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

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.

Important Topics of Class 10 Science Chapter 9 Light Reflection and Refraction

Benefits of Learning with Class 10 Science Light Chapter Class 10 Important Questions

• Light Chapter Class 10 Important Questions help students understand the core concepts of light, such as reflection, refraction, and image formation, ensuring a strong foundation for further studies.

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Conclusion 

The important questions for Class 10 Science Chapter 9, "Light: Reflection and Refraction," provided by Vedantu, offer students a comprehensive way to strengthen their understanding of key concepts like reflection, refraction, and image formation by mirrors and lenses. The light class 10 extra questions, designed in alignment with the CBSE exam pattern, help students practise effectively and clear any doubts related to the chapter. Regular revision and practice using these important questions can significantly build confidence and improve exam performance. Download the FREE PDF from Vedantu for thorough exam preparation.

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