Science Class- 10
Class 10 - Chapter 9: Light – Reflection and Refraction
Q1. Which one of the following materials cannot be used to make a lens?
Answer: (d) Clay.
A lens must be made of a transparent material that allows light to pass through. Clay is opaque and 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 the position of the object be?
Answer: (d) Between the pole of the mirror and its principal focus.
When an object is placed between the pole (P) and focus (F) 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?
Answer: (b) At twice the focal length.
When an object is placed at 2F of a convex lens, the image is formed at 2F on the other side. The image is real, inverted and of the same size as the object.
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
Answer: (a) Both concave.
According to the sign convention:
- A concave mirror has a negative focal length.
- A concave lens has a negative focal length.
Therefore, both the mirror and lens are concave.
Q5. 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 always produce virtual and erect images.
Q6. Which of the following lenses would you prefer to use while reading small letters found in a dictionary?
Answer: (c) A convex lens of focal length 5 cm.
A convex lens with a short focal length provides greater magnification and is therefore 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?
For a concave mirror to form an erect image, the object must be placed between the pole (P) and the principal focus (F).
Range of Object Distance:
0 cm < Object Distance < 15 cm
Nature of Image:
- Virtual
- Erect
- Magnified
The image formed is larger than the object.
Q8. Name the type of mirror used in the following situations.
(a) Headlights of a car: Concave mirror
Reason: It produces a strong parallel beam of light.
(b) Side/rear-view mirror of a vehicle: Convex mirror
Reason: It provides a wider field of view and forms erect images.
(c) Solar furnace: Concave mirror
Reason: It converges sunlight at the focus to produce high temperatures.
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.
Yes, the lens will produce a complete image.
Each part of a lens forms the entire image of the object.
Covering half of the lens only reduces the brightness of the image, but the complete image is still formed.
Experimental Verification:
- Form an image using a convex lens and screen.
- Cover half the lens with black paper.
- Observe that the complete image is still visible but appears dimmer.
Q10. An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Find the position, size and nature of the image formed.
Given:
| Quantity | Value |
|---|---|
| Object Height (ho) | 5 cm |
| Object Distance (u) | -25 cm |
| Focal Length (f) | +10 cm |
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 = 16.7 cm
Magnification:
m = v/u = 16.7/(-25) = -0.67
hi = m × ho
hi = -0.67 × 5 = -3.35 cm
Result:
| Quantity | Value |
|---|---|
| Image Distance | 16.7 cm |
| Image Height | 3.35 cm |
| Nature | Real, inverted and diminished |
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?
Given:
| Quantity | Value |
|---|---|
| Focal Length (f) | -15 cm |
| Image Distance (v) | -10 cm |
Lens Formula:
1/f = 1/v - 1/u
1/(-15) = 1/(-10) - 1/u
1/u = -1/10 + 1/15
1/u = -1/30
u = -30 cm
Answer: The object is placed 30 cm in front of the lens.
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.
Given:
| Quantity | Value |
|---|---|
| Object Distance (u) | -10 cm |
| Focal Length (f) | +15 cm |
Mirror Formula:
1/f = 1/v + 1/u
1/15 = 1/v - 1/10
1/v = 1/15 + 1/10
1/v = 1/6
v = 6 cm
Result:
The image is formed 6 cm behind the mirror.
The image is virtual, erect and diminished.
Q13. The magnification produced by a plane mirror is +1. What does this mean?
Magnification (m) = +1 means:
- The image is virtual and erect.
- The image size is equal to the object size.
- The positive sign indicates that the image is erect.
Q14. 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:
| Quantity | Value |
|---|---|
| Object Height | 5 cm |
| Object Distance (u) | -20 cm |
| Radius of Curvature (R) | 30 cm |
Focal Length = R/2 = +15 cm
Mirror Formula:
1/15 = 1/v - 1/20
1/v = 1/15 + 1/20
1/v = 7/60
v = 8.57 cm
Magnification:
m = -v/u = -(8.57)/(-20) = 0.43
hi = 0.43 × 5 = 2.15 cm
Result:
| Quantity | Value |
|---|---|
| Image Position | 8.57 cm behind mirror |
| Image Height | 2.15 cm |
| Nature | Virtual, erect and diminished |
Q15. An object of size 7.0 cm is placed 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 nature and size of the image.
Given:
| Quantity | Value |
|---|---|
| Object Height | 7 cm |
| Object Distance (u) | -27 cm |
| Focal Length (f) | -18 cm |
Mirror Formula:
1/f = 1/v + 1/u
1/(-18) = 1/v - 1/27
1/v = -1/18 + 1/27
1/v = -1/54
v = -54 cm
The screen should be placed 54 cm in front of the mirror.
Magnification:
m = -v/u = -(-54)/(-27) = -2
hi = -2 × 7 = -14 cm
Nature:
- Real
- Inverted
- Magnified
Image Size: 14 cm
Q16. Find the focal length of a lens of power -2.0 D. What type of lens is this?
Given:
Power (P) = -2.0 D
Formula:
P = 1/f
f = 1/P
f = 1/(-2)
f = -0.5 m
f = -50 cm
Answer:
Focal length = -50 cm.
Since the focal length 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 converging or diverging?
Given:
Power (P) = +1.5 D
Formula:
f = 1/P
f = 1/1.5
f = 0.67 m
f = 66.7 cm
Answer:
Focal length = +66.7 cm.
The positive power indicates a convex (converging) lens.