NCERT Solutions For Class 10 Physics Chapter 9 Light- Reflection and Refraction: Download PDF for Free

NCERT Solutions for Class 10 Light - Reflection and Refraction: NCERT books are considered the best source for preparation of theoretical concepts and making the basics strong. The questions provided in NCERT textbooks at the end of each chapter and intext question are quite important for examination point of view and for better understanding of the concepts. It is highly recommended to follow the NCERT Textbook for the exam. 

In this article you will find all the NCERT intext and final exercise solutions for CBSE Class 10 Science Chapter 9, Light - Reflection and Refraction. All the class 10 Science NCERT solutions have been collated in PDF format which students may easily download free of cost. 

Read more: Refraction of Light

Check NCERT Solutions for CBSE Class 10 Science Physics Chapter 9 Light – Reflection and Refraction: 

Intext Solutions Page No. -142

Question 1: Define the principal focus of a concave mirror. 

Answer: The principal focus of a concave mirror is the point where all the parallel rays of light that hit the mirror converge or meet after reflecting off the mirror. It’s the point where the light rays come together and is located in front of the mirror, along its optical axis. For a concave mirror, this focus is real and can be found by tracing the paths of the reflected rays.

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

Answer: To find the focal length of a spherical mirror, you use a simple formula that relates the focal length to the radius of curvature. The formula is:

Focal length = Radius of Curvature/2 

In this case, the radius of curvature is 20 cm. So, you divide this number by 2:

Focal length = 20 cm / 2 = 10 cm 

Therefore, the focal length of the mirror is 10 cm.

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

Answer: A convex mirror can give an erect and enlarged image of an object. In particular, when the object is placed very close to a convex mirror, the image appears enlarged and upright.

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

Answer: We prefer a convex mirror as a rear-view mirror in vehicles for several reasons:

• Wider Field of View: Convex mirrors are curved outward, allowing them to capture a larger area behind the vehicle. This helps drivers see more of the road and the traffic behind them.
• Reduced Blind Spots: By providing a broader view, convex mirrors help reduce blind spots, making it easier for drivers to spot vehicles and other objects that might be in their path.
• Erect and Diminished Image: Convex mirrors produce erect (upright) images that are smaller than the actual size of the objects. This allows drivers to see more of the surroundings, although the objects appear smaller and thus are safer for quick glances.
• Minimized Distortion: Convex mirrors have less distortion and glare compared to flat or concave mirrors, making them safer and more practical for driving.

Intext Solutions Page No. -145

Question 1: Find the focal length of a convex mirror whose radius of curvature is 32 cm.

Answer: To find the focal length of a convex mirror, you use the formula that relates the focal length (f) to the radius of curvature (R)

f=R/2

For a convex mirror, the radius of curvature is considered negative because the mirror's surface curves away from the incoming light. So, if the radius of curvature (R) is 32 cm, the focal length (f) is calculated as follows:

f= -32 cm/2 = -16 cm

Therefore, the focal length of the convex mirror is −16 cm. The negative sign indicates that it is a convex mirror.

Question 2: A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located? 

Answer: To find where the image is located for a concave mirror that produces a three times magnified real image, we can use the magnification formula and the mirror formula.

Magnification Formula:

Magnification(M) = Image distance(v) / Object distance (u)

Here, the magnification M is 3 (since the image is three times magnified) and the object distance uuu is 10 cm. So:

3= v/-10

(The object distance uuu is taken as negative for concave mirrors.)

Solving for v (the image distance):

v=3 x (-10) = -30 cm

The negative sign indicates that the image is on the same side as the object, which is typical for real images in concave mirrors.

Image Location: The image is located 30 cm in front of the mirror.

So, the image is positioned 30 cm away from the mirror, on the same side as the object.

Intext Solutions Page No. -150

Question1: 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?

Answer: When a ray of light traveling in air enters water at an angle, it bends towards the normal. This happens because light travels more slowly in water than in air. According to Snell's Law, when light moves from a medium where it travels faster (air) to a medium where it travels slower (water), it bends towards the normal line (an imaginary line perpendicular to the surface at the point of incidence).

Question 2: Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 × 10⁸ m s–1.

Answer: To find the speed of light in glass, you can use the formula that relates the speed of light in a medium to its refractive index:

Speed of light in the medium= Speed of light in vacuum/ Refractive Index 

Given:

Speed of light in vacuum (c) = 3 × 10⁸ m s–1

Refractive index of glass (n) = 1.50

values into the formula:

Speed of light in glass = 3 × 10⁸ m s–1/ 1.50

Speed of light in glass = 2 × 10⁸ m/s

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

Answer: The refractive index of diamond being 2.42 means that light travels 2.42 times slower in diamond than it does in a vacuum. A refractive index of 2.42 indicates that diamond bends light more than materials with lower refractive indices. This shows that diamond is a very effective material at bending light, which is why diamonds sparkle so much.

Intext Solutions Page No. -158

Question 1: Define 1 dioptre of power of a lens. 

Answer: 1 dioptre (D) of power of a lens is defined as the power of a lens that has a focal length of 1 meter.

Question 2: 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. 

Answer: Given:

Image distance (v) = 50 cm

Image size (h') = Object size (h)

To find:

Object distance (u)

Lens power (P)

Formula:

Magnification (m) = v/u = h'/h

Lens formula: 1/v + 1/u = 1/f

Lens power (P) = 1/f (where f is the focal length in meters)

Solution:

1. From the magnification formula:
• Since h' = h, m = 1.
• So, 1 = v/u
• Therefore, u = v = 50 cm
2. Using the lens formula:
• 1/50 + 1/50 = 1/f
• 1/25 = 1/f
• f = 25 cm = 0.25 m
3. Calculating lens power:
• P = 1/f = 1/0.25 = 4 diopters

Therefore, the needle is placed 50 cm in front of the convex lens, and the power of the lens is 4 diopters.

Question 3: Find the power of a concave lens of focal length 2 m

Answer: Power of a lens is a measure of how strongly it can bend light. A concave lens is used to diverge or spread out light rays.

Formula:

Power (P) = 1 / focal length (f)

Given:

Focal length (f) = 2 meters

Calculation:

Power (P) = 1 / 2 = 0.5 diopters

So, the power of the concave lens is 0.5 diopters.

EXERCISES

Question1: Which one of the following materials cannot be used to make a lens?

(a) Water (b) Glass (c) Plastic (d) Clay

Answer: (d) Clay

Question 2: 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.

Answer: (d) Between the pole of the mirror and its principal focus.

Question 3: 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.

Answer: (b) At twice the focal length

Question 4: A spherical mirror and a thin spherical lens have each 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.

Answer: (d) the mirror is convex, but the lens is concave.

Question 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: (d) either plane or convex.

Question 6: 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.

Answer: (c) A convex lens of focal length 5 cm

Question 7: 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.

Answer: To obtain an erect image of an object using a concave mirror with a focal length of 15 cm, the object should be placed between the pole of the mirror (P) and its principal focus (F). Here's the detailed explanation:

Range of Distance of the Object

For a concave mirror, an erect image is formed when the object is placed between the pole (P) and the principal focus (F) of the mirror.

• Focal Length (f) = 15 cm (Concave mirror, so focal length is positive, f = +15 cm)
• Object Distance (u): The object distance must be less than the focal length to get an erect image.

Thus, the range of the object distance uuu should be:

0 < u < 15 cm

Nature of the Image

• Erect: The image formed will be erect.
• Virtual: The image will be virtual because it is formed on the same side as the object.
• Magnified: The image will be larger than the object.

Ray Diagram

Question 8: 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.

Answer: Here’s the type of mirror used in each of the given situations and the reasoning behind it:

(a) Headlights of a car:

• Type of Mirror: Concave Mirror
• Reason: A concave mirror is used in headlights because it can focus light rays into a parallel beam. This helps in directing a concentrated and bright beam of light forward, which improves visibility for the driver at night.

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

• Type of Mirror: Convex Mirror
• Reason: A convex mirror is used for side and rear-view mirrors because it provides a wider field of view. It allows drivers to see more of the road and other vehicles behind them, reducing blind spots. The image formed by a convex mirror is also erect, which is useful for safe driving.

(c) Solar Furnace:

• Type of Mirror: Concave Mirror

• Reason: A concave mirror is used in solar furnaces to concentrate sunlight onto a small area. By focusing the sunlight to a single point, the concave mirror increases the intensity of solar energy, which is essential for heating or generating power efficiently.

Question 9: 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.

Answer: When one-half of a convex lens is covered with black paper, the lens will still produce a complete image of the object, but the quality of the image may be affected. Here’s an experimental verification: To verify this, you can perform a simple experiment:

Materials Needed:

• A convex lens
• A black paper or tape
• An object to view (e.g., a printed text or a small object)
• A screen or white paper

Procedure:

1. Cover Half of the Lens:
• Cover one half of the convex lens with black paper or tape, ensuring that no light passes through that portion.
2. Setup the Experiment:
• Place the lens on a stand or holder so that it can be directed towards the object.
• Position the object in front of the lens at a suitable distance (e.g., at the lens's focal length or beyond it).
• Place a screen or white paper on the other side of the lens to capture the image.
3. Observe the Image:
• Look at the image formed on the screen or paper.

Observations:

• Complete Image: You should observe that a complete image of the object is formed. The image will be formed because the uncovered half of the lens is still able to focus light.
• Image Quality: The image might be dimmer or have uneven brightness since only part of the lens is contributing to the image formation. The area where the black paper is applied will not contribute to image formation, potentially causing shadows or less sharpness in that region.

Conclusion:

Covering half of a convex lens with black paper will still produce a complete image of the object, but the image may not be as bright or uniformly clear as it would be with the entire lens uncovered. This is because the uncovered half of the lens still performs the function of focusing light, though with reduced effectiveness.

Question 10: 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.

Answer: Using the lens formula:

• 1/f = 1/v + 1/u
• 1/10 = 1/v + 1/25
• 1/v = 1/10 - 1/25
• 1/v = 3/50
• v = 50/3 cm

Using the magnification formula:

• m = v/u = h'/h
• h' = m * h
• h' = (50/3) / 25 * 5
• h' = 10/3 cm

Nature of image:

• Since v is positive, the image is real.
• Since h' is positive and larger than h, the image is inverted and magnified.

Ray diagram:

The point of intersection of these refracted rays gives the location of the image. As you can see, the image is real, inverted, and magnified.

Question 11: 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.

Answer: Given: 

• Focal length (f) = -15 cm (negative for concave lens)
• Image distance (v) = -10 cm (negative for virtual image)

To find:

• Object distance (u)

Using the lens formula:

• 1/f = 1/v + 1/u
• 1/(-15) = 1/(-10) + 1/u
• 1/u = 1/15 - 1/10
• 1/u = -1/30
• u = -30 cm

Therefore, the object is placed 30 cm in front of the concave lens.

Ray diagram:

The point of intersection of these refracted rays gives the location of the image. As you can see, the image is virtual, erect, and diminished.

Question 12: 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.

Answer: Using the mirror formula:

1/f = 1/v + 1/u

1/15 = 1/v + 1/10

1/v = 1/15 - 1/10

1/v = -1/30

v = -30 cm

Therefore, the image is formed 30 cm behind the convex mirror.

Nature of image:

• Since the image distance is negative, the image is virtual.
• For a convex mirror, a virtual image is always erect and smaller than the object.

Conclusion: The image is formed 30 cm behind the convex mirror and is virtual and erect.

Question 13: The magnification produced by a plane mirror is +1. What does this mean?

Answer: A magnification of +1 for a plane mirror means that the image formed is:

• Same size as the object.
• Upright or erect (not inverted).

This is a characteristic of plane mirrors. They always produce virtual images that are the same size as the object and are oriented in the same direction.

Question 14: 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.

Answer: Given:

• Object height (h) = 5.0 cm

• Object distance (u) = 20 cm

• Radius of curvature (R) = 30 cm

To find:

• Image distance (v)

• Nature of image

• Image height (h')

Step 1: Calculate the focal length (f):

f = R/2 = 30 cm / 2 = 15 cm

Step 2: Use the mirror formula to find the image distance:

1/f = 1/v + 1/u

1/15 = 1/v + 1/20

1/v = 1/15 - 1/20

1/v = 1/60

v = 60 cm

Step 3: Determine the nature of the image:

• Since the image distance (v) is positive, the image is virtual.
• Convex mirrors always form virtual images.

Step 4: Calculate the magnification (m):

• m = -v/u = -60 cm / 20 cm = -3

Step 5: Find the image height:

• m = h'/h
• h' = m * h = -3 * 5.0 cm = -15 cm

Conclusion:

• The image is formed 60 cm behind the convex mirror.
• The image is virtual and erect.
• The image is diminished (smaller than the object) with a height of 15 cm.

Question 15: 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 focussed image can be obtained? Find the size and the nature of the image.

Answer: Image distance: Use the mirror formula: 1/f = 1/v + 1/u.

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

Solve for v: v = -54 cm

Screen placement: Place the screen 54 cm in front of the mirror.

Image nature:

• Since v is negative, the image is real.
• Since the magnification (m = -v/u) is -2, the image is inverted and larger than the object.

So, the image will be real, inverted, and twice as big as the object.

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

Answer: Think of a lens like a magnifying glass. The stronger the magnifying glass, the more it can bend light. We call this strength the "power" of the lens.

Here's the simple rule:

Power = 1 / focal length

In your question:

Power = -2.0 D (D stands for diopters, which is the unit for lens power)

So, to find the focal length, we just flip this around:

Focal length = 1 / Power

Focal length = 1 / (-2.0) = -0.5 meters

A negative focal length means it's a concave lens. These lenses make objects appear smaller and farther away.

Question 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?

Answer: Think of a lens like a magnifying glass. The stronger the magnifying glass, the more it can bend light. We call this strength the "power" of the lens.

Here's a simple rule:

Power = 1 / focal length

In your question:

Power = +1.5 D (D stands for diopters, which is the unit for lens power)

So, to find the focal length, we just flip this around:

Focal length = 1 / Power

Focal length = 1 / (+1.5) = 0.67 meters

A positive power means it's a converging lens. These lenses are like magnifying glasses. They make objects look bigger and closer.

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These solutions follows the latest CBSE Class 10 Science syllabus 2024-25 and the updates NCERT textbook thus, do not include the deleted topics.

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