Light Class 10 NCERT Solutions – Complete Chapter Guide
Reflection and Mirrors
1. Define the principal focus of a concave mirror.
The principal focus of a concave mirror is the point on the principal axis where all parallel rays of light converging after reflection meet. It is a real focus for concave mirrors.
2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
The focal length (f) is half the radius of curvature (R):
f = R/2 = 20 cm/2 = 10 cm
f = R/2 = 20 cm/2 = 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 between the pole and the principal focus (i.e., within the focal length).
4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Convex mirrors are preferred because:
- They provide a wider field of view (diminished image shows more area)
- Always produce erect images
- Give virtual images that are safer for distance judgment
5. Find the focal length of a convex mirror whose radius of curvature is 32 cm.
For convex mirrors, focal length is negative by convention:
f = -R/2 = -32 cm/2 = -16 cm
f = -R/2 = -32 cm/2 = -16 cm
6. 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: m = -3 (real image has negative magnification), u = -10 cm
m = -v/u ⇒ -3 = -v/(-10) ⇒ v = -30 cm
The image is formed 30 cm in front of the mirror.
m = -v/u ⇒ -3 = -v/(-10) ⇒ v = -30 cm
The image is formed 30 cm in front of the mirror.
Refraction and Lenses
7. 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?
The light ray bends towards the normal because water is optically denser than air. When light enters a denser medium, its speed decreases causing it to bend towards the normal.
8. 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.
n = c/v ⇒ v = c/n = (3 × 10⁸ m/s)/1.50 = 2 × 10⁸ m/s
9. Find out, from Table 9.3, the medium having highest optical density. Also find the medium with lowest optical density.
| Medium | Refractive Index |
|---|---|
| Diamond | 2.42 (Highest) |
| Air | 1.0003 (Lowest) |
10. You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table 9.3.
Light travels fastest in the medium with lowest refractive index:
Answer: Water (n = 1.33)
| Medium | Refractive Index |
|---|---|
| Kerosene | 1.44 |
| Turpentine | 1.47 |
| Water | 1.33 |
11. The refractive index of diamond is 2.42. What is the meaning of this statement?
This means:
- The speed of light in diamond is 1/2.42 times its speed in vacuum
- Diamond bends light more than other common materials
- It indicates diamond’s high optical density
12. Define 1 dioptre of power of a lens.
1 dioptre is the power of a lens whose focal length is 1 meter.
P (in dioptres) = 1/f (in meters)
A lens with P = 1D has f = 1 m; P = 2D has f = 0.5 m
P (in dioptres) = 1/f (in meters)
A lens with P = 1D has f = 1 m; P = 2D has f = 0.5 m
13. 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 if the image is equal to the size of the object? Also, find the power of the lens.
When image size equals object size, u = 2f and v = 2f
Given v = 50 cm ⇒ 2f = 50 cm ⇒ f = 25 cm = 0.25 m
Power P = 1/f = 1/0.25 = +4 D
Object distance u = 2f = 50 cm
Given v = 50 cm ⇒ 2f = 50 cm ⇒ f = 25 cm = 0.25 m
Power P = 1/f = 1/0.25 = +4 D
Object distance u = 2f = 50 cm
14. Find the power of a concave lens of focal length 2 m.
For concave lens, f is negative:
P = 1/f = 1/(-2) = -0.5 D
P = 1/f = 1/(-2) = -0.5 D
Exercise Questions with 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 – Because it is opaque and cannot transmit light.
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?
Answer: Between the pole of the mirror and its principal focus.
(When object is within focal length of concave mirror)
(When object is within focal length of concave mirror)
3. Where should an object be placed in front of a convex lens to get a real image of the size of the object?
Answer: At twice the focal length (u = 2f)
This gives v = 2f with m = -1 (same size, real and inverted)
This gives v = 2f with m = -1 (same size, real and inverted)
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:
Negative focal length indicates:
- Concave mirror
- Concave (diverging) lens
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 and convex mirrors always produce erect images regardless of object distance.
Both plane and convex mirrors always produce erect images regardless of object distance.
6. Which lens would you prefer to use while reading small letters found in a dictionary?
Answer: (c) A convex lens of focal length 5 cm
Shorter focal length provides greater magnification for close objects.
Shorter focal length provides greater magnification for close objects.
7. We wish to obtain an erect image using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror?
Solution:
To get erect image from concave mirror, object must be within focal length:
0 < u < 15 cm
Nature: Virtual, erect and magnified
To get erect image from concave mirror, object must be within focal length:
0 < u < 15 cm
Nature: Virtual, erect and magnified
[Ray diagram showing object between pole and focus, virtual image behind mirror]
8. Name the type of mirror used in:
(a) Headlights of a car: Concave mirror – produces parallel beam when light source is at focus
(b) Side/rear-view mirror: Convex mirror – gives wider field of view
(c) Solar furnace: Concave mirror – concentrates sunlight at focus
(b) Side/rear-view mirror: Convex mirror – gives wider field of view
(c) Solar furnace: Concave mirror – concentrates sunlight at focus
9. One-half of a convex lens is covered with black paper. Will this lens produce a complete image?
Answer: Yes, but with reduced brightness.
Experiment shows:
Experiment shows:
- Image forms with same position and size
- Only half the light reaches the image
- Image appears dimmer but complete
10. An object 5 cm tall is held 25 cm away from a converging lens of focal length 10 cm. Find the position, size and nature of the image.
Solution:
Given: u = -25 cm, f = +10 cm, ho = 5 cm
Using lens formula:
1/v – 1/u = 1/f ⇒ 1/v = 1/10 – 1/25 = (5-2)/50 ⇒ v = 50/3 ≈ 16.67 cm
Magnification:
m = v/u = (50/3)/(-25) = -2/3
Image height:
hi = m × ho = (-2/3) × 5 ≈ -3.33 cm
Result: Real, inverted image of 3.33 cm at 16.67 cm on other side of lens.
Given: u = -25 cm, f = +10 cm, ho = 5 cm
Using lens formula:
1/v – 1/u = 1/f ⇒ 1/v = 1/10 – 1/25 = (5-2)/50 ⇒ v = 50/3 ≈ 16.67 cm
Magnification:
m = v/u = (50/3)/(-25) = -2/3
Image height:
hi = m × ho = (-2/3) × 5 ≈ -3.33 cm
Result: Real, inverted image of 3.33 cm at 16.67 cm on other side of lens.
11. A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed?
Solution:
For concave lens: f = -15 cm, v = -10 cm (virtual image)
1/v – 1/u = 1/f ⇒ -1/10 – 1/u = -1/15
⇒ -1/u = -1/15 + 1/10 = (-2+3)/30 ⇒ u = -30 cm
Object is 30 cm from lens.
For concave lens: f = -15 cm, v = -10 cm (virtual image)
1/v – 1/u = 1/f ⇒ -1/10 – 1/u = -1/15
⇒ -1/u = -1/15 + 1/10 = (-2+3)/30 ⇒ u = -30 cm
Object is 30 cm from lens.
[Ray diagram showing diverging rays from concave lens]
12. An object is placed 10 cm from a convex mirror of focal length 15 cm. Find position and nature of image.
Solution:
For convex mirror: f = +15 cm, u = -10 cm
1/v + 1/u = 1/f ⇒ 1/v = 1/15 – (-1/10) = (2+3)/30 ⇒ v = +6 cm
Magnification:
m = -v/u = -6/(-10) = 0.6
Result: Virtual, erect image at 6 cm behind mirror, diminished (0.6× size).
For convex mirror: f = +15 cm, u = -10 cm
1/v + 1/u = 1/f ⇒ 1/v = 1/15 – (-1/10) = (2+3)/30 ⇒ v = +6 cm
Magnification:
m = -v/u = -6/(-10) = 0.6
Result: Virtual, erect image at 6 cm behind mirror, diminished (0.6× size).
13. The magnification produced by a plane mirror is +1. What does this mean?
Answer:
- + sign indicates image is erect
- 1 indicates image size equals object size
- Image distance equals object distance
- Image is virtual and laterally inverted
14. An object 5.0 cm tall is placed 20 cm in front of a convex mirror of radius of curvature 30 cm. Find image position, nature and size.
Solution:
Given: R = 30 cm ⇒ f = R/2 = +15 cm, u = -20 cm, ho = 5 cm
Mirror formula:
1/v = 1/f – 1/u = 1/15 – (-1/20) = (4+3)/60 ⇒ v ≈ +8.57 cm
Magnification:
m = -v/u = -8.57/(-20) ≈ 0.428
Image height:
hi = m × ho ≈ 0.428 × 5 ≈ 2.14 cm
Result: Virtual, erect image of 2.14 cm at 8.57 cm behind mirror.
Given: R = 30 cm ⇒ f = R/2 = +15 cm, u = -20 cm, ho = 5 cm
Mirror formula:
1/v = 1/f – 1/u = 1/15 – (-1/20) = (4+3)/60 ⇒ v ≈ +8.57 cm
Magnification:
m = -v/u = -8.57/(-20) ≈ 0.428
Image height:
hi = m × ho ≈ 0.428 × 5 ≈ 2.14 cm
Result: Virtual, erect image of 2.14 cm at 8.57 cm behind mirror.
15. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. Find image distance, size and nature.
Solution:
Given: f = -18 cm, u = -27 cm, ho = 7 cm
Mirror formula:
1/v = 1/f – 1/u = -1/18 – (-1/27) = (-3+2)/54 ⇒ v = -54 cm
Magnification:
m = -v/u = -(-54)/(-27) = -2
Image height:
hi = m × ho = -2 × 7 = -14 cm
Result: Real, inverted image of 14 cm at 54 cm in front of mirror.
Given: f = -18 cm, u = -27 cm, ho = 7 cm
Mirror formula:
1/v = 1/f – 1/u = -1/18 – (-1/27) = (-3+2)/54 ⇒ v = -54 cm
Magnification:
m = -v/u = -(-54)/(-27) = -2
Image height:
hi = m × ho = -2 × 7 = -14 cm
Result: Real, inverted image of 14 cm at 54 cm in front of mirror.
16. Find focal length of a lens of power -2.0 D. What type of lens is this?
Solution:
P = 1/f ⇒ f = 1/P = 1/(-2.0) = -0.5 m = -50 cm
Negative power indicates a concave (diverging) lens.
P = 1/f ⇒ f = 1/P = 1/(-2.0) = -0.5 m = -50 cm
Negative power indicates a concave (diverging) lens.
17. A doctor prescribed a corrective lens of power +1.5 D. Find focal length and lens type.
Solution:
f = 1/P = 1/+1.5 ≈ +0.666 m = +66.6 cm
Positive power indicates a convex (converging) lens used to correct hypermetropia (farsightedness).
f = 1/P = 1/+1.5 ≈ +0.666 m = +66.6 cm
Positive power indicates a convex (converging) lens used to correct hypermetropia (farsightedness).