NCERT Class 10 Science Solutions: Light – Reflection and Refraction
Question:
Which one of the following materials cannot be used to make a lens?
A. Water
B. Glass
C. Plastic
D. Clay
Concept in a Minute:
To form a lens, a material must be transparent and have a refractive index different from the surrounding medium (usually air). This difference in refractive index causes light to bend and converge or diverge, which is the principle behind how lenses work. Materials that are opaque or do not significantly refract light cannot be used to make lenses.
Explanation:
A lens is an optical device that transmits and refracts light, converging or diverging the light rays. For a material to be suitable for making a lens, it must be transparent so that light can pass through it. Additionally, it must have a refractive index that allows it to bend light effectively.
Let’s analyze the given options:
A. Water: Water is transparent and has a refractive index of approximately 1.33. This difference from the refractive index of air (approximately 1.00) allows water to bend light, and it is indeed used to make lenses (e.g., in some contact lenses or in laboratory experiments).
B. Glass: Glass is a common material for making lenses. It is transparent and has a refractive index typically ranging from 1.5 to 1.7, depending on the type of glass.
C. Plastic: Many types of plastics are transparent and can be molded into lens shapes. They have varying refractive indices, making them suitable for producing lenses, especially in eyeglasses and cameras.
D. Clay: Clay, when fired to become pottery or ceramic, is generally opaque. Light cannot pass through it. Therefore, it cannot be used to make a lens, as transparency is a fundamental requirement for a lens.
Conclusion: Since clay is opaque, it cannot be used to make a lens.
Question:
No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be ______.
A. Plane
B. Concave
C. Convex
D. Either plane or convex
Concept in a Minute:
Understanding the types of mirrors (plane, concave, convex) and the nature (erect, inverted, virtual, real) and location of images formed by them.
Explanation:
A plane mirror always forms a virtual and erect image that is the same size as the object and located as far behind the mirror as the object is in front.
A concave mirror can form both real and inverted images, or virtual and erect images, depending on the object’s position. If the object is placed between the pole and the principal focus, the image is virtual, erect, and magnified. However, at most other positions, the image is inverted.
A convex mirror always forms a virtual, erect, and diminished image, regardless of the object’s position.
In this question, the key information is that the image appears erect *no matter how far you stand*.
If the mirror were concave, the image would be erect only when the object is placed between the pole and the principal focus. At other distances, the image would be inverted.
If the mirror were plane, the image would always be erect.
If the mirror were convex, the image would always be erect.
Since the question states that the image appears erect *no matter how far you stand*, both a plane mirror and a convex mirror fit this description. Therefore, the mirror is likely to be either plane or convex.
The final answer is $\boxed{D}$.
Question:
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.
Concept in a Minute:
Ray diagrams for concave mirrors and the nature of image formed for different object positions. Specifically, understanding when a concave mirror forms a virtual, erect, and magnified image.
Explanation:
A concave mirror forms a virtual, erect, and magnified image only when the object is placed between the pole (P) of the mirror and its principal focus (F). Let’s analyze the options:
A. Between the principal focus and the centre of curvature (F and C): In this case, the image formed is real, inverted, and magnified.
B. At the centre of curvature (C): When the object is at C, the image formed is real, inverted, and of the same size.
C. Beyond the centre of curvature (Beyond C): When the object is beyond C, the image formed is real, inverted, and diminished.
D. Between the pole of the mirror and its principal focus (Between P and F): When the object is placed between P and F, the rays diverge after reflection, and their virtual extensions meet behind the mirror, forming a virtual, erect, and magnified image. This matches the observation in the question.
Therefore, the correct position of the object is between the pole of the mirror and its principal focus.
The final answer is $\boxed{D}$.
Question:
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.
Concept in a Minute:
This question involves the use of the lens formula and the magnification formula for convex lenses. The key concepts are:
1. Lens Formula: Relates the focal length (f) of a lens to the object distance (u) and the image distance (v): 1/f = 1/v – 1/u.
2. Magnification (m) of a lens: Defines the ratio of the image height (h’) to the object height (h) and also the ratio of the image distance (v) to the object distance (u): m = h’/h = v/u. For real images, magnification is negative.
3. Sign Conventions: Understanding and applying the standard sign conventions for lenses is crucial (object is always placed to the left of the lens, so u is negative; real images are formed on the right, so v is positive; focal length of a convex lens is positive).
Explanation:
We are given that a convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. This means the image distance, v = +50 cm (positive because it’s a real image formed on the other side of the lens).
We are also told that the image is equal to the size of the object. This means the magnification (m) is -1 (negative because the image is inverted, and 1 because the size is equal).
Using the magnification formula, m = v/u:
-1 = +50 cm / u
Solving for u:
u = -50 cm
The negative sign indicates that the needle (object) is placed 50 cm in front of the convex lens.
Now, to find the power of the lens, we first need to calculate its focal length (f) using the lens formula: 1/f = 1/v – 1/u.
1/f = 1/(+50 cm) – 1/(-50 cm)
1/f = 1/50 cm + 1/50 cm
1/f = 2/50 cm
1/f = 1/25 cm
f = 25 cm
To calculate the power of the lens, we use the formula P = 1/f, where f is in meters.
f = 25 cm = 0.25 m
P = 1 / 0.25 m
P = 4 D (Diopters)
Therefore, the needle is placed 50 cm in front of the convex lens, and the power of the lens is +4 Diopters.
Question:
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.
Concept in a Minute:
Mirror Formula and Magnification for Convex Mirrors. The mirror formula relates object distance (u), image distance (v), and focal length (f) as 1/f = 1/v + 1/u. Magnification (m) relates image height (h’) and object height (h) as m = h’/h = -v/u. For a convex mirror, the focal length (f) is positive. Object distance (u) is always negative. Image formed by a convex mirror is always virtual, erect, and diminished, located behind the mirror.
Explanation:
Given:
Object distance, u = -10 cm (object is placed in front of the mirror, so it’s negative by convention)
Focal length of the convex mirror, f = +15 cm (focal length of a convex mirror is positive)
We need to find:
Position of the image (v)
Nature of the image (virtual/real, erect/inverted, diminished/enlarged)
Using the mirror formula:
1/f = 1/v + 1/u
Substitute the given values:
1/15 = 1/v + 1/(-10)
1/15 = 1/v – 1/10
Rearrange the formula to solve for 1/v:
1/v = 1/15 + 1/10
Find a common denominator for the fractions on the right side, which is 30:
1/v = (2/30) + (3/30)
1/v = 5/30
Now, solve for v by taking the reciprocal of both sides:
v = 30/5
v = +6 cm
The positive sign for v indicates that the image is formed behind the mirror.
To determine the nature of the image, we can calculate the magnification (m):
m = -v/u
Substitute the values of v and u:
m = -(+6 cm) / (-10 cm)
m = -6 / -10
m = +0.6
The magnification is positive, which means the image is erect.
Since the magnitude of magnification (|m| = 0.6) is less than 1, the image is diminished (smaller than the object).
As the image is formed behind the mirror (v is positive) and is erect (m is positive), the image is virtual.
Therefore, the position of the image is 6 cm behind the mirror, and the nature of the image is virtual, erect, and diminished.
Question:
A concave mirror produces three times magnified (enlarged) real image of object placed at 10 cm in front of it. Where is the image located?
Concept in a Minute:
Magnification (m) of a mirror is the ratio of the image height (hi) to the object height (ho) and also related to the image distance (v) and object distance (u) as m = hi/ho = -v/u. For a real image formed by a concave mirror, the magnification is negative. The mirror formula relates the focal length (f) of a mirror to the object distance (u) and image distance (v) as 1/f = 1/v + 1/u.
Explanation:
Given:
Magnification (m) = -3 (since it’s a real and magnified image)
Object distance (u) = -10 cm (object is placed in front of the mirror, so it’s negative by convention)
We know that magnification m = -v/u.
Substituting the given values:
-3 = -v / (-10 cm)
-3 = v / 10 cm
v = -3 * 10 cm
v = -30 cm
The image is located 30 cm in front of the mirror.
Detailed Solution Structure:
1. Identify the type of mirror: Concave mirror.
2. Understand the nature of the image: Real and magnified.
3. Recall the formula for magnification of a mirror: m = hi/ho = -v/u.
4. Determine the sign of magnification for a real image: For a real image, magnification is negative. Therefore, m = -3.
5. Identify the given object distance (u) and its sign convention: Object is placed in front, so u = -10 cm.
6. Use the magnification formula to find the image distance (v):
* m = -v/u
* Substitute known values: -3 = -v / (-10 cm)
* Solve for v.
7. State the final answer with the correct sign and unit, indicating the location of the image relative to the mirror.
Question:
Name the type of mirror used in the following situation:
Headlights of a car
Support your answer with reason.
Concept in a Minute:
The question asks about the type of mirror used in car headlights and requires a justification for the choice. This involves understanding the properties of different types of mirrors (plane, concave, convex) and how they are used to focus or diverge light. Specifically, the concept of converging or diverging light rays and the formation of images by mirrors are crucial.
Explanation:
The type of mirror used in the headlights of a car is a concave mirror.
Reason:
Headlights of a car are designed to produce a strong, parallel beam of light that can illuminate a distant road. A concave mirror is used for this purpose because it has the ability to converge parallel light rays to a focal point, or conversely, to take light from a source placed at its focal point and reflect it as a parallel beam. In a car headlight, the bulb (light source) is placed at the focal point of the concave reflector. When light emanates from the focal point, it strikes the concave mirror and is reflected as a beam of parallel rays. This parallel beam travels a long distance without much divergence, effectively illuminating the road ahead. A convex mirror would diverge light, and a plane mirror would not be able to concentrate the light effectively into a beam.
Question:
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?
Concept in a Minute:
The relationship between the power of a lens and its focal length is inverse. Power (P) is measured in diopters (D) and focal length (f) is measured in meters (m). The formula is P = 1/f. A positive power indicates a converging lens, and a negative power indicates a diverging lens.
Explanation:
The power of the corrective lens is given as +1.5 D.
We know that the power of a lens (P) is the reciprocal of its focal length (f) in meters.
So, the formula is P = 1/f.
To find the focal length, we can rearrange the formula to f = 1/P.
Given P = +1.5 D,
f = 1 / (+1.5) meters.
Calculating the value:
f = 1 / 1.5 = 1 / (3/2) = 2/3 meters.
To express this in centimeters, we multiply by 100:
f = (2/3) * 100 cm = 200/3 cm.
f ≈ 66.67 cm.
Since the power of the lens is positive (+1.5 D), the prescribed lens is a converging lens. Converging lenses are also known as convex lenses and are used to correct hypermetropia (farsightedness).
Therefore, the focal length of the lens is approximately 66.67 cm, and the prescribed lens is converging.
Question:
The refractive index of diamond is 2.42. What is the meaning of this statement?
Concept in a Minute:
The refractive index of a material quantifies how much light bends when it enters that material from a vacuum. It is defined as the ratio of the speed of light in a vacuum to the speed of light in the material. A higher refractive index means light travels slower in the material and bends more.
Explanation:
The statement “The refractive index of diamond is 2.42” means that the speed of light in diamond is 2.42 times slower than the speed of light in a vacuum. Mathematically, this is expressed as:
Refractive index (n) = Speed of light in vacuum (c) / Speed of light in the medium (v)
In this case, n = 2.42.
So, 2.42 = c / v.
This implies that v = c / 2.42.
Therefore, light travels at approximately 1/2.42 times the speed of light in a vacuum when it passes through diamond. This significant slowing down of light is also responsible for diamond’s high brilliance and sparkle because it causes light to bend (refract) at a large angle as it enters and exits the gem.
Question:
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?
Concept in a Minute:
Refraction of light: The bending of a light ray when it passes from one medium to another medium with a different optical density.
Optical density: A measure of how much a medium slows down light. Denser media have higher optical density.
Normal: An imaginary line perpendicular to the surface at the point where the light ray strikes.
Speed of light: Light travels faster in rarer media (like air) and slower in denser media (like water).
Explanation:
When a ray of light travels from a rarer medium (like air, which has lower optical density) to a denser medium (like water, which has higher optical density) obliquely, it bends towards the normal. This bending occurs because the speed of light decreases as it enters the denser medium. The part of the wavefront that enters the water first slows down, while the part still in air continues at its original speed. This difference in speed causes the light ray to change direction and bend towards the normal.
Question:
Name the type of mirror used in the following situation:
Side/rear-view mirror of a vehicle
Support your answer with reason.
Concept in a Minute:
Concave mirrors converge light, forming magnified or diminished, real or virtual images. Convex mirrors diverge light, always forming diminished, virtual, and erect images. The field of view is the area visible to the observer.
Explanation:
The type of mirror used as a side/rear-view mirror of a vehicle is a convex mirror.
Reason: Convex mirrors are used because they provide a wider field of view compared to plane mirrors. This is because convex mirrors diverge the light rays falling on them, causing them to spread out. As a result, the image formed by a convex mirror is always diminished (smaller than the object), virtual (cannot be projected on a screen), and erect (upright). The diminished nature of the image allows more objects to be seen within the mirror’s reflection, thus enhancing the driver’s ability to see traffic behind and to the sides of the vehicle. This wider field of view significantly improves safety by reducing blind spots.
Question:
Find the focal length of a lens of power −2.0 D. What type of lens is this?
Concept in a Minute:
The power of a lens is defined as the reciprocal of its focal length in meters. Power is measured in diopters (D). A positive power indicates a converging lens (convex lens), and a negative power indicates a diverging lens (concave lens).
Explanation:
The power (P) of a lens is related to its focal length (f) by the formula:
P = 1/f
where P is in diopters (D) and f is in meters (m).
Given the power of the lens, P = -2.0 D.
To find the focal length, we can rearrange the formula:
f = 1/P
Substituting the given value of P:
f = 1 / (-2.0 D)
f = -0.5 meters
To express the focal length in centimeters, we multiply by 100:
f = -0.5 m * 100 cm/m
f = -50 cm
The sign of the focal length indicates the type of lens. A negative focal length signifies a diverging lens. Therefore, this is a concave lens.
Answer:
The focal length of the lens is -0.5 m or -50 cm. This is a concave lens.
Question:
Name the type of mirror used in the following situation:
Solar furnace
Support your answer with reason.
Concept in a Minute:
Mirrors can be concave, convex, or plane. Concave mirrors converge parallel light rays to a focal point, concentrating energy. Convex mirrors diverge light rays and form virtual, diminished images. Plane mirrors form virtual, same-sized images. Solar furnaces utilize concentrated solar energy.
Explanation:
A concave mirror is used in a solar furnace.
Reason: Solar furnaces are designed to concentrate sunlight to generate high temperatures. Concave mirrors have the property of converging parallel rays of light (like sunlight) to a single point called the focal point. By placing the object to be heated (e.g., food or materials) at the focal point of a large concave mirror, a large amount of solar energy is focused onto a small area, leading to a significant increase in temperature, making it suitable for heating or cooking.
Question:
You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table.
| Material medium |
Refractive index | Material medium | Refractive index |
| Air | 1.0003 | Canada Balsam | 1.53 |
| Ice | 1.31 | – | – |
| Water | 1.33 | Rock salt | 1.54 |
| Alcohol | 1.36 | – | – |
| Kerosene | 1.44 | Carbon disulphide | 1.63 |
| Fused quartz |
1.46 | Dense flint glass |
1.65 |
| Turpentine oil | 1.47 | Ruby | 1.71 |
| Benzene | 1.50 | Sapphire | 1.77 |
| Crown glass |
1.52 | Diamond | 2.42 |
Concept in a Minute:
The speed of light in a medium is inversely proportional to the refractive index of that medium. A lower refractive index means light travels faster, and a higher refractive index means light travels slower.
Explanation:
The question asks in which of the given materials (kerosene, turpentine, and water) light travels fastest. We need to find the refractive indices of these materials from the provided table.
From the table:
Refractive index of kerosene = 1.44
Refractive index of turpentine oil = 1.47
Refractive index of water = 1.33
To determine where light travels fastest, we need to identify the material with the lowest refractive index. Comparing the values:
1.33 (Water) < 1.44 (Kerosene) < 1.47 (Turpentine oil)
Since water has the lowest refractive index among the three, light will travel fastest in water.
Question:
Find out, from Table, the medium having highest optical density. Also find the medium with lowest optical density.
| Material medium | Refractive index | Material medium | Refractive index |
| Air | 1.0003 | Canada Balsam | 1.53 |
| Ice | 1.31 | – | – |
| Water | 1.33 | Rock salt | 1.54 |
| Alcohol | 1.36 | – | – |
| Kerosene | 1.44 | Carbon disulphide | 1.63 |
| Fused quartz | 1.46 | Dense flint glass | 1.65 |
| Turpentine oil | 1.47 | Ruby | 1.71 |
| Benzene | 1.50 | Sapphire | 1.77 |
| Crown glass | 1.52 | Diamond | .42 |
Concept in a Minute:
Optical density is directly related to the refractive index of a medium. A higher refractive index indicates a higher optical density, meaning light travels slower in that medium. Conversely, a lower refractive index indicates a lower optical density.
Explanation:
To find the medium with the highest optical density, we need to identify the medium with the highest refractive index from the provided table. Looking at the table, the highest refractive index listed is 2.42 for Diamond. Therefore, Diamond has the highest optical density.
To find the medium with the lowest optical density, we need to identify the medium with the lowest refractive index from the provided table. Looking at the table, the lowest refractive index listed is 1.0003 for Air. Therefore, Air has the lowest optical density.
Question:
Define 1 dioptre of power of a lens.
Concept in a Minute:
The power of a lens is defined as the reciprocal of its focal length in meters. Power is measured in dioptres (D). A lens with a shorter focal length has a higher power, meaning it can bend light more strongly.
Explanation:
The power (P) of a lens is given by the formula:
P = 1 / f
where ‘f’ is the focal length of the lens measured in meters.
1 dioptre of power means that the lens has a focal length of 1 meter.
So, if P = 1 D, then f = 1 / 1 D = 1 meter.
This means a lens with a power of 1 dioptre converges or diverges parallel rays of light at a distance of 1 meter from its optical center. A positive power (e.g., +1 D) indicates a converging lens (convex lens), and a negative power (e.g., -1 D) indicates a diverging lens (concave lens).
Question:
Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Concept in a Minute:
The relationship between focal length (f) and radius of curvature (R) for a spherical mirror is given by the formula f = R/2. For a convex mirror, both the focal length and radius of curvature are considered positive.
Explanation:
The question asks to find the focal length of a convex mirror given its radius of curvature.
We are given the radius of curvature, R = 32 cm.
For a spherical mirror, the focal length (f) is half of its radius of curvature (R). This relationship is expressed by the formula:
f = R / 2
Since the mirror is a convex mirror, its focal length and radius of curvature are positive.
Substituting the given value of R into the formula:
f = 32 cm / 2
f = 16 cm
Therefore, the focal length of the convex mirror is 16 cm.
Question:
Find the power of a concave lens of focal length 2m.
Concept in a Minute:
The power of a lens is the reciprocal of its focal length, measured in meters. The unit of power is diopter (D). For a concave lens, the focal length is considered negative.
Explanation:
The power (P) of a lens is defined by the formula:
P = 1/f
where ‘f’ is the focal length of the lens in meters.
Given:
Focal length of the concave lens, f = 2m
Since it is a concave lens, its focal length is negative. Therefore, we should consider the focal length as f = -2m.
Now, we can calculate the power of the lens:
P = 1 / f
P = 1 / (-2m)
P = -0.5 D
So, the power of the concave lens is -0.5 diopters. The negative sign indicates that it is a concave lens.
Final Answer:
The power of the concave lens is -0.5 D.
Question:
Name the mirror that can give an erect and enlarged image of an object.
Concept in a Minute:
Types of mirrors and their image formation properties. Specifically, understanding the characteristics of images formed by concave and convex mirrors.
Explanation:
Mirrors can be classified into two main types: plane mirrors, concave mirrors, and convex mirrors.
Plane mirrors always form erect and virtual images of the same size as the object.
Convex mirrors always form erect, virtual, and diminished images.
Concave mirrors can form both real and virtual images. When the object is placed between the pole and the principal focus of a concave mirror, it forms an image that is erect, virtual, and enlarged.
Therefore, the mirror that can give an erect and enlarged image of an object is a concave mirror.
Question:
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−1.
Concept in a Minute:
The concept required to solve this question is the definition of refractive index of a medium. The refractive index (n) of a medium is defined as the ratio of the speed of light in vacuum (c) to the speed of light in that medium (v). Mathematically, it is expressed as n = c/v.
Explanation:
We are given the refractive index of glass, n = 1.50, and the speed of light in vacuum, c = 3 × 10⁸ m s⁻¹. We need to find the speed of light in glass, v.
Using the formula n = c/v, we can rearrange it to solve for v:
v = c/n
Substitute the given values into the formula:
v = (3 × 10⁸ m s⁻¹) / 1.50
Now, perform the division:
v = 2 × 10⁸ m s⁻¹
Therefore, the speed of light in the glass is 2 × 10⁸ m s⁻¹.
Question:
Define the principal focus of a concave mirror.
Concept in a Minute:
The question is about defining the principal focus of a concave mirror. This requires understanding how light rays behave when they strike a concave mirror and where they converge after reflection. Key concepts include reflection, concave mirror properties, and the path of parallel rays.
Explanation:
The principal focus of a concave mirror is the point on its principal axis where parallel rays of light, incident on the mirror, converge after reflection. Alternatively, it is the point from which rays of light, originating from it, become parallel to the principal axis after reflection. For a concave mirror, this point is real and lies in front of the mirror.
Question:
The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Concept in a Minute:
The focal length of a spherical mirror is half of its radius of curvature. This is a fundamental relationship for all spherical mirrors (both concave and convex).
Explanation:
For a spherical mirror, the focal length (f) is directly proportional to its radius of curvature (R). The relationship is given by the formula:
f = R / 2
In this question, the radius of curvature (R) of the spherical mirror is given as 20 cm.
To find the focal length (f), we substitute the given value of R into the formula:
f = 20 cm / 2
f = 10 cm
Therefore, the focal length of the spherical mirror is 10 cm.
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