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

NCERT Solutions For Class 10 Science Chapter 9 Light-Reflection and Refraction, this chapter covers the concepts of light, focusing on the phenomena of reflection and refraction using the straight-line propagation of light. These fundamental ideas are essential for understanding various optical phenomena observed in nature. NCERT Solutions For Class 10 Science Chapter 9 explores the behavior of light with spherical mirrors and its refraction, highlighting their practical applications in real-life scenarios.

1.0Download Class 10 Science Chapter 9 Light Reflection and Refraction NCERT Solutions: Free PDF

NCERT Solutions For Class 10 Science Chapter 9 Light- Reflection and Refraction .Practicing these solutions will help students develop a strong foundation in Physics and gain clarity on how to approach related problems effectively, ultimately aiding in securing good scores in board exams. For a detailed understanding, students can download the NCERT Class 10 Science Chapter 9 PDF solution below, curated by ALLEN’s experts.

2.0NCERT Solutions Class 10 Science Chapter 9 Light Reflection and Refraction : Overview

Before getting into the details of NCERT Solutions Class 10 Science Chapter 9 Light Reflection and Refraction, let's take a brief look at the topics and subtopics covered in this chapter of the NCERT Class 10 Science book.:

Topics covered in this chapter:

• Reflection of Light
• Spherical Mirrors
• Image Formation by Spherical Mirrors
• Representation of Images Formed by Spherical Mirrors Using Ray Diagram 
• Mirror Formula and Magnification
• Refraction of Light 
• Power of Lens 

3.0Reflection of Light

Reflection occurs when light strikes a surface and bounces back into the same medium. The laws of reflection state that the angle of incidence equals the angle of reflection.

Also Read Optic Reflection of Light

4.0Spherical Mirrors

• Concave Mirrors: Used in headlights and searchlights.
• Convex Mirrors: Common in rearview mirrors.

Also Read Reflection From Spherical Mirrors

5.0Some Basic Terms related to Spherical Mirrors

6.0Sign Conventions for Spherical Mirrors

7.0Mirror Formula

• The relationship between the object distance(u),the image distance(v) and the focal lengthf is given by Mirror Formula.

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

8.0Magnification

• The ratio of height image $h_2$ to the height of object $h_1$ is called magnification or linear magnification.

$m=\frac{h_2}{h_1}$

• Magnification is also related to object distance and image distance.

$m=\frac{h_2}{h_1}=-\frac{v}{u}$

• Magnification also expressed as

$m=\frac{f}{f-u}=\frac{f-v}{f}$

9.0Refraction of Light

• The phenomenon of change in path of light when it passes from one medium to another is called Refraction.

Also Read Optic Refraction of Light

Laws of Refraction

1. The incident ray, the refracted ray, and the normal at the point of incidence are all contained within a single plane.

$\frac{Sini}{Sinr}=Constant(\mu )$

10.0Lens Formula and Magnification

The relationship between object distance(u) ,image distance(v) and the focal length(f)

Is given by $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

Magnification-The ratio of height of image $(h_2)$ to the height of object $(h_1)$.

$m=\frac{h_2}{h_1}=+\frac{v}{u}$

$m=\frac{f}{f+u}=\frac{f-v}{f}$

11.0Power of Lens

• It is the extent of convergence or divergence of light rays. Falling on it. A lens of short focal length bends the light rays more through large angles.
• The power of a lens is defined as the inverse of its focal length.

$P=\frac{1}{F}$

• Units is Dioptre(D)
• For convex lens power is positive and for concave lens power is negative.

12.0Class 10 Science Chapter 9 Light Reflection and Refraction Part 1 NCERT: Detailed Solutions

1. Define the principal focus of a concave mirror.

Solution:

The principal focus of a concave mirror is a point on the principal axis, at which the incident rays parallel to the principal axis, after reflection, actually meet.

2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?

Solution:

Given, radius of curvature, R = 20 cm

Focal length, f = (1/2) × radius of curvature

or f = (1/2) × R = (1/2) × 20 = 10 cm

3. Name a mirror that can give an erect and enlarged image of an object.

Solution:

Concave mirror

4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?

Solution:

A convex mirror is preferred as a rear-view mirror because it always gives an erect, though diminished image. Also, it has a wider field of view as it is curved outwards. Thus, a convex mirror enables the driver to view a much larger area than would be possible with a plane mirror.

5. Find the focal length of a convex mirror whose radius of curvature is 32 cm.

Solution:

Given, radius of curvature, R = +32 cm

Focal length, f = (1/2) × radius of curvature

or f = (1/2) × R = (1/2) × (+32) = **+16 cm**

6. A concave mirror produces a three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?**

Solution:

Given, magnification, m = -3 (negative sign is taken as the image is real);

object distance, u = -10 cm

Now, m = -v/u or v = -m × u

or v = -(-3) × (-10) = **-30 cm**

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

(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

Solution:

Option (d) is correct. When an object is placed between the pole and principal focus of a concave mirror, the image formed is virtual, erect, and larger than the object.

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

(a) plane

(b) concave

(c) convex

(d) either plane or convex

Solution:

Option (d) is correct. A convex mirror always gives a virtual and erect image of a smaller size than the object placed in front of it. Similarly, a plane mirror will always give a virtual and erect image of the same size as that of the object placed in front of it. Therefore, the given mirror could be either plane or convex.

9. We wish to obtain an erect image of an object, using a concave mirror of focal length 15 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.

Solution:

The object should be placed between the pole and the focus to get an erect image. Thus, the range of distance from the mirror will be between 0 to 15 cm (or less than 15 cm), i.e., the range is 0 cm - 15 cm.

(Ray diagram would be included here in a visual format in a Word document)

10. Name the type of mirror used in the following situations

(a) Headlights of a car.

(b) Side/rear-view mirror of a vehicle.

(c) Solar furnace.

Support your answer with reason.

Solution

(a) Concave mirror.

A concave mirror is used in the headlights of a vehicle because it produces a powerful and almost parallel beam of light when the light source is placed at its principal focus.

A bulb placed at the focus of a concave mirror produces a strong, almost parallel beam.

(b) Convex mirror.

A convex mirror is used in the rear-view mirror of a vehicle. A convex mirror gives a virtual, erect, and diminished image of the object placed in front of it. Because of this, it has a wide field of view. It enables the driver to see most of the traffic behind him.

A convex mirror has a wide field of view (c) Concave mirror.

Concave mirrors are converging mirrors. That is why they are used to construct solar furnaces. Concave mirrors converge the parallel light incident on them at a single point which is called principal focus. Hence, they can be used to produce a large amount of heat at that point.

A solar furnace placed at the focus of a concave mirror.

11. An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.

Solution:

Given, object distance, u = -10 cm;

focal length, f = +15 cm;

image distance, v = ? ; magnification, m = ?

Mirror formula:

1/v + 1/u = 1/f

or 1/v + 1/(-10) = 1/(+15)

or 1/v = 1/10 + 1/15 = (3 + 2)/30 = 5/30 = 1/6

or v = +6 cm

Now, magnification,

m = -v/u = -(+6)/(-10) = +0.6

The image is located 6 cm behind the mirror; it is a virtual, erect, and diminished image.

12. The magnification produced by a plane mirror is +1. What does this mean?

Solution:

Magnification for a plane mirror is +1. The positive sign means the image is virtual and erect. The numerical value (magnitude) '1' means that the size of the image is exactly equal to the size of the object.

13. An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature, and size.

Solution:

Given, object distance, u = -20 cm;

radius of curvature, R = +30 cm;

height of object, h1 = +5 cm;

image distance, v = ? ; magnification, m = ?; height of image, h2 = ?

Focal length, f = R/2 = (+30)/2 = +15 cm

Mirror formula:

1/v + 1/u = 1/f

or 1/v + 1/(-20) = 1/(+15)

or 1/v = 1/20 + 1/15 = (3 + 4)/60 = 7/60

or v = +60/7 cm = +8.57 cm

Now, magnification,

m = -v/u = -(+60/7)/(-20) = +3/7

or m = +0.428

Also, m = h2/h1 or +3/7 = h2/(+5)

or h2 = +(15/7) = +2.14 cm

The image is located at a distance of 8.57 cm behind the mirror; it is virtual, erect, and diminished with a size of 2.14 cm.

14. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 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.

Solution:

Given, object height, h1 = +7 cm;

object distance, u = -27 cm;

focal length, f = -18 cm;

image distance, v = ? ; image height, h2 = ?

Mirror formula:

1/v + 1/u = 1/f

or 1/v + 1/(-27) = 1/(-18)

or 1/v = 1/27 - 1/18 = (2 - 3)/54 = -1/54

or v = -54 cm

Now, magnification,

m = -v/u = -(-54)/(-27) = -2

Also, m = h2/h1 or -2 = h2/(+7)

or h2 = -14 cm

The image is located at a distance of 54 cm in front of the mirror; it is real, inverted, and magnified with a size of 14 cm.

13.0Class 10 Science Chapter 9 Light Reflection and Refraction Part 2 NCERT : Detailed Solutions

1. A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?

Solution:

The light ray bends towards the normal. This is because it enters from a rarer medium (air) to a denser medium (water).

2. Light enters from air to glass having a refractive index of 1.50. What is the speed of light in the glass? The speed of light in a vacuum is 3 × 10⁸ ms⁻¹.

Solution:

We know that the absolute refractive index (n) of a medium is given by:

n = c / v  or  v = c / n

where:

c = speed of light in vacuum

v = speed of light in the medium

Thus, the speed of light in glass (v_glass) is:

v_glass = c / n_glass

v_glass = (3 × 10⁸ m/s) / 1.5

v_glass = 2 × 10⁸ m/s

3. Find out, from Table given below, the medium having highest optical density. Also find the medium with lowest optical density.

Solution

Among the media given in the above table, diamond has highest optical density as its refractive index is maximum among the given media. Air has least optical density as its refractive index is minimum among the given media.

4. You are given kerosene, turpentine and water. Using the table given in question 3, find in which of these does the light travel fastest.

Solution:

The light travels fastest in water as its refractive index is least (v ∝ 1/n) among the three given substances.

5. The refractive index of diamond is 2.42. What is the meaning of this statement?

Solution:

We know that the absolute refractive index (n) of a medium is given by:

n = c/v  or  v = c/n  i.e., v ∝ 1/n

Since the refractive index of diamond is 2.42, this suggests that the speed of light in diamond will reduce by a factor of 2.42 compared to its speed in air.

6. Define 1 dioptre of power of a lens.

Solution:

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

7. 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 the object? Also, find the power of the lens.

Solution:

Given, v = +50 cm (positive sign is taken because the real image in a convex lens is formed on the right side).

Since the image is real and the size of the image is equal to the size of the object, the magnification (m) = -1.

Now, m = -v/u  or  -1 = -(+50)/u

or u = -v = -(+50) = **-50 cm**

By the lens equation:

1/v - 1/u = 1/f

1/(+50) - 1/(-50) = 1/f

1/f = 2/50 = 1/25

or f = +25 cm = +0.25 m

Power, P = 1/f = 1/(+0.25) = +4 Diopter

8. Find the power of a concave lens of focal length 2 m.

Solution:

Given, focal length = -2 m (focal length of a concave lens is negative).

Power, P = 1/f = 1/(-2.0)

or P = -0.5 Diopter

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

(a) Water

(b)Glass

(c) Plastic

(d) Clay

Solution

(d) Clay is not transparent; therefore, it cannot be used for making lens.

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

(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.

Solution

When an object is placed at 2 F1​ in front of a convex lens, its image is formed at 2 F2​ on the other side of the lens. The image formed is real, inverted, and of the same size as the object.

Thus, option (b) is correct i.e., the object should be placed at twice the focal length from the optical centre of the lens.

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

(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

Solution

Option (a) is correct i.e., a concave mirror as well as concave lens both have negative focal length.

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

(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

Solution

Option (c) is correct.

For reading small letters in a dictionary, a convex lens of small focal length is used as it is kept quite close to the page of the dictionary.

13. 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.

Solution

The convex lens will form a complete image of an object, even if its one half is covered with black paper. Only the brightness of the image will be reduced, in this case it will be half of the brightness of the original image. It can be understood by the following two cases:

(a) Let the upper half of the lens be covered and an object be placed between the optical centre and the focus F1​. The light rays from the object falling on the lower half of the lens form a virtual, erect and magnified image [see figure(a)]

(b) Let the lower half of the lens be covered and an object be placed between optical centre and the focus F1​. The light rays from the object falling on the upper half of the lens form a virtual, erect and magnified image [see figure(b)]

14. 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 the nature of the image formed.

Solution:

Given, object height (h₁) = +5 cm; object distance (u) = -25 cm; focal length (f) = +10 cm.

By the lens equation:

1/v - 1/u = 1/f

1/v - 1/(-25) = 1/(+10)

1/v + 1/25 = 1/10

1/v = 1/10 - 1/25 = (5 - 2) / 50 = 3/50

Therefore, image distance (v) = +50 / 3 = +16.67 cm

Magnification (m) = +v/u = +(50/3) / (-25) = -2/3

Now, magnification (m) = h₂/h₁

-2/3 = h₂ / 5

Therefore, image height (h₂) = -2/3 × 5 = -10/3 = -3.33 cm

15. A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.

Solution:

Given, focal length (f) = -15 cm, image distance (v) = -10 cm (negative sign is taken because the image formed in a concave lens is always on the same side as the object).

By the lens equation:

1/v - 1/u = 1/f

1/(-10) - 1/u = 1/(-15)

-1/10 + 1/15 = 1/u

1/u = (-3 + 2) / 30 = -1/30

Therefore, object distance (u) = -30 cm

16. Find the focal length of a lens of power -2.0 D. What type of lens is this?

Solution:

Given, power, P = -2.0 D

Power, P = 1/f

or f = 1/P = 1/(-2) = -0.5 m

= -50 cm

The lens is a concave (or diverging) lens.

17. A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Solution:

Given, power, P = +1.5 D

Power, P = 1/f

or f = 1/P = 1/(+1.5) = 10/15 = +2/3 m

or f = +0.66 m = +66 cm

The lens is a convex (or converging) lens.

14.0Benefits of Light Reflection and Refraction 

• Foundation for Higher Studies: It forms the basis for understanding more complex concepts in Physics, which are covered in higher classes.
• Improved Problem-Solving Skills: Learning about the behavior of light through reflection and refraction enhances analytical and problem-solving abilities.
• Practical Understanding: The concepts are directly applicable to everyday phenomena, such as the formation of images by mirrors and lenses, which helps in understanding the real-world application of Physics.
• Preparation for Competitive Exams: A strong grasp of this chapter is crucial for competitive exams like NTSE, Olympiads, and other entrance tests where basic Physics concepts are tested.
• Enhanced Conceptual Clarity: Light Reflection and Refraction chapter helps in building a clear understanding of the nature and properties of light, which is essential for mastering more advanced topics in optics.
• Boosts Academic Performance: A thorough understanding of this chapter can lead to better performance in school exams, as it is an important part of the curriculum.
• Stimulates Curiosity: It encourages students to explore and experiment with light, fostering a scientific temperament and curiosity about natural phenomena.