CBSE Class 10 Important Questions on Science Chapter 9 - Light Reflection and Refraction 2024-25

The following list of important topics covered under the chapter on Light Reflection and Refraction has been provided for students so that they can take a glance at the major concepts and read them through before diving into the important questions.

• Reflection of Light

• Spherical Mirrors

• Image Formation by Spherical Mirrors

• Sign Convention for Spherical Mirrors

• Mirror Formula and Magnification

• Refraction of Light

• Refraction through a Rectangular Glass Slab

• Refractive Index

• Refraction by Spherical Lenses

• Image Formation by Lenses and Their Ray Diagrams

• Sign Convention for Spherical Lenses

• Lens Formula and Magnification

• Power of a Lens

Study Important Questions for Class 10 Science Chapter 9 - Light Reflection and Refraction

Long Answer Questions (5 Marks)

1. A convex lens has a focal length of \[10\] cm. At what distance from the lens should the object be placed so that it forms a real and inverted image $20$ cm. away from the lens? What would be the size of the image formed if the object is $2$ cm high? With the help of a ray, the diagram shows the formation of the image by the lens in this case?

Ans: Given, the focal length of the convex lens,

$\Rightarrow f=+10$ cm

Also, given the image formed is real and inverted with the image distance as $20$ cm.

$\Rightarrow v=+20$ cm

From the lens formula, we have: 

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

$\Rightarrow \dfrac{1}{10}=\dfrac{1}{20}-\dfrac{1}{u}$

$\Rightarrow \dfrac{1}{u}=\dfrac{1}{20}-\dfrac{1}{10}$

$\Rightarrow \dfrac{1}{u}=\dfrac{-1}{20}$

$\Rightarrow u=-20$ cm.

$\therefore $ The object is placed at a distance of $20$ cm.

Magnification is given as,

$\Rightarrow m=-\dfrac{v}{u}$

$\Rightarrow m=-\dfrac{20}{\left( -20 \right)}$

$\Rightarrow m=+1$

Also, magnification is given by, $m=\dfrac{Height\text{ of the image}}{Height\text{ of the object}}$.

$\therefore m=\dfrac{Height\text{ of the image}}{2}$

$\Rightarrow 1=\dfrac{Height\text{ of the image}}{2}$

$\Rightarrow Height\text{ of the image}=2$ cm.

Thus, the image is of the same size as that of the object and it is real and inverted.

The ray diagram representing the formation of the image by the lens in this case is:

It is observed that the image is formed at $2{F_2}$ with the object placed at $2{F_1}$.

2. Draw a ray diagram to show the use of a convex lens for the formation of images having the following characteristics. 

a. Real & inverted and diminished 

Ans: A ray diagram representing a real & inverted and diminished image is given below.

b. Virtual, erect & magnified.

Ans: A ray diagram representing a virtual, erect, and magnified image is given below.

3. 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 objects? Also, find the power of the lens.

Ans: Given, the image is real and inverted at a distance of $50$ cm.

$\Rightarrow v=+50$ cm

Also given, Height of image $=$ Height of object

We know, magnification is given as,

$\Rightarrow m=-\dfrac{v}{u}$

Also, magnification is given by, $m=\dfrac{Height\text{ of the image}}{Height\text{ of the object}}$.

$\therefore -\dfrac{v}{u}=\dfrac{Height\text{ of the image}}{Height\text{ of the object}}$

Since, Height of image $=$ Height of object and $v=+50$ cm

$\Rightarrow -\dfrac{50}{u}=1$

$\Rightarrow u=-50$ cm

From the lens formula, we have: 

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

$\Rightarrow \dfrac{1}{f}=\dfrac{1}{50}-\dfrac{1}{-50}$

$\Rightarrow f=25$ cm

$\Rightarrow f=0.25$ m

The power of a lens is given by,

$\Rightarrow P=\dfrac{1}{f}$

$\Rightarrow P=\dfrac{1}{0.25}$  m

$\Rightarrow P=+4$ D

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

Ans: If we cover one-half of a convex lens with black paper it is observed that it produces a complete image of the object.

Representing the given situation with a ray diagram.

Adjust the position of the burning candle such that its image is formed on the screen placed on the other side of the lens.

We get a full image of the burning candle on the screen.

Thus, from the above observation, we conclude that the formation of an image does not depend on the length of the lens, but the brightness of the image formed on the screen depends since the number of rays passing through the lens gets reduced on covering half the lens with black paper.

5. 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.

Ans: Given focal length of the lens, $f=+10$ cm.

Object distance from the converging lens is given to be $25$ cm and the height of the object ($ho$) is given to be $5$ cm.

Thus,

$\Rightarrow u=-25$ cm

$\Rightarrow ho=5$ cm

From the lens formula, we have: 

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

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

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

$\Rightarrow v=\dfrac{50}{3}$ cm

$\Rightarrow v=16.7$ cm

Thus, we conclude that the image formed is real and inverted and is formed at a distance of $16.7$ cm on the other side of the lens.

We know, magnification is given as,

$\Rightarrow m=-\dfrac{v}{u}$

Also, magnification is given by, $m=\dfrac{Height\text{ of the image}}{Height\text{ of the object}}=\dfrac{hi}{ho}$.

$\therefore -\dfrac{v}{u}=\dfrac{hi}{ho}$

$\Rightarrow \dfrac{16.7}{-25}=\dfrac{hi}{5}$

$\Rightarrow hi=-\dfrac{10}{3}$ cm

Hence, the image is diminished.

The ray diagram representing the above situation is drawn below.

6. A convex lens of focal length $15$ cm formed an image $10$ cm from the lens. How far is the object placed from the lens? Draw the ray diagram.

Ans: Given focal length of the lens, $f=15$ cm.

Image distance from the converging lens is given to be $10$ cm.

Thus,

$\Rightarrow v=-10$ cm

We observe that the image distance is less than the focal length, hence the image formed would be virtual and erect.

From the lens formula, we have: 

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

$\Rightarrow \dfrac{1}{15}=\dfrac{1}{-10}-\dfrac{1}{u}$

$\Rightarrow \dfrac{1}{u}=-\dfrac{1}{10}-\dfrac{1}{15}$

$\Rightarrow u=-6$ cm

The ray diagram representing the above situation is drawn below.

Light: Reflection and Refraction - Important Questions and Solutions Summary

light rays, mirrors, lenses, prisms, etc., is the advanced version of what you As mentioned earlier, the chapter related to light, reflection, and refraction of have studied in the previous classes. In Class 10, you will have to focus on the theoretical part, as well as, the use of these theories to answer questions.

After you have completed answering the basic questions in the exercise, you might need a question bank to learn more about this chapter. This is where Light Reflection and Refraction Class 10 Important Questions with Answers can be used perfectly. These questions will intellectually challenge your knowledge and foundation of concepts based on this topic. You can answer on your own and check whether you know the right answer or not.

Class 10 Light Reflection and Refraction Important Questions have been developed by the expert teachers to provide a strong idea of how questions are set in the exams. All these questions are developed to define how theoretical concepts are used to solve conceptual problems. You will gain a better insight into the chapters and strengthen your foundation accordingly.

Benefits of Studying the CBSE Class 10 Important Questions on Science Chapter 9 - Light Reflection and Refraction

The Important Questions for Class 10 Light Reflection and Refraction have been developed based on the basic and advanced concepts you have learned in your science textbook. Let us check the benefits of studying these questions first.

Thinking Out of the Box

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Getting to Know Possible Important Questions of Chapter 9 Class 10 Science

By studying Class 10th Science Chapter 9 Important Questions, you will get to know possible questions that might come in the board exams. All these questions are developed by experienced science teachers. They are well-aware of the type of questions asked in the CBSE exam. Hence, studying these questions will help you learn about different modes and patterns.

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The Ch 9 Science Class 10 Important Questions will also come with a proper solution. The solutions will have the perfect answers to all these important questions. You will get another set of questions and answers to follow and find out the best approaches to score well in the exams. These approaches are simplified and designed by following the CBSE guidelines.

Mastering Solving Numerical Questions

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