Chapter 9 - Light Reflection and Refraction Class 10 Notes (Science)
Chapter 9 Light Reflection and Refraction Class 10 Notes (Science)
Updated Solution 2024-2025 Updated Solution 2024-2025
NCERT Solutions for Science, Chapter 9 Light Reflection and Refraction Class 10 Notes, Question/Answers, Activity, Experiment’s & Projects
Chapter 9 Light Reflection and Refraction
Activity 9.1
- Take a large shining spoon. Try to view your face in its curved surface.
- Do you get the image? Is it smaller or larger?
Ans: Yes, you get the image. On the convex surface (outward curve), the image will be smaller and inverted. On the concave surface (inward curve), the image will be larger and could be magnified, especially if the spoon is held close.
- Move the spoon slowly away from your face. Observe the image. How does it change?
Ans: When you move the spoon slowly away from your face, the image formed on the convex (outward curve) surface will change in the following ways:
- Size: The image will become larger as you increase the distance between the spoon and your face.
- Clarity: The image may become less distinct or blurry at first, as the light rays reflecting from the curved surface start to spread out.
This happens because the convex surface acts like a converging mirror, and as you move further away, the light rays reflecting off the spoon are spread over a larger area, altering the image’s appearance.
- Reverse the spoon and repeat the Activity. How does the image look like now?
Ans: When you reverse the spoon and use the concave (inward curve) surface, the image will appear larger and upright, especially if the spoon is held close to your face. The concave surface acts like a diverging mirror, magnifying the reflected image when viewed from a close distance. If the spoon is moved further away, the image will gradually shrink and could eventually become inverted.
- Compare the characteristics of the image on the two surfaces.
Ans: The characteristics of the images formed on the convex and concave surfaces of the spoon are as follows:
1. Convex Surface (Outward Curve):
- Image size: The image appears smaller than the actual object.
- Image orientation: The image is inverted (upside down).
- Type of mirror: The convex surface acts as a converging mirror, which reduces the size of the reflected image as the distance between the spoon and the object increases.
- Distance effect: As the spoon is moved farther from the face, the image becomes larger but less distinct.
2. Concave Surface (Inward Curve):
- Image size: The image can appear larger and magnified, especially when the spoon is held close to the face.
- Image orientation: The image appears upright.
- Type of mirror: The concave surface acts as a diverging mirror when viewed up close, magnifying the reflected image. As the distance increases, the image shrinks and may eventually become inverted.
- Distance effect: When the spoon is close to the face, the image appears larger and upright, but as it is moved farther away, the image size decreases and may flip upside down.
In summary, the convex surface creates a smaller, inverted image, while the concave surface provides a larger, upright image when viewed closely.
Activity 9.2
CAUTION: Do not look at the Sun directly or even into a mirror reflecting sunlight. It may damage your eyes.
- Hold a concave mirror in your hand and direct its reflecting surface towards the Sun.
Ans: When you hold a concave mirror with its shiny, curved side facing the Sun, the mirror will reflect sunlight. The curved shape causes the light to focus at a single point, known as the focal point. If you bring a piece of paper or another object to this point, you can see the sunlight concentrated there, potentially creating a small, bright spot. This happens because the concave surface bends the light rays inward, intensifying the light.
- Direct the light reflected by the mirror on to a sheet of paper held close to the mirror.
Ans: To direct the light reflected by a mirror onto a sheet of paper, simply hold the paper in front of the mirror. The mirror will bounce light off its surface, and you can adjust the angle of the paper until the reflected light shines onto it. The paper will show a bright spot where the light focuses.
- Move the sheet of paper back and forth gradually until you find on the paper sheet a bright, sharp spot of light.
Ans: When you move the sheet of paper back and forth, you are adjusting the distance between the paper and the focal point of the concave mirror. The bright, sharp spot of light appears when the paper is placed exactly at the focal point where the mirror focuses the light. At this point, the light rays converge and form a concentrated, bright spot.
As you move the paper away or closer from the focal point, the sharpness and brightness of the spot will change, becoming either larger and more diffused or more concentrated. The correct position for a bright, sharp spot is at the focal point of the mirror, where the light converges to its smallest, most intense point.
- Hold the mirror and the paper in the same position for a few minutes. What do you observe? Why?
Ans: When you hold the mirror and paper in the same position for a few minutes, you will likely observe the paper starting to heat up and possibly burn or scorch. This happens because the concave mirror is focusing the sunlight onto a single point on the paper. The concentration of light at this focal point increases the intensity of the sunlight, which raises the temperature at that spot. Over time, the concentrated energy can cause the paper to heat up to the point where it starts to burn or char.
Activity 9.3
You have already learnt a way of determining the focal length of a concave mirror. In Activity 9.2, you have seen that the sharp bright spot of light you got on the paper is, in fact, the image of the Sun. It was a tiny, real, inverted image. You got the approximate focal length of the concave mirror by measuring the distance of the image from the mirror.
- Take a concave mirror. Find out its approximate focal length in the way described above. Note down the value of focal length. (You can also find it out by obtaining image of a distant object on a sheet of paper.)
- Mark a line on a Table with a chalk. Place the concave mirror on a stand. Place the stand over the line such that its pole lies over the line.
- Draw with a chalk two more lines parallel to the previous line such that the distance between any two successive lines is equal to the focal length of the mirror. These lines will now correspond to the positions of the points P, F and C, respectively. Remember – For a spherical mirror of small aperture, the principal focus F lies mid-way between the pole P and the center of curvature C.
- Keep a bright object, say a burning candle, at a position far beyond C. Place a paper screen and move it in front of the mirror till you obtain a sharp bright image of the candle flame on it.
- Observe the image carefully. Note down its nature, position and relative size with respect to the object size.
- Repeat the activity by placing the candle – (a) just beyond C, (b) at C, (c) between F and C, (d) at F, and (e) between P and F.
- In one of the cases, you may not get the image on the screen. Identify the position of the object in such a case. Then, look for its virtual image in the mirror itself.
- Note down and tabulate your observations.
Ans: Activity 9.3
Q1. Find the approximate focal length of the concave mirror:
Ans 1: To determine the focal length of a concave mirror, you can use the method of obtaining a sharp image of the Sun on a sheet of paper. Measure the distance from the mirror’s surface to the sharp image of the Sun. This distance gives the focal length of the concave mirror. Alternatively, you can place the mirror and focus it on a distant object (like a building or tree), and measure the distance from the mirror to the sharp image. The approximate focal length can be determined from this setup.
Q2. Mark a line on the Table and position the concave mirror:
Ans 2
- Mark a straight line on the table using chalk.
- Place the concave mirror on the stand such that its pole (the central point of the mirror) lies on the marked line.
- Draw two more lines parallel to the original line, with the distance between any two successive lines equal to the focal length of the mirror.
- These lines will represent the points PPP (pole), FFF (focal point), and CCC (center of curvature). The focal point FFF lies exactly halfway between PPP and CCC.
Q 3. Position the object (candle) at different locations:
Ans 3: Now, place a bright object (such as a burning candle) in front of the concave mirror at the following positions:
- (a) Object just beyond C:
- The image will be real, inverted, and diminished in size.
- The image will be formed between FFF and CCC.
- (b) Object at C:
- The image will be real, inverted, and of the same size as the object.
- The image will be formed at the same position as the object (at CCC).
- (c) Object between F and C:
- The image will be real, inverted, and magnified.
- The image will form beyond CCC.
- (d) Object at F:
- The image will form at infinity.
- It will be highly magnified and parallel rays.
- (e) Object between P and F:
- The image will be virtual, upright, and magnified.
- It will appear to be behind the mirror.
Q 4. In one of the cases, you may not get the image on the screen:
Ans 4: When the object is placed between the pole (P) and the focal point (F), the image formed is virtual. This image cannot be obtained on a screen since it is formed behind the mirror. Instead, it can be observed directly in the mirror, appearing upright and magnified.
Q 5. Tabulate your observations:
Ans 5:
| Object Position | Image Nature | Image Position | Image Size |
| Beyond CCC | Real, Inverted | Between FFF and CCC | Diminished |
| At CCC | Real, Inverted | At CCC | Same size as object |
| Between FFF and CCC | Real, Inverted | Beyond CCC | Magnified |
| At FFF | At infinity | At infinity | Magnified |
| Between PPP and FFF | Virtual, Upright | Behind the mirror | Magnified |
These steps will help you complete Activity 9.3 and understand the behavior of the concave mirror for different object positions.
Activity 9.4
Q 1. Draw neat ray diagrams for each position of the object shown in Table 9.1.
Ans 1: (a). At Infinity
(b). Object is beyond ‘C’
(c). Object is at ‘C’
(d). Object is in between C and F
(e). Object is at F
(f). between P and F object is in
Q 2. You may take any two of the rays mentioned in the previous section for locating the image.
Ans 2: To locate the image formed by a concave mirror, you can use any two of the following rays commonly mentioned in ray diagrams:
- Parallel Ray:
- A ray parallel to the principal axis of the mirror reflects through the focal point (F).
- This ray helps identify the direction of the reflected rays.
- Focal Ray:
- A ray passing through the focal point (F) before hitting the mirror reflects parallel to the principal axis.
- This ray helps define the path of the reflected rays.
By drawing these two rays and determining their point of intersection, you can locate the position of the image formed by the concave mirror.
Q 3. Compare your diagram with those given in Fig. 9.7
Ans 3: To compare your image with these ray diagrams, I can walk through the specific types of image formation shown:
- Case (a): The object is at infinity. The rays converge at the focal point (F), forming a highly diminished and real image.
- Case (b): The object is located beyond the center of curvature (C). The image is real, inverted, and reduced in size, formed between the focal point (F) and the center of curvature (C).
- Case (c): The object is at the center of curvature. The image formed is real, inverted, and equal in size to the object, at the center of curvature (C).
- Case (d): The object is between the center of curvature (C) and the focal point (F). The image is real, inverted, and magnified, and it forms beyond the center of curvature (C).
- Case (e): The object is at the focal point (F). The rays are parallel and do not converge, so no real image is formed. The image would be at infinity.
- Case (f): The object is between the focal point (F) and the mirror’s surface. The image formed is virtual, upright, and magnified, appearing behind the mirror.
Q 4. Describe the nature, position and relative size of the image formed in each case. Tabulate the results in a convenient format.
Object Position | Nature of Image | Position of Image | Relative Size of Image |
At Infinity | Real, Inverted, Diminished | At the focal point (F) | Highly diminished, point image |
Beyond Center of Curvature (C) | Real, Inverted | Between F and C | Smaller than the object, reduced |
At Center of Curvature (C) | Real, Inverted | At the center of curvature (C) | Same size as the object |
Between Center of Curvature (C) and Focal Point (F) | Real, Inverted | Beyond C | Magnified, larger than the object |
At Focal Point (F) | No real image (Rays parallel after reflection, no convergence) | At infinity | Not definable (parallel rays) |
Between Focal Point (F) and Mirror | Virtual, Upright | Behind the mirror | Magnified, larger than the object |
Ans 4: Here’s a summarized description of the nature, position, and relative size of the image formed in each case, organized in a convenient tabular format:
Explanation:
- Real and Inverted: The image is formed on the same side of the mirror as the object, and it is upside down (inverted).
- Virtual and Upright: The image is formed behind the mirror and is upright, appearing larger than the object.
- Diminished, Same Size, magnified: Refers to how the image compares in size with the object. If the image is smaller, it is diminished; if it is larger, it is magnified. If the size is equal, it’s the same size as the object.
This table helps clarify the relationships between object position and image characteristics for a concave mirror.
Activity 9.5
- Take a convex mirror. Hold it in one hand.
- Hold a pencil in the upright position in the other hand.
- Observe the image of the pencil in the mirror. Is the image erect or inverted? Is it diminished or enlarged?
Ans: The image of the pencil in the convex mirror is erect and diminished. Convex mirrors always form upright (erect) images that are smaller (diminished) compared to the actual object.
- Move the pencil away from the mirror slowly. Does the image become smaller or larger?
Ans: As the pencil is moved away from the convex mirror, the image remains smaller. Convex mirrors always form diminished images regardless of the object’s distance, so the size of the image does not increase.
- Repeat this Activity carefully. State whether the image will move closer to or farther away from the focus as the object is moved away from the mirror?
Ans: As stated in the activity, the image formed by a convex mirror will move closer to the focus as the object is moved farther away from the mirror. This means that as you increase the distance between the object and the mirror, the image will shift closer to the focal point, although it will still remain diminished and upright.
Activity 9.6
- Observe the image of a distant object, say a distant tree, in a plane mirror.
Ans: When you look at the reflection of a distant object like a tree in a plane mirror, you will observe an image that is formed behind the mirror. The image appears to be the same size as the actual tree, and it will be upright (not inverted), as a plane mirror forms a virtual image that has the same shape and size as the object.
- Could you see a full-length image?
Ans: Yes, you can see a full-length image of the tree in a plane mirror as long as the mirror is tall enough to reflect the entire tree. The image in a plane mirror will be of the same size as the object, and since it is a virtual image, it will appear behind the mirror at the same distance as the tree is in front of it. If the mirror’s size is small or too short, it will only reflect a portion of the tree, not the full length.
- Try with plane mirrors of different sizes. Did you see the entire object in the image?
Ans: If the plane mirror is small, you may not be able to see the entire object at once. A larger mirror allows for a wider field of view, enabling you to see the full-length image of the distant tree. The size of the mirror affects how much of the object can be reflected.
- Repeat this Activity with a concave mirror. Did the mirror show full length image of the object?
Ans: A concave mirror can produce different types of images depending on the object’s distance from the mirror. If the tree is far away, the image formed will be inverted and diminished in size. The concave mirror may not show a full-length image of the object because it reduces the size of distant objects. However, if the tree is placed closer, the image may appear larger and upright.
- Now try using a convex mirror. Did you succeed? Explain your observations with reason.
Ans: Using a convex mirror, you will always see a diminished, virtual, and upright image of the distant tree. The image will not be full-length, and it will appear smaller than the actual tree. Convex mirrors diverge light rays, making the image smaller and giving a wider field of view, but it can’t produce a full-length image of a distant object.
Questions
Q 1. Define the principal focus of a concave mirror.
Ans 1: The principal focus of a concave mirror is the point where parallel rays of light that are reflected from the mirror converge (come together). It is located in front of the mirror, along the mirror’s principal axis. When light rays hit the concave surface, they bounce off and meet at this focus. The distance between the mirror’s surface and the focus is called the focal length.
Q 2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Ans 2:
The focal length of a spherical mirror is related to its radius of curvature by the formula: f = R / 2
Where:
f is the focal length
R is the radius of curvature
Given that the radius of curvature R = 20 cm, we can calculate the focal length as: f = 20 / 2 = 10 cm
Therefore, the focal length of the spherical mirror is 10 cm.
Q 3. Name a mirror that can give an erect and enlarged image of an object.
Ans 3: A concave mirror can give an erect and enlarged image of an object. This happens when the object is placed very close to the mirror, within its focal length. The image formed appears larger and upright.
Q 4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Ans 4: We prefer a convex mirror as a rear-view mirror in vehicles because it gives a wider field of view. The curved shape of a convex mirror spreads out light, allowing the driver to see more of the road and surroundings behind them. It also makes objects appear smaller and farther away, which helps to reduce blind spots and improve safety while driving.
Questions
Q 1. Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Ans 1:
To find the focal length of a convex mirror, we use the formula: f = R / 2
Where:
f is the focal length
R is the radius of curvature
Given that the radius of curvature R = 32 cm, substitute the value into the formula: f = 32 / 2 = 16 cm
Therefore, the focal length of the convex mirror is 16 cm.
Q 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?
Ans 2: A concave mirror forms a real, magnified image when an object is placed in front of it. In this case, the object is 10 cm away, and the image is three times larger. To find where the image is located, we can use the magnification formula for mirrors:
Magnification (M) = − (Image distance (v)) / (Object distance (u))
Given that the magnification is 3 (enlarged real image), and the object distance u = − 10 cm (negative because it’s in front of the mirror), we can rearrange the formula to solve for the image distance v:
3=−10−v
This simplifies to: v = 30 cm
So, the image is located 30 cm in front of the mirror.
Activity 9.7
- Place a coin at the bottom of a bucket filled with water.
- With your eye to a side above water, try to pick up the coin in one go. Did you succeed in picking up the coin?
Ans: You likely did not succeed in picking up the coin in one go. This is because light bends (refracts) when it passes from water to air, making the coin appear to be in a different position than it actually is. Your perception of where the coin is is distorted due to the refraction of light.
- Repeat the Activity. Why did you not succeed in doing it in one go?
Ans: The reason for not succeeding is that when you look at the coin from above the water’s surface, the light rays are refracted (bent) as they move from water to air, altering the apparent position of the coin. The coin seems to be at a different depth than it actually is, making it harder to pick it up directly with your hand.
- Ask your friends to do this. Compare your experience with theirs.
Ans: Your friends will likely experience the same outcome because the optical illusion caused by the refraction of light is a universal phenomenon. The bending of light causes the coin to appear to be at a shallower depth, making it difficult for anyone to pick it up accurately from above the water.
Activity 9.8
- Place a large shallow bowl on a Table and put a coin in it.
- Move away slowly from the bowl. Stop when the coin just disappears from your sight.
- Ask a friend to pour water gently into the bowl without disturbing the coin.
- Keep looking for the coin from your position. Does the coin become visible again from your position? How could this happen?
Ans: Yes, the coin may become visible again from your position. When water is added to the bowl, it changes the way light travels and bends (refraction). The light rays from the coin may bend differently as they pass through the water, making the coin visible from your original position, even if it was previously out of sight.
This happens due to the refraction of light. When the water is poured into the bowl, the light rays passing through the water bend (refract) differently than when the bowl was empty. This bending of light can cause the coin, which was previously out of sight due to the angle of vision, to become visible again. The water alters the way light travels from the coin to your eyes, allowing you to see it even from a distance.
Activity 9.9
- Draw a thick straight line in ink, over a sheet of white paper placed on a Table.
- Place a glass slab over the line in such a way that one of its edges makes an angle with the line.
- Look at the portion of the line under the slab from the sides. What do you observe? Does the line under the glass slab appear to be bent at the edges?
Ans: When you observe the line under the glass slab from the sides, you will notice that the line appears to be bent at the edges. This happens because of refraction. The light rays traveling through the air and glass slab bend when they pass from one medium (air) to another (glass), causing the appearance of bending at the boundary between the two mediums. The glass slab alters the path of light, making the line under it appear shifted or bent.
- Next, place the glass slab such that it is normal to the line. What do you observe now? Does the part of the line under the glass slab appear bent?
Ans: When the glass slab is placed normal (perpendicular) to the line, the line under the slab will not appear bent. This is because the light rays passing through the slab will not change direction at an angle when entering or exiting the glass. Since the angle of incidence is 0 degrees, there is no refraction happening at the boundary, and the line will appear straight beneath the slab.
- Look at the line from the top of the glass slab. Does the part of the line, beneath the slab, appear to be raised? Why does this happen?
Ans: Yes, the part of the line beneath the glass slab will appear raised when viewed from above. This phenomenon is known as the “apparent rise” and occurs due to the refractive index of the glass slab. The light rays from the line are bent at the interface between the air and the glass, causing the position of the line to seem elevated. Since light travels more slowly through the glass than through air, the visual effect of the line appears as if it has shifted upward.
Chapter 9 – Light Reflection and Refraction Class 10 Notes, Question/Answer, Activity & Projects
Updated Solution 2024-2025
This complete solution is prepared as per the latest syllabus of 2024-25. If you have any further queries, feel free to ask!
Activity 9.10
- Fix a sheet of white paper on a drawing board using drawing pins.
- Place a rectangular glass slab over the sheet in the middle.
- Draw the outline of the slab with a pencil. Let us name the outline as ABCD.
- Take four identical pins.
- Fix two pins, say E and F, vertically such that the line joining the pins is inclined to the edge AB.
- Look for the images of the pins E and F through the opposite edge. Fix two other pins, say G and H, such that these pins and the images of E and F lie on a straight line.
- Remove the pins and the slab.
- Join the positions of tip of the pins E and F and produce the line up to AB. Let EF meet AB at O. Similarly, join the positions of tip of the pins G and H and produce it up to the edge CD. Let HG meet CD at O′.
- Join O and O′. Also produce EF up to P, as shown by a dotted line in Fig. 9.10.
Fig 9.10 Refraction of light through a rectangular glass slab
Ans: This activity demonstrates the refraction of light through a glass s
lab, highlighting the lateral displacement of light when it passes through a medium with a different refractive index. Here is a detailed explanation of the steps and their significance:
Materials Needed:
- White sheet of paper
- Drawing board
- Rectangular glass slab
- Four identical pins
- Drawing pins
- Pencil
- Ruler
Procedure:
1. Setup the Sheet and Glass Slab:
- Fix the white paper on a drawing board using drawing pins.
- Place the glass slab at the center of the paper and trace its outline using a pencil. Label the outline as ABCD.
2. Place the Initial Pins:
- Fix two pins (E and F) vertically on the paper. Ensure the line joining the pins is inclined (not parallel) to the edge AB of the slab.
3. View Through Opposite Edge:
- Observe the images of the pins (E and F) through the opposite edge CD of the slab.
4. Align the Second Set of Pins:
- Fix two more pins (G and H) such that these pins and the images of E and F appear in a straight line when viewed through edge CD.
5. Remove the Slab and Pins:
- Carefully remove the pins and the glass slab, ensuring the pinholes remain visible.
6. Draw the Lines:
- Join the pinholes of E and F: Extend the line to meet edge AB at a point O.
- Join the pinholes of G and H: Extend the line to meet edge CD at a point O’.
7. Connect O and O’:
- Draw a straight-line joining points O and O’.
- Extend the line joining E and F beyond O to point P (as shown in the diagram).
Observations:
- The refracted ray inside the slab (line joining O and O’) is parallel to the incident ray (EF) but displaced laterally.
- This lateral displacement is a characteristic of refraction through a glass slab.
Concepts Illustrated:
- Refraction of Light:
Light bends at the boundaries (air-glass and glass-air) due to the change in refractive index. - Lateral Displacement:
The incident ray (EF) is shifted laterally after passing through the slab but remains parallel to itself. - Straight-Line Path:
The pins G and H align with the images of E and F, confirming that light travels in straight lines in uniform media.
Questions
Q 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?
Ans 1: This happens because water is denser than air, which slows down the speed of light. As light moves from a less dense medium (air) to a denser medium (water), its path bends closer to the normal due to a change in its speed. This phenomenon is called refraction.
Q2. Light enters from air into glass having a refractive index of 1.50. What is the speed of light in the glass?
(The speed of light in vacuum is 3 × 10⁸ m/s.)
Ans 2:
The speed of light in glass can be found using the formula: v = c / n
Where:
• v = speed of light in the medium (glass)
• c = speed of light in vacuum = 3 × 10⁸ m/s
• n = refractive index of glass = 1.50
Substitute the values:
v = (3 × 10⁸) / 1.50
v = 2 × 10⁸ m/s
Therefore, the speed of light in glass is 2 × 10⁸ m/s.
Q 3. Find out, from Table 9.3, the medium having highest optical density. Also find the medium with lowest optical density.
Ans 3: Highest optical density: Diamond (Refractive Index = 2.42)
Lowest optical density: Air (Refractive Index = 1.0003)
Q 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.
Ans 4: The speed of light is inversely proportional to the refractive index of a medium. Among kerosene (n = 1.44), turpentine oil (n = 1.47), and water (n = 1.33), water has the lowest refractive index. Hence, light travels fastest in water.
Q 5. The refractive index of diamond is 2.42. What is the meaning of this statement?
Ans 5: The refractive index of diamond being 2.42 means that light travels 2.42 times slower in diamond than in a vacuum. This also indicates diamond’s high ability to bend (refract) light entering it.
Activity 9.11
CAUTION: Do not look at the Sun directly or through a lens while doing this Activity or otherwise. You may damage your eyes if you do so.
- Hold a convex lens in your hand. Direct it towards the Sun.
- Focus the light from the Sun on a sheet of paper. Obtain a sharp bright image of the Sun.
- Hold the paper and the lens in the same position for a while. Keep observing the paper. What happened? Why? Recall your experience in Activity 9.2.
Ans: When you hold the paper and lens in the same position, the sharp and bright image of the Sun on the paper will eventually disappear or become dimmer. This happens because the convex lens focuses the Sun’s light on the paper, but over time, the lens will start to heat up the paper, which can cause it to burn or absorb the light. Additionally, if the position of the lens or the paper shifts slightly, the focal point will change, making the image blur and fade.
Recall your experience in Activity 9.2. In Activity 9.2, it’s likely that you had an experience related to focusing light through a lens or observing the effects of a focused beam of light. This would provide a basis for understanding how a convex lens works to concentrate light and how this concentrated light can generate heat, potentially causing burns or fading of the image. The comparison between this activity and Activity 9.2 would involve understanding the relationship between light, lenses, and heat.
Activity 9.12
- Take a convex lens. Find its approximate focal length in a way described in Activity 9.11.
- Draw five parallel straight lines, using chalk, on a long Table such that the distance between the successive lines is equal to the focal length of the lens.
- Place the lens on a lens stand. Place it on the central line such that the optical center of the lens lies just over the line.
- The two lines on either side of the lens correspond to F and 2F of the lens respectively. Mark them with appropriate letters such as 2f₁, f₁, f₂, and 2f₂ respectively.
- Place a burning candle, far beyond 2f₁ to the left. Obtain a clear sharp image on a screen on the opposite side of the lens.
- Note down the nature, position and relative size of the image. Repeat this Activity by placing object Just behind 2f₁, Between f₁ and 2f₁, At f₁ Between f₁ and O. Note down and tabulate your observations.
Ans: Observations for Different Object Positions:
Object Position | Image Nature | Image Position | Relative Size |
Beyond 2f₁ | Real, Inverted | Between f₂ and 2f₂ | Reduced |
Between f₁ and 2f₁ | Real, Inverted | Beyond 2f₂ | Larger |
At f₁ | Real, Inverted | At 2f₂ | Same size as object |
Between f₁ and O | Virtual, Upright | Same side as the object | Larger |
Explanation of Results:
- When the object is between f₁ and 2f₁: The image is real, inverted, and larger than the object. It forms beyond 2f₂, showing magnification.
- When the object is at f₁: The image will form at 2f₂, with the size equal to that of the object.
- When the object is between f₁ and O: The image formed will be virtual, upright, and magnified. This is a typical result when the object is within the focal length of a convex lens.
Activity 9.13
- Take a concave lens. Place it on a lens stand.
- Place a burning candle on one side of the lens.
- Look through the lens from the other side and observe the image. Try to get the image on a screen, if possible. If not, observe the image directly through the lens.
- Note down the nature, relative size and approximate position of the image.
Ans: Nature, Relative Size, and Approximate Position of the Image:
- Nature of the Image: The image formed by a concave lens is always virtual, upright, and diminished.
- Relative Size: The image is smaller than the object (candle).
- Approximate Position: The image appears to be formed on the same side as the object (candle) but closer to the lens. It is formed at a position where light rays diverge after passing through the lens.
- Move the candle away from the lens. Note the change in the size of the image. What happens to the size of the image when the candle is placed too far away from the lens
Ans: Change in the Size of the Image as the Candle Moves Away from the Lens:
- As the candle moves farther away from the concave lens, the size of the image becomes smaller.
- When the candle is placed too far away from the lens (at an infinite distance), the image becomes very small and appears to be at the focal point of the lens. This is because the rays diverge less as they pass through the lens, and the virtual image formed appears very close to the lens.
Questions
Q 1. Define 1 dioptre of power of a lens.
Ans 1: One dioptre (1 D) is the unit used to measure the power of a lens. It is defined as the power of a lens whose focal length is 1 meter. In simpler terms, if a lens can focus light at a distance of 1 meter, it has a power of 1 dioptre. The formula to calculate lens power is: Power (D) = 1 / Focal length (m)
So, a lens with a shorter focal length has more power, and a lens with a longer focal length has less power.
Q 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.
Ans 2: To form an image that is equal in size to the object (needle) and inverted, the object must be placed at twice the focal length of the convex lens (i.e., at the 2f position). This is because, at this point, the lens produces a real, inverted image of the same size.
Given that the image is formed 50 cm away from the lens, this distance represents the image distance (v = 50 cm). For the image to be equal in size to the object, the object distance (u) must also be 50 cm, since the image and object are equidistant from the lens at this point.
To find the power of the lens, we use the lens formula: 1/𝑓 = 1/𝑣 − 1/𝑢
Here, v = 50 and u = − 50 cm (since the object is placed on the opposite side of the light direction).
Substitute these values: 1/f = 1/50 – 1/(-50) = 1/50 + 1/50 = 2/50
Now, solve for f: f = 50 / 2 = 25 cm
Now, to find the power of the lens (P), we use the formula: P = 1 / f (In meters)
Since f = 25 cm = 0.25 m: P = 1 / 0.25 = 4 diopters (D)
So, the needle is placed 50 cm in front of the lens, and the power of the lens is 4 diopters.
Q 3. Find the power of a concave lens of focal length 2 m.
Ans 3: To find the power of a concave lens, we use the formula: P = 1 / f
where:
- P is the power of the lens in diopters (D),
- f is the focal length in meters.
For a concave lens, the focal length is negative. Given that the focal length f = − 2 meters, the power of the lens is: P = 1 / (-2) = -0.5 diopters (D)
So, the power of the concave lens is -0.5 diopters.
Exercise
Q 1. Which one of the following materials cannot be used to make a lens?
(a). Water
(b). Glass
(c). Plastic
(d). Clay
Ans 1: (d). Clay: Clay cannot be used to make a lens because it does not have the necessary optical properties, such as clarity and uniform refractive index, that materials like water, glass, and plastic possess, which are used in the manufacture of lenses.
Q 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 center of curvature
(b). At the center of curvature
(c). Beyond the center of curvature
(d). Between the pole of the mirror and its principal focus.
Ans: 2 (d). Between the pole of the mirror and its principal focus:
Explanation: For a concave mirror, a virtual, erect, and magnified image is formed when the object is placed between the pole of the mirror (P) and its principal focus (F). In this case, the image appears behind the mirror, is virtual, upright, and larger than the object.
Q 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 center of the lens and its principal focus.
Ans 3: (b). At twice the focal length
Explanation: For a convex lens, when an object is placed at twice the focal length (2f), the image formed is real, inverted, and of the same size as the object. This is known as the condition for producing a real image of the same size as the object.
Q 4. A spherical mirror and a thin spherical lens have each a focal length of–15cm. The mirror and the lens are likely to be:
(a). both concaves.
(b). both convex.
(c). the mirror is concave and the lens is convex.
(d). the mirror is convex, but the lens is concave.
Ans 4: (a). both concaves.
Explanation:
- A spherical mirror with a negative focal length (−15 cm) indicates it is concave, as concave mirrors have negative focal lengths.
- A thin spherical lens with a negative focal length (−15 cm) indicates it is a concave lens, as concave lenses also have negative focal lengths.
Q 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.
Ans 5: (d). either plane or convex.
Q 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.
Ans 6: (c) A convex lens of focal length 5 cm, as it provides significant magnification, making it easier to read small letters.
Q 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.
Ans 7: To obtain an erect image using a concave mirror with a focal length of 15 cm, the object must be placed between the focal point and the mirror. This means the object distance (denoted as u) should be greater than the focal length (15 cm), but not at infinity.
Range of Object Distance:
- Object distance (u): The object should be between 15 cm and infinity from the mirror.
Nature of the Image:
- The image will be virtual (it cannot be projected on a screen) and erect (upright).
- It will be larger than the object.
Ray Diagram:
- The object is placed beyond the focal point but not at infinity.
- Rays coming from the object reflect off the concave mirror and diverge.
- The diverging rays appear to come from a point behind the mirror, forming a virtual, erect, and magnified image.
This is typical for an object placed between the focal point and the mirror, resulting in a virtual, magnified, and upright image.
Q 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.
Ans 8: Here are the types of mirrors used in the given situations, along with reasons:
(a) Headlights of a car: Concave mirror
- A concave mirror is used in headlights because it can focus light rays from the bulb into a parallel beam. This allows the headlights to project a bright, concentrated light in a specific direction, improving visibility on the road.
(b) Side/rear-view mirror of a vehicle: Convex mirror
- Convex mirrors are used for side and rear-view mirrors because they provide a wider field of view. The diverging rays from the objects in the mirror allow the driver to see a larger area behind and around the vehicle, enhancing safety by reducing blind spots.
(c) Solar furnace: Concave mirror
- A concave mirror is used in solar furnaces to focus sunlight onto a small area. The mirror collects and reflects the sun’s rays to a single focal point, concentrating the energy and generating high temperatures for industrial or scientific purposes.
Q 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.
Ans 9: Yes, the convex lens will still produce a complete image of the object even if one-half is covered with black paper. The reason is that a lens works based on the bending of light rays passing through it. When half of the lens is covered, only half of the light is being refracted, but the remaining uncovered part can still focus the light and form an image.
Experimental Verification:
- Set up an object (like a small light source) in front of the convex lens.
- Cover one-half of the lens with black paper.
- Observe the image formed on the screen.
Observation:
- A complete image will still form, but it may appear dimmer or have reduced brightness because only half of the light is being refracted.
- The shape and size of the image will remain the same, but it could be less intense compared to the image formed by an uncovered lens.
Explanation:
- The convex lens focuses light by bending rays that pass through it. Even with half the lens covered, the uncovered half still does its job in focusing light to form a complete image, although with less intensity.
Q 10. An object 5 cm in length is held 25 cm away from a converging lens of focal length10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Ans 10: Use the lens formula to find the image distance (v):
The lens formula is: 1/f = 1/v – 1/u
Given:
Focal length (f) = 10 cm
Object distance (u) = -25 cm (negative sign convention)
Substitute values into the lens formula: 1/10 = 1/v – 1/(-25)
→ 1/10 = 1/v + 1/25
Isolate 1/v: 1/v = 1/10 – 1/25
Find common denominator (50): 1/v = (5 – 2)/50 = 3/50
Solve for v: v = 50/3 ≈ 16.67 cm
Conclusion:
The image is formed 16.67 cm to the right of the lens (positive v indicates a real image).
Find the size of the image using the magnification formula:
The magnification (M) is given by: M = Image height / Object height = Image distance / Object distance
Substitute the known values: M = 16.67 / (-25) = -0.67
Thus, the image height is: Image height = M × Object height = − 0.67 × 5cm = − 3.33 cm
The negative sign indicates that the image is inverted (upside-down)
Nature of the image:
- The image is real because it is formed on the right side of the lens (positive image distance).
- The image is inverted because the magnification is negative.
- The image is diminished because the magnification is less than 1.
Q 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.
Ans 11: To solve this, we use the lens formula for concave lenses: 1/f = 1/v – 1/u
Where:
- f is the focal length of the lens,
- v is the image distance,
- u is the object distance.
Given:
- Focal length f = − 15 cm (negative for concave lens),
- Image distance v = − 10cm (negative for virtual image).
Substitute these values into the lens formula: 1/(-15) = 1/(-10) – 1/u
Now, solve for u (object distance): 1/u = 1/(-10) – 1/(-15)
1/u = (-3 + -2) / 30 = -5 / 30
u = -6 cm
Thus, the object is placed 6 cm in front of the lens.
Ray Diagram:
Q 12. An object is placed at a distance of 10 cm from a convex mirror of focal length15 cm. Find the position and nature of the image.
Ans 12: To find the position and nature of the image formed by a convex mirror, we can use the mirror formula: 1/f = 1/v + 1/u
Where:
- f is the focal length of the mirror,
- v is the image distance (which we need to find),
- u is the object distance.
Given:
- Focal length f = + 15cm (positive for convex mirrors),
- Object distance u = − 10cm (negative since the object is placed in front of the mirror).
Now, using the mirror formula: 1/15 = 1/v – 1/(-10)
Solving for v: 1/v = 1/15 + 1/10 = 2/30 + 3/30 = 5/30 = 1/6
Thus, v = 6 cm
Nature of the Image:
- The image is formed at a distance of 6 cm behind the mirror (virtual and erect).
- The image is diminished (smaller than the object).
So, the image is virtual, erect, and diminished, formed at a distance of 6 cm behind the convex mirror.
Q 13. The magnification produced by a plane mirror is +1. What does this mean?
Ans 13: A magnification of +1 produced by a plane mirror means that the image formed is the same size as the object. The positive sign indicates that the image is upright and virtual, meaning it cannot be projected on a screen but appears behind the mirror.
Q 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.
Ans 14: Find the image position (v) using the mirror formula: 1/f = 1/v + 1/u
Substitute the values: 1/15 = 1/v + 1/(-20)
Solving for v: 1/v = 1/15 + 1/20
1/v = (4 + 3) / 60 = 7 / 60
v = 60 / 7 = 8.57 cm
The image is formed 8.57 cm behind the mirror (positive sign indicates a virtual image).
Find the size of the image (hᵢ) using magnification:
Magnification (M) = hᵢ / hₒ = – v / u
Substitute the values: M = – 8.57 / (-20) = 0.429
Now, calculate the image size:
hᵢ = M × hₒ = 0.429 × 5.0 = 2.15 cm
Summary:
Position of the image: 8.57 cm behind the mirror
Nature of the image: Virtual, upright, and smaller than the object
Size of the image: 2.15 cm
Q 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 focused image can be obtained? Find the size and the nature of the image.
Ans 15: We use the mirror formula: 1/f = 1/v + 1/u
Where:
• f = –18 cm (concave mirror → focal length is negative)
• u = –27 cm (object in front of the mirror → negative)
• v = ? (image distance)
Step 1: Find v (Image Distance)
1/v = 1/f – 1/u
1/v = 1/(–18) – 1/(–27)
1/v = –1/18 + 1/27
1/v = (–3 + 2) / 54 = –1/54
v = –54 cm
The image is formed 54 cm in front of the mirror, so the screen should be placed 54 cm in front of the mirror.
Step 2: Find the Size of the Image
Magnification (M) = –v / u
M = –(–54) / (–27) = 2
Image size = M × object size = 2 × 7 = 14 cm
Final Answer:
Distance of the screen: 54 cm in front of the mirror
Size of the image: 14 cm
Nature of the image: Real and inverted
Q 16. Find the focal length of a lens of power – 2.0 D. What type of lens is this?
Ans 16: The power of a lens (P) is related to its focal length (f) by the formula: P = 100 / f (when f is in cm)
Given:
P = –2.0 D
Rearranging the formula:
f = 100 / P = 100 / (–2.0) = –50 cm
The negative sign indicates that the lens is diverging, which means it is a concave lens.
Final Answer:
Focal length: –50 cm (or –0.5 m)
Type of lens: Concave lens
Q 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?
Ans 17: To find the focal length (f) of the lens, use the formula: P = 1 / f
(Where P is in diopters (D) and f is in meters)
Given: P = +1.5 D
Rearranging the formula: f = 1 / P = 1 / 1.5 = 0.67 m (or 67 cm)
Since the power is positive, the lens is converging, i.e., a convex lens.
Final Answer:
Focal length: 0.67 meters (67 cm)
Type of lens: Converging (Convex) lens
Chapter 9 – Light Reflection and Refraction Class 10 Notes, Question/Answer, Activity & Projects
Updated Solution 2024-2025
This complete solution is prepared as per the latest syllabus of 2024-25. If you have any further queries, feel free to ask! 