NCERT Solutions for Class 10 Science Chapter 9 Light Reflection and Refraction
NCERT Solutions for Class 10 Science Chapter 9, Light: Reflection and Refraction explains the properties of light that allows us to see the world around us. Through this chapter, we answer questions such as why we can see our reflection in a mirror and why objects appear to be bent when viewed through water. NCERT solution for Class 10 Science chapter 9 will help to understand how light behaves when it bounces back from surfaces and bends while passing through different materials.
The Class 10 Science Chapter 9 of the NCERT Solutions prepared by experts at Careers360 provides detailed solutions to all the questions. You will also find the answers to the in-text questions that guide students to form a strong foundation of the chapter. The article also contains Higher Order Thinking Skills(HOTS) questions, which encourages reasoning and prepares students for higher studies.
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NCERT Solutions for Class 10 Science Chapter 9: Download Solution PDF
Students who wish to access the NCERT class 10 science chapter 9 solutions can click on the given link below to download the complete solution in PDF. It is useful for easy offline access and anytime revision.Download Solution PDF
NCERT Solutions for Class 10 Science Chapter 9: Intext Questions
In this section of NCERT class 10 chapter 9 solution, first you will find the complete question and answer of intext question of the NCERT textbook. It is followed by the complete exercise solutions of the chapter.
Topic 9.2.2 Representation of Images Formed by Spherical Mirrors Using Ray Diagrams
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.
F represents the focal length
Q.2 The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Answer:
$
\text { focal length }=\frac{\text { Radius }}{2}
$
So,
$
\text { focal length }=\frac{20}{2}=10 \mathrm{~cm}
$
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 mirrors usually give a virtual and erect image.
The concave mirror gives a virtual and enlarged image only when the object is between the pole and the 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 increasing 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.
Topic 9.2.4 Mirror formula 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 $(f) \times 2$
So,
$
f=\frac{R}{2}
$
putting the given values,
$
f=\frac{30}{2}=15 \mathrm{~cm}
$
Hence Focus of the convex mirror is 15 cm.
Answer:
Here Given Magnification:
m= -3 (real image)
Given magnification $\mathrm{m}=3$
and object distance $u=10 \mathrm{~cm}$ and we know,
Magnification $\mathrm{m}=\frac{-\mathrm{v}}{\mathrm{u}}$
$
\begin{aligned}
& -\mathrm{v}=3 \times-10 \\
& \mathrm{v}=30 \mathrm{~cm}
\end{aligned}
$
Thus the image is formed at a distance of 30 cm.
The image is 30 cm in front of the mirror.
Topic 9.3 Refraction of light
Answer:
As we know when the light goes from a rare medium to a denser medium the light bends towards the normal. So when the light goes from air to water it will bend toward the normal.
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,
$
\begin{aligned}
& 1.5=\frac{3 \times 10^8}{v} \\
& v=\frac{3 \times 10^8}{1.5} \\
& v=2 \times 10^8 \mathrm{~m} / \mathrm{s}
\end{aligned}
$
Hence the speed of the light in the glass is $2 \times 10^8 \mathrm{~m} / \mathrm{s}$.
| 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 |
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.
| 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 a diamond is 2.42, it means light is travelling from a 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.
Topic 9.3.8 Power of a Lens
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.
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 the image formed is inverted and the size image is equal to the size of the object, magnification (m) = -1.
magnification $(m)=-1$.
$
\begin{aligned}
& m=\frac{v}{u} \\
& -1=\frac{v}{u} \\
& u=-v \\
& u=-50 \mathrm{~cm}
\end{aligned}
$
The needle is placed 50 cm in front of the lens
Using the lens formula we have
$
\begin{aligned}
& \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 \mathrm{~cm}
\end{aligned}
$
Power of the lens P is given by
$
\begin{aligned}
P & =\frac{1}{f} \\
P & =\frac{1}{25 \times 10^{-2}} \\
P & =4 D
\end{aligned}
$
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)
The power of the lens P is given by
$\begin{aligned} & P=\frac{1}{f} \\ & P=\frac{1}{-2} \\ & P=-0.5 \mathrm{D}\end{aligned}$
The Power of the lens is - 0.5 Dioptre.
NCERT Class 10 Science Chapter 9: Exercise Questions
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.
(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:
The position of the object should be between the pole of the mirror and its principal focus.
(d) is the correct answer.
(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:
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
(b) is the correct answer.
(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.
Answer:
The mirror and the lens are both concave.
(a) is the correct answer.
Q 5. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
(a) only plane.
(b) only concave.
(c) only convex.
(d) either plane or convex.
Answer:
The mirror is likely to be either plane or convex.
(d) is the correct answer.
(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 a 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.
Answer:
$
\begin{aligned}
\frac{1}{f} & =\frac{1}{v}+\frac{1}{u} \\
v & =\frac{f u}{u-f}
\end{aligned}
$
For the image to be erect it has to be virtual and therefore as per convention $v$ has to be positive
$
\begin{aligned}
& \mathrm{v}>0 \\
& \frac{f u}{u-f}>0 \\
& \mathrm{u}<0 \\
& \mathrm{f}<0 \\
& \mathrm{fu}>0 \\
& \mathrm{u}-\mathrm{f}>0 \\
& \mathrm{u}>\mathrm{f}
\end{aligned}
$
Therefore, the object must be placed between the pole and the focus of the mirror. i.e $-15 \mathrm{~cm}<\mathrm{u}<0$
The image formed would be virtual and larger than the object.
Q 8. (a) Name the type of mirror used in the following situations.
- Headlights of a car.
Support your answer with reason.
Answer:
A concave mirror is used for the 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.
- Side/rear-view mirror of a vehicle
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.
- Solar furnace.
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.
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 the 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: The size of the image is the same but the brightness reduces
Answer:
Object distance u = -25 cm.
Focal length = 10 cm
Let image distance be v
$
\begin{aligned}
& \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 \mathrm{~cm}
\end{aligned}
$
The position of the image is 16.66 cm on the other side of the lens.
Object size $\mathrm{O}=5 \mathrm{~cm}$.
Let the image size be I
Magnification is m
$
\begin{aligned}
m & =\frac{v}{u} \\
\frac{I}{O} & =\frac{16.66}{-25} \\
I & =5 \times-\frac{16.66}{25} \\
I & =-3.33 \mathrm{~cm}
\end{aligned}
$
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
Answer:
Focal length, f = -15 cm
Image distance, v = -10 cm ( As in the 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
$
\begin{aligned}
& \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 \mathrm{~cm}
\end{aligned}
$
The object is placed at a distance of 30 cm from the lens.
Answer:
Object distance, u = -10 cm
Focal length, f = 15 cm
Let the image distance be v
As per the mirror formula
$
\begin{aligned}
& \frac{1}{f_1}=\frac{1}{v}+\frac{1}{u} \\
& \frac{1}{15}=\frac{1}{v}+\frac{1}{-10} \\
& v=6 \mathrm{~cm} \\
& \text { Magnification }=-\frac{v}{u} \\
& =-\frac{6}{-30} \\
& =0.2
\end{aligned}
$
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.
Answer:
The radius of curvature, R = 30 cm
Focal length, f = R/2 = 15 cm
Object distance, u = -20 cm
Let the image distance be $v$
$
\begin{aligned}
& \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\
& \frac{1}{15}=\frac{1}{v}+\frac{1}{-20} \\
& v=8.57 \mathrm{~cm}
\end{aligned}
$
Since v is positive image is formed behind the mirror.
$
\begin{aligned}
& \text { Magnification }=-\frac{v}{u} \\
& =-\frac{8.57}{-20} \\
& =0.428
\end{aligned}
$
Object size $=5 \mathrm{~cm}$
$
\begin{aligned}
& \text { Image size }=\text { Object size } \times \text { Magnification } \\
& =5 \times 0.428 \\
& =2.14 \mathrm{~cm}
\end{aligned}
$
A virtual, erect and diminished image of size 2.14 cm would be formed 8.57 cm behind the mirror.
Answer:
Object distance, u = -27 cm
Focal length, f = -18 cm
Let the image distance be v
$
\begin{aligned}
& \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\
& \frac{1}{-18}=\frac{1}{v}+\frac{1}{-27} \\
& v=-54 \mathrm{~cm}
\end{aligned}
$
The negative sign shows the image is formed in front of the mirror and is real.
Object size $=7 \mathrm{~cm}$
$
\begin{aligned}
& \text { Image size }=\text { magnification } \times \text { Object size } \\
& =-\frac{v}{u} \times O \\
& =-\frac{-54}{-27} \times 7 \\
& =-14 \mathrm{~cm}
\end{aligned}
$
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:
$\begin{aligned} & f=\frac{1}{P} \\ & f=\frac{1}{-2} \\ & f=-0.5 \mathrm{~m} \\ & f=-50 \mathrm{~cm}\end{aligned}$
The lens is concave because it's focal length is negative and is equal to -50 cm.
Answer:
$\begin{aligned} & f=\frac{1}{P} \\ & f=\frac{1}{1.5} \\ & f=0.6667 \mathrm{~m} \\ & f=66.67 \mathrm{~cm}\end{aligned}$
The less is convex because its focal length is positive and therefore is converging.
Class 10 Science NCERT Chapter 9: Higher Order Thinking Skills (HOTS) Questions
Q1. There is a point object and a plane mirror. If a mirror is moved by 10cm away from the object then the image will move by?
Answer:
Image formation from plane mirror -
1) Distance of object from mirror = Distance of image from the mirror.
2) Line joining a point object and its image is normal to the reflecting surface
3) The size of the image is the same as that of the object.
4) For a real object the image is virtual and for a virtual object, the image is real.
After moving the mirror by 10cm the object distance = x +10 cm
Image distance from m = x+ 10cm
Movement of the image from m
= (x+20) -x = 20 cm
Hence, the answer is 20cm
Q2. A light ray is an incident on a glass sphere of refractive index $\mu=\sqrt{3}$ at an angle of incidence 600 as shown the total deviation after two refractions is:
Answer:
At point $A$ :
$
\begin{aligned}
& \text { 1. } \sin 60^{\circ}=\sqrt{3} \sin r \\
& \Rightarrow r=30^{\circ}
\end{aligned}
$
from symmetry
$
r^{\prime}=r=30^0
$
Apply Snell's law at B.
$
\begin{aligned}
& \text { 1. } \sin e=\sqrt{3} \sin r^{\prime}=\frac{\sqrt{3}}{2} \\
& \Rightarrow e=60^{\circ} \\
& \delta_1=60^{\circ}-30^0=30^0 \\
& \delta_2=e-r^1=60^{\circ}-30^{\circ}=30^{\circ} \\
& \therefore \text { total deviation }=60^{\circ}
\end{aligned}
$
Q3. An object is at a distance of 10 cm from a mirror, and an image of the object is at a distance of 30 cm from the mirror from the same side as the object. Then the nature of the mirror and its power is -
Answer:
Optical power of a mirror -
$
P=-\frac{1}{f}
$
- wherein
$f=$ focal length with a sign and is in meters unit of power = diopter
$
\begin{aligned}
& \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\
& u=-10 \mathrm{~cm}, \mathrm{v}=-30 \mathrm{~cm} \\
& \frac{1}{f}=\frac{-1}{10}-\frac{1}{30}=-\frac{4}{30} \text { or } \mathrm{f}=-7.5 \mathrm{~cm}
\end{aligned}
$
$\therefore$ mirror is concave
$
p=\frac{-100}{f} D=\frac{100}{7.5} D=\frac{40}{3} D
$
Q4. An object is at a distance of 10 cm from a mirror and the image of the object is at a distance of 30 cm from the mirror on the same side as the object. The nature of the mirror and its focal length are:
Answer:
Sign Convention
1) All distances are measured from the pole.
2) Distance measured in the direction of incident rays is taken as positive.
3) Distance measured in the direction opposite to that of incident rays is taken as negative.
4) Distance above the principal axis as positive and below the principal axis as negative.
$u=-10 \mathrm{~cm}$
$v=-30 \mathrm{~cm}$ from sign convention
from mirror formula
$
\begin{aligned}
& \frac{1}{f}=\frac{1}{v}+\frac{1}{u}=-\frac{1}{30}-\frac{1}{10}=\frac{-4}{30} \\
& \mathrm{f}=-7.5 \mathrm{~cm}
\end{aligned}
$
Since focal length is negative hence its mirror concave.
NCERT Solutions for Class 10 Physics Chapter 9: Topics
The following are the topics covered in the NCERT Solutions of Class 10 Physics Light- Reflection and Refraction.
9.1 Reflection of Light
9.2 Spherical Mirrors
9.2.1 Image Formation by Spherical Mirrors
9.2.2 Representation of Images Formed by Spherical Mirrors Using Ray Diagrams
9.2.3 Sign Convention for Reflection by Spherical Mirrors
9.2.4 Mirror Formula and Magnification
9.3 Refraction of Light
9.3.1 Refraction through a Rectangular Glass Slab
9.3.2 The Refractive Index
9.3.3 Refraction by Spherical Lenses
9.3.4 Image Formation by Lenses
9.3.5 Image Formation in Lenses Using Ray Diagrams
9.3.6 Sign Convention for Spherical Lenses
9.3.7 Lens Formula and Magnification
9.3.8 Power of a Lens
Class 10 Science Light- Reflection and Refraction: Important Formulae
Some of the important formulae of NCERT Class 10 Chapter 9 are given below:
Mirror Formula (for Spherical Mirrors)
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
where,
v= Image distance
u= Object distance
f= Focal length of the mirror
Magnification for Mirrors
$\begin{aligned}
&m=\frac{\text { height of the image }\left(h^{\prime}\right)}{\text { height of the } \operatorname{object}(h)}\\
&\text { or, in terms of distances: }\\
&m=\frac{-v}{u}
\end{aligned}$
- If m is positive, the image is virtual and erect.
- If m is negative, the image is real and inverted.
Lens Formula (for Convex and Concave Lenses)
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
Refractive Index
$n_{21}=\frac{\sin i}{\sin r}=\frac{v_1}{v_2}$
where,
$n_{21}=$ Refractive index of medium 2 with respect to medium 1.
$i=$ Angle of incidence (angle between the incident ray and the normal).
$r=$ Angle of refraction (angle between the refracted ray and the normal).
$v_1=$ Speed of light in medium 1.
$v_2=$ Speed of light in medium 2.
Absolute Refractive Index
$n=\frac{c}{v}$
where,
$n=$ Absolute refractive index of the medium
$c=$ Speed of light in vacuum $\left(3.0 \times 10^8 \mathrm{~m} / \mathrm{s}\right)$
$v=$ Speed of light in the given medium
Power of a Lens
$P=\frac{1}{f}$
where:
$\begin{aligned} & P=\text { Power of the lens (in Dioptres (D)) } \\ & f=\text { Focal length of the lens (in meters) }\end{aligned}$
$P_{\mathrm{net}}=P_1+P_2+P_3+\ldots$
Approach to Solve Questions of Light Reflection and Refraction Class 10
How and when to solve the exercise problem is a very curious questions which arise in every student's mind, so here you will get some of the pointers which will help to find answers to these questions to score good marks in this chapter of class 10.
- Laws of Reflection & Refraction – Learn both laws for reflection and refraction.
- Mirror Formula & Magnification – Learn the mirror formula and magnification, use correct sign conventions, which helps in solving numericals problems
- Ray Diagrams – Practice formation of images for concave, convex lenses and mirrors using Ray Diagrams.
- Lens Formula & Power of Lens – Apply the lens formula and power of lens, to indentify the nature of lens, whether it is converging or diverging.
- Sign Conventions – Learn the sign convention rules for lens which helps in solving numerical problems
- Real-life Applications – Connect what you learn to everyday things like mirrors, magnifying glasses, and eyeglasses. It makes the concepts easier to understand and remember.
NCERT Solutions for Class 10 Science Chapter-wise
Frequently Asked Questions (FAQs)
Q: How can I get the PDF of NCERT solutions for Class 10 Chapter 9 pdf?
A:
The NCERT solutions class 10 Science chapter 9 PDF can be accessed using the download button. By downloading, you can view the solutions offline.
Q: What are the real-life applications of reflection and refraction?
A:
Reflection Applications: Used in vehicle mirrors, periscopes, telescopes, and makeup mirrors.
Refraction Applications: Found in eyeglasses, cameras, optical fibers, prisms, and mirages.
Q: How to solve ray diagram questions in Class 10 Science Chapter 9?
A:
To solve ray diagram questions, identify the type of mirror/lens, follow the standard rules for drawing rays, locate the image where rays intersect, and determine its nature (real/virtual, magnified/diminished). Practice different cases for better understanding.
Q: What is reflection and refraction in Chapter 9 of Class 10 Science?
A:
Reflection is the bouncing back of light when it strikes a smooth surface, following the laws of reflection. Refraction is the bending of light as it passes from one medium to another due to a change in speed.
Q: What are the important topics of Class 10 Science Chapter 9 Light Reflection and Refraction?
A:
Centre of Curvature, Principal Axis, Principal Focus, Focal Length, Images Formed by Spherical Mirrors, Mirror formula & Magnification, Refractive index, Refraction & Laws of refraction, Lens Formula & Magnification, Power of a lens.
Q: What is the weightage of Chapter 9 Light Reflection and Refraction in CBSE class 10 board final exam?
A:
6-7 marks. All the formulas and raydiagrams of the chapter are important. To get more problems of Light Reflection and Refraction refer to CBSE previous year papers, NCERT book and NCERT exemplar for Class 10.
Q: What are the benefits of referring the Caerres360 NCERT Solutions for Class 10 Science Chapter 9?
A:
Advantages of using Careers360 NCERT Solutions for reflection of light class 10:
The in-depth explanation for better understanding of concepts.
Aligned with the CBSE syllabus.
Authentic and precise for improving problem-solving skills.
Free PDF downloads for easy access by students.
Q: What are the topics and question count of NCERT Solutions for Class 10 Science Light Reflection and Refraction?
A:
Topics in NCERT Solutions for class 10 light reflection and refraction:
Light Reflection
Spherical Mirrors
Image Formation with Spherical Mirrors
Ray Diagrams for Mirror Image Representation (4 questions)
Mirror Formula and Magnification (2 questions)
Light Refraction (5 questions)
Lens Power (3 questions)
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