# 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

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

F represents the focal length

$focal\:length=\frac{Radius }{2}$

So,

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

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

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.

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

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.

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

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.

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   $2\times10^8m/s$.

 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

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.

 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

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

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

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

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.

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

Clay cannot be used to make a lens.

The position of the object should be between the pole of the mirror and its principal focus.

(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.

An object should be placed at twice the focal length in front of a convex lens to get a real image of the size of the object

(a) only plane.
(b) only concave.
(c) only convex.
(d) either plane or convex.

The mirror is likely to be either plane or convex.

(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.

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

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

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.

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.

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.

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

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

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

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.

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.

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

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.

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.

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

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

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

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

• 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.