NCERT Solutions - Class 10 Science Chapter 9: Light – Reflection and Refraction
Intext Questions (Page No. 142)
Q1. Define the principal focus of a concave mirror.
Answer -
The principal focus of a concave mirror is the point on its principal axis where light rays parallel to the principal axis converge after reflection from the mirror.
The principal focus of a concave mirror is the point on its principal axis where light rays parallel to the principal axis converge after reflection from the mirror.
Q2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Answer -
The focal length (f) of a spherical mirror is half of its radius of curvature (R).
f = R / 2 = 20 cm / 2 = 10 cm
The focal length (f) of a spherical mirror is half of its radius of curvature (R).
f = R / 2 = 20 cm / 2 = 10 cm
Q3. Name a mirror that can give an erect and enlarged image of an object.
Answer -
A concave mirror can give an erect and enlarged image of an object when the object is placed between the pole and the focus of the mirror.
A concave mirror can give an erect and enlarged image of an object when the object is placed between the pole and the focus of the mirror.
Q4. 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 because:
• It always forms an erect and diminished image.
• It provides a wider field of view compared to a plane or concave mirror.
This helps the driver see a larger area of the road behind.
We prefer a convex mirror as a rear-view mirror in vehicles because:
• It always forms an erect and diminished image.
• It provides a wider field of view compared to a plane or concave mirror.
This helps the driver see a larger area of the road behind.
Intext Questions (Page No. 145)
Q1. Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer -
The relation between focal length (f) and radius of curvature (R) is:
f = R / 2
Given R = 32 cm
⇒ f = 32 / 2 = 16 cm
So, the focal length of the convex mirror is +16 cm (positive sign as focal length of a convex mirror is positive).
The relation between focal length (f) and radius of curvature (R) is:
f = R / 2
Given R = 32 cm
⇒ f = 32 / 2 = 16 cm
So, the focal length of the convex mirror is +16 cm (positive sign as focal length of a convex mirror is positive).
Q2. 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 -
Given:
Magnification (m) = –3 (real image has negative magnification)
Object distance (u) = –10 cm (object is in front of the mirror)
We know,
m = v / u
⇒ –3 = v / (–10)
⇒ v = 30 cm
So, the image is formed at a distance of 30 cm in front of the mirror.
Given:
Magnification (m) = –3 (real image has negative magnification)
Object distance (u) = –10 cm (object is in front of the mirror)
We know,
m = v / u
⇒ –3 = v / (–10)
⇒ v = 30 cm
So, the image is formed at a distance of 30 cm in front of the mirror.
Intext Questions (Page No. 150)
Q1. 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 travels from a rarer medium (air) to a denser medium (water), it slows down and bends towards the normal. This bending occurs because the speed of light is lower in water than in air.
When a ray of light travels from a rarer medium (air) to a denser medium (water), it slows down and bends towards the normal. This bending occurs because the speed of light is lower in water than in air.
Q2. 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 -
Given:
Refractive index (n) = 1.50
Speed of light in vacuum (c) = 3 × 108 m/s
Refractive index, n = c / v
⇒ v = c / n = (3 × 108) / 1.50 = 2 × 108 m/s
So, the speed of light in glass is 2 × 108 m/s.
Given:
Refractive index (n) = 1.50
Speed of light in vacuum (c) = 3 × 108 m/s
Refractive index, n = c / v
⇒ v = c / n = (3 × 108) / 1.50 = 2 × 108 m/s
So, the speed of light in glass is 2 × 108 m/s.
Q3. Find out, from Table 9.3, the medium having highest optical density. Also find the medium with lowest optical density.
Answer -
From Table 9.3:
Highest optical density: Diamond (Refractive index = 2.42)
Lowest optical density: Air (Refractive index = 1.0003)
From Table 9.3:
Highest optical density: Diamond (Refractive index = 2.42)
Lowest optical density: Air (Refractive index = 1.0003)
Q4. You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table 9.3.
Answer -
Light travels fastest in the medium with the lowest refractive index.
From Table 9.3:
Refractive indices:
Water = 1.33
Kerosene ≈ 1.44
Turpentine ≈ 1.47
Therefore, light travels fastest in water.
Light travels fastest in the medium with the lowest refractive index.
From Table 9.3:
Refractive indices:
Water = 1.33
Kerosene ≈ 1.44
Turpentine ≈ 1.47
Therefore, light travels fastest in water.
Q5. The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer -
This means that the speed of light in diamond is 2.42 times slower than its speed in vacuum.
In other words, the ratio of the speed of light in vacuum to the speed of light in diamond is 2.42.
This means that the speed of light in diamond is 2.42 times slower than its speed in vacuum.
In other words, the ratio of the speed of light in vacuum to the speed of light in diamond is 2.42.
Intext Questions (Page No. 158)
Q1. Define 1 dioptre of power of a lens.
Answer -
1 dioptre is the power of a lens whose focal length is 1 metre.
Definition: A lens has a power of 1 dioptre if it forms a focal point at a distance of 1 metre from the lens.
Unit: The SI unit of power of a lens is dioptre (D).
1 dioptre is the power of a lens whose focal length is 1 metre.
Definition: A lens has a power of 1 dioptre if it forms a focal point at a distance of 1 metre from the lens.
Unit: The SI unit of power of a lens is dioptre (D).
Q2. 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 -
When a convex lens forms a real, inverted image of the same size as the object, the object must be placed at 2F (twice the focal length).
Given: Image distance (v) = +50 cm
So, object distance (u) = –50 cm (since object is placed at 2F on the same side)
Using lens formula:
1/f = 1/v – 1/u = 1/50 – (–1/50) = 1/50 + 1/50 = 2/50 = 1/25
⇒ f = 25 cm = 0.25 m
Power (P) = 100/f (in cm) = 100/25 = +4 D
When a convex lens forms a real, inverted image of the same size as the object, the object must be placed at 2F (twice the focal length).
Given: Image distance (v) = +50 cm
So, object distance (u) = –50 cm (since object is placed at 2F on the same side)
Using lens formula:
1/f = 1/v – 1/u = 1/50 – (–1/50) = 1/50 + 1/50 = 2/50 = 1/25
⇒ f = 25 cm = 0.25 m
Power (P) = 100/f (in cm) = 100/25 = +4 D
Q3. Find the power of a concave lens of focal length 2 m.
Answer -
Given: Focal length (f) = –2 m (concave lens)
Power (P) = 100 / f (in cm) = 100 / (–200) = –0.5 D
So, the power of the concave lens is –0.5 dioptre.
Given: Focal length (f) = –2 m (concave lens)
Power (P) = 100 / f (in cm) = 100 / (–200) = –0.5 D
So, the power of the concave lens is –0.5 dioptre.
Exercises
Q1. Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Answer -
Correct option: (d) Clay
Clay is an opaque material and does not allow light to pass through, so it cannot be used to make a lens.
Correct option: (d) Clay
Clay is an opaque material and does not allow light to pass through, so it cannot be used to make a lens.
Q2. 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
(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 -
Correct option: (d) Between the pole of the mirror and its principal focus
When an object is placed between the pole and the focus of a concave mirror, the image formed is virtual, erect, and magnified.
Correct option: (d) Between the pole of the mirror and its principal focus
When an object is placed between the pole and the focus of a concave mirror, the image formed is virtual, erect, and magnified.
Q3. Where should an object be placed in front of a convex lens to get a real image of the same size as the object?
(a) At the focus
(b) At twice the focal length
(c) At infinity
(d) Between optical centre and focus
(a) At the focus
(b) At twice the focal length
(c) At infinity
(d) Between optical centre and focus
Answer -
Correct option: (b) At twice the focal length
When the object is placed at 2F (twice the focal length) in front of a convex lens, a real and same-sized image is formed at 2F on the other side.
Correct option: (b) At twice the focal length
When the object is placed at 2F (twice the focal length) in front of a convex lens, a real and same-sized image is formed at 2F on the other side.
Q4. 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
(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 -
Correct option: (a) Both concave
Negative focal length is a property of concave mirrors and concave lenses. Hence, both must be concave.
Correct option: (a) Both concave
Negative focal length is a property of concave mirrors and concave lenses. Hence, both must be concave.
Q5. 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
(a) Only plane
(b) Only concave
(c) Only convex
(d) Either plane or convex
Answer -
Correct option: (d) Either plane or convex
Both plane and convex mirrors always form virtual and erect images, regardless of the object's position.
Correct option: (d) Either plane or convex
Both plane and convex mirrors always form virtual and erect images, regardless of the object's position.
Q6. 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
(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 -
Correct option: (c) A convex lens of focal length 5 cm
A convex lens of shorter focal length acts as a magnifying glass and is suitable for reading small letters.
Correct option: (c) A convex lens of focal length 5 cm
A convex lens of shorter focal length acts as a magnifying glass and is suitable for reading small letters.
Q7. 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 -
The object should be placed between the pole and the focus of the concave mirror, i.e., within 15 cm.
The image formed is virtual, erect, and magnified.
[Ray diagram is unfortunately not available now.]
The object should be placed between the pole and the focus of the concave mirror, i.e., within 15 cm.
The image formed is virtual, erect, and magnified.
[Ray diagram is unfortunately not available now.]
Q8. 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.
(a) Headlights of a car.
(b) Side/rear-view mirror of a vehicle.
(c) Solar furnace.
Support your answer with reason.
Answer -
(a) Concave mirror: It converges light to a point and is used in headlights to focus the beam.
(b) Convex mirror: It gives a wider field of view, suitable for rear-view mirrors.
(c) Concave mirror: It concentrates sunlight to a point to generate heat.
(a) Concave mirror: It converges light to a point and is used in headlights to focus the beam.
(b) Convex mirror: It gives a wider field of view, suitable for rear-view mirrors.
(c) Concave mirror: It concentrates sunlight to a point to generate heat.
Q9. 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 -
Yes, the lens will still form a complete image because every part of the lens contributes to image formation. However, the brightness of the image will be reduced due to less light passing through.
Yes, the lens will still form a complete image because every part of the lens contributes to image formation. However, the brightness of the image will be reduced due to less light passing through.
Q10. 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 -
Given:
Object height (h) = 5 cm
Object distance (u) = –25 cm (object is on the left side of the lens, so negative)
Focal length (f) = +10 cm (converging lens, so positive)
To Find:
Image distance (v), image height (h′), nature of the image
Using the lens formula:
1/f = 1/v – 1/u
⇒ 1/10 = 1/v – (–1/25)
⇒ 1/10 = 1/v + 1/25
⇒ 1/v = 1/10 – 1/25
⇒ (LCM = 50) ⇒ 1/v = (5 – 2)/50 = 3/50
⇒ v = 50/3 ≈ 16.67 cm
Image position:
The image is formed at approximately 16.67 cm on the other side of the lens (positive sign).
Magnification (m):
m = v/u = 16.67 / (–25) ≈ –0.67
Image height (h′):
h′ = m × h = –0.67 × 5 = –3.33 cm
Nature of the image:
– Since magnification is negative, the image is real and inverted.
– The image height is 3.33 cm, so the image is smaller than the object.
Note: The ray diagram is not available in this text-based format, but it can be drawn by placing the object beyond 2F on the left of a convex lens. The image will form between F and 2F on the right side, real and inverted.
Given:
Object height (h) = 5 cm
Object distance (u) = –25 cm (object is on the left side of the lens, so negative)
Focal length (f) = +10 cm (converging lens, so positive)
To Find:
Image distance (v), image height (h′), nature of the image
Using the lens formula:
1/f = 1/v – 1/u
⇒ 1/10 = 1/v – (–1/25)
⇒ 1/10 = 1/v + 1/25
⇒ 1/v = 1/10 – 1/25
⇒ (LCM = 50) ⇒ 1/v = (5 – 2)/50 = 3/50
⇒ v = 50/3 ≈ 16.67 cm
Image position:
The image is formed at approximately 16.67 cm on the other side of the lens (positive sign).
Magnification (m):
m = v/u = 16.67 / (–25) ≈ –0.67
Image height (h′):
h′ = m × h = –0.67 × 5 = –3.33 cm
Nature of the image:
– Since magnification is negative, the image is real and inverted.
– The image height is 3.33 cm, so the image is smaller than the object.
Note: The ray diagram is not available in this text-based format, but it can be drawn by placing the object beyond 2F on the left of a convex lens. The image will form between F and 2F on the right side, real and inverted.
Q11. 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 (concave lens)
image distance (v) = –10 cm (image on same side of lens, virtual)
To Find: Object distance (u)
Using lens formula:
1/f = 1/v – 1/u
⇒ 1/–15 = 1/–10 – 1/u
⇒ –1/15 = –1/10 – 1/u
⇒ 1/u = –1/10 + 1/15 = (–3 + 2)/30 = –1/30
⇒ u = –30 cm
Conclusion:
The object is placed 30 cm in front of the lens.
Nature of the image:
Since the image distance is negative, the image is virtual, erect, and diminished.
Note: The ray diagram is not available in this text format but can be drawn by placing the object in front of the concave lens and forming a virtual image closer to the lens on the same side.
Given:
focal length (f) = –15 cm (concave lens)
image distance (v) = –10 cm (image on same side of lens, virtual)
To Find: Object distance (u)
Using lens formula:
1/f = 1/v – 1/u
⇒ 1/–15 = 1/–10 – 1/u
⇒ –1/15 = –1/10 – 1/u
⇒ 1/u = –1/10 + 1/15 = (–3 + 2)/30 = –1/30
⇒ u = –30 cm
Conclusion:
The object is placed 30 cm in front of the lens.
Nature of the image:
Since the image distance is negative, the image is virtual, erect, and diminished.
Note: The ray diagram is not available in this text format but can be drawn by placing the object in front of the concave lens and forming a virtual image closer to the lens on the same side.
Q12. 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 -
Given:
Object distance (u) = –10 cm
Focal length (f) = +15 cm (convex mirror)
To Find: Image distance (v)
Using mirror formula:
1/f = 1/v + 1/u
⇒ 1/15 = 1/v – 1/10
⇒ 1/v = 1/15 + 1/10 = (2 + 3)/30 = 5/30 = 1/6
⇒ v = 6 cm
Conclusion:
The image is formed 6 cm behind the mirror.
Nature of the image:
The image is virtual, erect, and smaller than the object.
Note: Ray diagram not available in text format.
Given:
Object distance (u) = –10 cm
Focal length (f) = +15 cm (convex mirror)
To Find: Image distance (v)
Using mirror formula:
1/f = 1/v + 1/u
⇒ 1/15 = 1/v – 1/10
⇒ 1/v = 1/15 + 1/10 = (2 + 3)/30 = 5/30 = 1/6
⇒ v = 6 cm
Conclusion:
The image is formed 6 cm behind the mirror.
Nature of the image:
The image is virtual, erect, and smaller than the object.
Note: Ray diagram not available in text format.
Q13. The magnification produced by a plane mirror is +1. What does this mean?
Answer -
It means the image is of the same size as the object, virtual and erect.
It means the image is of the same size as the object, virtual and erect.
Q14. An object 5 cm is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position, nature and size of the image.
Answer -
Given:
Object height (h) = 5 cm
Object distance (u) = –20 cm
Radius of curvature (R) = 30 cm ⇒ f = R/2 = +15 cm
Using mirror formula:
1/f = 1/v + 1/u
⇒ 1/15 = 1/v – 1/20
⇒ 1/v = 1/15 + 1/20 = (4 + 3)/60 = 7/60
⇒ v = 60/7 ≈ 8.57 cm
Magnification:
m = v/u = 8.57 / (–20) ≈ –0.43
Image height:
h′ = m × h = –0.43 × 5 = –2.15 cm
Conclusion:
Image is formed 8.57 cm behind the mirror. It is virtual, erect, and smaller than the object.
Note: Diagram not available in this format.
Given:
Object height (h) = 5 cm
Object distance (u) = –20 cm
Radius of curvature (R) = 30 cm ⇒ f = R/2 = +15 cm
Using mirror formula:
1/f = 1/v + 1/u
⇒ 1/15 = 1/v – 1/20
⇒ 1/v = 1/15 + 1/20 = (4 + 3)/60 = 7/60
⇒ v = 60/7 ≈ 8.57 cm
Magnification:
m = v/u = 8.57 / (–20) ≈ –0.43
Image height:
h′ = m × h = –0.43 × 5 = –2.15 cm
Conclusion:
Image is formed 8.57 cm behind the mirror. It is virtual, erect, and smaller than the object.
Note: Diagram not available in this format.
Q15. 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.
Answer -
Given:
Object height (h) = 7.0 cm
Object distance (u) = –27 cm
Focal length (f) = –18 cm
Using mirror formula:
1/f = 1/v + 1/u
⇒ 1/–18 = 1/v – 1/27
⇒ 1/v = –1/18 + 1/27 = (–3 + 2)/54 = –1/54
⇒ v = –54 cm
Magnification:
m = v/u = –54 / –27 = 2
Image height:
h′ = m × h = 2 × 7.0 = 14.0 cm
Conclusion:
Screen should be placed at 54 cm in front of the mirror.
Image is real, inverted, and magnified.
Note: Ray diagram not included here.
Given:
Object height (h) = 7.0 cm
Object distance (u) = –27 cm
Focal length (f) = –18 cm
Using mirror formula:
1/f = 1/v + 1/u
⇒ 1/–18 = 1/v – 1/27
⇒ 1/v = –1/18 + 1/27 = (–3 + 2)/54 = –1/54
⇒ v = –54 cm
Magnification:
m = v/u = –54 / –27 = 2
Image height:
h′ = m × h = 2 × 7.0 = 14.0 cm
Conclusion:
Screen should be placed at 54 cm in front of the mirror.
Image is real, inverted, and magnified.
Note: Ray diagram not included here.
Q16. Find the focal length of a lens of power –2.0 D. What type of lens is this?
Answer -
Given:
Power (P) = –2.0 D
Using formula:
P = 100/f
⇒ f = 100 / P = 100 / (–2.0) = –50 cm = –0.5 m
Conclusion:
The focal length is –50 cm.
Since the power is negative, it is a concave (diverging) lens.
Given:
Power (P) = –2.0 D
Using formula:
P = 100/f
⇒ f = 100 / P = 100 / (–2.0) = –50 cm = –0.5 m
Conclusion:
The focal length is –50 cm.
Since the power is negative, it is a concave (diverging) lens.
Q17. 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 -
Given:
Power (P) = +1.5 D
Using formula:
P = 100/f
⇒ f = 100 / P = 100 / 1.5 ≈ 66.67 cm = 0.667 m
Conclusion:
The focal length of the lens is approximately 66.67 cm.
Since the power is positive, the lens is converging (convex).
Given:
Power (P) = +1.5 D
Using formula:
P = 100/f
⇒ f = 100 / P = 100 / 1.5 ≈ 66.67 cm = 0.667 m
Conclusion:
The focal length of the lens is approximately 66.67 cm.
Since the power is positive, the lens is converging (convex).