Get summaries, questions, answers, solutions, notes, extras, PDF and guides for Chapter 9 Light: Reflection and Refraction: Class 10 Science textbook, which is part of the syllabus for students studying under SEBA (Assam Board), NBSE (Nagaland Board), TBSE (Tripura Board), CBSE (Central Board), MBOSE (Meghalaya Board), BSEM (Manipur Board), WBBSE (West Bengal Board), and all other boards following the NCERT books. These solutions, however, should only be treated as references and can be modified/changed.
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Summary
We see things because light bounces off them and enters our eyes. This bouncing is called reflection. Light usually travels in straight paths. When light hits a smooth surface like a mirror, it follows certain rules. The angle at which light hits the surface is the same as the angle at which it bounces off.
Flat mirrors, known as plane mirrors, create images that are upright and the same size as the object. These images appear behind the mirror. They are called virtual, which means they cannot be formed on a screen. These images are also flipped left to right.
Curved mirrors are called spherical mirrors. A concave mirror curves inwards, like the inside of a spoon. It can form different kinds of images. Sometimes the image is real, meaning it can be projected onto a screen, and it might be upside down. Other times, the image is virtual and upright. Concave mirrors are used in torches. A convex mirror curves outwards, like the back of a spoon. It always forms images that are virtual, upright, and smaller than the object. Cars use convex mirrors as side-view mirrors because they offer a wider view.
Spherical mirrors have a few important points: the pole (the center of the mirror’s surface), the center of curvature (the center of the sphere from which the mirror is a part), the principal axis (a line passing through the pole and center of curvature), and the principal focus (a point where parallel light rays meet or appear to meet after reflection). The distance from the pole to the principal focus is the focal length. A formula, 1/v + 1/u = 1/f, connects the object distance (u), the image distance (v), and the focal length (f). Magnification tells us if the image is larger or smaller than the object.
Light also bends when it travels from one material to another, for example, from air to water. This bending is called refraction. Refraction occurs because the speed of light changes as it moves through different materials. When light enters a denser material where it slows down, it bends towards an imaginary line perpendicular to the surface. When it enters a less dense material where it speeds up, it bends away from this line. This is why a straw in a glass of water appears bent.
Lenses also work because of refraction. A convex lens is thicker in the middle and causes light rays to come together, or converge. It can form both real and virtual images. A concave lens is thinner in the middle and causes light rays to spread out, or diverge. It always forms images that are virtual, upright, and smaller. Lenses also have a principal focus and a focal length. The lens formula is 1/v – 1/u = 1/f. The power of a lens indicates how much it can bend light. The unit for the power of a lens is the dioptre.
Textbook solutions
Intext Questions and Answers I
1. Define the principal focus of a concave mirror.
Answer: When a number of rays parallel to the principal axis are falling on a concave mirror, the reflected rays meet/intersect at a point on the principal axis of the mirror. This point is called the principal focus of the concave mirror.
2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Answer: For spherical mirrors of small apertures, the radius of curvature is found to be equal to twice the focal length. This is expressed as R = 2f.
Radius of curvature (R) = 20 cm
Focal length (f) = R/2
= 20 cm / 2
= 10 cm
3. 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 (P) and the principal focus (F) of the mirror.
4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Answer: Convex mirrors are preferred as rear-view mirrors in vehicles because they always give an erect, though diminished, image. Also, they have a wider field of view as they are curved outwards. Thus, convex mirrors enable the driver to view a much larger area than would be possible with a plane mirror.
Intext Questions and Answers II
1. Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer: Relation between radius of curvature and focal length of a spherical mirror: R = 2 f
Given R = 32 cm,
∴ f = R / 2
= 32 cm / 2
= 16 cm.
As the mirror is convex, f is taken as positive: f = +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?
Answer: Magnification produced by a spherical mirror: m = – v / u
For a three-times enlarged real image, m = –3 (real ⇒ inverted ⇒ negative magnification).
Object distance (u) = –10 cm (by convention, real object distances are negative).
Substitute into m formula:
∴ –3 = – v / (–10)
⇒ –3 = v / 10
⇒ v = –30 cm
The negative sign for v indicates the image lies on the same side as the object (in front of the mirror), and is real and inverted.
Location of image: 30 cm in front of the concave mirror.
Intext Questions and Answers III
1. 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 to a denser medium, it slows down and bends towards the normal. Since water is optically denser than air, a ray of light travelling in air that enters obliquely into water will bend towards the normal.
2. 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 x 10⁸ m s⁻¹.
Answer: v = c / n
Here, c = 3 × 10⁸ m s⁻¹ (speed of light in vacuum)
n = 1.50 (refractive index of glass)
Substitute the known values:
v = (3 × 10⁸ m s⁻¹) / 1.50
⇒ v = 3 × 10⁸ m s⁻¹ / 1.5
⇒ v = 2 × 10⁸ m s⁻¹
3. Find out, from Table 9.3, the medium having highest optical density. Also find the medium with lowest optical density.
| Material Medium | Refractive Index |
|---|---|
| Air | 1.0003 |
| Ice | 1.31 |
| Water | 1.33 |
| Alcohol | 1.36 |
| Kerosene | 1.44 |
| Fused quartz | 1.46 |
| Turpentine oil | 1.47 |
| Benzene | 1.50 |
| Crown glass | 1.52 |
| Canada Balsam | 1.53 |
| Rock salt | 1.54 |
| Carbon disulphide | 1.63 |
| Dense flint glass | 1.65 |
| Ruby | 1.71 |
| Sapphire | 1.77 |
| Diamond | 2.42 |
Answer: From Table 9.3, the medium having the highest optical density is Diamond, with a refractive index of 2.42. The medium with the lowest optical density listed in Table 9.3 is Air, with a refractive index of 1.0003.
4. 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: According to Table 9.3, the refractive indices are:
Water: 1.33
Kerosene: 1.44
Turpentine oil: 1.47
The speed of light is higher in a rarer medium, which has a lower refractive index. Comparing the refractive indices, water has the lowest refractive index (1.33). Therefore, light travels fastest in water among the given substances.
5. The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer: The statement that the refractive index of diamond is 2.42 means that the ratio of the speed of light in air (or vacuum) to the speed of light in diamond is equal to 2.42.
Intext Questions and Answers IV
1. Define 1 dioptre of power of a lens.
Answer: 1 dioptre is the power of a lens whose focal length is 1 metre. The SI unit of power of a lens is ‘dioptre’, and it is denoted by the letter D.
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.
Answer: m = –1
⇒ – v / u = –1
⇒ v / u = 1
⇒ v = u
v = 50 cm (real image distance).
∴ u = 50 cm
1 / f = 1 / v + 1 / u
⇒ 1 / f = 1 / 50 + 1 / 50
⇒ 1 /f = 2 / 50
⇒ 1 / f = 1 / 25
∴ f = 25 cm
P = 1 / f (in metres)
⇒ P = 1 / 0.25 m
⇒ P = +4 D
The needle must be placed 50 cm in front of the convex lens. The power of the lens is +4 D.
3. Find the power of a concave lens of focal length 2 m.
Answer: The power of a lens be given by: P = 1 / f (in metres)
Here,
f = – 2 m
P = 1 / (– 2)
⇒ P = – 0.5 dioptres
Exercise Questions and Answers
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
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) Between the pole of the mirror and its principal focus.
3. Where should an object be placed in front of a convex lens to get a real image of the size of the object?
(a) At the principal focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre of the lens and its principal focus.
Answer: (b) At twice the focal length
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
(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.
5. 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.
Answer: (d) either plane or convex.
6. 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.
Answer: (c) A convex lens of focal length 5 cm.
7. 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: To obtain an erect image using a concave mirror of focal length 15 cm, the object should be placed at a distance less than 15 cm from the mirror, i.e., between the pole (P) and the principal focus (F) of the mirror.
The nature of the image formed will be virtual and erect.
The image will be larger than the object (enlarged).
8. 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.
Answer: (a) Headlights of a car: Concave mirrors are used.
Reason: Concave mirrors are used in vehicle headlights to get powerful parallel beams of light when the light source is placed at their principal focus.
(b) Side/rear-view mirror of a vehicle: Convex mirrors are used.
Reason: Convex mirrors are preferred because they always give an erect, though diminished, image. Also, they have a wider field of view as they are curved outwards, enabling the driver to view a much larger area of the traffic behind him/her to facilitate safe driving.
(c) Solar furnace: Large concave mirrors are used.
Reason: Large concave mirrors are used to concentrate sunlight to produce heat in solar furnaces. The sunlight is converged at the focus of the mirror, producing a high temperature.
9. 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, if one-half of a convex lens is covered with a black paper, the lens will still produce a complete image of the object.
Explanation: Each small portion of an extended object acts like a point source, and an infinite number of rays originate from each of these points. To form the image of a point, rays of light from that point, after refraction from the lens, intersect. Even if half of the lens is covered, light rays from all parts of the object can still pass through the uncovered portion of the lens and converge to form a complete image at the same position. However, the brightness or intensity of the image will be reduced because the number of rays forming the image is reduced by half.
Experiment: One-half of a convex lens covered with black paper is taken. An object is placed infront of it, say beyond 2F1 It produces a diminished, real and inverted image between F2 and 2F2. Experiment is repeated by placing object at different positions. In all cases image is formed.
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.
Answer: Given:
Object height, h = +5 cm
Object distance, u = -25 cm
Focal length of converging (convex) lens, f = +10 cm
We need to find the image distance (v), image height (h’), and nature of the image.
Using the lens formula:
1/v – 1/u = 1/f
⇒ 1/v – 1/(-25 cm) = 1/(10 cm)
⇒ 1/v + 1/25 = 1/10
⇒ 1/v = 1/10 – 1/25
⇒ 1/v = (5 – 2)/50
⇒ 1/v = 3/50
⇒ v = 50/3 cm
⇒ v = +16.67 cm
The image is formed at a distance of 16.67 cm on the other side of the lens.
Magnification,
m = h’/h
⇒ m = v/u
⇒ m = (+16.67 cm) / (-25 cm)
⇒ m = -0.67
Now, for the image size:
h’ = m x h
⇒ h’ = (-2/3) x 5 cm
⇒ h’ = -10/3 cm
⇒ h’ = -3.33 cm
So, the position of the image is 16.67 cm on the other side of the lens.
The size of the image is 3.33 cm.
The nature of the image is real, inverted, and diminished.
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? Draw the ray diagram.
Answer: Given,
Focal length of concave lens, f = -15 cm
Image distance, v = -10 cm (A concave lens always forms a virtual image on the same side of the lens as the object)
We need to find the object distance (u).
Using the lens formula:
1/v – 1/u = 1/f
⇒ 1/(-10 cm) – 1/u = 1/(-15 cm)
⇒ -1/10 – 1/u = -1/15
⇒ -1/u = -1/15 + 1/10
⇒ -1/u = (-2 + 3)/30
⇒ -1/u = 1/30
⇒ u = -30 cm
Thus, the object is placed 30 cm from the lens on the same side as the image.
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.
Answer: Given:
Object distance, u = -10 cm
Focal length of convex mirror, f = +15 cm
We need to find the image distance (v) and the nature of the image.
Using the mirror formula:
1/v + 1/u = 1/f
⇒ 1/v + 1/(-10 cm) = 1/(15 cm)
⇒ 1/v – 1/10 = 1/15
⇒ 1/v = 1/15 + 1/10
⇒ 1/v = (2 + 3)/30
⇒ 1/v = 5/30 = 1/6
⇒ v = +6 cm
The image is formed at a distance of 6 cm behind the mirror.
Magnification,
m = -v/u
⇒ m = -(+6 cm) / (-10 cm)
⇒ m = 6/10
⇒ m = +0.6
So, the position of the image is 6 cm behind the convex mirror.
The nature of the image is virtual, erect, and diminished.
13. The magnification produced by a plane mirror is +1. What does this mean?
Answer: If the magnification produced by a plane mirror is +1, it means:
The positive sign (+) indicates that the image formed is virtual and erect.
The value ‘1’ indicates that the size of the image is equal to the size of the object (h’ = h).
Therefore, a magnification of +1 for a plane mirror signifies that the image is virtual, erect, and 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.
Answer: Given:
Object height, h = +5.0 cm
Object distance, u = -20 cm
Radius of curvature of convex mirror, R = +30 cm
Focal length, f = R/2 = +30 cm / 2 = +15 cm
We need to find the image distance (v), nature of the image, and image height (h’).
Using the mirror formula:
⇒ 1/v + 1/u = 1/f
⇒ 1/v + 1/(-20 cm) = 1/(15 cm)
⇒ 1/v – 1/20 = 1/15
⇒ 1/v = 1/15 + 1/20
⇒ 1/v = (4 + 3)/60
⇒ 1/v = 7/60
⇒ v = +60/7 cm
⇒ v = +8.57 cm
The image is formed at a distance of 60/7 cm (or approximately 8.57 cm) behind the mirror.
Magnification, m = -v/u = h’/h
⇒ m = -(+60/7 cm) / (-20 cm)
⇒ m = (60/7) x (1/20)
⇒ m = 3/7
⇒ m = +0.43
Now, for the image size:
h’ = m x h
⇒ h’ = (3/7) x 5.0 cm
⇒ h’ = 15/7 cm
⇒ h’ = +2.14 cm
So, the position of the image is 60/7 cm (approx. 8.57 cm) behind the mirror.
The nature of the image is virtual, erect, and diminished.
The size of the image is 15/7 cm (approx. 2.14 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, so that a sharp focussed 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 of concave mirror, f = -18 cm
We need to find the image distance (v) (where the screen should be placed), image height (h’), and nature of the image.
Using the mirror formula:
1/v + 1/u = 1/f
⇒ 1/v + 1/(-27 cm) = 1/(-18 cm)
⇒ 1/v – 1/27 = -1/18
⇒ 1/v = 1/27 – 1/18
⇒ 1/v = (2 – 3)/54
⇒ 1/v = -1/54
⇒ v = -54 cm
A screen should be placed at a distance of 54 cm in front of the mirror to obtain a sharp focused image. The negative sign indicates the image is formed on the same side as the object, which is where real images are formed by a concave mirror.
Magnification, m = -v/u = h’/h
⇒ m = -(-54 cm) / (-27 cm)
⇒ m = 54 / (-27) = -2
Now, for the image size:
h’ = m x h
⇒ h’ = (-2) x 7.0 cm
⇒ h’ = -14.0 cm
So, the screen should be placed 54 cm in front of the mirror.
The size of the image is 14.0 cm.
The nature of the image is real, inverted, and enlarged.
16. Find the focal length of a lens of power – 2.0 D. What type of lens is this?
Answer: Given:
Power of the lens, P = -2.0 D
The relationship between power and focal length (f, in metres) is P = 1/f.
So, f = 1/P
⇒ f = 1 / (-2.0 D)
⇒ f = -0.5 metres
⇒ f = -50 cm
The focal length of the lens is -0.5 m or -50 cm.
Since the power of the lens is negative (and consequently the focal length is negative), the lens is a concave lens.
17. 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 of the corrective lens, P = +1.5 D
The focal length (f, in metres) is given by f = 1/P.
f = 1 / (+1.5 D)
⇒ f = 1 / (3/2) m
⇒ f = 2/3 m
⇒ f = +0.67 metres or +66.67 cm
The focal length of the lens is +2/3 m (or approximately +66.67 cm).
Since the power of the lens is positive, the lens is a convex lens. A convex lens is a converging lens.
Therefore, the prescribed lens is converging.
Extras
Additional MCQs (Knowledge Based)
1. What phenomenon describes the tendency of light to bend around very small opaque objects and not travel in a straight line?
A. Reflection
B. Refraction
C. Diffraction
D. Dispersion
Answer: C. Diffraction
45. If two thin lenses of powers P₁ and P₂ are placed in contact, the net power of the combination is:
A. P₁P₂
B. P₁/P₂
C. P₁ + P₂
D. P₁ – P₂
Answer: C. P₁ + P₂
Additional MCQs (Competency Based)
1. Assertion (A): When light travels from air into a glass slab, its path bends.
Reason (R): The speed of light changes as it enters a different transparent medium.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R does not explain A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (a) Both A and R are true and R is the correct explanation of A.
28. Sequence the steps to determine the focal length of a concave mirror using a distant object (like the sun):
(i) Measure the distance between the mirror and the paper.
(ii) Hold a concave mirror facing the distant object.
(iii) Move a sheet of paper in front of the mirror until a sharp, bright spot is formed.
(iv) The measured distance is the approximate focal length.
(a) (ii) → (iii) → (i) → (iv)
(b) (iii) → (ii) → (iv) → (i)
(c) (ii) → (i) → (iii) → (iv)
(d) (i) → (iii) → (ii) → (iv)
Answer: (a) (ii) → (iii) → (i) → (iv)
Additional Questions and Answers
1. What makes things visible to our eyes?
Answer: During the day, sunlight helps us to see objects. An object reflects light that falls on it. This reflected light, when received by our eyes, enables us to see things.
39. Define 1 dioptre of power.
Answer: 1 dioptre is the power of a lens whose focal length is 1 metre.
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