TEXT BOOK ANSWER OF CLASS 10 SCIENCE CHAPTER 9 LIGHT - REFLECTION AND REFRACTION
- Get link
- X
- Other Apps
CLASS 10 SCIENCE
CHAPTER -9 REFLECTION AND REFRACTION
Video explanation of this chapter is available on our Youtube channel :
Class 10 Science Ch - 9 Light : Reflection and Refraction
In-text Questions and Answers
Section 9.2 – Spherical Mirrors
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 rays parallel to the principal axis meet after reflection from the mirror. It is represented by F.
Q2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Given:
R = 20 cm
Using:
R = 2f
Therefore,
f = R/2 = 20/2 = 10 cm
Answer: 10 cm
Q3. Name a mirror that can give an erect and enlarged image of an object.
Answer:
A concave mirror can produce an erect and enlarged image when the object is placed between the pole (P) and focus (F). The image is virtual and erect.
Q4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Answer:
A convex mirror is preferred because:
- It always forms an erect and diminished image.
- It provides a wider field of view.
- It allows the driver to see a larger area of traffic behind the vehicle.
In-text Questions – Mirror Formula
Q1. Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Given:
R = +32 cm
Using:
f = R/2
f = 32/2 = +16 cm
Answer: +16 cm
Q2. A concave mirror produces three times magnified real image of an object placed at 10 cm in front of it. Where is the image located?
Given:
u = −10 cm
m = −3
For a mirror:
m = −v/u
Therefore,
−3 = −v/(−10)
So,
v = −30 cm
Answer: The image is formed 30 cm in front of the mirror. Since the image is real, it is inverted and enlarged.
In-text Questions – Refraction
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:
The ray bends towards the normal because water is optically denser than air. When light travels from a rarer medium to a denser medium, its speed decreases and it bends towards the normal.
Q2. Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass?
Given:
c = 3 × 10⁸ m/s
n = 1.50
Using:
n = c/v
Therefore,
v = c/n
v = (3 × 10⁸)/1.5
v = 2 × 10⁸ m/s
Answer: 2 × 10⁸ m/s
Q3. Find the medium having the highest and lowest optical density.
From the table:
- Highest optical density: Diamond, refractive index = 2.42
- Lowest optical density: Air, refractive index = 1.0003
Answer: Highest – Diamond; Lowest – Air.
Q4. You are given kerosene, turpentine and water. In which of these does light travel fastest?
Refractive indices:
- Water = 1.33
- Kerosene = 1.44
- Turpentine oil = 1.47
Light travels faster in the medium having the lower refractive index.
Answer: Water
Q5. The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer:
It means that the ratio of the speed of light in air/vacuum to the speed of light in diamond is 2.42.
In other words, light travels about 2.42 times faster in air/vacuum than in diamond.
In-text Questions – Lenses
Q1. Define 1 dioptre of power of a lens.
Answer:
One dioptre is the power of a lens whose focal length is 1 metre.
1 D = 1 m⁻¹
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 if the image is equal in size to the object? Also find the power of the lens.
For a convex lens, an object placed at 2F₁ forms a real, inverted image of the same size at 2F₂.
Therefore:
2f = 50 cm
f = 25 cm = 0.25 m
Power:
P = 1/f
P = 1/0.25 = +4 D
Answer:
- Object distance = 50 cm
- Focal length = 25 cm
- Power = +4 D
Q3. Find the power of a concave lens of focal length 2 m.
For a concave lens:
f = −2 m
P = 1/f
P = 1/(−2)
P = −0.5 D
Answer: −0.5 D
EXERCISE ANSWER
1. Which material cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Answer:
(d) Clay
A lens must be made from a transparent material so that light can pass through it.
2. Concave mirror gives a virtual, erect and enlarged image. Where is the object?
(a) Between F and C
(b) At C
(c) Beyond C
(d) Between P and F
Answer:
(d) Between the pole and its principal focus
A concave mirror produces a virtual, erect and enlarged image when the object lies between P and F.
3. Where should an object be placed in front of a convex lens to obtain a real image of the same size?
(a) At F
(b) At twice the focal length
(c) At infinity
(d) Between O and F
Answer:
(b) At twice the focal length (2F₁)
At 2F₁, a convex lens forms a real, inverted image of the same size at 2F₂.
4. A spherical mirror and a thin spherical lens each have focal length −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:
(a) Both concave
- A concave mirror has negative focal length.
- A concave lens also has negative focal length.
5. No matter how far you stand from a mirror, your image is 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
A plane mirror always forms an erect image, while a convex mirror also always forms an erect image.
6. Which lens is preferred for reading small letters in a dictionary?
(a) Convex lens, f = 50 cm
(b) Concave lens, f = 50 cm
(c) Convex lens, f = 5 cm
(d) Concave lens, f = 5 cm
Answer:
(c) A convex lens of focal length 5 cm
A convex lens with a short focal length has greater magnifying power and can be used as a magnifying glass.
7. Erect image using a concave mirror of focal length 15 cm
Given:
f = 15 cm
For a concave mirror to produce an erect image, the object must be placed between P and F.
Therefore, the object should be placed at a distance:
Less than 15 cm from the mirror.
Nature of image:
- Virtual
- Erect
- Enlarged
- Formed behind the mirror
Ray diagram: Draw one ray parallel to the principal axis and another ray towards the pole; extend the reflected rays backward to obtain the virtual image behind the mirror.
8. Name the mirror used in the following situations.
(a) Headlights of a car
Concave mirror
It produces powerful parallel beams of light when the source is placed near its focus.
(b) Side/rear-view mirror
Convex mirror
It gives an erect, diminished image and provides a wider field of view.
(c) Solar furnace
Concave mirror
It concentrates sunlight at a point and produces a large amount of heat.
9. One-half of a convex lens is covered with black paper. Will it produce a complete image?
Answer: Yes.
Covering half of the lens does not prevent the formation of the complete image. The uncovered portion of the lens forms the complete image, although the brightness of the image decreases.
Observation:
- Complete image is formed.
- Image becomes less bright because less light passes through the lens.
10. Object of 5 cm is placed 25 cm from a converging lens of focal length 10 cm.
Given:
h = +5 cm
u = −25 cm
f = +10 cm
Using lens formula:
1/v − 1/u = 1/f
1/v − 1/(−25) = 1/10
1/v + 1/25 = 1/10
1/v = 1/10 − 1/25
1/v = 3/50
Therefore:
v = 50/3 = 16.67 cm
Magnification:
m = v/u
m = 16.67/(−25)
m ≈ −0.667
h′ = m × h
h′ = −0.667 × 5
h′ ≈ −3.33 cm
Answer:
- Image distance = 16.67 cm
- Image size = 3.33 cm
- Nature = Real, inverted and diminished
- Magnification ≈ −0.67
11. A concave lens of focal length 15 cm forms an image 10 cm from the lens.
Given:
f = −15 cm
v = −10 cm
Using:
1/v − 1/u = 1/f
−1/10 − 1/u = −1/15
−1/u = −1/15 + 1/10
−1/u = 1/30
Therefore:
u = −30 cm
Answer:
The object is placed 30 cm in front of the lens.
The image is virtual, erect and diminished.
12. Object is 10 cm from a convex mirror of focal length 15 cm.
Given:
u = −10 cm
f = +15 cm
Mirror formula:
1/v + 1/u = 1/f
1/v − 1/10 = 1/15
1/v = 1/15 + 1/10
1/v = 1/6
Therefore:
v = +6 cm
Answer:
- Image distance = 6 cm behind the mirror
- Nature = Virtual and erect
- Size = Diminished
13. Magnification produced by a plane mirror is +1. What does this mean?
Answer:
Magnification +1 means:
- The image is the same size as the object.
- The positive sign indicates that the image is erect and virtual.
14. Object of 5 cm is placed 20 cm in front of a convex mirror of radius 30 cm.
Given:
h = 5 cm
u = −20 cm
R = +30 cm
Focal length:
f = R/2 = 30/2 = +15 cm
Mirror formula:
1/v + 1/u = 1/f
1/v − 1/20 = 1/15
1/v = 1/15 + 1/20
1/v = 7/60
Therefore:
v = 60/7 = +8.57 cm
Magnification:
m = −v/u
m = −(8.57/−20)
m = +0.4285
Image height:
h′ = m × h
h′ = 0.4285 × 5
h′ ≈ 2.14 cm
Answer:
- Image distance = 8.57 cm behind the mirror
- Image size = 2.14 cm
- Nature = Virtual, erect and diminished
15. Object of 7 cm is placed 27 cm in front of a concave mirror of focal length 18 cm.
Given:
h = +7 cm
u = −27 cm
f = −18 cm
Mirror formula:
1/v + 1/u = 1/f
1/v − 1/27 = −1/18
1/v = −1/18 + 1/27
1/v = −1/54
Therefore:
v = −54 cm
The screen should be placed 54 cm in front of the mirror.
Magnification:
m = −v/u
m = −(−54/−27)
m = −2
Image height:
h′ = m × h
h′ = −2 × 7
h′ = −14 cm
Answer:
- Image distance = 54 cm in front of mirror
- Image size = 14 cm
- Nature = Real, inverted and enlarged
- Magnification = −2
16. Find the focal length of a lens of power −2.0 D.
Given:
P = −2 D
Using:
P = 1/f
Therefore:
f = 1/P
f = 1/(−2)
f = −0.5 m
or
f = −50 cm
Since focal length is negative, it is a concave lens.
Answer:
Focal length = −0.5 m or −50 cm; type = concave lens.
17. A doctor prescribes a corrective lens of power +1.5 D.
Given:
P = +1.5 D
Using:
P = 1/f
f = 1/1.5
f = +0.667 m
or approximately
f = +66.7 cm
Since the focal length is positive, the lens is a convex (converging) lens.
Answer:
- Focal length = +0.67 m or +66.7 cm
- Type = Convex/converging lens
- Get link
- X
- Other Apps
Comments
Post a Comment