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

 

NCERT solutions for class 10 science chapter 10 Light Reflection and Refraction: Physics can be one of the toughest subjects to understand. Chapter 10 Light Reflection and Refraction is one of those but it is a scoring one. With the help of solutions for NCERT class 10 science chapter 10 Light and Reflection and Refraction you will able to understand all the interesting questions related to reflections and different types of lenses, mirrors, etc. Also, you will get clarity about the concepts of lenses that form images of objects, reflecting surfaces of all types and the laws of reflection. If you have any problem in answering the questions or you are not getting the correct answers then don't worry, CBSE NCERT solutions for class 10 science chapter 10 Light Reflection and Refraction is there for you to help. For other subject solutions, you can check here NCERT solutions.

Here are the important Topics of NCERT class 10 science chapter 10 - Light Reflection and Refraction mentioned below:

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 Reflection of Light

    10.3.1 Reflection through a Rectangular Glass Slab

    10.3.2 The Refractive Index

    10.3.3 Reflection 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 Lenses Formula and Magnification

    10.3.8 Power of a lens

NCERT solutions for class 10 science chapter 10 Light Reflection and Refraction

Topic 10.2 Spherical Mirrors

Q.1   Define the principal focus of a concave mirror.

Answer:

It is the point on the principal axis where a beam of light parallel to the principal axis after reflection actually meets.

concave mirror

F represents the focal length

Q.2   The radius of curvature of a spherical mirror is 20 cm. What is its focal length?

Answer:

 focal\:length=\frac{Radius }{2}

So,

focal\:length=\frac{20}{2}=10cm

Hence the Focal length of the spherical mirror is 10 cm.

Q.3   Name a mirror that can give an erect and enlarged image of an object.

Answer:

Convex mirror usually gives a virtual and erect image.

The concave mirror gives a virtual and enlarged image only when the object is between pole and focus.

Q.4  Why do we prefer a convex mirror as a rear-view mirror in vehicles?

Answer:

Convex mirrors are preferred as a rearview mirror in vehicles as: 

i) It forms an erect (Rigidly upright or straight.) image of an object hence object becomes easily identified. 

ii) It forms a diminished image of the object thus increases the field of vision.

iii) An object that is far away from us is seen closer which helps us to take early decisions while driving.

CBSE NCERT solutions for class 10 science chapter 10 Light Reflection and Refraction 

Topic 10.2.4 Mirror forumula and magnification

Q.1   Find the focal length of a convex mirror whose radius of curvature is 32 cm.

Answer:

As we know for the convex mirror,

Radius (R) = Focus (f) X 2

So,

f=\frac{R}{2}

putting the given values,

f=\frac{30}{2}=15cm

Hence Focus of the convex mirror is 15 cm.

Q. 2    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:

Here Given Magnification:

m= -3 (real image)

The object distance :

u =-10cm

Now As we know,

m=-\frac{v}{u}

v = - m\times u

v = - (-3)\times(-10)

v = -30cm

Hence,

The image is 30 cm in front of the mirror.

NCERT textbook solutions for class 10 science chapter 10 Light Reflection and Refraction

Topic 10.3 Refraction of light

Q.1  A ray of light traveling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?

Answer:

AS we know that when the light goes from rare medium to denser medium the light bends towards the normal. So when the light goes from air to water it will bend toward the normal. 

Q. 2  Light enters from air to glass having a refractive index 1.50. What is the speed of light in the glass? The speed of light in a vacuum is 3 \times 108\: m\: s^1.

Answer:

As we know, from the definition of the refractive index that 

 Refractive Index :

 \mu=\frac{c}{v}

Where = Velocity of light in the medium and 

            c = Speed of light in vacuum

So putting those given values we get,

1.5 = \frac{3\times10^ {8}}{v}

v=\frac{3\times10^8}{1.5}

v= 2\times10^8m/s

Hence the speed of the light in the glass is   2\times10^8m/s.

Q.3  Find out, from Table 10.3, the medium having highest optical density. Also, find the medium with the lowest optical density.

 

Material Medium Refractive Index Material Medium Refractive Index

Air

Ice

water

Alcohol

Kerosene

Fused Quartz

Turpentine oil

Benzene

Crown Glass

 

1.0003

1.31

1.33

1.36

1.44

1.46

1.47

1.50

1.52

Canada Balsam

Rock Salt

Carbon-Disulphide

Dense Flint Glass

Ruby

Sapphire

Diamond

1.53

1.54

1.63

1.65

1.71

1.77

2.42

 

Answer:

As we know optical density is the tendency to hold(absorb)  the light. So,

more refractive index = more absorbing power = more optical density.

It can be observed from table 10.3 that diamond and air respectively have the highest and lowest refractive index. Therefore, diamond has the highest optical density and air has the lowest optical density.

Q.4  You are given kerosene, turpentine, and water. In which of these does the light travel fastest? Use the information given in Table 10.3

Material Medium Refractive Index Material Medium Refractive Index

Air

Ice

water

Alcohol

Kerosene

Fused Quartz

Turpentine oil

Benzene

Crown Glass

 

1.0003

1.31

1.33

1.36

1.44

1.46

1.47

1.50

1.52

Canada Balsam

Rock Salt

Carbon-Disulphide

Dense Flint Glass

Ruby

Sapphire

Diamond

1.53

1.54

1.63

1.65

1.71

1.77

2.42

Answer:

As we can see from the table:

Refractive index of kerosene = 1.44

Refractive index of terbutaline = 1.47

Refractive index of water = 1.33

We know from the definition of refractive index, that the speed of light is higher in a medium with the lower refractive index.

So, the light travels fastest in water relative to kerosene and turpentine. 

speed of light---> water > kerosene > turpentine

Q.5  The refractive index of diamond is 2.42. What is the meaning of this statement?

Answer:

Refractive Index shows the comparison of light speeds in two mediums. Light may travel from rarer to denser medium or from denser to rarer medium. When we say the refractive index of diamond is 2.42, it means light is traveling from rarer to a denser medium, and the speed of light in the air (vacuum) is 2.42 times the speed of light in a diamond. On the other hand, if the light travels from denser to rarer medium, that is, from diamond to air the refractive index will be the reciprocal of 2.42.

Solutions for NCERT class 10 science chapter 10 Light Reflection and Refraction 

Topic 10.3.8 Power of Lenses

 Q.1  Define 1 dioptre of power of a lens.

Answer:

A lens whose focal length is 1 metre is said to have the power of 1 dioptre.

Q.2   A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.

Answer:

Since the image formed by the lens is real and inverted it is formed on the side opposite to the one where the object is placed.

Image position v = 50 cm

Let the object position be u.

Since image formed is inverted and the size image is equal to the size of the object, magnification (m) = -1.

\\m=\frac{v}{u}\\ -1=\frac{v}{u}\\ u=-v\\ u=-50\ cm

The needle is placed 50 cm in front of the lens

Using lens formula we have

\\\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\\ \frac{1}{f}=\frac{1}{50}-\frac{1}{-50}\\ \frac{1}{f}=\frac{2}{50}\\ f=25\ cm

Power of the lens P is given by

\\P=\frac{1}{f}\\ P=\frac{1}{25\times 10^{-2}}\\ P=4\ D

The power of the lens P is 4 Dioptre.

Q.3  Find the power of a concave lens of focal length 2 m.

Answer:

The focal length of the lens is f = -2 m. (The focal length of a concave lens is negative)

Power of the lens P is given by

\\P=\frac{1}{f}\\ P=\frac{1}{-2}\\ P=-0.5\ D

The Power of the lens is - 0.5 Dioptre.

NCERT solutions for class 10 science chapter 10 Light Reflection and Refraction: Exercise solutions

Q1.     Which one of the following materials cannot be used to make a lens?


 (a) Water     (b) Glass     (c) Plastic     (d) Clay

 

Answer:

Clay cannot be used to make a lens.

(d) is the correct answer.

Q 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:

I would prefer to use a convex lens of focal length 5 cm while reading small letters found in a dictionary because a convex lens gives magnified image if the object is in between focus and radius of curvature and the magnification will be high for shorter focal length

(c) is the correct answer.

Q 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:

\\\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ v=\frac{fu}{u-f}

 For the image to be erect it has to be virtual and therefore as per convention v has to be positive

v > 0

\frac{fu}{u-f}>0

u < 0

f < 0

fu > 0

u - f > 0

u > f

 Therefore the object must be placed between the pole and focus of the mirror.

 i.e -15 cm < u < 0

 The image formed would be virtual and larger than the object.

 

ray diagram

Q 8.(a)  Name the type of mirror used in the following situations.

 Support your answer with reason.

Answer:

A concave mirror is used for headlights of a car as they can produce parallel beams of high intensity which can travel through large distances if the source of light is placed at the focus of the mirror.

Q 8.b) Name the type of mirror used in the following situations.

Support your answer with reason.

Answer:

The convex mirror would be used as a side/rear-view mirror of a vehicle as the image produced will be erect, virtual and diminished. A convex mirror also has a larger field of view as compared to a plane mirror of the same size.

Q 8.c)  Name the type of mirror used in the following situations.

Support your answer with reason.

Answer:

The concave mirror would be used in a solar furnace to concentrate the incident rays from the sun on a solar panel.

Q 9.   One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.

Answer:

The lens would produce a complete image. Only the brightness of the image would be diminished.

 Verification by experiment:

 Apparatus required: a convex lens, a screen and a candlelight

 procedure:

1. Place screen behind a lens placed vertically

2. Move the candle to obtain a clear full-length image of a candle on the screen

3. Cover half of the lens with the black paper without disturbing the position of the lens 

4. Note the observations

 Observation: Size of the image is the same but the brightness reduces

Q 10.   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.

Answer:

Object distance u = -25 cm.

Focal length = 10 cm

Let image distance be v

\\\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ \frac{1}{10}=\frac{1}{v}+\frac{1}{-25}\\ \frac{1}{v}=\frac{1}{10}+\frac{1}{25}\\ v=16. 66\ cm

The position of the image is 16.66 cm on the other side of the lens.

Object size O = 5 cm.

Let the image size be I

Magnification is m

\\m=\frac{v}{u}\\ \frac{I}{O}=\frac{16.66}{-25}\\ I=5\times -\frac{16.66}{25}\\ I=-3.33\ cm

The nature of the image is real and its size is -3.33 cm.

The formation of the image is shown in the following ray diagram

Ray diagram

 

Q 11.   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.

Answer:

Focal length, f = -15 cm

Image distance, v = -10 cm ( As in case of the concave lens image is formed on the same side as the object is placed)

Let the object distance be u. 

As per the lens formula

\\\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\\ \frac{1}{u}=\frac{1}{v}-\frac{1}{f}\\ \frac{1}{u}=\frac{1}{-10}-\frac{1}{-25}\\ v=-30\ cm

 The object is placed at a distance of 30 cm from the lens.

Ray diagram

Q 12.   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.

Answer:

Object distance, u = -10 cm

Focal length, f = 15 cm

Let the image distance be v

As per the mirror formula

\\\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ \frac{1}{15}=\frac{1}{v}+\frac{1}{-10}\\ v=6\ cm

\\Magnification=-\frac{v}{u}\\ =-\frac{6}{-30}\\ =0.2

 The image formed is virtual, erect, diminished and is formed 6 cm behind of the mirror.

Q 13.  The magnification produced by a plane mirror is +1. What does this mean?

Answer:

It means the image formed is virtual, erect and of the same size as the object.

Q 14.   An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature, and size.

Answer:

Radius of curvature, R = 30 cm

Focal length, f = R/2 = 15 cm

Object distance, u = -20 cm

Let the image distance be v 

\\\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ \frac{1}{15}=\frac{1}{v}+\frac{1}{-20}\\ v=8.57\ cm

Since v is positive image is formed behind the mirror.

\\Magnification=-\frac{v}{u}\\ =-\frac{8.57}{-20}\\ =0.428

Object size = 5 cm

\\Image \ size= Object\ size \times Magnification\\ =5\times 0.428\\ =2.14\ cm

A virtual, erect and diminished image of size 2.14 cm would be formed 8.57 cm behind the mirror.

Q 15.   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 focussed image can be obtained? Find the size and nature of the image.

Answer:

Object distance, u = -27 cm

Focal length, f = -18 cm

Let the image distance be v

As per the mirror formula

\\\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\ \frac{1}{-18}=\frac{1}{v}+\frac{1}{-27}\\ v=-54\ cm

The negative sign shows the image is formed in front of the mirror and is real.

Object size = 7 cm

\\Image\ size=magnification\times Object\ size\\ =-\frac{v}{u}\times O\\ =-\frac{-54}{-27}\times 7\\ =-14\ cm

A real, inverted and magnified image of size 14 cm is formed in front of the mirror.

Q 16.   Find the focal length of a lens of power – 2.0 D. What type of lens is this?

Answer:

\\f=\frac{1}{P}\\ f=\frac{1}{-2}\\ f=-0.5\ m\\ f=-50\ cm

The lens is concave because it's focal length is negative and is equal to -50 cm.

Q 17.  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?

Answer:

\\f=\frac{1}{P}\\ f=\frac{1}{1.5}\\ f=0.6667\ m\\ f=66.67\ cm

The less is convex because its focal length is positive and therefore is converging.

NCERT Solutions for Class 10 Science - Chapter wise

Chapter No.

Chapter Name

Chapter 1

CBSE NCERT solutions for class 10 science chapter 1 Chemical Reactions and Equations

Chapter 2

NCERT free solutions for class 10 science chapter 2 Acids, Bases, and Salts

Chapter 3

NCERT textbook solutions for class 10 science chapter 3 Metals and Non-Metals

Chapter 4

Solutions for NCERT class 10 science chapter 4 Carbon and Its Compounds

Chapter 5

CBSE NCERT solutions for class 10 science chapter 5 Periodic Classification of Elements

Chapter 6

NCERT textbook solutions for class 10 chapter 6 Life Processes

Chapter 7

NCERT free solutions for science class 10 chapter 7 Control and Coordination

Chapter 8

Solutions for NCERT class 10 science class 8 How do Organisms Reproduce?

Chapter 9

CBSE NCERT solutions for class 10 science chapter 9 Heredity and Evolution

Chapter 10

NCERT solutions for class 10 science chapter 10 Light Reflection and Refraction

Chapter 11

NCERT solutions for class 10 science chapter 11 The Human Eye and The Colorful World

Chapter 12

CBSE NCERT solutions for class 10 science chapter 12 Electricity

Chapter 13

Solutions for NCERT class 10 science chapter 13 Magnetic Effects of Electric Current

Chapter 14

NCERT textbook solutions for class 10 science chapter 14 Sources of Energy

Chapter 15

NCERT free solutions for class 10 science chapter 15 Our Environment

Chapter 16

CBSE NCERT solutions for class 10 science chapter 16 Sustainable Management of Natural Resources

Benefits NCERT solutions for class 10 science chapter 10 Light Reflection and Refraction

  • The solutions provided in this article are easy to understand.
  • NCERT solutions provided here for chapter 10 Light Reflection and Refraction can help you to score well in the CBSE board exam for class 10 science.
  • You can also find exercise questions and additional exercise questions in the same place.
  • Step-wise step solutions are provided by the experts if you have any doubt, you can ask them directly.
 

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