

Sign Conventions of Spherical Mirror

• All the distances are measured from the pole of the mirror as the origin.
• Distances measured in the direction of incident rays are taken as positive.
• Distances measured opposite to the direction of incident rays are taken as negative.
• Distances measured upward and perpendicular to the principal axis are taken as positive.
• Distances measured downward and perpendicular to the principal axis are taken as negative. \(\frac { 1 }{ f } =\frac { 1 }{ v } +\frac { 1 }{ u }\) …where f, v and u are focal length, image distance, object distance

Linear Magnification: This is the ratio of the height of the image to the height of the object. \(m=\frac { { h }^{ ‘ } }{ h }\) …where m = magnification, h = height of image, h’ = height of object

Use of Convex Mirror: Convex mirror used as rear view mirror in vehicles, as shop security mirrors, etc.

Refraction of Light: The bending of light at the interface of two different mediums is called Refraction of light.

• If the velocity of light in medium is more, then medium is called optical rarer. Example, air or vacuum is more optical rarer.
• If the velocity of light in medium is less, then medium is called optical denser. Example, glass is more denser than air.

Refractive Index: It represents the amount or extent of bending of light when it passes from one medium to another. There are two types of refractive index

• Relative refractive index and
• Absolute refractive index.

Refractive index of medium with respect to other medium is called Relative Refractive Index. Refractive index of medium 1 with respect to medium 2 = \(\frac { Speed\quad of\quad light\quad in\quad medium\quad 2(V2) }{ Speed\quad of\quad light\quad in\quad medium\quad 1(V1) }\)

Refractive index of medium with respect to air or vacuum is called Absolute Refractive Index. Absolute refractive index of medium (m) = \(\frac { Speed of light in air(c) }{ Speed of light in medium (Vm) }\)

Incident ray: It is incoming ray on the refracting surface.

Refracted ray: It is an outgoing ray from the refracting surface.

An angle of incidence (i): It is the angle between incident rays and perpendicular line (normal) at the point of incidence.

An angle of refraction (r): It is the angle between refracted rays and perpendicular line (normal) at the point of incidence.

Law of Refraction: According to this law

• “The incident ray, refracted ray and normal at the point of incidence all lie in the same plane.”
• “The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant.” \(\frac { sin\quad i }{ sin\quad r }\) = constant (µ)

Lens: The transparent refracting medium bounded by two surfaces in which at least one surface is curved is called lens. Lenses are mainly two type

• Convex lens and
• Concave lens.

Center of Curvature: The centres of two spheres, of which lens is part is called the centre of curvature.

Radii of Curvature: The radii of spheres, of which lens is part is called radius of curvature.

Principal Axis: The line joining the centres of curvature of two surfaces of lens is called principal axis.

Optical Center: It is a special point on the principal axis. Light incident on the optical centre passes through the lens without deviation.

Principal Focus: The point on the principal axis at which all incident rays parallel to the principal axis converge or appear to diverge after refraction through the lens.

Special Rays for Image Formation by Lens:

• An incident ray, parallel to the principal axis, after refraction passes through (or appears to come from), second focus of the lens.
• An incident ray, passing through the optical center of the lens, goes undeviated from the lens.
• An incident ray, passing through the (first) principal focus of the lens, or directed toward it, becomes parallel to the principal axis after refraction through lens.

Use of Lens: In photographic cameras, magnifying glass, microscope, telescope, the human eye.

1 . Light travels in a straight line.

2. Light gets reflected when it falls on polished surfaces; like mirrors.

3. Light suffers refraction when it travels from one medium to another.

4. There is a change in the wavelengths!light when it moves from one medium into another.

5. The bouncing back of light when it strikes a smooth or polished surface is called reflection of light. Reflection is of two types; Specular or regular and Diffuse or irregular reflection.

6. The angle of incidence is equal to the angle of reflection. Mathematically, we have ∠i = ∠r.

7. The image is as far behind the mirror as the object is in front.

8. The image is unmagnified, virtual and erect.

9. The image has right-left reversal.

10. Focal length of a plane mirror is infinity.

11. Power of a plane mirror is zero.

12. If a plane mirror is turned by an angle, the reflected ray turns by 2θ.

13. The least size of a plane mirror to view an object is equal to half the size of the object.

14. Pole (Vertex): The central point of a mirror is called its pole.

15. Centre of curvature : The centre of the sphere of which the mirror is a part is called the centre of curvature. It is denoted by C.

16. Radius of curvature : The radius of the sphere of which the mirror is a part is called the radius of curvature. It is denoted by R.

17. Principal axis : The straight line passing through the pole and the centre of curvature of the mirror is called the principal axis.

18. Principal focus : It is a point on the principal axis at which the rays parallel to the principal axis meet after reflection or seem to come from. For a concave mirror, the focus lies in front of the mirror and for a convex mirror, it lies behind the mirror. In short, a concave mirror has a real focus while aconvex mirror has a virtual focus.

19. Focal plane : A plane, drawn perpendicular to the principal axis and passing through the principal focus.

20. Focal length : The distance between the pole and the focus is called the focal length. It is represented by f. The focal length is half the radius of curvature.

21. Aperture: The size of the mirror is called its aperture. It is also defined as the effective diameter of the light reflecting area of the mirror.

22. Real image : When the rays of light, after reflection from a mirror, actually meet at a point, then the image formed by these rays is said to be real. Real images can be obtained on a screen.

23. Virtual image: When the rays of light, after reflection from a mirror, appear to meet at a point, then the image formed by these rays is said to be virtual. Virtual images can’t be obtained on a screen.

24. The following rays are used while drawing ray diagrams to find the position of an image :

• A ray of light parallel to the principal axis after reflection passes through the focus. (1)
• A ray of light passing through the focus after reflection becomes parallel to the principal axis. (2)

25. For mirrors, the following results hold : u is – ve, if the object is in front of the mirror. (Real object) u is + ve, if the object is behind the mirror. (Virtual object) v is – ve, if the image is in front of the mirror. (Real image) vis +ve, if the image is behind the mirror. (Virtual image) Focal length of a concave mirror is taken as – ve. Focal length of a convex mirror is taken as +ve.

26. When the image formed by a spherical mirror is real, it is also inverted and is on the same side of the mirror as the object. Since both v and u are negative, the magnification is negative.

27. When the image formed by a spherical mirror is virtual, it is also erect and is on the other side of the mirror as the object. In this case, u is – ve and v is + ve , therefore, m is positive.

28. The expression for the mirror formula is 1/u+1/v = 1/f

30. If m is positive, the image is erect w.r.t the object and if m is negative, the image is inverted w.r.t. the object.

32. The bending of light when it travels from one medium into another is called refraction of light

34. As light travels from ,one medium to another, the frequency of light does not change.

35. Light refracts because it has different speeds in different media.

36. The refraction of light obeys the following two laws :

• The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.

38. The most familiar and widely used optical device is the lens. A lens is an optical system with two refracting surfaces. The simplest lens has two spherical surfaces close enough together that we can neglect the distance between them. Such a lens is called a thin lens. The two common types of lenses are Converging lens or Convex lens, Diverging lens or Concave lens.

39. It should be noted that, if the above lenses are surrounded by .a material with a refractive index greater than that of the lens, the convex lens gets converted into a concave lens and vice-versa.

40. Any lens that is thicker at its centre than at its edges is a converging lens with positive f, and any lens that is thicker at its edges than at the centre is a diverging lens with negative f.

42.Principal axis: Since the lens contains two spherical surfaces, therefore, it has two centres of curvatures. The line joining these centres and passing through the optical centre is called principal axis.

43. Aperture: The effective width of a lens through which refraction takes place is called the aperture.

44. Focus and Focal Length : If a beam of light moving parallel to the principal axis of a convex lens is incident on it, the rays converge or meet at a point on the principal axis. This point F is called the focus. The distance CF is called the focal length. If a beam of light moving parallel to the principal axis is incident on a concave lens, the beam of light diverges. If these diverged rays are produced backward, they meet at a point F on the principal . axis. The transmitted rays appear to come from this point. This point F is called the focus and distance CF is called the focal length.

• All rays parallel to the principal axis after refraction pass through the principal focus or seem to come from it.
• A ray of light passing through the focus after refraction becomes parallel to the principal axis.
• A ray of light passing through the optical centre of the lens after refraction passes undeviated.

46.  A convex and a concave lens can be supposed to be made-up of prisms.

49. New Cartesian sign conventions :

• All distances, object distance (u), image distance (v) and focal length f are measured from the optical centre.
• The distances measured in the direction of incident ray are taken as positive and distances measured against the direction of incident ray are taken as negative.
• All distances (heights) of objects and images above principal axis are taken as positive and those below the principal axis are taken as negative.

50. For the two lenses, the sign conventions take the form

• u is- ve, if the object is in front of the lens. (Real object)
• u is +ve, if the object is virtual.
• v is – ve, if the image is on the same side as that of the object. (Virtual image )
• v is +ve, if the image is real.
• Focal length of a concave lens is taken as – ve.
• Focal length of a convex lens is taken as +ve.

51. Lens formula for convex lens 1/v-1/u = 1/f

53. If the magnification of a lens is negative, then the image formed is inverted and real.

54. If the magnification of a lens is positive, then the image formed is erect and virtual.

55. Power is defined as the reciprocal of the focal length. Power is measured in dioptre.

NCERT Notes for Class 10 Science

• Chapter 1 Chemical Reactions and Equations Class 10 Notes
• Chapter 2 Acids Bases and Salts Class 10 Notes
• Chapter 3 Metals and Non-metals Class 10 Notes
• Chapter 4 Carbon and its Compounds Class 10 Notes
• Chapter 5 Periodic Classification of Elements Class 10 Notes
• Chapter 6 Life Processes Class 10 Notes
• Chapter 7 Control and Coordination Class 10 Notes
• Chapter 8 How do Organisms Reproduce Class 10 Notes
• Chapter 9 Heredity and Evolution Class 10 Notes
• Chapter 10 Light Reflection and Refraction Class 10 Notes
• Chapter 11 Human Eye and Colourful World Class 10 Notes
• Chapter 12 Electricity Class 10 Notes
• Chapter 13 Magnetic Effects of Electric Current Class 10 Notes
• Chapter 14 Sources of Energy Class 10 Notes
• Chapter 15 Our Environment Class 10 Notes
• Chapter 16 Management of Natural Resources Class 10 Notes

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NCERT Solutions for Class 10 Science Chapter 10 The Human Eye and the Colourful World

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• Chapter 11 Human Eye And Colourful World

NCERT Class 10 Science Chapter 10 Question and Answers - FREE PDF Download

Vedantu provides NCERT Solutions for class 10 science chapter 10, a key resource for students who want to navigate the concepts of Science at the 10th-grade level. Students will learn about the concepts related to this chapter, such as the power of accommodation, defects of vision, refraction of light, Atmospheric refraction, scattering of light, etc.

Download the FREE PDF of Class 10 NCERT Solutions for The Human Eye and The Colourful World, which are prepared by Vedantu master teachers. Solutions are prepared according to the CBSE Science class 10 syllabus and exam pattern.

Glance on NCERT Solutions Chapter 10 Science Class 10 Human Eye and Colourful World

Human eye class 10 NCERT Solutions emphasise the understanding of the human eye working along with different functions performed by the human eye.

It covers a thorough explanation of the human eye parts, along with its structural depiction.

In human eye class 10 questions and answers, the phrases least distance of distinct vision and cataract are introduced.

Students will learn about visual problems such as myopia, hypermetropia, and presbyopia and how to repair them. These three flaws are explained using well-labelled graphics.

Class 10 science ch 10 NCERT solutions go through the many elements of the human eye, as well as their properties and functions.

The cornea, pupil, optic nerve, retina, iris, eye lens, and ciliary muscles are all part of it. Students can practise human eye diagrams, visual abnormalities, and their repair.

Class 10 answers the human eye and the colourful world question, which teaches students why different light colours bend at different angles as they travel through a prism.

Chapter Human Eye and Colourful World is based on the structure of the human eye, the functioning of each eye part, and methods to treat vision defects. In the annual examination, this unit holds a weightage of 4 marks.

Access NCERT Solutions For Class 10 Science Chapter 10 The Human Eye and The Colourful World

1. What is meant by the power of accommodation of the eye?

Ans: When the ciliary muscles are relaxed, the eye lens becomes thin. This results in the increase of the focal length, and distant objects are clearly visible to the eyes. 

To see the nearby objects clearly, the ciliary muscles contract which makes the eye lens thicker. This results in reduction of focal length of the eye lens, and the nearby objects become visible to the eyes.

Therefore, the focal length can be adjusted by the human eye lens to view both distant and nearby objects on the retina. This ability of the eye refers to the power of accommodation of the eye.

2. A person with a myopic eye cannot see objects beyond \[1.2\] m distinctly. What should be the type of corrective lens used to restore proper vision?

Ans: The person can clearly see the objects which are near, but he cannot see the objects beyond \[1.2m\] . This arises as the image of an object beyond \[1.2m\] is obtained in front of the retina and not at the retina, as displayed in the figure given below.

To correct this defect of vision, a concave lens should be used. The image will be brought back to the retina by the concave lens as shown in the given figure.

3. What is the far point and near point of the human eye with normal vision?

Ans: The near point of the eye refers to the minimum distance of the object from the eye that can be seen clearly without any strain to the eye. This distance is \[25\] cm, for a normal human eye.

The far point of the eye refers to the maximum distance to which the eye can clearly see the objects, without getting any strain to the eye. The far point of the normal human eye is at an infinite distance.

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: A student has difficulty in reading the blackboard while sitting in the last row. It means that he is unable to see distant objects clearly.

The student is suffering from myopia. This defect can be corrected using a concave lens.

1. 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

Ans: The correct answer is (b) accommodation, 

The focal length of the eye lens is changed by the human eye to see objects situated at various distances from the eye. This is possible due to the power of accommodation of the eye lens.

2. The human eye forms the image of an object at its

Ans: (d) The image of an object is formed by the human eye at its retina.

3. The least distance of distinct vision for a young adult with normal vision is about

b) \[2.5cm\]

c) \[25cm\]

d) \[2.5m\]

Ans: (c) The least distance of distinct vision refers to the minimum distance to see a clear and distinct image. This distance is \[25cm\] for a young adult with normal vision.

4. The change in focal length of an eye lens is caused by the action of the

c) Ciliary muscles

Ans: (c) The curvature of the eye lens is changed by the relaxation or contraction of ciliary muscles.

The focal length of the eyes is changed due to the change in curvature of the eye lens. Therefore, the change in focal length of an eye lens occurs due to the action of ciliary muscles.

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 

(a) distant vision?

Ans:  The power \[P\]of a lens of focal length \[f\] is given by the relation

\[P=\frac{1}{f}\]  (\[f\]in metres)

Power of the lens used for correcting distant vision \[=-5.5D\]

Focal length of the required lens, \[f=\frac{1}{P}\]

\[\Rightarrow f=\frac{1}{-5.5}=-0.181m\]

The focal length of the lens for correcting distant vision is \[-0.181\]m.

(b) Near vision? 

Ans : Power of the lens used for correcting near vision \[=+1.5D\] 

\[\Rightarrow f=\frac{1}{1.5}=+0.667\]

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?

Ans: The person is suffering from an eye defect called myopia. In this defect, the image of an object is formed in front of the retina. Hence, a concave lens is used to correct this defect of vision.

Object distance, \[u=\,infinity=\infty \]

Image distance, \[v=\,-80cm\]

Focal length \[=f\]

According to the lens formula,

\[\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\]

\[\Rightarrow -\frac{1}{80}-\frac{1}{\infty }=\frac{1}{f}\]

\[\Rightarrow \frac{1}{f}=-\frac{1}{80}\]

\[\Rightarrow f=-80cm\] or \[-0.8m\]

Power of the lens and the focus can be given as:

\[P=\frac{1}{f}\] ,( \[f\]is in metres)

\[\Rightarrow P=\frac{1}{-0.8}=-1.25D\]

Thus, a concave lens of power \[-1.25D\] is required by the person to correct his defect.

7. 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 25cm.

Ans: A person who suffers from hypermetropia can see distinct objects clearly but faces difficulty in seeing nearby objects clearly. This happens because the eye lens focuses the incoming divergent rays beyond the retina.

This defect of vision can be fixed using a convex lens. A convex lens of a suitable power converges the incoming light in such a manner that the image is formed on the retina, as shown in the given figure.

The person will be able to clearly see the object kept at 25cm (near point of the

normal eye), if the image of the object is formed at his near point, which is given as \[1m\].

Object distance, \[u=-25cm\] 

Image distance, \[v=-1m=-100cm\]

Focal length, \[f\]

Using the lens formula, 

\[\Rightarrow -\frac{1}{100}-\frac{1}{-25}=\frac{1}{f}\]

\[\Rightarrow \frac{1}{f}=\frac{1}{25}-\frac{1}{100}\]

\[\Rightarrow \frac{1}{f}=\frac{4-1}{100}\]

\[\Rightarrow f=\frac{100}{3}cm=33.33cm=0.33m\]

Power, 

\[P=\frac{1}{f}\] (\[f\] is in meters)

\[\Rightarrow P=\frac{1}{0.33}\]

\[\Rightarrow P=+3D\]

Thus, a convex lens of power +3.0 D is needed to rectify the defect.

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

Ans: The size of eyes cannot increase or decrease because of which the image distance remains constant.

When the object’s distance from the eye is increased, the image distance in the eye does not change. The increase in the object distance is balanced by the change in the focal length of the eye lens. The focal length of the eye changes in such a manner that the image is always formed at the retina of the eye.

10. Why do stars twinkle?

Ans: Stars emit their own light and they shine due to the atmospheric refraction of light.

Stars are situated very far away from the earth. Therefore, they are considered as point sources of light. When the light coming from stars enters the earth’s atmosphere, it gets refracted at different levels because of the variation in air density present at different levels of the atmosphere. 

When the star light refracted by the atmosphere comes more towards the surface of earth, it appears brighter than when it comes less towards the earth. Clearly, it appears as if the stars are twinkling at night.

11. Explain why the planets do not twinkle?

Ans: Planets do not twinkle because they appear larger in size than the stars as they are comparatively closer to earth. Planets are a collection of many point-sized sources of light. The various regions of these planets produce either brighter or dimmer effect in such a way that the resultant of brighter and dimmer effect is zero.

Clearly, the twinkling effects of the planets are nullified and that is why they do not twinkle.

12. Why does the Sun appear reddish early in the morning?

Ans: During sunrise, the light rays from the sun travel a greater distance in the earth’s atmosphere before reaching our eyes. In this journey, the shorter wavelengths of lights are spread out and only the longer wavelengths reach our eyes. 

Since blue colour has a short wavelength and red colour has a long wavelength, red colour can reach our eyes after the atmospheric scattering of light. Thus, the sun appears to be reddish early in the morning.

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

Ans: Since there is no atmosphere in outer space that can scatter the sunlight, the sky appears dark instead of blue to an astronaut. 

Since the sunlight is not scattered in space, no scattered light reaches the astronaut’s eyes and the sky appears black to them instead of blue.

Topics Covered In Class 10 Science Chapter 10: The Human Eye and The Colourful World

Benefits of the Human Eye Class 10 NCERT Solutions

Class 10 Science ch 10 NCERT solutions help students to appear for the Class 10 Science exam with a better plan by keeping all the vital points in mind.  Some of the benefits of considering NCERT Solutions for Class 10 Chapter 10 include:

After analysing the human eye class 10 questions and answers , students can design their study material in a well-structured way. They get new ideas to shape their study materials.

T he human eye and the colourful world question answers are comprehensive and prepared by a team of Science experts. 

C lass 10, the human eye and the colourful world question answers It covers a thorough explanation of the human eye parts and its structural depiction. 

H uman eye class 10 NCERT Solutions emphasise the understanding of the human eye working along with different functions performed by the human eye. 

In human eye class 10 questions and answers, the phrases least distance of distinct vision and cataract are introduced. 

The students can finish their homework confidently once they follow Chapter 10 Science Class 10 NCERT Solutions.

The solutions are designed after a lot of research on the particular topic of the human eye and the colourful world class 10 , and thus help students to build a strong foundation on every topic.

Study Materials For Class 10 Science Chapter 10

Class 10 science ch 10 NCERT solutions by Vedantu are designed with the main aim of helping students focus on the chapter's important concepts. Every question answer is explained briefly and in an easy-to-understand language to help students improve their conceptual knowledge. The solutions also contain various shortcut techniques which can be used to remember the concepts effectively. Students can refer to the solutions after reading the complete chapter to improve their understanding.

FREE PDF Links for Other Chapter-wise NCERT Solutions Class 10 Science

You can also access chapter-wise NCERT Solutions for Class 10 Science from the links below and kick-start your preparation for Class 10 Board exams.

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FAQs on NCERT Solutions for Class 10 Science Chapter 10 The Human Eye and the Colourful World

1. Explain Defects of Vision and How to Correct Them from class 10 science chapter 10 question answers.

Myopia, hypermetropia, and presbyopia are three significant defects of vision. Myopia refers to nearsightedness due to which a person can see nearby objects. A person who has myopia cannot see far objects clearly. This defect could be resolved with the help of a concave lens. In hypermetropia, a person cannot see near objects clearly and is corrected by using a convex lens. In presbyopia, a person is difficult to see both nearby and far objects. It is treated with the use of bifocal lenses.

2. What are the Atmospheric Refraction Phenomena mentioned in the human eye and the colourful World Class 10?

The deflection of light in the atmosphere refers to the phenomena of atmospheric refraction. The change in the direction of propagation of radiation in traversing the atmosphere is known as atmospheric radiation. This refraction is due to light velocity through air. With the increased density, it decreases. It is responsible for various phenomena like delayed sunset, the twinkling of stars, advanced sunrise, and more. The concept of atmospheric refraction is crucial to understand other phenomena occurring in nature.

3. Where can I get the NCERT Solutions chapter 10 science class 10?

NCERT Solutions for Chapter 10 of Class 10 Science have been offered by various subject experts through the Vedantu app and website. These solutions have been specifically framed in a simple and easy-to-understand language so that the students do not face any difficulty in understanding the concepts being explained in the answers. Solutions for Chapter 10 of Class 10 Science NCERT can be accessed in PDF form free of cost from Vedantu.

4. What are the topics covered in the NCERT Solutions for class 10 the human eye and the colourful world question answers?

Chapter 10 of Class 10 Science NCERT is called “The Human Eye and Colourful World”. As the name of the chapter suggests, it talks about the workings of the human eye in detail. Topics covered in the chapter include different parts of the eye, their functions, and their structural depiction. Concepts like the accommodation of the human eye, visionary defects, and their correction are also a part of this chapter. The teaching process also includes various diagrams for various topics.

5. How are NCERT Solutions helpful for the human eye and the colourful world question answers?

Grasping all the concepts and being able to use and solve equations appropriately is important to ace Class 10 Science Chapter 10. Hence, referring to NCERT Solutions for Class 10 Science Chapter 10 can provide students with the required help in understanding how to answer questions that can be particularly difficult. These solutions are useful during self-study and students can use them to practice regularly for questions involving equations.

6. What are the important topics in NCERT Solutions for Chapter 10 of Class 10 Science?

Chapter 10: The Human Eye and The Colourful World consists of the following important topics:

Human Eye Structure

Scattering of Light

Defects in Vision and Corrections

Atmospheric Refractions

7. How do you study for class 10 science chapter 10 question answers?

By following the given below points, one can study Chapter 10 of Class 10 Science.

Make a list of the subtopics. For instance, if there are five subtopics, and you need one day for each, make sure you book five days for this chapter.

Make short and easy-to-read notes. 

Go through NCERT Solutions. 

Go through the diagrams and draw the important ones. 

Revise at regular intervals.

8. Mention the topics covered in human eye class 10 NCERT solutions.

The topics covered are:

Defects of Vision

Refraction of Light

Dispersion of Light

Atmospheric Refraction

Scattering of light

NCERT Solutions for Class 10 Science

Ncert solutions for class 10.

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RD Sharma Solutions , RS Aggarwal Solutions and NCERT Solutions

NCERT Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction

July 13, 2020 by Veerendra

These Solutions are part of NCERT Solutions for Class 10 Science . Here we have given NCERT Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction. Learn Insta provides you the Free PDF download of NCERT Solutions for Class 10 Science (Physics) Chapter 10 – Light Reflection and Refraction solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 10 – Light Reflection and Refraction Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

NCERT Questions

In Text Book Questions

Question 1. Define the principal focus of a concave mirror. Answer: A point on the principal axis where the parallel rays of light after reflecting from a concave mirror meet.

More Resources

• NCERT Solutions for Class 10 Science
• NCERT Exemplar Solutions for Class 10 Science
• HOTS Questions for Class 10 Science
• Value Based Questions in Science for Class 10
• Previous Year Question Papers for CBSE Class 10 Science

Question 3. Name a mirror that can give an erect and magnified image of an object. Answer: A concave mirror.

Question 4. Why do we prefer a convex mirror as a rear-view mirror in vehicles ? [CBSE (All India) 2007, 2011, 2012] Answer: This is because a convex mirror forms an erect and diminished (small in size) images of the objects behind the vehicle and hence the field of view behind the vehicle is increased.

Question 5. Find the focal length of a convex mirror whose radius of curvature is 32 cm. Answer: R = +32 cm. Therefore, f = R/2 = +32/2 = +16 cm. Thus, focal length of the convex mirror = +16 cm.

Question 6. A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located ? Answer: m – -3, But m = -v/u, so v = 3u u = -10 cm v = 3 (-10 cm) =-30 cm Thus, the image is located at a distance of 30 cm to the left side of the concave mirror.

Question 7. A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal ? Why ? Answer: The ray of light bends towards the normal because the speed of light decreases when it goes from air (rarer medium) into water (denser medium).

Question 9. You are given kerosene, turpentine and water. In which of these does the light travel faster ? Answer: We know, v = c/n. Refractive index (n) of water is 1.333, whereas refractive index of kerosene is 1.44 and that of turpentine is 1.47. As refractive index of water is least, so speed of light in water is more than in kerosene and turpentine. Hence, light travels faster in water.

Question 11. Define 1 dioptre of power of a lens. Answer: Power = I/f (in m). Power of a lens is 1 dioptre if focal length of the lens is 1 metre or 100 cm.

NCERT Chapter End Exercises

Question 1. Which one of the following materials cannot be tised to make a lens 1 (a) water (b) glass (c) plastic (d) clay. Answer: (d). This is because clay is opaque (i.e. light cannot pass through it).

Question 2. The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object ? (a) between the principal focus and the centre of curvature (b) at the centre of curvature (c) beyond the centre of curvature (d) between the pole of the mirror and its principal focus. Answer: (d).

Question 3. Where should an object be placed in front of a convex lens to get a real image of the size of the object ? (Bihar Board 2012) (a) at the principal focus of the lens (b) at twice the focal length (c) at infinity (d) between the optical centre of the lens and its principal focus. Answer: (b).

Question 4. A spherical mirror and a thin spherical lens have each a focal length of — 15 cm. The mirror and the lens are likely to be (a) both are concave (b) both are convex (c) the mirror is concave and the lens is convex (d) the mirror is convex but the lens is concave. Answer: (a).

Question 5. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be (a) plane only (b) concave only (c) convex only (d) either plane or convex. Answer: (d).

Question 6. Which of the following lenses would you prefer to use while reading small letters found in a dictionary ? (a) a convex lens of focal length 50 cm (b) a concave lens of focal length 50 cm (c) a convex lens of focal length 5 cm (d) a concave lens of focal length 5 cm. Answer: (c). Magnifying power of a reading glass (Convex lens) = 1/f.

Question 7. We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror ? What is the nature of the image ? Is the image larger or smaller than the object ? Draw a ray diagram to show the image formation in this case. Answer: A concave mirror produces an erect image if the object is placed between the pole and the focus of the concave mirror. Thus, object may be placed at any position whose distance is less than 15 cm from the concave mirror. The image is virtual and erect. The image is larger than the object. For a ray diagram, see figure 24.

Question 8. Name the type of mirror used in the following situations : (a) head lights of a car (b) side rear view mirror of a vehicle (c) solar furnace. Support your answer with reason. (CBSE 2012, 2013) Answer: (a) Concave mirror. When a bulb is placed at the focus of a concave mirror, then the beam of light from the bulb after reflection from the concave mirror goes as a parallel beam which lights up the front road. (b) Convex mirror. Image formed by a convex mirror is erect and small in size. The field of view behind the vehicle is large. (c) Concave mirror. Concave mirror focuses rays of light coming from the sun at its focus. So, the temperature at the focus is raised.

Practical Skills Based Questions (2 Marks)

Get 100 percent accurate NCERT Solutions for Class 10 Science Chapter 10 (Light Reflection and Refraction) explained by expert Science teachers. We provide step by step solutions for the questions given in Class 10 Science textbook as per CBSE Board guidelines from the latest NCERT book for Class 10 Science. The topics and sub-topics in  Chapter 10 Light Reflection and Refraction

• 10.1 REFLECTION OF LIGHT
• 10.2 SPHERICAL MIRRORS
• 10.2.1 Image Formation by Spherical Mirrors
• 10.2.2 Representation of Images Formed by Spherical Mirrors Using Ray Diagrams
• 10.2.3 Sign Convention for Reflection by Spherical Mirrors
• 10.2.4 Mirror Formula and Magnification
• 10.3 REFRACTION OF LIGHT
• 10.3.1 Refraction through a Rectangular Glass Slab
• 10.3.2 The Refractive Index
• 10.3.3 Refraction by Spherical Lenses
• 10.3.4 Image Formation by Lenses
• 10.3.5 Image Formation in Lenses Using Ray Diagrams
• 10.3.6 Sign Convention for Spherical Lenses
• 10.3.7 Lens Formula and Magnification
• 10.3.8 Power of a Lens.

We cover all exercises in the chapter given below:-

• EXERCISE 10.1 – 4 Questions with Solutions
• EXERCISE 10.2 – 2 Questions with Solutions
• EXERCISE 10.3 – 5 Questions with Solutions
• EXERCISE 10.4 – 3 Questions with Solutions
• EXERCISE 10.5 – 17 Questions with Solutions.

Download the free PDF of Chapter 10 Light Reflection and Refraction or save the solution images and take the print out to keep it handy for your exam preparation.

Hope given NCERT Solutions for Class 10 Science Chapter 10 are helpful to complete your science homework.

If you have any doubts, please comment below. Learn Insta try to provide online science tutoring for you.

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Light Reflection and Refraction Class 10 Numericals

Welcome to this page on Numericals on Light for Class 10 Science chapter 10. These questions will help you improve your ranks with the great collection of Numericals on Light Class 10 of 3 marks questions. These light reflection and refraction class 10 questions and answers are from various topics and formulas like

• Mirror Formula
• Lense Formula
• Refractive index etc.
• Light Class 10 Numericals

Answer 1. This occurs due to the, phenomenon of refraction of light.  Here,  the ray of light from the coin travels from a denser medium to a rarer medium . In this process  it bends away from the normal . The point from which the refracted rays appear to come gives the apparent position of the coin. As the rays appear to come from a point above the coin, so, the coin seems to be raised. 

Answer 2.   Given object size = 5 cm object distance from lens $u = -30$ cm focal length $f = 20$ cm,  We have to find $v = ?$ Using the lens formula $\frac{1}{f} = \frac{1}{v}-\frac{1}{u}$ We have, $ \frac{1}{v}  =  \frac{1}{u}  +  \frac{1}{f}  =  \frac{1}{-30} +  \frac{1}{20}  = \frac{ -2+3 }{ 60 }  =  \frac{1}{60}$   or, $\frac{1}{v}   =  \frac{1}{60}$ Thus  $v = 60cm$ . This is part (i) of the question magnification  $\frac{v}{u} = \frac{60}{-30} = -2 $  $Magnification \quad =  \quad \frac{Image  \quad  size}{Object  \quad size}$ $\frac{h_i }{h_o}=-2 $ $\frac{h_i}{5}=-2$ $h_{i}= -2 \times 5 = -10 \quad cm$ The image is real inverted and magnified.

Answer 3. (1) New Cartesian sign conventions for measuring the various distances in the ray diagrams (reflection by spherical mirrors) 1. All the distances in a ray diagram are measured from the pole of the spherical mirror. 2. The distances measured in the direction of incident light are taken as positive. 3. The distances measured in the direction opposite to the direction of incident light are taken as negative. 4. The heights measured upwards and perpendiculars to the principal axis of the mirror are taken as positive. 5. The heights measured downwards and perpendiculars to the principal axis of the mirror are taken as negative. Answer 3.  (2) Visit page https://physicscatalyst.com/Class10/light-reflection-and-refraction.php#spherical-mirrors

Answer 4 The following activity can be used to determine the principal focus and approximate focal length of a concave mirror.  (1) Place the mirror vertically on a table.  (2) Place an object (a pencil) in front of the mirror and move it towards and away from the mirror.  (3) Find the approximate spot where the reflection of the object changes from upright to upside down. This point is the focus of the mirror. Measure the distance between this point and the mirror. This length is the focal length of the mirror. 

Answer 5 Given that  $h_1 = 4.5 \quad cm,\quad u = - 12 cm, \quad f = 15 cm$ We have to find $ v = ?$ using the equation of the mirror formula $\frac{1}{f} = \frac{1}{v}+\frac{1}{u}$ we have, $\frac{1}{15} = \frac{1}{v}+\frac{1}{-12}$ $\frac{1}{15}+\frac{1}{12}=\frac{1}{v}$ By solving we get v= +6.6 cm. Now, magnification $m= \frac{h_2}{h_1}=-\frac{v}{u}$ Thus, $h_2 = - \frac{v}{u} \times h_1 = \frac{-6.6}{-12} \times 4.5 = +2.5 \quad cm$ Hence, magnification of image, $m=\frac{h_2}{h_1} = \frac{2.5}{4.5} = .56$ The height of the image is 2.5 cm. The positive sign indicates that the image is erect, virtual, and diminished.

$Magnification = \frac {h_i}{h_o}=-\frac {v}{u}$ For real image $m=-\frac {v}{u} =-2$ v=2u Now Using the mirror equation, $\frac {1}{v} + \frac {1}{u}=\frac {1}{f}$ $\frac {1}{2u} + \frac {1}{u} =\frac {1}{-30}$ u=-45 cms. which is between the focal length and the Curvature. For virtual image $m=-\frac {v}{u} =2$ v=-2u Now Using the mirror equation, $\frac {1}{v} + \frac {1}{u}=\frac {1}{f}$ $\frac {1}{-2u} + \frac {1}{u} =\frac {1}{-30}$ u=-15 cm which is between the focal length and the pole

Answer 7 Linear magnification is the ratio of the height of the image to the height of the object. It is represented by the letter $m$ $m=\frac{height \quad of \quad image}{ height \quad of \quad object} = \frac{h_i}{h_o}$ where, $h_i$ is the height of image and $h_o$ is the height of object. If the image formed is virtual and erect, then the magnification is positive. If the image formed is real and inverted, then the magnification is negative.

If the object is placed at 25 cm in front of the concave mirror having focal length 20 cm, this means that the object is placed in between focus and the centre of curvature of the mirror. If the object is placed in front of the center of curvature then the image will be formed beyond the centre of curvature.So the image formed is a real image. The nature of the image will be will be inverted and enlarged If the object is placed at 15 cm in front of the concave mirror having focal length 20 cm, this means that the object is placed in between focus and the pole of the mirror. An object placed between the pole and focus of a concave mirror forms a virtual image.The nature of the image will be will be erect and enlarged

Answer 9 Similarity :- Both produce Virtual Image .  dissimilarity :- Convex Mirror produces diminished image while plane mirror produce the Image of the same size as that of the object . 

Answer 10  snell' law of refraction is:- "the ratio of sine of angle of incident to sine of angle of refraction is always constant for given medium"  where, $\frac{{\sin i}}{{\sin r}} = constant = {n_{12}}$ where, $n_{12}$ = refractive index of medium 2 with respect to the medium 1. The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence , all lie in the same plane.

$h_o=5 cm$, f=10 cm , u=-15 cm ,v=? Using lens formula $\frac {1}{f}=\frac {1}{v} - \frac {1}{u}$ $\frac {1}{10} = \frac {1}{v}- \frac {1}{-15}$ or v=30 cm Now $m =\frac { v}{u} = -2$ Now $\frac {h_i}{h_o} = -2$ or $h_i= -10 cm$ So,image is real, inverted and enlarged

a. Refractive index of an optical medium is ratio speed of the light in air to the speed of the light in the optical medium.So the refractive index of Ruby is 1.71 means ratio of speed of light in air to the speed of light in ruby is equal to 1.71 b. $Refractive \; Index = \frac {speed \; of \; light \; in \; air}{speed \; of \; light \;in \;optical\; medium}$ $speed \; of \; light \;in \;optical\; medium = \frac {speed \; of \; light \; in \; air}{Refractive \; Index}$ So higher the refractive index, lower the speed So, Speed will be maximum in water and lowest in Sapphire Now $speed \; of \; light \;in Sapphire = \frac {speed \; of \; light \; in \; air}{Refractive \; Index\; of\; Sapphir}$ $= \frac {3 \times 10^8}{1.77} =1.69 \times 10^8 m/s$

Given that f=30cm , u= -20cm , v=? Using Lens formula $\frac {1}{f} = \frac {1}{v} - \frac {1}{u}$ $ \frac {1}{30}=\frac {1}{v} - \frac {1}{-20}$ v = -60 cm Hence the image is at a distance of 60 cm from the lens.The negative sign indicates it is on same side on lens as the object and it is a real image. Now the size can be obtained using the magnification formula $m=\frac {h_i}{h_o}=\frac {v}{u}$ $\frac {h_i}{10}= \frac {-60}{-20}$ $h_i=30cm$ Hence Position of image is 60 cm on same side of lens and image is 30 cm and it is erect image

Given that f=15cm , u= -10cm , v=? Using Lens formula , $\frac {1}{f} = \frac {1}{v} - \frac {1}{u}$ $ \frac {1}{15}=\frac {1}{v} - \frac {1}{-10}$ v = -30 cm Hence the image is at a distance of 30 cm from the lens.The negative sign indicates it is on same side on lens as the object and it is a real image. Now the size can be obtained using the magnification formula $m=\frac {h_i}{h_o}=\frac {v}{u}$ $\frac {h_i}{6}= \frac {-30}{-10}$ $h_i=18cm$ Hence Position of image is 30 cm on same side of lens and image is 18 cm and it is erect image

(i) $n_d = \frac {v_{vaccum}}{v_{diamond}}$ or $v_{diamond} = \frac {v_{vaccum}}{n_d} = 1.25 \times 10^8$ m/s (ii) $r_w$ < $r_{g}$ < $r_{c}$ (iii)(a)since speed is glass is less than water, Glass is optically densor (b)it will go normal. as angle of incidene is 0 ,so the angle of refraction will be zero (iii)for Glass $n_g = \frac {v_{vaccum}}{v_{glass}}$ $\frac {3}{2}= \frac {c}{2 \times 10^8}$ or $c=3 \times 10^8$ m/s. For water $n_w = \frac {v_{vaccum}}{v_{water}}$ $v_{water}= \frac {c}{4/3}$ or $c=2.25 \times 10^8$ m/s

Given $f_A=20 cm$,$f_B=15 cm$,$f_C=10 cm$,$u_1=30 cm$,$u_2=10 cm$,$u_3=20 cm$ a. For same size image, object should be placed at Center of Curvature So for Mirror A, Image will not of same size for these positions For Mirror B,Image will be of same size for position $u_1=30 cm$ For mirror C,Image will be of same size for position $u_3=20 cm$ b.We need enlarged and erect image for shaving. For erect and enlarged image, Object should be placed between Pole and Focus. Face would be generally kept at more than 10 cm distance, so mirrors A and B are suitable for shaving

Image formed at 10 cm in front of concave mirror,inverted,smaller than object.

(a) check this Refraction by Spherical Lenses (b) f= -30 cm; $h_o = 5cm$ ,v=-15 cm Using lens Formula, $\frac {1}{f} = \frac {1}{v} -\frac {1}{u}$ $\frac {1}{-30} = \frac {1}{-15} -\frac {1}{u}$ u=-30 cm Now $ \frac {v}{u}=\frac {h_i}{h_o}$ or $h_i = \frac {v \times h_o}{u} = 2.5 cm$

a. Since this is converging lens, it is a convex lens v=+100 cm Now Magnification for lens is given by $m= \frac {v}{u} =\frac {h_i}{h_o}$ Since the image is inverted $\frac {v}{u} =-2$ or u =-50 cm The object is placed at a distance 50 cm from the lens Now using Lens formula $\frac {1}{f} = \frac {1}{v} -\frac {1}{u}$ $\frac {1}{f} = \frac {1}{100} -\frac {1}{-50}$ or f=33.33 cm Now Power of the lens $power = \frac {1}{f(m} = 3D$

a. check this Sign convention for reflection by spherical mirrors b. Here u=-18 cm, m=1/3 Now $m= -\frac {v}{u}$ or v = 6 cm Now using Mirror Formula $\frac {1}{f} = \frac {1}{v} +\frac {1}{u}$ $\frac {1}{f} = \frac {1}{6} +\frac {1}{-18}$ f=9 cm. Hence a convex mirror

a. Yes b. u=-15 ,f=20 cm ,v=? Now using Lens formula $\frac {1}{f} = \frac {1}{v} -\frac {1}{u}$ $\frac {1}{20} = \frac {1}{v} -\frac {1}{-15}$ v=-60 cm Now $m= \frac {v}{u}=\frac {h_i}{h_o}$ or $h_i=16 cm$ So image is enlarged and erect

This Numericals on Light Class 10 Sciencewith answers is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail.

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• Light Reflection And Refraction

NCERT Exemplar Class 10 Science Solutions for Chapter 10 - Light Reflection and Refraction

Ncert exemplar solutions class 10 science chapter 10 – free pdf download.

NCERT Exemplar Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction will boost your confidence in attempting the board exam by providing you with different types of questions that can be asked on the topic of Reflection and Refraction. In Chapter 10, students will understand the phenomena of reflection and refraction of light, and at the same time, they will learn about the basic concepts related to some optical phenomena occurring in nature. Students will also understand the reflection of light by spherical mirrors and refraction of light and other concepts like the mirror formula, refractive index, lens formula and more.

We are providing free NCERT Exemplar Solutions for Class 10 Science Chapter 10 here for the students to practise well and obtain good marks in the board exam. These exemplars will facilitate easy learning, and students will be able to get clarity on all the concepts and topics given in this chapter. Besides, these Exemplar Solutions that are given here act as a very useful reference tool which will help them get familiar with all the important chapter topics, practice questions, complete revisions, and ultimately, be well prepared to write the final exams.

Students can take a quick look at Class 10 Science Chapter 10 NCERT Exemplar by downloading the solutions PDF from the link below.

Download the PDF of NCERT Exemplar for Class 10 Science Chapter 10 – Light Reflection and Refraction

Access Answers to NCERT Exemplar Class 10 Science Chapter 10 – Light Reflection and Refraction

Multiple-choice questions.

1. Which of the following can make a parallel beam of light when light from a point source is incident on it?

(a) Concave mirror as well as convex lens

(b) Convex mirror as well as a concave lens

(c) Two plane mirrors placed at 90° to each other

(d) Concave mirror as well as a concave lens

The answer is (a) Concave mirror as well as convex lens

Explanation:

Emergent beams are parallel, and the image is produced at infinity when light rays from a point source are incident on a concave mirror and convex lens.

2. A 10 mm long awl pin is placed vertically in front of a concave mirror. A 5 mm long image of the awl pin is formed at 30 cm in front of the mirror. The focal length of this mirror is

(a) – 30 cm

(b) – 20 cm

(c) – 40 cm

(d) – 60 cm

The answer is (b) – 20 cm

Here, size of object = O = + 10.0 mm = + 1.0 cm (as, 1 cm = 10 mm)

Size of Image size = I = 5.0 mm = 0.5 cm

Image distance, v = − 30 cm  (as the image is real)

Let, object distance = u

Focal length, f =?

Magnification m= I(size of image)/O(size of image)

Magnification is given by m= -v/u

0.5/1=-30/u

Focal length is given by 1/f =1/v + 1/u

1/f=1/-30 + 1/60

3. Under which of the following conditions a concave mirror can form an image larger than the actual object?

(a) When the object is kept at a distance equal to its radius of curvature

(b) When an object is kept at a distance less than its focal length

(c) When an object is placed between the focus and centre of curvature

(d) When an object is kept at a distance greater than its radius of curvature

The answer is (c) When an object is placed between the focus and centre of curvature

When an object is placed between F and C, an enlarged image is formed beyond C.

4. Figure 10.1 shows a ray of light as it travels from medium A to medium B. Refractive index of the

medium B relative to medium A is

(a) √3 / √2

The answer is (a) 3 / 2

Refractive Index of B with respect to A

=sin i/sin r

=sin60/sin45

= ( √3 /2) /(1/ √2)

5. A light ray enters from medium A to medium B, as shown in Figure 10.2. The refractive

index of medium B relative to A will be

(a) greater than unity

(b) less than unity

(c) equal to unity

The answer is (a) greater than unity

6. Beams of light are incident through holes A and B and emerge out of the box through holes C and D, respectively, as shown in the Figure10.3. Which of the following could be inside the box?

(a) A rectangular glass slab

(b) A convex lens

(c) A concave lens

(d) A prism

The answer is (a) A rectangular glass slab

When incident rays fall perpendicularly on the point of incidence. A rectangular glass slab would refract and then re-refract it.

7. A beam of light is incident through the holes on side A and emerges out of the holes on the other face of the box, as shown in Figure 10.4. Which of the following could be inside the box?

(a) Concave lens

(b) Rectangular glass slab

(d) Convex lens

The answer is (d) Convex lens

Convex lenses are converging lenses. They have the ability to converge a parallel beam of light into a point.

8. Which of the following statements is true?

(a) A convex lens has 4 dioptre power having a focal length 0.25 m

(b) A convex lens has –4 dioptre power having a focal length 0.25 m

(c) A concave lens has 4 dioptre power having a focal length 0.25 m

(d) A concave lens has –4 dioptre power having a focal length 0.25 m

The answer is (a) A convex lens has 4 dioptre power having a focal length 0.25 m

A positive value for focal length indicates a convex lens.

9. Magnification produced by a rearview mirror fitted in vehicles

(a) is less than one

(b) is more than one

(c) is equal to one

(d) can be more than or less than one, depending upon the position of the object in front of it

The answer is (a) is less than one

A Convex mirror is used in the rearview mirror. The convex mirror always gives a smaller image. Hence, the magnification produced by the rear view mirror is always less than 1.

10. Rays from Sun converge at a point 15 cm in front of a concave mirror. Where should an object be placed, so that size of its image is equal to the size of the object?

(a) 15 cm in front of the mirror

(b) 30 cm in front of the mirror

(c) between 15 cm and 30 cm in front of the mirror

(d) more than 30 cm in front of the mirror

The answer is (d) more than 30 cm in front of the mirror

Here f= 15,

The radius of curvature is double the focal length

When an object is placed on c, its image is of the same size, inverted and is formed on c.

11. A full-length image of a distant tall building can definitely be seen by using

(a) a concave mirror

(b) a convex mirror

(c) a plane mirror

(d) both concave as well as plane mirror

The answer is (b) a convex mirror

The field of a convex mirror is more than any type of mirror. Hence the full-length size of the building can be seen by using a convex mirror.

12. In torches, searchlights and headlights of vehicles, the bulb is placed

(a) between the pole and the focus of the reflector

(b) very near to the focus of the reflector

(c) between the focus and centre of curvature of the reflector

(d) at the centre of curvature of the reflector

The answer is (b) very near to the focus of the reflector

Headlight reflectors and searchlights are in the shape of a concave mirror. When the source of light is placed at the focus, reflected light appears like a beam.

13. The laws of reflection hold good for

(a) plane mirror only

(b) concave mirror only

(c) convex mirror only

(d) all mirrors, irrespective of their shape

The answer is (d) all mirrors, irrespective of their shape

14. The path of a ray of light coming from air passing through a rectangular glass slab traced by four students are shown as A, B, C and D in Figure 10.5. Which one of them is correct?

The answer is b) B

Light bends towards normal when it passes from air to glass. Light bends away from normal when it passes from glass to air. This is appropriately shown in figure b).

15. You are given water, mustard oil, glycerine and kerosene. In which of these media a ray of light incident obliquely at the same angle would bend the most?

(a) Kerosene

(c) Mustard oil

(d) Glycerine

The answer is (d) Glycerine

Refractive indices

Kerosene-1.44

Mustard oil-1.46

Glycerine-1.47

Hence Glycerine is optically dense hence ray of light bends more with glycerine.

16. Which of the following ray diagrams is correct for the ray of light incident on a concave mirror as shown in Figure 10.6?

The answer is (d) Fig. D

In the case of the concave mirror, an incident ray is parallel to the principal axis and passes through F after reflection.

17. Which of the following ray diagrams is correct for the ray of light incident on a lens shown in Fig. 10.7?

(a) Fig. A.

(b) Fig. B.

(c) Fig. C.

(d) Fig. D.

The answer is (a) Fig. A.

In a convex lens, the incident ray passing through F goes parallel to the principal axis after refraction.

18. A child is standing in front of a magic mirror. She finds the image of her head bigger, the middle portion of her body of the same size and that of the legs smaller. The following is the order of combinations for the magic mirror from the top.

(a) Plane, convex and concave

(b) Convex, concave and plane

(c) Concave, plane and convex

(d) Convex, plane and concave

The answer is (c) Concave, plane and convex

When the object is between F and P of the concave mirror enlarged image is formed behind the mirror. Hence child can see her head bigger in a concave mirror. She can see her body size of the same size because the plane mirror gives an image of the original size. Convex mirror gives diminished images, and the baby’s legs appear smaller.

19. In which of the following, the image of an object placed at infinity will be highly diminished and point sized?

(a) Concave mirror only

(b) Convex mirror only

(c) Convex lens only

(d) Concave mirror, convex mirror, concave lens and convex lens

The answer is (d) Concave mirror, convex mirror, concave lens and convex lens

Short Answer Questions

20. Identify the device used as a spherical mirror or lens in the following cases, when the image formed is virtual and erect in each case.

(a) Object is placed between the device and its focus, and an image formed is enlarged and behind it.

(b) Object is placed between the focus and device, and an image formed is enlarged and on the same side as that of the object.

(c) Object is placed between infinity and device, an image formed is diminished and between focus and optical centre on the same side as that of the object.

(d) Object is placed between infinity and device, image formed is diminished and between pole and focus, behind it.

(a) Concave mirror

(b) Convex lens

(c) Concave lens

(d) Convex mirror

21. Why does a light ray incident on a rectangular glass slab immersed in any medium emerges parallel to itself? Explain using a diagram.

When a light ray enters a denser medium from the rarer medium, it bends towards the normal In this case extent of bending of the ray at the opposite parallel is the same. Hence emergent ray is parallel to the incident ray.

22. A pencil, when dipped in water in a glass tumbler, appears to be bent at the interface of air and water. Will the pencil appear to be bent to the same extent if, instead of water, we use liquids like kerosene or turpentine? Support your answer with reasons.

The bending of light here is a function of refraction. Refraction is dependent on refractive indices. Refractive indices of kerosene or turpentine would not be the same as water. Hence the degree of the bend would be different in different mediums.

23. How is the refractive index of a medium related to the speed of light? Obtain an expression for the refractive index of a medium with respect to another in terms of the speed of light in these two media.

Refractive Index can be seen as the factor by which the speed and the wavelength of the radiation are reduced with respect to their vacuum values.

w=civ (where n:refractive index,c=speed of light,v:velocity of light in that medium)

The refractive index of one medium in relation to a second medium is given by the ratio of the speed of light In the second medium to the speed of light in the first medium.

24. Refractive index of a diamond with respect to glass is 1.6, and the absolute refractive index of glass is 1.5. Find out the absolute refractive index of the diamond.

Absolute RI of diamond= 1.6

Absolute RI of glass= 1.5

Multiplying them we get 2.4

25. A convex lens of focal length 20 cm can produce a magnified virtual as well as real image. Is this a correct statement? If yes, where shall the object be placed in each case for obtaining these images?

When an object is placed in F and F2 of a convex lens, we get an inverted, enlarged and real image is formed beyond 2F2, which is on the other side of the lens. Hence we need to place the object between 20 and 40 cm of the lens.

When an object is placed between F and 0 of a convex lens, its enlarged, erect and virtual image is formed beyond FL i.e. on the same side of the lens. So for this, we need to place the object at a distance less than 20 cm from the lens.

26. Sudha finds out that the sharp image of the window pane of her science laboratory is formed at a distance of 15 cm from the lens. She now tries to focus on the building visible to her outside the window instead of the window pane without disturbing the lens. In which direction will she move the screen to obtain a sharp image of the building? What is the approximate focal length of this lens?

To obtain a clear image of the building, Sudha has to move the screen towards the lens. The focal length will be approximately 15 cm. The rays of light coming from a distant object such as a tree (or a distant building or electricity pole) can be considered to be parallel to each other. When parallel rays of light are incident on a convex lens, the rays, after refraction, converge at focus on the other side of the lens.

27. How are the power and focal length of a lens related? You are provided with two lenses of focal lengths 20 cm and 40 cm, respectively. Which lens will you use to obtain more convergent light?

The power of the lens is inversely proportional to the focal length of the lens. A lens with a focal length of 20 has more power than a lens with a focal length of 40 cm. A lens with higher power should be used to obtain more convergent light.

28. Under what condition in an arrangement of two plane mirrors, incident ray and reflected ray will always be parallel to each other, whatever may be the angle of incidence? Show the same with the help of a diagram.

If two plane mirrors are placed perpendicular to each other, then the incident ray and reflected ray will always be parallel to each other.

29. Draw a ray diagram showing the path of rays of light when it enters with oblique incidence

(i) from air into water, (ii) from water into air.

The speed of light decreases when it passes from a rarer medium to a denser medium, and also, light rays bend towards normal.

When a ray of light passes from a denser medium to a rarer medium-light, rays bend away from the normal.

Long Answer Questions

30. Draw ray diagrams showing the image formation by a concave mirror when an object is placed

(a) between pole and focus of the mirror

(b) between focus and centre of curvature of the mirror

(c) at the centre of curvature of the mirror

(d) a little beyond the centre of curvature of the mirror

(e) at infinity

31. Draw ray diagrams showing the image formation by a convex lens when an object is placed

(a) between optical centre and focus of the lens

(b) between focus and twice the focal length of the lens

(c) at twice the focal length of the lens

(d) at infinity (e) at the focus of the lens

32. Write laws of refraction. Explain the same with the help of a ray diagram, when a ray of light passes through a rectangular glass slab.

Laws of refraction

• Incident ray refracted ray and normal at the point of incidence lie in the same plane.
• The ratio of the sine of incidence and sine of refraction is constant for the given colour and pair of media.

• ABCD is a glass slab. EF is an incident ray which is incident on point 0 on the air-glass interface.

• NO is normal and _LEON = : I; which Is angle of incidence.

• NV is normally extended towards the glass slab and ..LV ’00 = Zri; which is the angle of refraction.

• 00′ is refracted ray from surface AB. It behaves like an incident rayon surface CD.

• the ray EF bends when it enters the slab to become 00′.

• MO’ and O’M’ are normal on surface CD.

• GH is the emergent ray.

• ZOO’ Al = Li,; which is the angle of Incidence at surface CD.

• Z.110′ H = Zr,; which is the angle of refraction at surface CD.

• It is observed that EF. NO and 00′ lie in the same plane: which is in accordance with the first law of refraction.

• It is also observed that EF II GH, which means the emergent ray is parallel to the incident ray. This happens because the degree of bend at opposite surfaces of the glass slab is the same.

33. Draw ray diagrams showing the image formation by a concave lens when an object is placed

(a) at the focus of the lens

(c) beyond twice the focal length of the lens

(a) The ray diagram when the object is placed at the focus of the concave lens:

(b) The ray diagram when the object is placed between focus and twice the length of focal length of the lens:

(c) Ray diagram when the object is beyond twice the focal length of the concave lens:

34. Draw ray diagrams showing the image formation by a convex mirror when an object is placed

(a) at infinity

(b) at a finite distance from the mirror

a) At Infinity

b) At infinite distance from the mirror

35. The image of a candle flame formed by a lens is obtained on a screen placed on the other side of the lens. If the image is three times the size of the flame and the distance between the lens and the image is 80 cm, at what distance should the candle be placed from the lens? What is the nature of the image at a distance of 80 cm and the lens?

As the image is obtained on the screen, it is real.

Magnification , m = –3 ,

u = –80/3 cm .

1/f = 1/v – 1/u

=1/80 + 3/80

1/f = 1/20cm

f = 20 cm .

The lens is convex and the image formed at 80 cm from the lens is real and inverted.

36. Size of the image of an object by a mirror having a focal length of 20 cm is observed to be reduced to 1/3rd of its size. At what distance has the object been placed from the mirror? What is the nature of the image and the mirror?

Using = 1/v+1/u=1/f

Calculate u;u = – 80 cm.

The image is real and inverted. The mirror is concave.

37. Define the power of a lens. What is its unit? One student uses a lens with a focal length of 50 cm and another of –50 cm. What is the nature of the lens and its power used by each of them?

The degree of convergence and divergence provided by a lens is called the power of the lens. The unit of power of the lens is Diopter D.

The focal length of the lens used by the first student is in positive; hence, it is a convex lens. The lens of the second student is a concave lens.

p=1/f =1/0.5 =2

Power of lens (first student) =+2

Power of lens (second student) = ).2

38. A student-focussed the image of a candle flame on a white screen using a convex lens. He noted down the position of the candle screen and the lens as under

Position of candle = 12.0 cm

Position of convex lens = 50.0 cm

Position of the screen = 88.0 cm

i) What is the focal length of the convex lens?

ii) Where will the image be formed if he shifts the candle towards the lens at a position of 31.0 cm?

iii) What will be the nature of the image formed if he further shifts the candle towards the lens?

iv) Draw a ray diagram to show the formation of the image in case (iii), as said above.

Position of the candle flame = 12.0cm

Position of the lens= 50.0 cm

Position of the screen=88.0 cm

i) u= 50-12= 38 cm.

Image distance v= 88-50= 38cm

Focal length =1/v – 1/u= 1/f

ii) Object distance u= 50-31= 19 cm

Object distance = focal length

Hence the image is formed at infinity.

If he further shifts the candle towards the lens. The object comes between F and 0. In this case. The image is virtual, enlarged and erect and is formed on the same side of the lens.

Topics Covered in NCERT Exemplar Class 10 Light Reflection and Refraction:

• Reflection of Light
• Spherical Mirrors
• Image Formation by Spherical Mirrors
• Representation of Images Formed by Spherical Mirrors Using Ray Diagrams Ex
• Sign Convention for Reflection by Spherical Mirrors
• Mirror Formula and Magnification
• Refraction through a Rectangular Glass Slab
• The Refractive Index
• Refraction by Spherical Lenses
• Image Formation by Lenses
• Image Formation in Lenses Using Ray Diagrams
• Sign Convention for Spherical Lenses
• Lens Formula and Magnification
• Power of a Lens.

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NCERT Solutions for Chapter 10 Light Reflection and Refraction Class 10 Science

NCERT Solutions for Class 10 Science Chapter 10 Light – Reflection and Refraction- This article includes free NCERT Solutions for Class 10 Science Chapter 10 Light – Reflection and Refraction, developed by the top Biology experts at GFG, according to the latest CBSE Syllabus 2023-24 and guidelines. The solutions to all the exercises in NCERT Class 10 Science Chapter 10 Light – Reflection and Refraction have been collectively covered in NCERT Solutions for Class 10 .

Also, to learn the basic concepts of NCERT Class 10 Chapter 10 Light – Reflection and Refraction from scratch, check out Class 10 NCERT Science Chapter 10 Light – Reflection and Refraction .

NCERT Solutions for Class 10 Science Chapter 10 Light- Reflection and Refraction: Exercise

Q1. which one of the following materials cannot be used to make a lens (a) water (b) glass (c) plastic (d) clay.

We cannot use clay to make lens because clay is an opaque material and light cannot pass through clay. Thus we cannot see anything through a clay lens. Thus option (d) clay is the correct answer.

Q2. The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object? (a) Between the principal focus and the centre of curvature (b) At the centre of curvature (c) Beyond the centre of curvature (d) Between the pole of the mirror and its principal focus.

Concave image forms a virtual, erect and larger image when the object is between the principal focus and pole of the mirror. Thus option (d) is the correct answer.

Q3. Where should an object be placed in front of a convex lens to get a real image of the size of the object? (a) At the principal focus of the lens (b) At twice the focal length (c) At infinity (d) Between the optical centre of the lens and its principal focus.

To get a real image of the object through a convex lens, we need to place the object at a distance equal to radius of curvature of the lens which is twice the focal lengh. Thus option (b) is the correct answer.

Q4. A spherical mirror and a thin spherical lens have a focal length of -15 cm. The mirror and the lens are likely to be (a) both concave (b) both convex (c) the mirror is concave, and the lens is convex (d) the mirror is convex, but the lens is concave

The focal length given here is negative. Negative focal length is a property of concave mirror. Thus the mirror and lens both are concave. Thus option (a) is the correct answer.

Q5. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be (a) plane (b) concave (c) convex (d) either plane or convex.

A plane and convex mirror always form an erect image. Thus the mirror is likely to be either plane or convex. Thus option (d) is the correct answer.

Q6. Which of the following lenses would you prefer to use while reading small letters found in a dictionary? (a) A convex lens of focal length 50 cm (b) A concave lens of focal length 50 cm (c) A convex lens of focal length 5 cm (d) A concave lens of focal length 5 cm

A mirror with shorter focal length has a higher magnification power and a convex lens makes an erect image when object is between the mirror and its focal length. Thus we will make use of a convex mirror with focal length of 5cm. Thus option (c) is the correct answer.

Q7. We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.

A concave mirror will form an erect image of an object when the distance of object from the mirror is less than its focal length. Thus we will place the image between 0 to 15 cm. The image thus formed is virtual, erect, and enlarged (larger than the object). The ray diagram is shown below:

Q8. Name the type of mirror used in the following situations. (a) Headlights of a car (b) Side/rear-view mirror of a vehicle (c) Solar furnace Support your answer with a reason.

(a) A concave mirror is used in the headlights of car because when the light source is placed at the focus of the concave mirror, it produces a powerful parallel beam of light which helps to see the objects on the front of car clearly to a large distance. (b) Side/ rear view mirror of a vehicle is a convex mirror because we need to see a large amount of area and vehicles behind us while driving. A convex mirror is used because a convex mirror forms a virtual and diminished image of the objects which provide us with a smaller view of the large area. (c) Solar furnace makes use of concave mirror because the sun’s rays coming from infinity are converged by the concave mirror at its focus which concentrates the energy at a particular point to produce a high temperature.

Q9. One-half of a convex lens is covered with black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.

Yes, the lens will produce a complete image of the object but the image will be reduced in intensity. This is because to form an image, we need only two rays of light coming from an object. As the upper portion of lens is not covered, at least two rays can pass through upper part of lens and form the image. This can be verified by an experiment. Experiment: To show that a half lens can produce a complete image of an object Requirements: Convex lens, Candle, matchstick, Two black sheets of paper, Table Procedure: Fix the convex lens on the table. Now light a candle with matchstick and place it on one side of the lens. Turn off the lights to ensure that the room is dark Take a sheet of paper and obtain the image of the candle on the sheet by moving it. Mark the positions of the candle, lens and image in this case. Now, cover one half of the lens with a black sheet. Repeat the above experiment. Observations: A complete image of candle is obtained on the sheet but it is less bright as compared to previous image. Thus, we conclude that a half lens can form a complete image of an object.

Q10. An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and nature of the image formed.

Given u = -25 cm, h = 5cm, f = 10 cm, h’ = ? Using lens formula: 1/f = 1/v – 1/u 1/10 = 1/v – (-1/25) 1/v = 1/10 – 1/25 1/v = 15/250 = 3/50 v = 50/3 cm = 16.67 cm We know that h/h’ = v/u 5/h’ = 50/3*(-25) h’ = 250/(-75) = -10/3 h’ = 3.33 cm Negative sign indicates that the image is inverted and diminished in size as h’ < h. It is formed at a distance of 16.67 cm behind the lens. The ray diagram is shown below:

Q11. A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.

Given u = ?, f = -15 cm, v = -10 cm Using lens formula: 1/f = 1/v – 1/u -1/15 = -1/10 – 1/u 1/u = -1/10 + 1/15 1/u = -5/150 u = -30 cm Thus, the object is placed at a distance of 30 cm in front of the lens. The ray diagram is shown below:

Q12. An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.

Given u = -10 cm, f = 15 cm Using mirror formula : 1/f = 1/v + 1/u 1/15 = 1/v + (-1/10) 1/v = 1/15 + 1/10 1/v = 25/250 v = 6 cm Magnification = -v/u = -6/-10 = 0.6 < 1 Thus a diminished image is formed at a distance 6 cm behind the mirror.

Q13. The magnification produced by a plane mirror is +1. What does this mean?

As magnification = h’/h. A magnification of 1 means that the height of image is same as the height of object and a positive sign indicates that the image is virtual and erect.

Q14. An object 5 cm is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position, nature and size of the image.

Given h = 5 cm, u = -20 cm, R = 30 cm As R = 2f 30 = 2f f = 15 cm Using mirror formula: 1/f = 1/v + 1/u 1/15 = 1/v – 1/20 1/v = 1/15 + 1/20 1/v = 35/300 v = 300/35 = 8.57 cm Magnification m= -v/u = -8.57/-20 = 0.428 m = h’/h 0.428 = h’/5 h’ = 2.14 cm As m < 1, v is positive and h’ < h, a virtual , erect and diminished image of size 2.14 cm is formed.

Q15. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharply focused image can be obtained? Find the size and nature of the image.

Given h = 7, u = -27 cm, f = -18 cm, v = ? Using mirror formula: 1/f = 1/v + 1/u -1/18 = 1/v – 1/27 1/v = -1/18 + 1/27 1/v = -1/54 v = -54 cm Magnification m= -v/u = 54/-27 = -2 As m > 1 and has a negative sign, a real, inverted and magnified image of double the size of the object is formed at a distance of 54 cm in front of the mirror.

Q16. Find the focal length of a lens of power -2.0 D. What type of lens is this?

Given P = -2.0 D We know that 1/f = P 1/f = -2 f = -1/2 = -0.5 m = -50cm As the focal length is negative, it is a concave lens.

Q17. A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Given P = 1.5 D We know that 1/f = P 1/f = 1.5 f = 1/1.5 = 0.667m Thus the focal length is 0.667 m. As the focal length is positive, it is a convex lens which is converging lens.

Important Topics Discussed in NCERT Class 10 Science Chapter 10 Light – Reflection and Refraction

The topics covered in Chapter 10 ‘Light-Reflection and Refraction’ chapter in Class 10 NCERT Science are:

• Reflection of Light
• Spherical Mirrors
• Refraction of Light
• Image Formation by Spherical Lens
• Combination of Lenses
• Power of Lenses
• Dispersion of Light
• Atmospheric Refraction

Key Features of NCERT Solutions for Class 10 Science Chapter 10 Light – Reflection and Refraction

• NCERT solutions are created by a team of professionals a GFG, with the intention to benefit students.
• These solutions are very accurate and comprehensive, which can help students prepare for any academic as well as competitive exam.
• All the solutions provided are in a step-by-step format for better understanding.

Also, Check:

• NCERT Solutions for Class 10 Science Chapter 1 Chemical Reactions and Equations
• NCERT Solutions for Class 10 Science Chapter 10 Light – Reflection and Refraction
• NCERT Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current
• NCERT Solutions for Class 10 Science Chapter 10 The Human Eye And The Colourful World

NCERT Solutions for Class 10 Science Chapter 10 Light-Reflection and Refraction – FAQs

1. what is reflection.

Reflection is the phenomenon where light, sound, or other waves bounce back when they encounter a surface that does not absorb them, obeying the law of reflection.

2. State the Law of reflection?

According to the law of reflection, the angle of incidence is equal to angle of reflection. It also states that the ray of incidence, normal to the plane of mirror, and the reflected ray all lie in the same plane.

3. What do you mean by irregular reflection?

Irregular reflection is a type of reflection in which the incident light reflects in different directions when it reflects from a rough or irregular surface.

4. What do you Mean by Regular Reflection?

Regular reflection is a type of reflection in which the incident light reflects at the same angle of incidence with the normal when it falls on a smooth surface or a plane mirror.

5. What is Refraction?

Refraction is the bending of light as it passes from one medium to another with a different optical density, causing the light to change direction and possibly speed.

6. What is the key Difference between Reflection and Refraction?

The key difference between reflection and refraction is that reflection involves the bouncing back of waves when they hit a surface, while refraction is the bending of waves as they pass from one medium to another due to a change in their speed and direction.

7. How does a Light Bend when it Passes from Rarer to Denser Medium?

Light bends towards the normal when it passes from rarer to denser medium.

8. What do you Mean by Focal Length of a Mirror?

Focal length of a mirror is defined as the distance between the pole of a mirror and its principal focus.

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Light – Reflection and Refraction Worksheet Class 10 PDF Download

In Class 10 Physics, there are important topics in Light – Reflection and Refraction on which students should practise questions to develop a strong grip. The Light – Reflection and Refraction Worksheet is a great way to solve various questions on Light – Reflection and Refraction. A regular solving of questions can improve students' speed and accuracy to give answers. 

Practising questions from the Light – Reflection and Refraction worksheet is an amazing way to have a strong foundation in the topic. It is considered as a great study tool because it helps students to identify all types of questions (easy, moderate, complex). During the Class 10 CBSE board exam preparation, identification of all types of questions can help students to score good marks in Class 10 Physics board exam. 

Light – Reflection and Refraction Worksheet PDF

Selfstudys provides Light – Reflection and Refraction worksheet PDF so that students can download it and practise the questions given. Portable Document Format (PDF) is a digital copy of the Light – Reflection and Refraction and electric fields worksheet answers. Through this digital copy, students can access these questions from anywhere. Through solving questions, students can get a brief idea about those questions that can be asked in the CBSE Class 10 Physics exam. This PDF form of worksheet can be helpful for all the board students: CBSE, KVS, UP, MP etc as all the boards follow NCERT books. 

Marking Scheme of the Light – Reflection and Refraction

With the help of Light – Reflection and Refraction worksheet, students would be able to identify the marking scheme for Class 10 Physics chapter Light – Reflection and Refraction. According to the marking scheme for each question, students would be able to give the accurate answers. Accuracy is very important while giving answers as accordingly Class 10 marks can be improved. Getting good marks can help students in selecting their desired field after 10th.

Through a marking scheme, students would be able to know the marks for each question. According to these marks, students can identify important topics in the chapter Light – Reflection and Refraction and can study accordingly. Studying according to the marks provided can help students to score well in the chapter. Marking scheme in the Light – Reflection and Refraction worksheet PDF can also help students to give answers in a comprehensive way. 

Light – Reflection and Refraction Worksheet with Answers

Solutions for these worksheets are also given in the website so that students don’t face any difficulty in searching answers. Answer is the response or reply given by students to each and every question. With the help of Light – Reflection and Refraction worksheet answers, students can go through it whenever they are in doubt. These answers can be very helpful throughout the preparation for Class 10 CBSE Physics exam. According to these answers students can be well prepared for the Class 10 Physics board exam. According to these answers, students can also write the answers in a comprehensive way. 

How to Download Light – Reflection and Refraction Worksheet?

To have the access to the Light – Reflection and Refraction worksheet, students can go through the given steps:

• Open Selfstudys website. 

• Bring the arrow towards CBSE and select it from the navigation bar. 

• A drop-down menu will appear, select KVS NCERT CBSE Worksheet from the list. 

• A new page will appear, select Class 10th from the list of classes.

• Select Physics from the list of subjects. 

• Again a new page will appear, select Light – Reflection and Refraction Assignment- 1. 

Features of Light – Reflection and Refraction Worksheet

Worksheet is a sheet which includes questions to practise so that students can easily understand all concepts. This is one of the most important features, apart from this. Some more features are: 

• Created by : Light – Reflection and Refraction worksheet are personally created by experts or teachers. These worksheets are created with proper research of the chapter Light – Reflection and Refraction. 
• Variety of questions : It contains all types of questions: one mark question, two mark question, three mark question, five mark question, value based questions. 
• Solutions : Answers to all these are available in the Light – Reflection and Refraction worksheet PDF. Through the answers students can solve their doubts then and there. These answers are comprehensive so that students can understand the concepts easily. 
• Explanation : Answers for 3 marks and 5 marks are explained in detail so that students don’t face any difficulty in giving these answers in the Class 10 Physics board exam. These explanations can help students to score good marks in Class 10 Physics board exam.
• Convenience : It is very helpful for students to practise on their electronic device as it can be very convenient compared to the traditional method. In the coming years, technology based worksheets will be followed more compared to traditional worksheets. By going through the Selfstudys website, students can access the Light – Reflection and Refraction worksheet PDF. 

Benefits of Light – Reflection and Refraction Worksheet

Through the worksheet, students can also understand complex topics in the chapter. This is one of the most crucial benefits of solving a Light – Reflection and Refraction worksheet PDF Class 10. Other than this there are more benefits:

• Improvisation of Grades : Through solving worksheets, students can improve their grades/marks in the chapter Light – Reflection and Refraction. As it is important to get a 90+ marks in Class 10 Physics board exam. 
• Tracking of performance : Students can also track their performance through solving questions. Through the tracking of performance students can easily improvise themselves. 
• Logical Reasoning : Solving Light – Reflection and Refraction worksheet can improve students' logical reasoning. Logical reasoning skills can be helpful for students in their further career.
•  Strong Foundation : Solving Light – Reflection and Refraction and electric fields worksheet answers can help to form a strong foundation for the chapter Light – Reflection and Refraction. Strong foundation can help students to perform well in further chapters in Class 10 CBSE Physics textbook. 
• Enhanced Learning - By solving Light – Reflection and Refraction problems worksheet regularly can enhance students' learning. Students learning is very important as they can understand all concepts quickly. 

Why is the Worksheet on Light – Reflection and Refraction for Class 10 Important? 

It is very important for Class 10 students to solve worksheets on Light – Reflection and Refraction. Some of the importance are clearly discussed below:

•  Speed and Accuracy : Regular solving of questions from the worksheet can improve a student's speed and accuracy of answers. Speed and accuracy of all answers can help students to answer correctly with proper speed in the Class 10 Physics board exam. 
• Clarification : Electrostatic Force worksheet can help students to improve the clarity in all the topics included in the chapter. 
•  Guide : Light – Reflection and Refraction worksheet can be helpful for both teachers and students. Teachers can guide their students according to the answers given by them. Students can analyse themselves and can improve accordingly. Basically worksheets act as guides to teachers and students. 
• Systematic Order - Questions in the worksheet are systematically arranged so that students don’t face any difficulty in searching it. This systematic order of questions can help students select the questions correctly. Selection of questions is very important during Class 10 Physics exam as accordingly students can score well. 
• Learning Process - Regular solving of Light – Reflection and Refraction problems worksheet can enhance the learning process so that students can score well in Class 10 board exam. 

When Should A Class 10 Student Use a Worksheet on Light – Reflection and Refraction?

Students of Class 10 should use the Light – Reflection and Refraction worksheet after understanding each and every topic in that chapter. Through solving questions in the worksheet, students can have a better understanding of the chapter. Better understanding of the chapter can help students to score well in Light – Reflection and Refraction. Regular solving of questions can also help students to create a strong foundation for the chapter Light – Reflection and Refraction. A strong foundation of the chapter Light – Reflection and Refraction can help students in understanding further chapters.

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All Activities Class 10 Science Chapter 10 | Class 10 Chapter 10 Science All Actvities

All activities class 10 science chapter 10 light-reflection and refraction.

In this article, we will discuss all activities and solutions of class 10 Science  Chapter 10 Light- Reflection and Refraction . You will study the complete explanation and conclusion of all activities.  These activities are based  on  NCERT Class 10 Science book .

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CBSE Class 10 Science Chapter 10 Light MCQ Assignment

• Last modified on: 1 year ago
• Reading Time: 12 Minutes

MULTIPLE CHOICE QUESTIONS

1. Focal length of plane mirror is (a) at infinity          (b) zero (c) negative             (d) none of these

2. Image formed by plane mirror is (a) real and erect             (b) real an inverted (c) virtual and erect        (d) virtual and inverted

3. A concave mirror gives, real, inverted and same size image if the object is placed (a) at F                 (b) at infinity (c) at C                 (d) beyond C

4. Power of a lens is –40, its focal length is (a) 4 m                     (b) – 40 cm (c) – 0.25 m             (d) – 25 m.

5. A concave mirror gives virtual, erect and enlarged image if the object is placed: (a) at infinity                      (b) between F and C (c) between P and F          (d) at F.

6. The mirror that always gives virtual and erect image of the object but image of smaller size than the size of the object is (a) Plane mirror           (b) Concave mirror (c) Convex mirror        (d) none of these

7. All the distances in case of spherical mirror are measured in relation to (a) object to image               (b) the pole of the mirror (c) the focus of the mirror (d) the image to the object.

8. The radius of curvature and focal length of a concave mirror are (a) positive                     (b) negative (c) both                           (d) none of these

9. The object distance in both concave as well as convex mirror is (a) negative                     (b) positive (c) zero                             (d) none of these

10. The ratio of the speed of light in vacuum to that in a medium is known as (a) magnification             (b) refraction (c) refractive index          (d) Snell’s law

11. In optics an object which has higher refractive index is called (a) optically rarer             (b) optically denser (c) optical density             (d) refractive index

12. The optical phenomena, twinkling of stars, is due to (a) atmospheric reflection          (b) total reflection (c) atmospheric refraction          (d) total refraction

13. Convex lens focus a real, point sized image at focus, the object is placed (a) at focus                 (b) between F and 2f (c) at infinity             (d) at 2f

14. The unit of power of lens is (a) metre              (b) centimeter (c) diopter            (d) m–1

15. The radius of curvature of a mirror is 20 cm the focal length is (a) 20 cm               (b) 10 cm (c) 40 cm               (d) 5 cm

16. The refractive indices of some media are given below:

In which of these is the speed of light minimum and maximum, respectively. (a) X-minimum, W-maximum               (b) Z-minimum, W-maximum (c) W-minimum, X-maximum               (d) X-minimum, Z-maximum

17. The power of a lens is + 1.6 D. The nature of lens is (a) Convex lens                            (b) Concave lens (c) both concave an convex      (d) none of these

18. An incident ray makes 60° angle with the surface of the plane mirror, the angle of its refraction is (a) 60°                (b) 90° (c) 30°                (d) 0°

19. The angle of reflection in the given figure is

(a) 90°            (b) 180° (c) 0°               (d) 30°

20. A mirror that has very wide field view is (a) concave            (b) convex (c) plane                 (d) none of these

21. If the object is placed at focus of a concave mirror, the image is formed at (a) infinity                             (b) focus (c) centre of curvature        (d) between F and O.

1. (a)        2. (c)          3. (c)         4. (c)         5. (c)        6. (c)      7. (b)      8. (b)      9. (a)     

10. (c) 11. (b)    12. (c)       13. (c)      14. (c)      15. (b)    16. (c)    17. (a)     18. (c) 

19. (c)    20. (b) 21. (a)

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We have provided below free printable Class 10 Physics Assignments for Download in PDF. The Assignments have been designed based on the latest NCERT Book for Class 10 Physics . These Assignments for Grade 10 Physics cover all important topics which can come in your standard 10 tests and examinations. Free printable Assignments for CBSE Class 10 Physics , school and class assignments, and practice test papers have been designed by our highly experienced class 10 faculty. You can free download CBSE NCERT printable Assignments for Physics Class 10 with solutions and answers. All Assignments and test sheets have been prepared by expert teachers as per the latest Syllabus in Physics Class 10. Students can click on the links below and download all Pdf Assignments for Physics class 10 for free. All latest Kendriya Vidyalaya Class 10 Physics Assignments with Answers and test papers are given below.

Physics Class 10 Assignments Pdf Download

We have provided below the biggest collection of free CBSE NCERT KVS Assignments for Class 10 Physics . Students and teachers can download and save all free Physics assignments in Pdf for grade 10th. Our expert faculty have covered Class 10 important questions and answers for Physics as per the latest syllabus for the current academic year. All test papers and question banks for Class 10 Physics and CBSE Assignments for Physics Class 10 will be really helpful for standard 10th students to prepare for the class tests and school examinations. Class 10th students can easily free download in Pdf all printable practice worksheets given below.

Topicwise Assignments for Class 10 Physics Download in Pdf

Advantages of Class 10 Physics Assignments

• As we have the best and largest collection of Physics assignments for Grade 10, you will be able to easily get full list of solved important questions which can come in your examinations.
• Students will be able to go through all important and critical topics given in your CBSE Physics textbooks for Class 10 .
• All Physics assignments for Class 10 have been designed with answers. Students should solve them yourself and then compare with the solutions provided by us.
• Class 10 Students studying in per CBSE, NCERT and KVS schools will be able to free download all Physics chapter wise worksheets and assignments for free in Pdf
• Class 10 Physics question bank will help to improve subject understanding which will help to get better rank in exams

Frequently Asked Questions by Class 10 Physics students

At https://www.cbsencertsolutions.com, we have provided the biggest database of free assignments for Physics Class 10 which you can download in Pdf

We provide here Standard 10 Physics chapter-wise assignments which can be easily downloaded in Pdf format for free.

You can click on the links above and get assignments for Physics in Grade 10, all topic-wise question banks with solutions have been provided here. You can click on the links to download in Pdf.

We have provided here topic-wise Physics Grade 10 question banks, revision notes and questions for all difficult topics, and other study material.

We have provided the best collection of question bank and practice tests for Class 10 for all subjects. You can download them all and use them offline without the internet.

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• CBSE Class 10 NCERT Solutions

NCERT Solutions for Class 10 Science Chapter 10 The Human Eye and the Colourful World: Download PDF For Free

Ncert solutions class 10 science chapter 10 the human eye and the colourful world: students get here detailed and accurate solutions for all the exercise and intext questions for ncert class 10 chapter 10 the human eye and the colourful world. .

NCERT Solutions for Class 10 Science Chapter 10: Both intext and exercise questions given in the NCERT Books are of great importance for students. It is essential that students are aware of the accurate answers to the questions. This article gives detailed answers for NCERT Class 10 Science Chapter 10 The Human Eye and the Colourful World. 

Questions given in the NCERT books are always considered very important as not only do they help you to analyse your understanding of the concepts involved in a chapter but also in preparing for the examinations. Students should read the chapters carefully along with attempting the questions given.

Also Check: Difference Between Pound and Kilogram: Major Differences Between The Units

NCERT Class 10 Science The Human Eye and the Colourful World Solutions 

Intext questions and solutions page no. 164.

Q. What is meant by power of accommodation of the eye? 

Sol. Power of the accommodation of the eye is the ability of an eye to adjust itself in different conditions. For example-When the ciliary muscles are relaxed, the eye lens becomes thin, leading to an increase in focal length. Due to this, distant objects are clearly visible to the eyes. 

In case of nearby objects, the ciliary muscles contract making the eye lens thicker. Hence, the focal length of the eye lens decreases and nearby objects become visible to the eyes. Hence, the human eye lens is able to adjust its focal length to view both distant and nearby objects on the retina.

Q. 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?

  Sol. The person is able to see nearby objects clearly, but he is unable to see objects beyond 1.2 m. This happens because the image of an object beyond 1.2 m is formed in front of the retina and not at the retina.

To correct this defect of vision, use of a concave lens is required. The concave lens will bring the image back to the retina.

Q. What is the far point and near point of the human eye with normal vision?

Sol. The far point of the normal human eye is infinity. The far point of the eye is the maximum distance to which the eye can see the objects clearly. 

The near point of the eye is the minimum distance of the object from the eye, which can be seen distinctly without strain. For a normal human eye, this distance is 25 cm. 

Q. 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?

Sol. He is unable to see distant objects clearly. He is suffering from myopia. This defect can be corrected by using a concave lens. 

NCERT Class 10 Science Chapter 10 Exercise Questions

Q. The human eye can focus on 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. 

Sol. (b) Accommodation 

Q. The human eye forms the image of an object at its

 (a) cornea. (b) iris. (c) pupil. (d) retina

Sol. (d) Retina 

Q. 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

Sol. (c) 25 cm

Q.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

Sol. (c ) ciliary muscles  

Q. 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?

Sol . (i) Power of the lens used for correcting distant vision = −5.5 D

f=1/P=−1/5.5=−0.181m

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 

P=1/f 

f=1/P=1/1-5=0.667m

The focal length of the lens for correcting near vision is 0.667 m.

Q. 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? 

Sol. 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

1/v−1/u=1/f

1/f=−1/80−1/∞

P=1/−0.8=−1.25D

A concave lens of power −1.25 D is required by the person to correct his defect.

Q. 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. 

Sol. A person suffering from hypermetropia can see distinct objects clearly but faces difficulty in seeing nearby objects clearly. It happens because the eye lens focuses the incoming divergent rays beyond the retina. This defect of vision is corrected by using a convex lens. 

Object distance, u= −25 cm

Image distance, v= −1 m = −100 m

Using the lens formula, 

1/f=−1/100+1/25

P=1/0.33=3D

A convex lens of power +3.0 D is required to correct the defect.

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

Sol. A normal eye is unable to clearly see objects placed closer than 25 cm because the ciliary muscles of eyes are unable to contract beyond a certain limit and the object appears blurred and produces strain in the eyes.

Also Check: Law Of Conservation Of Mass: Definition, Formula And More

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

Sol. 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.

Q. Why do stars twinkle?

Sol. Stars are considered as point sources of light. They emit their own light. They twinkle due to the refraction of light by the atmosphere. When the light coming from stars enters the earth’s atmosphere, it gets refracted at different levels. When the star light refracted by the atmosphere comes more towards us, it appears brighter. Hence, it appears as if the stars are twinkling at night.

Q. Explain why the planets do not twinkle.

Sol. Planets appear larger in size than the stars as they are closer to earth. The different parts of planets produce either brighter or dimmer effects in such a way that the average of brighter and dimmer effect is zero. Hence, the twinkling effects of the planets are nullified and they do not twinkle.

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

Sol. The sky appears dark instead of blue to an astronaut because of the absence of atmosphere in outer space. Atmosphere can scatter the sunlight. As the sunlight is not scattered, no scattered light reaches the eyes of the astronauts and the sky appears black to them. 

CBSE Video Courses for Class 10 Students

Class 10 students can study effectively for the exams with the help of video courses prepared by the subject matter experts. These video courses will explain the concepts in a simple and interactive manner which will help learners to understand clearly. 

CBSE Class 10 Video Courses 

Also, check

CBSE Class 10 Science Lab Manual PDF

  CBSE Class 10 Science syllabus 2024-25

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BSEB Bihar Board Exam 2025: Registration For Class 10 Begins, Check Steps To Apply

The Bihar School Examination Board (BSEB) has also recently released the dummy registration card for Bihar Board Class 10th and 12th.

Students aspiring to appear for the Bihar Board exam 2025 can apply by visiting the official website, secondary.biharboardonline.com.

BSEB Bihar Board Exam 2025: Steps To Register

• Visit the official website, secondary.biharboardonline.com
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The official website states: "Commencement of Online Submission of Application Form for Secondary Annual Exam 2025 is from September 11, 2024, to September 2, 2024. The last date for application fee payment is September 24, 2024."

Bihar Board Class 10 Exam 2025: Important Points To Consider

• The name, subject, gender, marital status, and caste category of the student, mother, and father should be filled in the application form with complete accuracy to avoid any errors in the mark sheet issued after the examination
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• The student should enter only a valid mobile number and email ID in the prescribed columns of the application form
• One mobile number can be used to fill the registration application form for only one student

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