Courses
Explore all Vedantu courses by class or target exam, starting at ₹1350
Long Term Courses
Full Year Courses Starting @ just Rs 9000
One-to-one LIVE classes
Learn one-to-one with a teacher for a personalised experience
Courses for Kids
Find courses by class
Starting @ ₹1,350
Find courses by target
Starting @ ₹1,350
Free study material
Get class-wise, author-wise, & board-wise free study material for exam preparation
Online Tuition
Get class-wise, subject-wise, & location-wise online tuition for exam preparation

NCERT Solutions for Class 10 Science Chapter 9 Light Reflection and Refraction

ffImage
Last updated date: 26th Jun 2024
Total views: 535.8k
Views today: 14.35k
MVSAT offline centres Dec 2023

NCERT Solutions for Class 10 Science Chapter 9 Light Reflection and Refraction - FREE PDF Download

NCERT Solutions for class 10 science chapter 9 Light Reflection and Refraction, provided by Vedantu, is a key resource for students who want to navigate the concepts of Science at the 10th-grade level. Students can go through the light reflection and refraction class 10 questions and answers PDF in which they will learn about the concepts of reflection and refraction.

toc-symbol
Table of Content
1. NCERT Solutions for Class 10 Science Chapter 9 Light Reflection and Refraction - FREE PDF Download
2. Glance on NCERT Solutions for Class 10 Science Chapter 9 Light Reflection and Refraction
3. Access NCERT Solutions for Class 10 Science Chapter 9 – Light Reflection and Refraction
4. Topics Covered In Class 10 Science Chapter 9 Light Reflection and Refraction
5. Key Features of Light Reflection and Refraction Class 10 NCERT Solutions
6. Important Study Material Links for Class 10 Science Chapter 9
7. FREE PDF Links for Other Chapter-wise NCERT Solutions Class 10 Science 
8. Related Important Links for Class 10 Science
FAQs


The advantage of NCERT Solutions ch 9 science class 10 is its alignment with the latest NCERT Science syllabus. Students can learn about fundamental concepts of light. The solutions closely follow the CBSE Science class 10 syllabus and exam pattern. You can access the FREE PDF for light class 10 NCERT  solutions from this page. 


Glance on NCERT Solutions for Class 10 Science Chapter 9 Light Reflection and Refraction

  • Light reflection and refraction class 10 embarked on a fascinating journey to understand the fundamental concepts of reflection and refraction mechanisms and their various aspects. 

  • Students can access the light reflection and refraction class 10 questions and answers PDF to resolve all their doubts related to this chapter.

  • It is quite necessary to have at least a basic knowledge of Science, which may help you in the future. Light class 10 NCERT solutions are structured in a comprehensive and presentable way to provide students with complete coverage of the topics and scientific terms, which can help students score well in their exams and gain perspective towards career goals. 

  • Opting for ch 9 science class 10 can help a student ensure excellent exam performance. This free PDF contains solutions to all problems in light reflection and refraction class 10. 

  • Class 10 Science Chapter 9 is mostly about the reflection and refraction of light. The main topics included spherical mirrors, magnification, reflection, refraction, and lens power. Light Reflection and Refraction Class 10 Solutions include diagrams, logical problems, and detailed content in simple words to simplify simplification.

  • Reflection of Light Class 10 refers to light waves bouncing off the surface of objects that cannot absorb energy produced by radiation. For example, the surface of a pool will display the clarity of surrounding objects when still. Whereas all the rays would be scattered once a rock is thrown inside, preventing systematic light reflection.

Popular Vedantu Learning Centres Near You
centre-image
Mithanpura, Muzaffarpur
location-imgVedantu Learning Centre, 2nd Floor, Ugra Tara Complex, Club Rd, opposite Grand Mall, Mahammadpur Kazi, Mithanpura, Muzaffarpur, Bihar 842002
Visit Centre
centre-image
Anna Nagar, Chennai
location-imgVedantu Learning Centre, Plot No. Y - 217, Plot No 4617, 2nd Ave, Y Block, Anna Nagar, Chennai, Tamil Nadu 600040
Visit Centre
centre-image
Velachery, Chennai
location-imgVedantu Learning Centre, 3rd Floor, ASV Crown Plaza, No.391, Velachery - Tambaram Main Rd, Velachery, Chennai, Tamil Nadu 600042
Visit Centre
centre-image
Tambaram, Chennai
location-imgShree Gugans School CBSE, 54/5, School road, Selaiyur, Tambaram, Chennai, Tamil Nadu 600073
Visit Centre
centre-image
Avadi, Chennai
location-imgVedantu Learning Centre, Ayyappa Enterprises - No: 308 / A CTH Road Avadi, Chennai - 600054
Visit Centre
centre-image
Deeksha Vidyanagar, Bangalore
location-imgSri Venkateshwara Pre-University College, NH 7, Vidyanagar, Bengaluru International Airport Road, Bengaluru, Karnataka 562157
Visit Centre
View More
Competitive Exams after 12th Science
tp-imag
bottom-arrow
tp-imag
bottom-arrow
tp-imag
bottom-arrow
tp-imag
bottom-arrow
tp-imag
bottom-arrow
tp-imag
bottom-arrow
Watch videos on
NCERT Solutions for Class 10 Science Chapter 9 Light Reflection and Refraction
icon
Light Class 10 in One Shot Part -2 | Refraction CBSE Class 10 2023-24| Abhishek Sir @VedantuClass9_10_11
Vedantu 9&10
Subscribe
iconShare
11.2K likes
225K Views
2 years ago
Download Notes
yt video
Light Class 10 in One Shot | Reflection Part -1 | CBSE Class 10 | Abhishek Sir @VedantuClass9_10_11 ​
Vedantu 9&10
27.7K likes
648.7K Views
2 years ago
Play Quiz
Download Notes
yt video
Laws of Refraction and Refractive Index | Light L-11 |CBSE Class 10 Abhishek Sir @VedantuClass9_10_11
Vedantu 9&10
13.1K likes
297K Views
3 years ago
Download Notes
yt video
Refraction | Light L-10 | CBSE Class 10 Abhishek Sir @VedantuClass9_10_11
Vedantu 9&10
14.2K likes
303.6K Views
3 years ago
Play Quiz
Download Notes
See Moresee more youtube videos

Access NCERT Solutions for Class 10 Science Chapter 9 – Light Reflection and Refraction

1. What is the principal focus of a concave mirror? 

Ans: The light rays parallel to the principal axis converge at a point on the principal axis after reflecting from the concave mirror.

The point of convergence on the principal axis of a concave mirror is called the principal focus of a concave mirror. 


2. What is the focal length of a spherical mirror whose radius of curvature is $20cm$?

Ans: It is given that,

Radius of curvature of a spherical mirror, $R=20cm$ 

It is known that, 

Radius of curvature is twice the focal length.

$R=2f$ 

$\Rightarrow f=R/2$ 

$\Rightarrow f=\frac{20}{2}$

$\Rightarrow f=10cm$ 

Therefore, the focal length of a spherical mirror with the radius of curvature equal to $20cm$ is $f=10cm$.


3. Which mirror gives an erect and enlarged image of an object? 

Ans: If the object is placed between pole and the principal focus of a concave mirror the image formed is virtual, erect and enlarged.

Erect and enlarged images are not possible in case of convex or plane mirrors.


4. Why is convex mirror preferred as a rear-view mirror in vehicles?

Ans: If the objects are placed in front of a convex mirror, then the image formed is an erect and diminished image.

In vehicles we need erect images and we need to see as many areas as possible behind the vehicles.

So, a convex mirror is preferred as a rear-view mirror in vehicles.


5. What is the focal length of a convex mirror whose radius of curvature is $32cm$?

Ans: It is given that,

Radius of curvature of a convex mirror, $R=32cm$ 

It is known that, 

Radius of curvature is twice the focal length.

$R=2f$ 

$\Rightarrow f=R/2$ 

$\Rightarrow f=\frac{32}{2}$

$\Rightarrow f=16cm$ 

Therefore, the focal length of a convex mirror with the radius of curvature equal to $32cm$ is $f=16cm$.


6. Find the location of image for a concave mirror that produces three times magnified (enlarged) real image of the object placed at $10cm$ in front of it.

Ans: It is given that,

Distance of object in front of mirror, $u=-10cm$ (negative sign due to the location of the object in front of the mirror)

Distance of image from mirror, $v=?$

It is known that,

Magnification of a spherical mirror,$m=\frac{{{h}_{i}}}{{{h}_{o}}}=-\frac{v}{u}$ 

where,

${{h}_{i}}$ is the height of the image

${{h}_{o}}$ is the height of the object

Let, ${{h}_{o}}=h$ 

It is given that three times the enlarged real image of the object is produced.

So, ${{h}_{i}}=-3h$ ($-$ due to real image formation)

$\Rightarrow m=-\frac{3h}{h}=-\frac{v}{u}$ 

$\Rightarrow 3=\frac{v}{-10}$ 

$\Rightarrow v=-30cm$(negative sign due to the formation of inverted image)

Therefore, the location of image from the mirror is at a distance of $30cm$ and the nature of the image is inverted.


7. When a light ray travelling in the air enters obliquely into the water, how does it bend towards the normal or away from the normal? State reason.

Ans: When a light ray is travelling from the rarer medium to denser medium, it refracts towards the normal.

Here, the light ray bends towards the normal because the light ray is moving from air(rarer) to water(denser) medium. 


8. Find the speed of light in glass if the light enters from air to glass having a refractive index of $1.50$. Take the speed of light in vacuum, $c=3\times {{10}^{8}}m{{s}^{-1}}$ .

Ans:  It is given that,

Refractive index of glass,$\mu =1.50$ 

Speed of the light in vacuum, $c=3\times {{10}^{8}}m{{s}^{-1}}$

Speed of the light in glass,${{v}_{g}}=?$ 

It is known that,

$\mu =\frac{c}{{{v}_{g}}}$ 

$\Rightarrow 1.5=\frac{3\times {{10}^{8}}m{{s}^{-1}}}{{{v}_{g}}}$ 

$\Rightarrow {{v}_{g}}=\frac{3\times {{10}^{8}}}{1.5}$ 

$\Rightarrow {{v}_{g}}=2\times {{10}^{8}}m{{s}^{-1}}$

Therefore, the speed of light in glass is ${{v}_{g}}=2\times {{10}^{8}}m{{s}^{-1}}$ .


9. From the table, find the medium having the highest optical density and the lowest optical density.

Material medium

Refractive Index

Material medium

Refractive Index

Air

1.0003

Crown glass

1.52

Ice

1.31

Canada Balsam

1.53

Water

1.33

Rock salt

1.54

Alcohol

1.36

Carbon disulphide

1.63

Kerosene

1.44

Dense flint glass

1.65

Fused quartz

1.46

Ruby

1.71

Turpentine oil

1.47

Sapphire

1.77

Benzene

1.50

Diamond

2.42


Ans: To find the materials of highest and lowest optical densities check for its refractive index. The one with the highest refractive index will have the highest optical density and the one with lowest refractive index will have the lowest optical density.

The highest optical density is for Diamond i.e., $\mu =2.42$

The lowest optical density is for Air i.e.,$\mu =1.0003$


10. Among kerosene, turpentine oil and water in which medium does the light travel fastest? Refer to the table for refractive index.

Material Medium

Refractive Index

Material Medium

Refractive Index

Air

1.0003

Crown Glass

1.52

Ice

1.31

Canada Balsam

1.53

Water

1.33

Rock Salt

1.54

Alcohol

1.36

Carbon Disulphide

1.63

Kerosene

1.44

Dense Flint Glass

1.65

Fused Quartz

1.46

Ruby

1.71

Turpentine Oil

1.47

Sapphire

1.77

Benzene

1.50

Diamond

2.42


Ans: It is known that,

Absolute refractive index,$\mu =\frac{c}{{{v}_{m}}}$ 

where,

$c$ is the speed of light in vacuum$=3\times {{10}^{8}}m{{s}^{-1}}$ 

${{v}_{m}}$ is the speed of light in medium

$\Rightarrow \mu \propto \frac{1}{{{v}_{m}}}$ ($c$ is constant)

To compare the speed of light we compare the refractive index. Speed of light and refractive index are inversely proportional i.e., if the refractive index is more than the speed of light is less.

The order of refractive index among kerosene, turpentine oil and water is${{\mu }_{\ker osene}}>{{\mu }_{turpentineoil}}>{{\mu }_{water}}$ .

The order of speed of light becomes ${{\mu }_{\ker osene}}<{{\mu }_{turpentineoil}}<{{\mu }_{water}}$ .

Therefore, the speed of light is fastest in water.


11. What is the meaning of the statement “The refractive index of diamond is $2.42$”?

Ans: It is known that,

Refractive index of a medium,$\mu =\frac{c}{{{v}_{m}}}$ 

where,

$c$ is the speed of light in vacuum

${{v}_{m}}$ is the speed of light in medium

$2.42=\frac{c}{{{v}_{d}}}$

“The refractive index of diamond is $2.42$” means that the speed of light in vacuum is $2.42$ times the speed of light in diamond or the speed of light in diamond is $\frac{1}{2.42}$ times the speed of light in vacuum.


12. What is $1$ dioptre of power of a lens?

Ans: When $f$ is the focal length in metres of a lens with power $P$ then$P=\frac{1}{f(metres)}$ .

Dioptre is the S.I. unit of power of a lens and is denoted by $D$.

The power of a lens of focal length $1$ metre is defined as $1$ dioptre.


13. Image formed by a convex lens is real and inverted and at a distance of $50cm$from the lens. Find the position of the needle in front of the convex lens when the image is equal to the size of the object. Also, calculate the power of the lens.

Ans: It is given that,

Distance of image from convex lens,$v=50cm$

Distance of object in front of lens,$u=?$

The image formed is real and inverted. So, the magnification of the lens is $-1$.


Self-created

Image source: Self-created

It is known that, 

Magnification of a convex lens,$m=\frac{v}{u}$ 

$\Rightarrow -1=\frac{v}{u}$ 

$\Rightarrow -1=\frac{50}{u}$ 

$\Rightarrow u=-50cm$ 

Lens formula:$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ 

$f$ is the focal length of the lens.

$\Rightarrow \frac{1}{f}=\frac{1}{50}-\frac{1}{(-50)}$ 

$\Rightarrow \frac{1}{f}=\frac{1}{50}+\frac{1}{50}$ 

$\Rightarrow \frac{1}{f}=\frac{2}{50}$ 

$\Rightarrow f=\frac{50}{2}$ 

$\Rightarrow f=25cm=0.25m$ 

It is known that,

Power of the lens,$P=\frac{1}{f(metres)}$ 

$\Rightarrow P=\frac{1}{(+0.25)}$ 

$\Rightarrow P=+4D$ 

Therefore, the object distance from the lens is $u=-50cm$ and power of the lens is $P=+4D$.


14. What is the power of a concave lens of focal length$2m$?

Ans: It is given that,

Focal length of a concave lens is,$f=-2m$ 

Power of the concave lens,$P=\frac{1}{f(metres)}$ 

$\Rightarrow P=\frac{1}{(-2)}$ 

$\Rightarrow P=-0.5D$ 

Therefore, the power of a concave lens is $P=-0.5D$.


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

  1. Water

  2. Glass

  3. Plastic

  4. Clay

Ans: d) Clay can’t be used to make a lens because it is opaque.


16. The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should the position of the object be?

  1. Between the principal focus and the centre of curvature

  2. At the centre of curvature

  3. Beyond the centre of curvature

  4. Between the pole of the mirror and its principal focus.

Ans: d) The object is placed between the pole of the mirror and its principal focus when the image formed is virtual, erect and larger than the object in the concave mirror. 


concave mirror is observed


17. Where should an object be placed in front of a convex lens to get a real image of the size of the object? 

  1. At the principal focus of the lens 

  2. At twice the focal length 

  3. At infinity 

  4. Between the optical centre of the lens and its principal focus

Ans: b) An object should be placed at a distance of twice the focal length of a convex lens to get a real image of the size of the object. 


When object is placed at twice the focal length

18. A spherical mirror and a thin spherical lens have each a focal length of $-15cm$ . The mirror and the lens are likely to be

  1. Both concave 

  2. Both convex 

  3. The mirror is concave and the lens is convex 

  4. The mirror is convex, but the lens is concave 

Ans: a) For a concave lens the primary focus is on the same side as the object and is negative. In the case of a concave mirror, the focus is in front of the mirror and negative. Therefore, the mirror and lens are likely to be concave.


19. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be 

  1. Plane 

  2. Concave 

  3. Convex 

  4. Either plane or convex

Ans: d) Erect images are produced by both plane and convex mirrors for objects at any position.


20. Which of the following lenses would you prefer to use while reading small letters found in a dictionary?

  1. A convex lens of focal length \[\mathbf{50}\text{ }\mathbf{cm}\]

  2. A concave lens of focal length \[\mathbf{50}\text{ }\mathbf{cm}\]

  3. A convex lens of focal length \[\mathbf{5}\text{ }\mathbf{cm}\]

  4. A concave lens of focal length \[\mathbf{5}\text{ }\mathbf{cm}\]

Ans: a) When the object is placed between focus and optic centre, magnified and erect images are formed in a convex lens. So, while reading small letters a convex lens is preferred.

21. We wish to obtain an erect image of an object, using a concave mirror of focal length \[\mathbf{15}\text{ }\mathbf{cm}\]. 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.

Ans: To obtain an erect image in a concave mirror the object should be placed between Focus and the Optic centre.

Here, the focal length of the concave mirror is given as $15cm$.

Therefore, the range of distance of the object from the mirror is from $0cm$ to$15cm$.

The nature of the image is virtual.

The image is larger than the object. 


ray diagram to show the image formation

A virtual, erect and magnified image is formed.


22. Name the type of mirror used in the following situations and support your answer with reason.

  1. Headlights of a car

Ans: In the headlights of a car, a concave mirror is used. Because in concave mirrors a parallel beam of light is produced if the bulb is placed at the focus.

  1. Side/Rear-View Mirror of a Vehicle

Ans: In a side/rear-view mirror of a vehicle , a convex mirror is used. Because when objects are placed in front of the convex mirror, erect and diminished images are formed which gives a wider field of view.

  1. Solar Furnace

Ans: In solar furnaces, Concave mirrors are used. They converge sunlight to a point and produce high temperatures because of their converging properties.


23. 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.

Ans:  Yes, the lens produces a complete image of the object with less intensity.

Consider the following two cases:


when lower half of lens is covered with black paper

In the first case the lower half of the lens is covered with black paper. Light rays coming from the object are refracted only from the upper half and the image is formed, whereas in the lower half the light rays are blocked.


when upper half of lens is covered with black paper

In the second case the upper half of the lens is covered with black paper. Light rays coming from the object are refracted only from the lower half and the image is formed, whereas in the upper half the light rays are blocked.

Therefore, change in intensity of the image is observed i.e., the intensity of the image is less and the complete image is formed. 


24. An object of \[\mathbf{5}\text{ }\mathbf{cm}\] in length is held \[\mathbf{25}\text{ }\mathbf{cm}\] away from a converging lens of focal length \[\mathbf{10}\text{ }\mathbf{cm}\] . Draw the ray diagram and find the position, size and the nature of the image formed.

Ans: It is given that,

Height of the object,${{h}_{o}}=5cm$

Distance of object in front of lens,$u=-25cm$

Distance of image from lens,$v=?$

Focal length of the lens,$f=+10cm$ 

From lens formula:$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ 

$\Rightarrow \frac{1}{v}=\frac{1}{f}+\frac{1}{u}$ 

$\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{25}$ 

$\Rightarrow \frac{1}{v}=\frac{25-10}{250}$ 

$\Rightarrow \frac{1}{v}=\frac{15}{250}$ 

$\Rightarrow v=\frac{250}{15}$ 

$\Rightarrow v=16.66cm$ 

The positive value of $v$ indicates that the image and the object are on opposite sides.

It is known that,

Magnification ,$m=\frac{{{h}_{i}}}{{{h}_{o}}}=-\frac{v}{u}$ 

Height of the image, ${{h}_{i}}=?$ 

$\Rightarrow m=\frac{{{h}_{i}}}{5}=-\frac{16.66}{25}=-0.66$ 

$\Rightarrow {{h}_{i}}=-0.66\times 5$ 

$\Rightarrow {{h}_{i}}=-3.3cm$


convex lens

As the magnification is $-0.66$, the negative sign indicates that the object is inverted and less than $1$ indicates that the image is smaller than the object.

Therefore, the position of the image is $16.66cm$ from the lens. The height of the object is $3.3cm$. The nature of the image is real, inverted and diminished.


25. A concave lens of focal length \[\mathbf{15}\text{ }\mathbf{cm}\] forms an image \[\mathbf{10}\text{ }\mathbf{cm}\] from the lens. Find the distance of an object from the lens? Draw the ray diagram.

Ans: It is given that, 

Focal length of the lens,$f=-15cm$ 

Distance of image from lens,$v=-10cm$

Distance of object in front of lens,$u=?$

From lens formula:$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ 

$\Rightarrow \frac{1}{u}=\frac{1}{v}-\frac{1}{f}$ 

$\Rightarrow \frac{1}{u}=\frac{1}{(-10)}-\frac{1}{(-15)}$ 

$\Rightarrow \frac{1}{u}=\frac{-1}{10}+\frac{1}{15}$ 

$\Rightarrow \frac{1}{u}=\frac{-5}{150}$ 

$\Rightarrow u=-\frac{150}{5}$ 

$\Rightarrow u=-30cm$ 


concave lens

Thus, the object is at a distance of $30cm$ from the lens.


26. An object is placed at a distance of \[\mathbf{10}\text{ }\mathbf{cm}\] from a convex mirror of focal length \[\mathbf{15}\text{ }\mathbf{cm}\] . Find the position and nature of the image. 

Ans: It is given that,

Focal length of the convex mirror,$f=+15cm$ 

Distance of object in front of convex mirror,$u=-10cm$

Distance of image from convex mirror,$v=?$

From mirror formula:$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ 

$\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}$ 

$\Rightarrow \frac{1}{v}=\frac{1}{15}-\frac{1}{(-10)}$ 

$\Rightarrow \frac{1}{v}=\frac{1}{15}+\frac{1}{10}$ 

$\Rightarrow \frac{1}{v}=\frac{25}{150}$ 

$\Rightarrow v=\frac{150}{25}$ 

$\Rightarrow v=6cm$ 

The positive value of $v$ indicates that the image is formed behind the mirror and virtual.

It is known that,

Magnification ,$m=\frac{{{h}_{i}}}{{{h}_{o}}}=-\frac{v}{u}$ 

$\Rightarrow m=\frac{-6}{-10}=+0.6$ 

The positive magnification indicates that the image is erect. As magnification is less than $1$ it indicates that the image is smaller than the object.

Therefore, the position of image is $6cm$ behind the mirror. Nature of image is virtual, erect and diminished.


27. What does “The magnification produced by a plane mirror is \[+1\] ” mean?

Ans: It is known that,

Magnification produced by a plane mirror is,$m=\frac{{{h}_{i}}}{{{h}_{o}}}=-\frac{v}{u}$ 

where,

${{h}_{i}}$ is the height of the image

${{h}_{o}}$ is the height of the object

$u$ is the distance of the object in front of lens

$v$ is the distance of image from lens

Magnification produced by a plane mirror is $+1$ means that ${{h}_{i}}={{h}_{o}}$ that is the size of the image is the same as the size of the object. The positive size indicates that the image is erect.

Therefore, magnification equal to $+1$ means that the size of the image is the same as object and erect.


28. An object \[\mathbf{5}\text{ }\mathbf{cm}\] in length is placed at a distance of \[\mathbf{20}\text{ }\mathbf{cm}\]in front of a convex mirror of radius of curvature\[\mathbf{30}\text{ }\mathbf{cm}\] . Find the position of the image, its nature and size. 

Ans: It is given that,

Distance of object in front of the mirror,$u=-20cm$

Distance of image from the mirror,$v=?$

The radius of curvature of the mirror,$R=30cm$

The focal length of the mirror,$f=?$ 

It is known that,

Radius of curvature is equal to twice the focal length.

$\Rightarrow R=2f$ 

$\Rightarrow 30=2f$ 

$f=\frac{30}{2}=15cm$ 

From mirror formula:$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ 

$\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}$ 

$\Rightarrow \frac{1}{v}=\frac{1}{15}-\frac{1}{(-20)}$ 

$\Rightarrow \frac{1}{v}=\frac{1}{15}+\frac{1}{20}$

$\Rightarrow \frac{1}{v}=\frac{20+15}{300}$ 

$\Rightarrow \frac{1}{v}=\frac{35}{300}$ 

$\Rightarrow v=\frac{60}{7}$ 

$\Rightarrow v=8.57cm$ 

The positive value of $v$ indicates that the image is formed behind the mirror.

It is known that,

Magnification ,$m=\frac{{{h}_{i}}}{{{h}_{o}}}=-\frac{v}{u}$ 

Height of the object, ${{h}_{o}}=5cm$

Height of the image, ${{h}_{i}}=?$ 

$\Rightarrow {{h}_{i}}=-\frac{v}{u}\times {{h}_{o}}$ 

$\Rightarrow {{h}_{i}}=-\frac{8.57}{(-20)}\times 5$ 

$\Rightarrow {{h}_{i}}=2.14cm$

Therefore, the image is formed at a distance of \[8.57cm\]behind the mirror. The nature of the image is virtual, erect and diminished.


29. An object of size \[\mathbf{7}.\mathbf{0}\text{ }\mathbf{cm}\] is placed at \[\mathbf{27}\text{ }\mathbf{cm}\] in front of a concave mirror of focal length \[\mathbf{18}\text{ }\mathbf{cm}\] . At what distance from the mirror should a screen be placed, so that a sharp focused image can be obtained? Find the size and the nature of the image.

Ans: It is given that,

Distance of object in front of mirror,$u=-27cm$

Distance of image from the mirror,$v=?$

Focal length of the mirror,$f=-18cm$ 

From mirror formula:$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ 

$\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}$ 

$\Rightarrow \frac{1}{v}=\frac{1}{-18}-\frac{1}{(-27)}$ 

$\Rightarrow \frac{1}{v}=-\frac{1}{18}+\frac{1}{27}$

$\Rightarrow \frac{1}{v}=\frac{-3+2}{54}$ 

$\Rightarrow \frac{1}{v}=-\frac{1}{54}$ 

$\Rightarrow v=-54cm$ 

The negative value of $v$ indicates that the screen should be placed at a distance of  $54cm$ in front of the mirror and the image is real.

It is known that,

Magnification ,$m=\frac{{{h}_{i}}}{{{h}_{o}}}=-\frac{v}{u}$ 

Height of the object, ${{h}_{o}}=7cm$

Height of the image, ${{h}_{i}}=?$ 

$\Rightarrow {{h}_{i}}=-\frac{v}{u}\times {{h}_{o}}$ 

$\Rightarrow {{h}_{i}}=-\frac{-54}{(-27)}\times 7$

$\Rightarrow {{h}_{i}}=-14cm$

The height of image is \[14cm\].

Therefore, the image is formed at a distance of \[54cm\]in front of the mirror. Nature of image is real, inverted and enlarged.


30. Find the focal length of a lens of power\[-\mathbf{2}.\mathbf{0}\text{ }\mathbf{D}\] . What type of lens is this?

Ans: It is given that,

Power of a lens,$P=-2.0D$ 

Focal length of a lens, $f=?$ 

Power of a lens,$P=\frac{1}{f(metres)}$ 

$\Rightarrow -2=\frac{1}{f}$ 

$\Rightarrow f=-\frac{1}{2}=-0.5m$ 

Negative $f$ indicates concave lens.

Therefore, the focal length of lens is $f=-0.5m$and the lens is concave.


31. A doctor has prescribed a corrective lens of power \[+\mathbf{1}.\mathbf{5}\text{ }\mathbf{D}\]. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Ans: It is given that,

Power of a lens,$P=+1.5D$ 

Focal length of a lens, $f=?$ 

Power of a lens,$P=\frac{1}{f(metres)}$ 

$\Rightarrow 1.5=\frac{1}{f}$ 

$\Rightarrow f=\frac{1}{1.5}=0.66m$ 

Positive $f$ indicates a convex lens.

Therefore, the focal length of lens is $f=0.66m$ and the lens prescribed is a diverging lens.


Topics Covered In Class 10 Science Chapter 9 Light Reflection and Refraction

List of Topics Covered in Science Chapter 9 Class 10

Topics

Subtopics

Reflection of Light


Spherical Mirrors

Image formation by spherical mirrors, Representation using Ray Diagrams, Sign Convention, Mirror Formulas and Magnification

Refraction of Light

Reflection through a rectangular glass slab, The Refractive Index, Refraction by spherical lenses, Image formation using Ray diagrams, Sign convention, Lens formula, and magnification. Power of  a Lens


Key Features of Light Reflection and Refraction Class 10 NCERT Solutions

Class 10th science chapter 9 question answer offers great opportunities for students to achieve maximum marks. To do so, students must have a good understanding of the concepts of this chapter, which is duly sufficient by NCERT Solutions for Class 10 Science Chapter 9. The main benefits of these NCERT Solutions are given below:

  • Light reflection and refraction class 10 questions and answers PDF are given by faculties who are specialists in this field and have excellent experience. The solutions they created are exact and will help you get good test marks.

  • The solutions to various questions in this chapter are clarified with charts and viable diagrams with regard to an easy understanding of students.

  • Step-by-step derivation and application of the mirror formula for concave and convex mirrors.

  • The solutions are designed and created keeping in mind one of the main objectives of helping students in obtaining high marks.

  • Detailed steps and diagrams explain light's refraction through a glass slab and the concept of lateral displacement.

  • Explanation of the working principles of optical instruments such as microscopes and telescopes.

  • The solutions are available for free download in PDF format, allowing students to access them offline at their convenience. The user-friendly format ensures easy navigation through the document.


Important Study Material Links for Class 10 Science Chapter 9


Conclusion

Class 10 science chapter 9 question-answer solutions provide students with simple and detailed definitions and explanations of each concept covered in the chapter. Therefore, it is highly recommended that students download and refer to our comprehensive and expert-curated light class 10 NCERT solutions to get a gist of the chapter before the exam and to know how to answer the questions in the exam. Students can also refer to our plethora of other study resources related to this chapter, which are available for free on our website and mobile app.


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

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



Related Important Links for Class 10 Science

FAQs on NCERT Solutions for Class 10 Science Chapter 9 Light Reflection and Refraction

1. Why do Vehicles have Convex-Shaped rear-view mirrors?

As discussed in class 10th science chapter 9 question answer,  a convex mirror is used as a rearview mirror in vehicles as it gives a more extensive field of view, which causes the driver to see the vast majority of the traffic behind him. Convex mirrors consistently structure an erect, virtual, and decreased picture of the things set in front of it.

2. What is Meant by 1 Dioptre of Power of Lens from light reflection and refraction class 10 questions and answers PDF?

Dioptre is the SI unit of intensity of focal point and is denoted by the letter D. 1 dioptre can be characterized as the intensity of a focal point/power of a lens of central length 1 metre.

3. What are the concepts covered in Chapter 9, Science Light class 10 NCERT solutions?

The concepts included in this chapter are:

Reflection of Light

  • Spherical Mirrors

  • Image Formation by Spherical Mirrors

  • Representation of Images Formed by Spherical Mirrors Using Ray Diagrams

  • Sign Convention for Reflection by Spherical Mirrors

  • Mirror Formula and Magnification

Refraction of Light

  • Refraction through a Rectangular Glass Slab

  • The Refractive Index

  • Refraction by Spherical Mirrors

  • Image Formation by Lenses

  • Image Formation by Lenses Using Ray Diagrams

  • Sign Convention for Spherical Lenses

  • Lens Formula and Magnification

  • Power of a Lens

4. Where can I get the NCERT Solutions of class 10 science chapter 9 question answer?

Students will find the NCERT Solutions of Chapter 9 of Class 10 Science on Vedantu's website.

  • Visit the page for NCERT Solutions for Chapter 9 of Class 10 Science.

  • This link will take you to Vedantu's official webpage (vedantu.com).

  • There you will find the PDF file of the NCERT Solution of Chapter 10 of Class 10 Science.

  • Click on the option of “Download PDF” which is at the top of the PDF file.

  • The file will get downloaded free of cost after clicking the option.

You can also download solutions from the Vedantu app.

5. What do you understand about the refraction of light? Also, state the laws of refraction of light from class 10 Science chapter 9. 

Refraction of light is the process in which the direction and speeds of light change when moving from one medium to another.


Laws of Refraction of Light:

  • The normal, the incident ray and the refracted ray 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 a constant for the light of a given colour and for the given pair of media. This law is known as Snell's law.

6. What is meant by the term magnification of spherical mirrors?

The magnification of spherical mirrors is defined as the ratio of the image's height to the object's height. It is denoted by ‘m’. This magnification is also described in terms of object and image distance. The image distance's negative ratio to the object's distance is known as magnification. The negative magnification shows that the images are real. The positive sign in the value of magnification tells that the image is virtual. Students can refer to light reflection and refraction class 10  NCERT Solutions to learn more about the chapter.

7. Write a short note on the power of a lens from ch 9 Science Class 10?

The power of a lens is defined as the ability of the lens to diverge or converge the light rays depending on the focal length. The short focal length of the concave lens causes higher divergence than the one with a longer focal length. The power of the lens is also defined as the reciprocal of its focal length. It is denoted by “P”. The SI unit of power of the lens is ‘dioptre’. 1 dioptre is described as the power of a lens whose focal length is 1 metre.