Light Reflection and Refraction solutions Class 10 Physics NCERT
30 October
Class 10 Science – Light: Reflection and Refraction (In-Text + Exercise Questions with Answers)
Welcome to Physics Xtreme – your destination for mastering physics the smart way! In this detailed and interactive blog post, you'll find complete, CBSE-focused, exam-ready answers for all the Intext and Exercise questions from the NCERT chapter Light – Reflection and Refraction. Let's make learning science fun, simple, and exciting! 🌟
Intext Questions – Page 142
1. Define the principal focus of a concave mirror.
The principal focus of a concave mirror is the point on the principal axis where light rays that are parallel to the principal axis converge after reflection. It lies between the pole and the center of curvature.
Pro Tip: The principal focus is always real for a concave mirror and virtual for a convex mirror.
2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
We know the relation:
Focal length (f) = Radius of Curvature (R) ÷ 2
So, f = 20 ÷ 2 = 10 cm
Hence, the focal length of the mirror is 10 cm.
Focal length (f) = Radius of Curvature (R) ÷ 2
So, f = 20 ÷ 2 = 10 cm
Hence, the focal length of the mirror is 10 cm.
3. Name a mirror that can give an erect and enlarged image of an object.
A concave mirror can produce an erect and enlarged image when the object is placed very close to the mirror, i.e., between the pole and the focus.
This is the reason why concave mirrors are used by dentists and makeup artists.
4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?
A convex mirror is preferred in vehicles because:
- It always forms an erect and diminished image.
- It offers a wider field of view.
- It enables the driver to see more vehicles and objects behind.
Hence, for safety and broader visibility, convex mirrors are ideal for rear-view applications.
Intext Questions – Page 145
1. Find the focal length of a convex mirror whose radius of curvature is 32 cm.
We use the formula:
f = R / 2
Here, R = 32 cm
⇒ f = 32 / 2 = 16 cm
Since it's a convex mirror, its focal length is taken as positive.
f = R / 2
Here, R = 32 cm
⇒ f = 32 / 2 = 16 cm
Since it's a convex mirror, its focal length is taken as positive.
∴ Focal length of the convex mirror is +16 cm.
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?
Given:
u = –10 cm (object in front of mirror)
m = –3 (real and inverted image ⇒ magnification is negative)
Using the magnification formula:
m = –v/u
⇒ –3 = –v / –10
⇒ v = –30 cm
u = –10 cm (object in front of mirror)
m = –3 (real and inverted image ⇒ magnification is negative)
Using the magnification formula:
m = –v/u
⇒ –3 = –v / –10
⇒ v = –30 cm
∴ The image is formed at a distance of 30 cm in front of the mirror (real and inverted).
Intext Questions – Page 158
1. Define 1 dioptre of power of a lens.
1 dioptre is the power of a lens whose focal length is 1 metre.
It is the SI unit of lens power and is denoted by the symbol "D".
It is the SI unit of lens power and is denoted by the symbol "D".
1 D = 1 m⁻¹
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.
Since the image is real, inverted, and of same size as the object:
⇒ Object is placed at 2F (i.e., twice the focal length)
Given: v = +50 cm (image is real ⇒ positive), u = –50 cm (object in front)
Using lens formula:
1/f = 1/v – 1/u
⇒ 1/f = 1/50 – (–1/50) = 2/50 = 1/25
⇒ f = 25 cm = 0.25 m
Now, Power (P) = 100/f (in cm)
⇒ P = 100 / 25 = +4 D
⇒ Object is placed at 2F (i.e., twice the focal length)
Given: v = +50 cm (image is real ⇒ positive), u = –50 cm (object in front)
Using lens formula:
1/f = 1/v – 1/u
⇒ 1/f = 1/50 – (–1/50) = 2/50 = 1/25
⇒ f = 25 cm = 0.25 m
Now, Power (P) = 100/f (in cm)
⇒ P = 100 / 25 = +4 D
∴ Object is at 50 cm, and power of the lens is +4 dioptres.
3. Find the power of a concave lens of focal length 2 m.
f = –2 m (concave lens has negative focal length)
Power (P) = 100 / f (in cm) = 100 / (–200) = –0.5 D
Power (P) = 100 / f (in cm) = 100 / (–200) = –0.5 D
∴ The power of the concave lens is –0.5 dioptres.
Exercise Questions – NCERT Solutions
1. Which one of the following materials cannot be used to make a lens?
(a) Water (b) Glass (c) Plastic (d) Clay
Answer: (d) Clay
Answer: (d) Clay
Clay is opaque and does not allow light to pass through. Lenses must be transparent to refract light properly.
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)
(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)
When the object is placed between the pole and focus of a concave mirror, it forms a virtual, erect, and magnified image.
3. Where should an object be placed in front of a convex lens to get a real image of the same size?
(a) At the principal focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre and focus
Answer: (b)
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre and focus
Answer: (b)
When an object is placed at 2F (twice the focal length), a convex lens forms a real, inverted image of the same size at 2F on the other side.
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
Answer: (a) both concave
Negative focal length indicates that both are concave in nature. Concave mirrors and concave lenses have negative focal lengths.
5. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
Answer: (d) either plane or convex
Both plane mirrors and convex mirrors form erect images regardless of object distance.
6. Which of the following lenses would you prefer to use while reading small letters in a dictionary?
Answer: (c) A convex lens of focal length 5 cm
A convex lens with a small focal length acts as a strong magnifying glass, ideal for reading small text.
7. We wish to obtain an erect image using a concave mirror of focal length 15 cm. What should be the range of object distances? What is the nature and size of the image? Draw a ray diagram.
- To get an erect image, object should be placed between the pole and focus (i.e., within 15 cm).
- The image will be virtual, erect, and magnified.
Ray Diagram:kindly refer text book for ray diagram. ray diagrams will be uploaded soon.
- The image will be virtual, erect, and magnified.
Ray Diagram:kindly refer text book for ray diagram. ray diagrams will be uploaded soon.
Such mirrors are used in makeup mirrors to see enlarged images.
8. Name the type of mirror used in the following situations. Support your answer with reasons.
| Situation | Type of Mirror | Reason |
|---|---|---|
| (a) Headlights of a car | Concave mirror | Focuses light rays to form a powerful beam |
| (b) Side/rear-view mirror of a vehicle | Convex mirror | Wider field of view and erect images |
| (c) Solar furnace | Concave mirror | Converges sunlight at a point to produce heat |
9. One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object?
Yes, the lens will still produce a complete image, but the image will be dimmer.
Light from all parts of the object can still pass through the uncovered portion of the lens to form a complete image.
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.
Given:
Object height, h = 5 cm
Object distance, u = –25 cm
Focal length, f = +10 cm (convex lens)
Using lens formula:
1/f = 1/v – 1/u
⇒ 1/10 = 1/v + 1/25
⇒ 1/v = 1/10 – 1/25 = (5 – 2)/50 = 3/50
⇒ v = 50/3 ≈ 16.67 cm
Magnification (m) = v/u = 16.67 / –25 ≈ –0.67
Image height = m × object height = –0.67 × 5 = –3.35 cm
Object height, h = 5 cm
Object distance, u = –25 cm
Focal length, f = +10 cm (convex lens)
Using lens formula:
1/f = 1/v – 1/u
⇒ 1/10 = 1/v + 1/25
⇒ 1/v = 1/10 – 1/25 = (5 – 2)/50 = 3/50
⇒ v = 50/3 ≈ 16.67 cm
Magnification (m) = v/u = 16.67 / –25 ≈ –0.67
Image height = m × object height = –0.67 × 5 = –3.35 cm
∴ The image is formed at 16.67 cm on the opposite side, inverted and smaller in size.
Ray diagram: for now kindly refer text book. ray diagrams will be uploaded soon...
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?
Given:
f = –15 cm, v = –10 cm (image formed on same side of object)
Using lens formula:
1/f = 1/v – 1/u
⇒ 1/–15 = 1/–10 – 1/u
⇒ –1/15 + 1/10 = –1/u
⇒ (–2 + 3)/30 = –1/u ⇒ 1/30 = –1/u
⇒ u = –30 cm
f = –15 cm, v = –10 cm (image formed on same side of object)
Using lens formula:
1/f = 1/v – 1/u
⇒ 1/–15 = 1/–10 – 1/u
⇒ –1/15 + 1/10 = –1/u
⇒ (–2 + 3)/30 = –1/u ⇒ 1/30 = –1/u
⇒ u = –30 cm
∴ Object is placed 30 cm in front of the lens.
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.
Given:
f = +15 cm, u = –10 cm
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
f = +15 cm, u = –10 cm
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
∴ The image is virtual, erect, and located 6 cm behind the mirror.
13. The magnification produced by a plane mirror is +1. What does this mean?
A magnification of +1 means the image is:
- Erect (positive sign)
- Same size as the object
Plane mirrors always form virtual, erect images of the same size as the object.
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.
Given:
Object height = 5 cm, u = –20 cm, R = 30 cm ⇒ f = +15 cm
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 ≈ 8.57 cm
Magnification (m) = v/u = 8.57 / –20 = –0.43
Image height = –0.43 × 5 ≈ –2.15 cm
Object height = 5 cm, u = –20 cm, R = 30 cm ⇒ f = +15 cm
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 ≈ 8.57 cm
Magnification (m) = v/u = 8.57 / –20 = –0.43
Image height = –0.43 × 5 ≈ –2.15 cm
∴ Image is virtual, erect, 8.57 cm behind the mirror, and smaller in size (2.15 cm).
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? Find size and nature of the image.
Given:
u = –27 cm, f = –18 cm
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 = v/u = –54 / –27 = 2
Image height = 2 × 7 = 14 cm
u = –27 cm, f = –18 cm
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 = v/u = –54 / –27 = 2
Image height = 2 × 7 = 14 cm
∴ The image is real, inverted, and formed 54 cm in front of the mirror. Screen should be placed at 54 cm.
16. Find the focal length of a lens of power –2.0 D. What type of lens is this?
f = 100 / P = 100 / –2 = –50 cm = –0.5 m
∴ The lens has a focal length of –50 cm and is a concave (diverging) lens.
17. A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the lens diverging or converging?
f = 100 / 1.5 = 66.67 cm = 0.666 m
∴ The lens is a convex (converging) lens with a focal length of approximately 66.67 cm.
Final Words
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