Class 8 Science Chapter 10 Light: Mirrors and Lenses (New Course)

Key Concept: Image Formation by Concave Mirrors

c) Real, inverted, and of the same size
[Solution Description]
When an object is placed at the center of curvature of a concave mirror, the image is formed at the same distance on the opposite side of the mirror. The characteristics are:
– The image is real because it is formed where light converges.
– It is inverted relative to the object.
– The size of the image is equal to the size of the object.
Therefore, the image formed is real, inverted, and of the same size as the object.

Key Concept: Image Characteristics of Convex Mirrors

b) Virtual, erect, and diminished
[Solution Description]
A convex mirror always forms a virtual image regardless of the object’s position. This image is also erect (upright) and diminished in size compared to the actual object. These properties arise because the reflected rays diverge, making them appear to originate from a point behind the mirror. Thus, the correct description is virtual, erect, and diminished.

Key Concept: Image Size Variation with Object Distance for Concave Mirror

a) From reduced to same size, then to enlarged
[Solution Description]
In a concave mirror, as the object moves from beyond the center of curvature towards the focal point, the characteristics of the image change as follows:
– Initially, when the object is beyond the center of curvature, the image is real, inverted, and reduced in size.
– As the object approaches the center of curvature, the image becomes the same size as the object.
– When the object is between the center of curvature and the focal point, the image is real, inverted, and enlarged.
Hence, the image size changes from reduced to enlarged as the object moves closer to the focal point.

Key Concept: Analytical Difficulty, Multiple Concepts

a) Real, inverted, same size
[Solution Description]
Placing the object at twice the focal length means $u = 2f = 40$ cm. Using the mirror formula:
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
Substituting $f = 20$ cm and $u = -40$ cm,
$\frac{1}{20} = \frac{1}{v} + \frac{1}{-40}$
Simplifying,
$\frac{1}{20} + \frac{1}{40} = \frac{1}{v} \implies \frac{2+1}{40} = \frac{1}{v} \implies v = \frac{40}{3} \approx 13.33 \text{ cm}$
At this point, the image is real, inverted, and of the same size as the object because it forms at the center of curvature.

Key Concept: Characteristics of Images in Concave and Convex Mirrors

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that a convex mirror always forms an erect and diminished image, which is true because no matter where the object is placed, the reflected rays appear to come from a point behind the mirror, making the image appear smaller and upright.
The reason claims that the focal point of a convex mirror is behind the mirror, causing light rays to diverge. This is indeed true as well. In a convex mirror, the focal length is negative, indicating that the focus lies on the back side of the mirror surface, which causes incoming parallel rays after reflection to spread out (diverge).
While both statements are true, the reason provided does not directly explain why images are always erect and diminished; it explains the nature of the reflections but doesn’t specifically link to the characteristic of the image being diminished or erect of its own accord.

Key Concept: Image Characteristics of Convex Mirrors

d) Erect and diminished
[Solution Description] Convex mirrors always form an image that is erect and smaller than the actual object, known as diminished. Even as the object moves further away, while the image size may slightly decrease, the image never becomes larger or inverted. This is due to the diverging nature of convex mirrors where parallel rays appear to originate from a focal point behind the mirror.

Key Concept: Concave and convex mirrors, image characteristics

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
Assertion: Concave mirrors can indeed form a real, inverted, and enlarged image of an object if the object is placed between the focal point ($f$) and the center of curvature ($C$) of the mirror. In this case, the image is formed beyond the center of curvature, is real, inverted, and larger than the object.
Reason: Convex mirrors have the characteristic of always producing virtual, erect, and diminished images because the reflected rays diverge. The image appears to be located behind the mirror, which explains its virtual nature, and it is always smaller than the actual object regardless of the object’s placement in front of the convex mirror.
Both the assertion and the reason are true, but the reason does not explain why concave mirrors specifically form real and enlarged images under certain conditions. Therefore, both statements are true, but they describe different behaviors of spherical mirrors without the reason explaining the assertion directly.

Key Concept: Image Characteristics of Concave Mirrors

b) The image becomes inverted and diminishes in size
[Solution Description] As per the characteristics of images formed by concave mirrors, when the object is close to the mirror, the image is erect and enlarged. However, as the object is moved farther away, the image becomes inverted. Initially, it remains enlarged but then starts to diminish in size. This change occurs due to the converging nature of concave mirrors, which affect image formation based on the object’s distance.

Key Concept: Multi-step Solution, Image Formation

a) Virtual, 6 cm behind
[Solution Description]
For a convex mirror, the focal length $f$ is positive. Using the mirror formula:
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
Here, $u = -10$ cm and $f = 15$ cm. Substituting these values,
$\frac{1}{15} = \frac{1}{v} + \frac{1}{-10}$
Simplifying,
$\frac{1}{15} + \frac{1}{10} = \frac{1}{v} \implies \frac{2+3}{30} = \frac{1}{v} \implies v = \frac{30}{5} = 6 \text{ cm}$
Since $v$ is positive (behind the mirror), the image is virtual, erect, and diminished.

Key Concept: Characteristics of Images Formed by Spherical Mirrors

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that when an object is positioned far from a concave mirror, the resulting image is both real and inverted. This statement is true because a concave mirror forms a real image on the same side as the object when the object is beyond the focal point, and such an image is inverted.
The reason correctly explains that as the object moves farther from a concave mirror, the size of the image decreases. Initially, if the object is at or near the focus, the image is large (enlarged). As the object keeps moving away, the image formed becomes smaller, eventually approaching a point size as the object reaches infinity.
Therefore, both statements are true, and the reasoning given is indeed the correct explanation of the assertion.

Key Concept: Comparison with Plane Mirrors

a) Always the same size and laterally inverted
[Solution Description] A plane mirror forms an image that is always erect and of the same size as the actual object. Unlike spherical mirrors (concave and convex), there are no changes in orientation or size as the object’s distance varies. Furthermore, lateral inversion is observed in images formed by all three types of mirrors: plane, concave, and convex.

Key Concept: Image Characteristics, Real-World Application

a) 6 cm
[Solution Description]
Using the magnification formula for mirrors, $m = -\frac{v}{u}$ where $m$ is magnification, $v$ is the image distance, and $u$ is the object distance. Given $m = 3$ (enlarged and erect implies positive magnification), we have:
$3 = -\frac{v}{u} \implies v = -3u$
The mirror formula is $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$. Substituting $f = 4$, we get:
$\frac{1}{4} = \frac{1}{-3u} + \frac{1}{u}$
Simplifying,
$\frac{1}{4} = \frac{1-3}{3u} \implies \frac{1}{4} = -\frac{2}{3u} \implies u = -6 \text{ cm}$
Therefore, the distance between the tooth and mirror is 6 cm.

Key Concept: Concept of Incident Ray

c) $60^\circ$
[Solution Description]
The angle of incidence is measured from the normal, not the surface. If the ray strikes the mirror at $30^\circ$ with the surface, then the angle of incidence ($i$) with the normal is $90^\circ – 30^\circ = 60^\circ$. According to the law of reflection, the angle of incidence is equal to the angle of reflection, so the angle of reflection ($r$) is also $60^\circ$. Therefore, the reflected ray makes an angle of $60^\circ$ with the normal.

Key Concept: Incident Ray, Normal to the Surface, Reflected Ray

a) $60^\circ$
[Solution Description]
To solve this problem, we need to understand how rotation affects the angles involved in reflection.
Initially, the angle of incidence $i$ is $45^\circ$. According to the laws of reflection, the angle of reflection $r$ is also $45^\circ$.
When the mirror is rotated by $15^\circ$, the normal to the surface moves by the same angle. This changes both the angle of incidence and the angle of reflection each by $15^\circ$:
New angle of incidence = $45^\circ – 15^\circ = 30^\circ$
New angle of reflection (because it equals the new angle of incidence due to the law of reflection) = $30^\circ$
The new angle between the incident ray and the reflected ray is given by:
$\text{Angle between incident and reflected rays} = i + r = 30^\circ + 30^\circ = 60^\circ$

Key Concept: Incident Ray, Normal to the Surface, Reflected Ray

c) 3 Reflections
[Solution Description]
Let’s analyze the path of the light beam within an equilateral triangle formed by three mirrors.
1. The light enters parallel to one side. As it hits the first mirror, it reflects according to the laws of reflection.
2. In an equilateral triangle with sides composed of mirrors, any ray entering parallel to one side undergoes multiple reflections until it eventually exits.
3. For an equilateral triangle, the light typically reflects off all three sides once before making its way out.
Thus, the number of reflections experienced by the light beam before exiting is 3.

Key Concept: Application of Laws of Reflection

c) Both angles are equal
[Solution Description] For any given situation involving reflection from a smooth surface, the angle of incidence (the angle made by the incident ray with the normal) must equal the angle of reflection (the angle made by the reflected ray with the normal). Thus, irrespective of the value of $\theta$, the two angles will remain equal as per the law of reflection.

Key Concept: Incident Ray, Normal to the Surface, Reflected Ray

b) $90^\circ$
[Solution Description]
For the light to complete a total deviation of $180^\circ$, consider the geometry created by the reflections.
When light reflects off two mirrors, the relationship between the angle of deviation ($D$), angle of incidence ($i$), and angle between the mirrors ($\theta$) can be expressed as:
$D = 360^\circ – 2\theta$
Given that the total deviation $D = 180^\circ$, substitute into the formula:
$180^\circ = 360^\circ – 2\theta$
Rearranging gives:
$2\theta = 360^\circ – 180^\circ$
$2\theta = 180^\circ$
Solving for $\theta$:
$\theta = \frac{180^\circ}{2} = 90^\circ$

Key Concept: Incident ray, normal to the surface, and reflected ray all lie in the same plane; Angle of incidence equals angle of reflection

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The laws of reflection state two fundamental principles: First, the incident ray, the normal, and the reflected ray must lie in the same plane. This ensures that no matter how the mirror is tilted, these components will remain coplanar due to the geometric nature of light reflection. Second, it specifies that the angle of incidence ($i$) is always equal to the angle of reflection ($r$), regardless of the type of mirror used, whether plane or curved.
In this case, both statements are true. The assertion correctly describes one aspect of the laws of reflection, focusing on the geometric arrangement of rays and normals during reflection. The reason complements this by explaining the consistency of angles involved in reflections across different mirrors, but it does not directly justify why the components lie in the same plane as described in the assertion. Hence, the correct option is b).

Key Concept: Reflection Laws

c) Assertion is true, but Reason is false.
[Solution Description] According to the first law of reflection, the incident ray, the normal at the point of incidence, and the reflected ray all lie in the same plane. This is a fundamental property of how light reflects off surfaces. The assertion is true because it describes one of the laws of reflection correctly.
However, the reason states that bending the surface or changing its orientation will affect this alignment. This part is incorrect because regardless of the curvature or orientation changes in the reflective surface, locally at the point of incidence, the incident ray, normal, and reflected ray will always lie in the same plane. The reason incorrectly attributes the effect on the entire plane as opposed to localised actions at the point of incidence.
Therefore, Assertion is true, but Reason is false.

Key Concept: Angle of Incidence and Reflection

a) Both angles are $0^\circ$
[Solution Description]
When a light ray approaches a mirror along the line parallel to the normal, it essentially means the light is hitting the mirror perpendicularly. In this scenario, both the angle of incidence and the angle of reflection are $0^\circ$, because they are measured with respect to the normal.

Key Concept: Concept of Plane Reflection

c) All three lie in the same plane
[Solution Description]
According to the first law of reflection, the incident ray, the normal, and the reflected ray all lie in the same plane. This geometric arrangement ensures that the behavior of light follows predictable laws when encountering reflective surfaces.

Key Concept: Incident ray, normal to the surface, and reflected ray lie in the same plane

d) The incident ray, normal, and reflected ray lie in the same plane
[Solution Description] According to the laws of reflection, the incident ray, the normal to the surface at the point of incidence, and the reflected ray all lie in the same plane. This principle is fundamental and applies universally to all reflective surfaces.

Key Concept: Angle of Incidence equals Angle of Reflection

c) It reflects at an angle of $30^\circ$
[Solution Description] According to the law of reflection which states that the angle of incidence is equal to the angle of reflection, if a light ray strikes a mirror at an angle of $30^\circ$ with respect to the normal, it will reflect off the mirror at the same angle of $30^\circ$.

Key Concept: Reflection Laws

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that the angle of incidence ($i$) is equal to the angle of reflection ($r$), which is a fundamental law of reflection applicable to all types of mirrors including plane and spherical mirrors. This statement is true.
The reason given is that the incident ray, the normal at the point of incidence, and the reflected ray all lie in the same plane. This is also a fundamental principle of the laws of reflection, pointing to the geometric nature of light interaction with reflective surfaces.
Both statements are correct. However, the second statement (Reason) does not directly explain why the first statement (Assertion) is true; it describes another aspect of the laws of reflection. Therefore, both Assertion and Reason are true but Reason is NOT the correct explanation of Assertion.

Key Concept: Applications of convex mirrors

c) They provide a wider field of view.
[Solution Description]
Convex mirrors provide a wider field of view than flat mirrors due to their outward curvature. This allows drivers to see more area behind and beside the vehicle, enhancing road safety.

Key Concept: Definition of convex mirror

b) It has a reflecting surface curved outwards.
[Solution Description]
A convex mirror has a reflecting surface that curves outward. In contrast, a concave mirror curves inward. This fundamental difference affects how they reflect light and form images.

Key Concept: Image Formation, Curvature of Convex Mirror

a) Virtual and reduced in size
[Solution Description]
To find the image characteristics formed by a convex mirror, we use the mirror formula:
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
where $f$ is the focal length, $v$ is the image distance, and $u$ is the object distance. Since this is a convex mirror, both the focal length and object distance are taken as positive numbers.
Given:
$f = 20 \, \text{cm}$
$u = 40 \, \text{cm}$
Substituting these values into the equation:
$\frac{1}{20} = \frac{1}{v} + \frac{1}{40}$
Solving for $v$:
First, calculate $\frac{1}{40}$:
$\frac{1}{40} = 0.025$
Now substitute back to solve for $\frac{1}{v}$:
$\frac{1}{20} – \frac{1}{40} = \frac{1}{v}$
Simplifying,
$0.05 – 0.025 = \frac{1}{v}$
So,
$\frac{1}{v} = 0.025$
Hence,
$v = 40 \, \text{cm}$
The image distance is positive, indicating that the image forms behind the mirror. The image will be virtual and diminished since convex mirrors always form such images relative to the object. Therefore, the image is half the size of the object.

Key Concept: Field of View Expansion, Safety Implementation

d) It expands the field of view, aiding in early detection of vehicles.
[Solution Description]
The radius of curvature ($R$) of a convex mirror determines its focal length ($f$). For any spherical mirror, the relationship between focal length and radius of curvature is given by:
$f = \frac{R}{2}$
For a convex mirror with $R = 50 \, \text{cm}$,
$f = \frac{50}{2} = 25 \, \text{cm}$
A convex mirror with this focal length provides a wide viewing angle due to its outward curvature. At sharp bends, this expanded field of view enables drivers to see approaching vehicles from the opposite direction, reducing blind spots and increasing reaction time to prevent accidents. It ensures safer navigation through tight turns and intersections by offering visibility of oncoming traffic well before entering the turn.

Key Concept: Convex Mirror Properties, Image Formation

c) Assertion is true, but Reason is false.
[Solution Description]
In a convex mirror, the reflecting surface curves outward, which causes it to diverge light rays rather than converge them, making the rays appear to originate from a focal point behind the mirror. This property results in the formation of virtual images that are diminished compared to the actual object. Therefore, convex mirrors indeed create smaller images. However, the reason given is incorrect because the wider field of view of a convex mirror is due to its divergent nature and not due to converging rays. Converging occurs in concave mirrors which focus rays at a point, unlike convex mirrors.

Key Concept: Convex Mirror Applications, Image Diminishment

d) They offer a larger field of view with diminished images.
[Solution Description]
Convex mirrors are used as rearview mirrors because they provide a wider field of view compared to flat or concave mirrors. This allows drivers to see more area behind them, which helps in making informed decisions while driving.
The property of forming diminished images ensures that objects appear smaller but cover a larger part of the field of view. Although the image is smaller, it gives an accurate representation of the surroundings, thus enhancing safety on the road.
Additionally, convex mirrors ensure that all images are erect and not inverted, contributing to easy interpretation of what is being seen without any confusion.

Key Concept: Applications of convex mirrors

b) They provide a wider field of view
[Solution Description] Convex mirrors provide a wider field of view compared to plane mirrors. They allow drivers to see more of the road and surrounding area, making them ideal for use as side-view mirrors in vehicles to monitor traffic behind.

Key Concept: Concept of image formation by convex mirrors

b) The image size decreases
[Solution Description] As per the properties of convex mirrors, they always form diminished images which are smaller than the actual object. When an object is moved further away from a convex mirror, the size of the image becomes even more diminished.

Key Concept: Convex Mirror: Curved outward, always forms diminished images

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that images formed by convex mirrors are smaller than the actual object, which is true as convex mirrors always form virtual and diminished images of objects placed in front of them. This occurs due to the way light reflects off the curved surface, causing the reflected rays to diverge.
The reason explains that convex mirrors have a wider field of view, making them ideal for use in vehicles. This statement is also true because the outward curve allows more area to be visible to the observer, which is why they are used on vehicle side mirrors to provide a better view of traffic behind.
However, while both statements are correct, having a wider field of view does not explain why the image is diminished. The diminishment is due to the nature of the reflection and divergence of light rays, not directly related to the wide field of view.
Therefore, both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

Key Concept: Characteristics of images formed by convex mirrors

c) The image is virtual and diminished.
[Solution Description]
Convex mirrors always form erect and diminished images regardless of the position of the object. This property makes them useful for applications like side-view mirrors on vehicles.

Key Concept: Properties of convex mirrors

b) The image is erect and diminished
[Solution Description] A convex mirror always forms an erect and diminished image. This means that the image will be upright and smaller than the actual object, no matter where the object is placed in front of the mirror.

Key Concept: Convex Mirror: Curved outward, always forms diminished images

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
Convex mirrors are widely used in vehicles as rear-view mirrors because they provide a wider field of view compared to concave or plane mirrors. The reason behind this application is that convex mirrors reflect light outwards due to their outward curve, causing light rays to diverge. This divergence leads to the formation of virtual, erect, and diminished images, which allows drivers to see a larger area behind them. Therefore, while both the assertion and the reason are true statements, the reason provided correctly explains why convex mirrors are preferred for use in rear-view applications in vehicles.

Key Concept: Angle Calculation for Reflection

c) 55°
[Solution Description]
The angle of incidence ($i$) is given as 35°. According to the law of reflection, the angle of reflection ($r$) is equal to the angle of incidence. Since these angles are measured from the normal, both $i$ and $r$ are 35°. The angle made by the reflected ray with the mirror surface will be complementary to the angle of reflection. Therefore, it is computed as:
$\text{Angle with mirror} = 90^\circ – r = 90^\circ – 35^\circ = 55^\circ$
Thus, the angle made by the reflected ray with the mirror surface is 55°.

Key Concept: Analyzing Ray Direction in Mirrors

c) The angle between the incident and reflected rays is 100°.
[Solution Description]
The angle of incidence ($i$) here is 50°, and according to the first law of reflection, the angle of reflection ($r$) is equal to this:
$r = i = 50^\circ$
Since the incident ray, normal, and reflected ray lie in the same plane, the reflected ray will also make a 50° angle with the normal. The direction of the ray changes such that it reflects symmetrically about the normal line.
Hence, after striking the mirror, the angle between the normal and the reflected ray remains the same but opposite in direction, confirming the reflection is symmetrical around the normal.

Key Concept: Application of Reflection Law

b) 90°
[Solution Description] The angle of incidence and the angle of reflection are both 45°. The angle between the incident ray and the reflected ray is the sum of these angles: $45^\circ + 45^\circ = 90^\circ$.

Key Concept: Law of Reflection

c) 30°
[Solution Description] According to the law of reflection, the angle of incidence equals the angle of reflection. Therefore, if $i = 30^\circ$, then $r = 30^\circ$.

Key Concept: Angle of incidence, Angle of reflection, Plane mirrors

b) 80°
[Solution Description]
Initially, the angle of incidence $i$ is 30°, thus the angle of reflection $r$ is also 30° due to the law of reflection.
When the mirror is rotated by 10° clockwise, the normal rotates along with the mirror. The new angle of incidence becomes $30° + 10° = 40°$. According to the law of reflection, the new angle of reflection is also 40°.
The total change in the direction of the reflected ray is twice the rotation angle of the mirror because it reflects symmetrically with respect to the incident ray. Therefore, the new angle between the reflected ray and the original incident ray is $2 \times 10° = 20°$.
Thus, the new angle between the reflected ray and the original incident ray is $60° + 20° = 80°$.

Key Concept: Incident ray, Reflected ray, Normal plane alignment

a) 100°
[Solution Description]
By definition, the incident ray, normal, and reflected ray must lie in the same plane for standard reflection. However, if they hypothetically lie in different planes, consider how they would normally interact under standard conditions.
For a typical setup:
The angle of incidence is given as 50°, so the angle of reflection is also 50°. The angle between the incident ray and the reflected ray is then calculated as $2 \times 50° = 100°$, considering both angles from the normal line.
Hypothetical scenarios cannot alter the mathematical rule of reflection; hence, even if positioned unusually, the theoretical minimum angle remains governed by this principle.

Key Concept: Reflection laws, Spherical mirrors, Angle measurement

b) 75°
[Solution Description]
For a spherical mirror, the normal at any point on the mirror is along the radius. Since the tangent at that point makes an angle of 60° with the principal axis, the normal (radius) makes an angle of $90° – 60° = 30°$ with the principal axis.
Given that the incident ray makes a 45° angle with the surface, the angle of incidence $i$ is the complementary angle made with the normal, which is $90° – 45° = 45°$. Hence, according to the law of reflection, the angle of reflection $r$ is also 45°.
Thus, the angle between the reflected ray and the principal axis is $30° + 45° = 75°$.

Key Concept: Laws of Reflection

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
In the law of reflection, it states that the angle of incidence ($i$) is equal to the angle of reflection ($r$). This law applies universally to all types of mirrors, including plane, concave, and convex mirrors. Therefore, the assertion is true.
The reason provided mentions that plane mirrors maintain the parallel nature of incident light rays when reflected. This statement is true as well. For instance, if multiple parallel beams of light hit a plane mirror, the reflected beams remain parallel. However, this explanation pertains specifically to the behavior of light concerning plane mirrors and does not explain why the angle of incidence equals the angle of reflection in general. Hence, while both statements are true, the reason given is not the correct explanation for the assertion made about the angles being equal across all mirror types.

Key Concept: Laws of Reflection: Angle of Incidence

c) Assertion is true, but Reason is false.
[Solution Description]
According to the law of reflection, the angle of incidence ($i$) is equal to the angle of reflection ($r$). This law applies universally to all reflective surfaces, such as plane mirrors and spherical mirrors, whether they are concave or convex.
Furthermore, one of the fundamental laws of reflection states that the incident ray, the normal to the surface at the point of incidence, and the reflected ray must all lie in the same plane. If they did not, it would violate the geometric requirement for calculating these angles correctly, as they rely on a common intersection within the same spatial plane. Thus, the Reason provided is false while the Assertion is true because while they discuss related topics, the Reason contradicts the known law about the planar alignment of rays.
Therefore, the correct response is that the Assertion is true, but the Reason is false.

Key Concept: Understanding Incident and Reflected Rays

c) The angle of reflection is 20°.
[Solution Description]
According to the laws of reflection, the angle of incidence is equal to the angle of reflection regardless of the type of mirror. Therefore, when the angle of incidence is 20°, the angle of reflection is also:
$r = i = 20^\circ$
So the angle of reflection is 20°.

Key Concept: Angle of incidence equals angle of reflection, Laws of reflection for different types of mirrors

Correct Answer option Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that for a concave mirror, the angle of incidence is always equal to the angle of reflection. According to the syllabus, it is known that one of the fundamental laws of reflection is that the angle of incidence ($i$) equals the angle of reflection ($r$) for any type of mirror. The reason correctly supports this by stating that the laws of reflection apply universally to all mirrors, including concave ones. Thus, both the assertion and the reason are true statements, and the reason provides the correct explanation for the assertion.

Key Concept: Definition of Terms

c) Normal
[Solution Description] The normal is defined as a line drawn at a right angle (90 degrees) to the surface of a plane mirror at the point where the incident ray strikes it.

Key Concept: Lenses and Their Image Formation

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion is true because a convex lens can indeed form both real and virtual images depending on the position of the object relative to the focal point. When an object is placed within the focal length of a convex lens, it forms a virtual, erect, and enlarged image on the same side as the object. If the object is beyond the focal length, the image formed is real and inverted on the opposite side of the lens.
The reason given is false because a convex lens does not always make objects appear smaller; instead, objects appear larger when they are within the focal length of the lens, which contradicts the statement in the reason. Hence, the correct response is: Assertion is true, but Reason is false.

Key Concept: Convex Lenses and Image Formation

Correct Answer option a) Erect and enlarged
[Solution Description]
When an object is placed at a small distance from a convex lens, according to the properties of lenses, it appears erect and enlarged through the lens. This is because a convex lens can converge light beams onto a point behind it, creating a magnified image if the object is very close.
Therefore, when you look through the lens from the opposite side, this creates an effect similar to that of a magnifying glass where the object appears larger and upright.

Key Concept: Concave Lenses and Image Formation

Correct Answer option a) Erect and diminished
[Solution Description]
A concave lens always forms an image that is erect and diminished in size compared to the actual object being viewed. Concave lenses are diverging lenses, meaning they spread out light rays passing through them. This results in images that are smaller than the real objects.
Since the nature of the lens is to make the light rays diverge, the virtual image formed is always on the same side as the object and appears smaller and upright.

Key Concept: Convergence and Divergence of Light, Types of Lenses

c) Virtual, Erect, and Diminished
[Solution Description]
Using the lens formula:
$\frac{1}{f} = \frac{1}{v} – \frac{1}{u}$
Insert the given values: $f = -15 \, \text{cm}, u = -10 \, \text{cm}$,
$\frac{1}{-15} = \frac{1}{v} + \frac{1}{10}$
Rearranging gives:
$\frac{1}{v} = \frac{1}{-15} – \frac{1}{10} = -\frac{2+3}{30} = -\frac{5}{30} = -\frac{1}{6}$
Therefore, $v = -6 \, \text{cm}$.
Since the image distance is negative, the image is virtual, erect, and diminished in size as it forms on the same side as the object.

Key Concept: Convex and Concave Lenses, Image Formation with Lenses

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that a convex lens can form an enlarged and erect image when the object is placed at a distance from the lens that is less than its focal length. This is true because for objects within the focal point of a convex lens, the image formed is virtual, erect, and magnified.
The reason claims that a concave lens always produces a virtual, erect, and diminished image irrespective of the position of the object. This is also true since a concave lens diverges light rays and thus cannot form real images; it always forms virtual, erect, and smaller images than the object.
While both statements are true, the reason provided does not explain why the convex lens forms an enlarged, erect image as described in the assertion. Therefore, the reason is not the correct explanation for the assertion.

Key Concept: Lenses and Their Image Formation

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that a convex lens can produce both magnified and reduced images based on the object’s distance from the lens. This is true because when an object is placed at different distances such as between the focal point and twice the focal length or beyond twice the focal length, the nature of the image changes. If the object is within the focal length, it produces a virtual, erect, and magnified image; if it’s beyond twice the focal length, the image is real, inverted, and reduced.
The reason claims that convex lenses always cause the image to appear larger due to convergence. While it is true that convex lenses converge light rays, they do not always make the image appear larger. Depending on the object’s position relative to the lens’s focal points, the image may be either magnified or reduced.
Therefore, while both statements are true individually, the reason does not correctly explain the assertion since the divergence or reduction in image size under certain conditions isn’t addressed by the reason.

Key Concept: Image Formation by Lenses

c) Erect and enlarged
[Solution Description] When an object is placed at a small distance from a convex lens, the image appears erect and enlarged. This characteristic makes convex lenses useful for magnification purposes like in magnifying glasses.

Key Concept: Lens Application in Daily Life

Correct Answer option a) Camera, Telescope, Microscope
[Solution Description]
Lenses are widely used in various optical instruments around us. Common examples include eyeglasses, cameras, telescopes, microscopes, and even our eyes. These devices use lenses to focus, magnify, or correct vision. Among the given options, a camera, telescope, and microscope all utilize lenses to perform their respective functions such as focusing light to take pictures or magnifying distant objects for observation.
Hence, these devices rely on the principles of lenses to operate effectively.

Key Concept: Applications of Lenses, Convergence and Divergence of Light

b) Magnification is -4, image is real and inverted
[Solution Description]
Start by finding the image distance $v$ using the lens formula:
$\frac{1}{f} = \frac{1}{v} – \frac{1}{u}$
With $f = 4 \, \text{cm}, u = -5 \, \text{cm}$,
$\frac{1}{4} = \frac{1}{v} + \frac{1}{5}$
Solving for $\frac{1}{v}$:
$\frac{1}{v} = \frac{1}{4} – \frac{1}{5} = \frac{5-4}{20} = \frac{1}{20}$
Hence, $v = 20 \, \text{cm}$.
Magnification $m$ is calculated by:
$m = \frac{v}{u} = \frac{20}{-5} = -4$
The negative sign indicates that the image is inverted and the magnitude shows it’s four times larger than the object.

Key Concept: Definition of Lens

c) A lens is a piece of transparent material with curved surfaces.
[Solution Description] A lens is defined as a piece of transparent material, usually made of glass or plastic, with curved surfaces. This curvature allows the lens to refract light rays and form images.

Key Concept: Types of Lenses

b) Convex lens
[Solution Description] A convex lens is characterized by being thicker in the middle compared to its edges. This shape causes light beams to converge when passing through it.

Key Concept: Convergence and Divergence of Light, Image Formation by Lenses

a) Real, Inverted, and Enlarged
[Solution Description]
The lens formula is given by:
$\frac{1}{f} = \frac{1}{v} – \frac{1}{u}$
Substituting the known values: $f = 20 \, \text{cm}, u = -30 \, \text{cm}$,
$\frac{1}{20} = \frac{1}{v} + \frac{1}{30}$
Solving for $\frac{1}{v}$:
$\frac{1}{v} = \frac{1}{20} – \frac{1}{30} = \frac{3-2}{60} = \frac{1}{60}$
Thus, $v = 60 \, \text{cm}$.
Since the image distance is positive, the image is real and formed on the opposite side of the lens. It will also be inverted.

Key Concept: Concave Mirrors, Reflecting Telescopes

b) 4 meters in front of the mirror
[Solution Description]
When light rays from a distant object such as a star hit the concave mirror, they are considered to be parallel due to the object’s immense distance. The concave mirror focuses these parallel rays to its focal point. The position where the image forms can be determined using the mirror formula:
$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$
For an object at infinity, $u = -\infty$, so:
$\frac{1}{f} = \frac{1}{-\infty} + \frac{1}{v} \implies \frac{1}{v} = \frac{1}{f} + 0$
Hence,
$v = f$
Given that the focal length $f = 4 \, \text{m}$, the image of the star will form 4 meters in front of the mirror. Thus, the image forms exactly at the focal point.

Key Concept: Lenses, Human Eye and Focus Adjustment

b) Lens becomes thinner
[Solution Description]
The human eye adjusts focus using a process called accommodation, facilitated by the ciliary muscles altering the lens’s shape. When viewing nearby objects, the ciliary muscles contract, causing the lens to thicken and increase its power to converge light more strongly.
However, when shifting focus to distant objects, the ciliary muscles relax, resulting in the lens becoming thinner and reducing its converging power. This thinned-out state is suitable for parallel rays from distant sources, aligning them onto the retina properly for clear vision.
Therefore, the correct description of the lens adjusting for distant objects involves it becoming thinner.

Key Concept: Uses of Spherical Mirrors and Lenses

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion is that concave mirrors are used in vehicle headlights to focus light. This is true because concave mirrors can collect and focus the light coming from the bulb, directing it into a beam that illuminates the road ahead.
The reason provided states that a concave mirror converges light beams to a focal point. This statement is also true. A concave mirror has the property of converging parallel light rays to a single focal point due to its curved reflective surface.
Both statements are true, and the reason correctly explains why concave mirrors are suitable for use in vehicle headlights. So, the correct answer is that both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

Key Concept: Uses of Concave Mirrors

c) Concave Mirror
[Solution Description]
Dentists use concave mirrors to get an enlarged view of teeth because concave mirrors can focus light and create a magnified image when the object is within its focal length.
When a dentist holds a concave mirror close to the teeth, it forms an enlarged and erect image which allows the dentist to inspect the teeth closely for any issues.

Key Concept: Applications of Lenses

d) They focus light to form clear images
[Solution Description] Lenses in telescopes are used to refract and focus light from distant objects, allowing the viewer to see clear and enlarged images of celestial bodies. The objective lens gathers light and creates an image, which the eyepiece lens magnifies for observation.

Key Concept: Uses of Convex Mirrors

c) Convex Mirror
[Solution Description]
Convex mirrors are used as side-view mirrors on vehicles. They provide an erect and diminished image that allows drivers to see a larger area behind them. This characteristic helps in reducing blind spots and ensures safer driving by giving a broader view of traffic approaching from behind.

Key Concept: Concave Mirrors, Telescopes

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that concave mirrors are utilized in telescopes for their capability to converge parallel light rays from distant astronomical bodies to a single focal point, enabling focused observation of the universe. This is indeed a true statement as concave mirrors are integral components of reflecting telescopes.
On the other hand, the reason provided talks about the characteristics of a convex mirror, which indeed forms a virtual, erect, and diminished image, but this property does not explain why concave mirrors are used in telescopes. Therefore, while both statements are true, the reason does not serve as a correct explanation for the assertion.

Key Concept: Uses of Lenses in Vision Correction

c) Convex Lens
[Solution Description]
Convex lenses are used in eyeglasses for people with farsightedness (hyperopia). These lenses converge the light rays before they enter the eye, focusing the light correctly onto the retina, thereby allowing a clear image to be formed.
This correction enables people to see near objects clearly, which they otherwise would not be able to due to their vision condition.

Key Concept: Uses of Concave Mirrors

c) Employed as reflectors in car headlights
[Solution Description] Concave mirrors can converge light rays to a focal point, making them ideal for applications requiring focused beams of light. One such application is in vehicle headlights, where they help direct and intensify the light beam projected onto the road.

Key Concept: Uses of Spherical Mirrors and Lenses

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion states that concave mirrors are used in telescopes to focus light from distant celestial bodies, which is true. This is because concave mirrors can gather and converge light to a focal point, making them ideal for observing distant objects in space. The reason, however, claims that convex mirrors are preferred for telescopes due to their wider field of view, which is false. Convex mirrors do indeed provide a wider field of view, but they are not used in telescopes for focusing purposes as they diverge light rather than focus it. Therefore, the Assertion is true, but the Reason is false.

Key Concept: Use of Convex Mirrors

c) To provide a wider view in vehicle side mirrors
[Solution Description] Convex mirrors are known for providing a wider field of view with erect and diminished images, which makes them suitable for scenarios where a comprehensive perspective is needed, such as side-view mirrors on vehicles. This allows drivers to see more traffic behind them compared to using flat or concave mirrors.

Key Concept: Convex Mirrors, Vehicle Side-View Mirrors

c) Divergence of reflected rays creating smaller images
[Solution Description]
Convex mirrors always form virtual, erect, and diminished images regardless of the position of the object. This is because they cause rays of light to diverge upon reflection, giving the appearance that the light comes from a point behind the mirror.
These properties give drivers a broader perspective by displaying a larger area beside and behind the vehicle than would be possible with a flat mirror. The image is reduced in size but remains upright, allowing for effective monitoring of traffic conditions.
The key principle here is the divergence of reflected rays, leading to a diminished yet erect image.

Key Concept: Imaginary Hollow Sphere Concept, Characteristics of Spherical Mirrors

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
In a concave mirror, when an object is placed between the focal point and the mirror, the reflected rays appear to diverge from a point on the same side as the object. When extended backward, these rays seem to converge at a point behind the mirror, resulting in an enlarged virtual image. This property is due to the geometry of the concave mirror, which forms part of an imaginary hollow sphere with the reflecting surface on the inside. Thus, while both statements are true, the reason given does not directly explain why the image is enlarged; it simply describes the construction of the mirror.
Therefore, the correct answer is: Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

Key Concept: Laws of Reflection, Spherical Mirror Types

b) Concave mirror; used in solar cookers for focusing sunlight
[Solution Description] The scenario describes parallel rays converging at a point after reflecting off the mirror, indicating the use of a concave mirror. Concave mirrors focus light to a point and are often used in optical devices like telescopes, vehicle headlights, or solar cookers where concentrating light is beneficial.

Key Concept: Spherical Mirrors, Practical Applications

c) They provide a wider field of view by forming erect, diminished images
[Solution Description] Convex mirrors are widely used in vehicle rearview mirrors due to their ability to provide a wider field of view. They always form images that are erect and diminished, allowing drivers to see more of the surroundings than with a flat or concave mirror. This property helps in ensuring safety by reducing blind spots.

Key Concept: Characteristics of Spherical Mirrors

c) Assertion is true, but Reason is false.
[Solution Description]
In a concave mirror, when the object is placed beyond the focal point, it produces real and inverted images. When the object is between the focal point and the mirror, it produces virtual and erect images. This ability to form different types of images underlies the statement that a concave mirror can produce both real and virtual images. Thus, Assertion is true.
On the other hand, the reason states that the reflective coating in a concave mirror is on the outer surface; however, this is incorrect. In reality, the reflective coating for a concave mirror is applied to the inner surface. Therefore, the Reason is false.
Since Assertion is true but Reason is false, option c) matches the solution.

Key Concept: Characteristics of Images Formed by Spherical Mirrors

c) Erect and enlarged
[Solution Description] When an object is placed very close to a concave mirror (between the pole and the focus), the image formed is erect and enlarged. This occurs because light rays from the object diverge after reflection and appear to come from a larger, upright virtual image located behind the mirror.

Key Concept: Spherical Mirror Characteristics, Image Formation

a) The image is virtual, upright, and magnified
[Solution Description] The situation describes an object placed between the pole and the focal point of a concave mirror. According to the mirror formula:
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
where $f$ is the focal length, $v$ is the image distance, and $u$ is the object distance. Substituting $f = 10$ cm and $u = -5$ cm:
$\frac{1}{10} = \frac{1}{v} + \frac{1}{-5}$
Simplifying:
$\frac{1}{v} = \frac{1}{10} + \frac{1}{5} = \frac{1}{10} + \frac{2}{10} = \frac{3}{10}$
Therefore, $v = \frac{10}{3} \approx 3.33$ cm. Since $v > 0$, the image is real. However, because it appears upright and magnified, we note it is also virtual and located on the same side as the object. This occurs with concave mirrors when the object is within the focal length.

Key Concept: Characteristics of Images Formed by Spherical Mirrors

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
To solve this question, we need to analyze both the assertion and reason separately and verify their correctness.
In the case of a **concave mirror**, it is known that:
– If an object is placed beyond the focal point, the image formed is **real** and **inverted**.
– If the object is within the focal length, the image appears **virtual**, **erect**, and **enlarged**.
Therefore, the assertion is true because a concave mirror indeed has the ability to form both real and virtual images based on where the object is situated in relation to the focal point.
Now examining the reason concerning a **convex mirror**:
– Regardless of how close or far the object is placed, the image is always **diminished**, **erect**, and **virtual** with respect to the object.
Hence, the reason is also true as it correctly describes the behavior of an image formed by a convex mirror.
However, the reason does not explain why a concave mirror can form different types of images; instead, it addresses the characteristics of a convex mirror. Thus, while both statements are true, the reason does not serve as the explanation for the assertion.

Key Concept: Characteristics of Images Formed by Concave Mirrors

c) Enlarged and erect
[Solution Description]
When an object is placed very close to a concave mirror, the rays reflecting from the mirror diverge and meet at a point behind the mirror. This causes the mirror to form an enlarged virtual image that appears upright.

Key Concept: Applications of Spherical Mirrors

c) Convex mirror
[Solution Description]
Convex mirrors are typically used for vehicle side-view mirrors because they provide an erect image, albeit diminished, allowing for a wider field of view, which helps drivers see more area behind them.

Key Concept: Laws of Reflection and Spherical Mirrors

c) The rays converge at a focal point
[Solution Description] When parallel rays of light strike a concave mirror, they are reflected inward and converge at a single point known as the focal point of the mirror. This property arises from the spherical shape of the mirror’s reflective surface, which causes all incident parallel rays to meet or intersect at this common point.

Key Concept: Laws of Reflection and Spherical Mirrors

c) Rays diverge after reflection
[Solution Description]
The laws of reflection state that the angle of incidence equals the angle of reflection. In the case of a convex mirror, the reflected rays diverge as if originating from a focal point behind the mirror. This gives the impression that the source of the light is located behind the mirror, resulting in a diminished image.

Key Concept: Practical Applications of Spherical Mirrors

d) They show a wider field of view
[Solution Description] Convex mirrors are used as side-view mirrors in vehicles because they can cover a wider field of view due to their outward bulging shape. They always form images that are erect and diminished, which allows drivers to see more area and traffic behind them while maintaining smaller image sizes for distant objects.

Key Concept: Concept: Convex and Concave Lenses

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion states that a convex lens can enlarge an object placed close to it, which is true. When an object is placed close to a convex lens, the lens converges light rays making the object appear larger.
The reason states that a concave lens is thicker in the middle compared to its edges, which is false. A concave lens is actually thinner in the middle and thicker at the edges. Therefore, the given reason is incorrect.
Thus, while the assertion is true, the reason provided is false. The correct answer is option c).

Key Concept: Image Characteristics Through Lenses, Concave and Convex Lens Behavior

a) Virtual, erect, 4.29 cm behind the concave lens
[Solution Description]
To determine the final image properties, we use the lens formula:
For the convex lens (focal length $f_1 = 10$ cm):
$\frac{1}{v_1} – \frac{1}{u} = \frac{1}{f_1}$
where $u = -15$ cm; solving,
$\frac{1}{v_1} + \frac{1}{15} = \frac{1}{10}$
$v_1 = \frac{150}{5} = 30\, \text{cm}$
This means the image from the convex lens forms 30 cm on the opposite side.
Now, consider this image as an object for the concave lens, which is 40 cm from the real image position of the first lens (since it is 10 cm behind the convex lens).
For the concave lens (focal length $f_2 = -5$ cm):
$\frac{1}{v_2} – \frac{1}{u_2} = \frac{1}{f_2}$
where $u_2 = -(40 – 10) = -30$ cm; solving,
$\frac{1}{v_2} + \frac{1}{30} = -\frac{1}{5}$
$v_2 = \frac{-30 \times 5}{-35} = \frac{150}{35} \approx 4.29\, \text{cm}$
The final image is virtual, erect, and located approximately 4.29 cm back from the concave lens.

Key Concept: Light Convergence and Image Formation, Convex Lens

a) Curvature of the lens
[Solution Description]
To solve this problem, we begin by understanding that a convex lens converges light rays to a point known as the focal point. The energy required to ignite the paper is given as 300 J. Since the lens focuses sunlight effectively at its focal point, we need to find the power concentration at this point.
Power can be calculated using the formula:
$P = \frac{\text{Energy}}{\text{Time}} = \frac{300\, \text{J}}{20\, \text{s}} = 15\, \text{W}$
Given that all the sunlight passing through the lens is concentrated at the focal point, the intensity of sunlight at this point should be enough to deliver this power output of 15 W in 20 seconds to accumulate 300 J.
The primary property of the lens contributing to this phenomenon is its ability to converge parallel incident light rays into a single point due to its shape and material, i.e., being a convex lens.
Therefore, the property responsible is the lens’s curvature and refractive index, facilitating convergence.

Key Concept: Lenses and Light Divergence and Convergence, Optics of Lenses

b) Remains slightly diverging
[Solution Description]
When a parallel beam of light encounters a concave lens first, the light diverges. After divergence, the light beams spread out as though they originate from a focal point located behind the concave lens.
As these diverged beams enter the convex lens, since the distance between the lenses is less than the sum of their focal lengths, the convex lens will attempt to reconverge these diverging rays towards another focal point. However, because the rays are already diverging upon entering the convex lens, the exiting rays will not converge significantly and may end up remaining parallel or slightly diverging, depending on the exact distances and focal lengths.
Thus, the final outcome depends on whether or how effectively the convex lens can counteract the initial divergence imparted by the concave lens. In typical scenarios under these conditions, light would exit slightly divergent or nearly parallel, indicating more dispersion than significant reconvergence.

Key Concept: Effects of Lens Curvature

a) Erect and enlarged
[Solution Description]
When an object is placed close to a convex lens and seen through it, the object appears erect and enlarged in size. As the distance between the object and the lens increases, the image characteristics change.

Key Concept: Convex Lens and Image Formation

b) Virtual and erect, 30 cm from the lens
[Solution Description]
Using the lens formula, $\frac{1}{f} = \frac{1}{v} – \frac{1}{u}$, where $f$ is the focal length, $v$ is the image distance, and $u$ is the object distance.
Given: $f = 15$ cm, $u = -10$ cm (object distance is taken as negative in lens formula).
Substituting the values into the lens formula:
$\frac{1}{15} = \frac{1}{v} + \frac{1}{10}$
Rearranging gives:
$\frac{1}{v} = \frac{1}{15} – \frac{1}{10} = \frac{2}{30} – \frac{3}{30} = -\frac{1}{30}$
Thus, $v = -30$ cm.
The negative sign indicates the image is formed on the same side as the object, meaning it is virtual and erect. Since $v$ indicates a greater distance than the object distance, the image is enlarged.

Key Concept: Concave Lens and Object Distance

b) Virtual and upright, 12 cm from the lens
[Solution Description]
Using the lens formula again, $\frac{1}{f} = \frac{1}{v} – \frac{1}{u}$,
Given: $f = -20$ cm (concave lenses have a negative focal length), $u = -30$ cm.
Substitute these into the lens formula:
$\frac{1}{-20} = \frac{1}{v} + \frac{1}{30}$
Rearranging gives:
$\frac{1}{v} = \frac{1}{-20} – \frac{1}{30} = -\frac{3}{60} – \frac{2}{60} = -\frac{5}{60} = -\frac{1}{12}$
Thus, $v = -12$ cm.
The negative value for $v$ indicates that the image is virtual and upright, appearing on the same side of the lens as the object. It is also diminished.

Key Concept: Light Convergence/Divergence by Lenses

a) Diverges the light beams
[Solution Description]
A concave lens diverges the light beams that pass through it. This means it spreads out the light rays, making them travel away from each other, hence it is also called a diverging lens.

Key Concept: What Is a Lens?

a) It is thicker at the middle than at the edges
[Solution Description]
A convex lens is characterized by being thicker in the middle compared to its edges. This shape allows it to converge light rays, which is why it is also known as a converging lens.

Key Concept: Converging and Diverging Lenses

c) A convex lens converges light; a concave lens diverges light
[Solution Description]
A convex lens (converging lens) focuses parallel rays of light to a point called the focus. Hence, it converges the light beams.
A concave lens (diverging lens) spreads out parallel rays of light such that they appear to originate from a single point called the focus. Therefore, it diverges the light beams.
Convex lenses are used to converge incoming light, while concave lenses disperse or diverge them.

Key Concept: Convex lens, Image formation

c) Assertion is true, but Reason is false.
[Solution Description]
For a convex lens, when an object is placed at a distance less than its focal length, the light rays diverge on passing through the lens and never meet on the opposite side. Instead, these divergent rays appear to meet on the same side of the lens as the object. Thus, a virtual image is formed that is both enlarged and erect.
The reason given states that convex lenses can only form real images by converging light rays, which is incorrect because they can also form virtual images when the object is within the focal point.
Therefore, while the assertion is true, the reason provided is false.

Key Concept: Lens properties and behavior

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that a convex lens can form both real and virtual images based on where the object is placed relative to the lens. This is true. For example, when an object is positioned beyond the focal point of a convex lens, a real and inverted image is formed. Conversely, if the object is within the focal length, the image appears virtual, upright, and enlarged.
The reason given is about concave lenses, which indeed always create virtual, erect, and smaller images regardless of where the object is located.
While both statements are true, the reason provided does not directly explain why a convex lens can produce both real and virtual images. Thus, the correct option is:

Key Concept: Representation of Light and Reflection

b) Normal
[Solution Description] The normal is the term used for the line drawn perpendicular to the surface of the mirror at the point of incidence. This line is crucial in measuring the angles of incidence and reflection since these angles are measured with respect to the normal.

Key Concept: Experiments, Basic Concepts of Reflection

c) 35°
[Solution Description] Initially, the angle of incidence is $25^\circ$, which means the angle of reflection is also $25^\circ$. When the mirror is rotated $10^\circ$ clockwise, the normal rotates as well. However, since the light ray’s direction remains unchanged, the new angle of incidence becomes $25^\circ + 10^\circ = 35^\circ$. By the law of reflection, the new angle of reflection is equal to the new angle of incidence, which is $35^\circ$.

Key Concept: Laws of Reflection

c) Assertion is true, but Reason is false.
[Solution Description]
According to the laws of reflection, the angle of incidence ($i$) is equal to the angle of reflection ($r$). This applies to all reflective surfaces, including plane mirrors. Therefore, the assertion that the angle of incidence is equal to the angle of reflection is true.
Regarding the reason, a plane mirror forms virtual images, not real ones. Real images are formed when light rays actually converge at a point after reflection or refraction. In contrast, the images seen in plane mirrors appear to be behind the mirror and cannot be projected onto a screen, hence they are virtual.
Since the statement about plane mirrors forming real images is false, the reason provided does not support the assertion. Thus, while the assertion is true, the reason is false.

Key Concept: Plane of Reflection

b) Same plane
[Solution Description] According to the laws of reflection, the incident ray, the normal, and the reflected ray all lie in the same plane. This ensures that the path of light remains predictable and consistent during the reflection process.

Key Concept: Basic Concepts of Reflection, Laws of Reflection

c) 100°
[Solution Description] The angle of incidence ($i$) is measured from the normal to the surface, not from the surface itself. Hence, if the ray makes an angle of $40^\circ$ with the surface, it makes $90^\circ – 40^\circ = 50^\circ$ with the normal. According to the law of reflection, the angle of reflection ($r$) equals the angle of incidence. Therefore, $r = 50^\circ$. The angle between the incident ray and the reflected ray is $i + r = 50^\circ + 50^\circ = 100^\circ$.

Key Concept: Laws of Reflection, Plane and Spherical Mirrors

c) Assertion is true, but Reason is false.
[Solution Description]
The assertion states that in a plane mirror, the angle of incidence ($i$) is always equal to the angle of reflection ($r$). This is one of the fundamental laws of reflection which applies perfectly to plane mirrors where $i = r$. Therefore, the assertion is true.
The reason suggests that light rays reflecting off convex mirrors always converge to a point on the principal axis. However, this statement is false. In reality, light rays reflecting from a convex mirror diverge as if they are coming from a focal point behind the mirror. Convex mirrors do not cause convergence but rather divergence.
Hence, the correct response is that the assertion is true, but the reason is false.

Key Concept: Application of Law of Reflection

b) 60°
[Solution Description]
According to the law of reflection, the angle of incidence ($i$) is equal to the angle of reflection ($r$). Given that the angle of incidence is 30°, we have:
$i = r = 30°$
The angle between the incident ray and the reflected ray can be found by considering the two angles together along the path of light. Therefore, this angle is given by:
Angle between incident ray and reflected ray $= i + r = 30° + 30° = 60°$
Thus, the angle between the incident ray and the reflected ray is 60°.

Key Concept: Effect of Incidence on Plane Mirror

c) 50°
[Solution Description]
Let the angle of incidence be $i$ and the angle of reflection also be $i$, since they are equal as per the law of reflection. If the total angle between the incident and reflected ray is 100°, then:
$i + i = 100°$
Solving for $i$ gives:
$2i = 100°$
$i = \frac{100°}{2} = 50°$
Therefore, the angle of incidence is 50°.

Key Concept: Understanding Ray Diagrams

d) It is perpendicular to the mirror’s surface.
[Solution Description]
The normal line in a ray diagram is defined as a line drawn perpendicular to the surface of the mirror at the point where the incident ray strikes the mirror. This means it forms a 90° angle with the mirror’s surface. Furthermore, the incident ray, the normal, and the reflected ray all lie in the same plane according to the laws of reflection.

Key Concept: Angles of Incidence and Reflection

a) They are equal
[Solution Description] According to the law of reflection, the angle of incidence is always equal to the angle of reflection when light reflects off a plane mirror. Mathematically, this can be expressed as $i = r$. Since both angles are equal, if the angle of incidence is 30 degrees, then the angle of reflection will also be 30 degrees.

Key Concept: Laws of Reflection, Practical Applications

c) The reflected rays converge at a point 15 cm in front of the mirror.
[Solution Description] For a concave mirror, parallel light rays that are incident along the principal axis converge at the focal point after reflection. This is due to the law of reflection where each incident ray hits the mirror surface and reflects such that the angle of incidence equals the angle of reflection. Given that the focal length is 15 cm, any parallel beam hitting the mirror will converge at this focal point, thus demonstrating this property of a concave mirror.

Key Concept: Angles of Incidence and Reflection

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
In the laws of reflection, the angle of incidence ($i$) is indeed equal to the angle of reflection ($r$), which applies to all mirrors, whether they are plane, concave, or convex. This makes the assertion true. However, the reason provided talks about the behavior of light in spherical mirrors (concave and convex). For concave mirrors, light rays parallel to the principal axis converge at a focal point after reflecting. For convex mirrors, such rays appear to diverge from a virtual focal point behind the mirror. This statement does not directly explain why the angles of incidence and reflection are equal; it describes the focusing properties of spherical mirrors. Therefore, while both statements are true, the reason is not the correct explanation of the assertion.

Key Concept: Image Formation by Concave Mirrors

c) Enlarged and virtual
[Solution Description]
When an object is placed very close to a concave mirror, the rays reflected from the mirror appear to diverge from a point behind the mirror. This creates an erect and virtual image. The image is also larger than the object.

Key Concept: Light Convergence in Concave Mirrors

a) They converge at a point
[Solution Description]
A concave mirror converges parallel light beams to a focal point in front of the mirror. This property makes concave mirrors useful for focusing light or heat energy onto a specific point.

Key Concept: Properties of Concave Mirrors, Image Formation

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
[Solution Description]
The assertion states that a concave mirror always forms an inverted image when the object is beyond its focal point. This is true because for objects placed beyond the focal point, the reflected rays converge and form a real and inverted image on the same side as the object.
The reason given is also true. It correctly explains that the nature of a concave mirror’s curved inward surface causes parallel incoming light rays to converge at the focal point after reflection.
Thus, both the assertion and reason are true, and the reason provides a correct explanation of the assertion. Therefore, the correct answer is option a).

Key Concept: Use of Concave Mirror

b) Solar furnace
[Solution Description] A concave mirror focuses parallel light rays to a point called the focus. This property is used effectively for concentrating sunlight to ignite a piece of paper.
When sunlight falls on a concave mirror, it reflects and converges at its focus, generating enough heat to burn paper.

Key Concept: Concave Mirror, Magnification, Object-Image Relationship

b) 45 cm in front of the mirror
[Solution Description]
The magnification $(m)$ produced by a mirror is given by:
$m = -\frac{v}{u}$
where $v$ is the image distance and $u$ is the object distance.
We are given that the magnification is 3 (since the image is three times larger), so:
$3 = -\frac{v}{-15}$
Solving for $v$ gives:
$v = 45 \text{ cm}$
Thus, the image is located 45 cm in front of the mirror.

Key Concept: Magnification by Concave Mirror

c) -1
[Solution Description] The magnification $m$ produced by a mirror is given by the formula:
$m = -\frac{v}{u}$
When the object is at the center of curvature, $u = -R$ and $v = -R$ because the image also forms at the center of curvature.
Thus,
$m = -\frac{-R}{-R} = -1$
The negative sign indicates that the image is inverted, and the magnitude being 1 means the size of the image is equal to the object.

Key Concept: Concave Mirror, Image Formation, Focal Point

a) Real, inverted image at 6.67 cm on the same side
[Solution Description]
To find the nature and position of the image formed by a concave mirror, we use the mirror formula:
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
where $f$ is the focal length, $v$ is the image distance, and $u$ is the object distance. Given:
\begin{align*}
f & = 10 \text{ cm}
u & = -20 \text{ cm}
\end{align*}
(Note: The object distance is negative as per the sign convention.)
Plugging in these values:
$\frac{1}{10} = \frac{1}{v} – \frac{1}{20}$
Simplifying,
$\frac{1}{v} = \frac{1}{10} + \frac{1}{20} = \frac{3}{20}$
Therefore,
$v = \frac{20}{3} \approx 6.67 \text{ cm}$
Since $v$ is positive, the image is real and inverted.

Key Concept: Concave Mirror: Image Formation

c) Assertion is true, but Reason is false.
[Solution Description]
When an object is placed very close to a concave mirror (within its focal length), the mirror forms an erect and enlarged image. This is because the light rays coming from the object converge after reflecting from the concave surface, producing a virtual image on the same side as the object which appears larger.
On the other hand, it is incorrect to say that concave mirrors always form inverted images regardless of the object’s position. The nature of the image depends on the distance of the object from the mirror. Only when the object is beyond the focal point does the image become real and inverted.
Hence, the Assertion is true but the Reason is false.

Key Concept: Image Formation by Concave Mirror

a) Real and inverted
[Solution Description] To determine the type of image formed, we use the mirror equation:
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
where $f = -15$ cm (since it’s a concave mirror and focal length is negative), $u = -10$ cm (object distance is taken as negative in mirror formula convention), and $v$ is the image distance.
Rearranging the equation to find $v$:
$\frac{1}{v} = \frac{1}{f} – \frac{1}{u}$
Substituting the known values:
$\frac{1}{v} = \frac{1}{-15} – \frac{1}{-10} = -\frac{2}{30} + \frac{3}{30} = \frac{1}{30}$
So,
$v = 30 \text{ cm}$
Since the value of $v$ is positive, the image formed is real and inverted.

Key Concept: Concave Mirror: Image Formation

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.
[Solution Description]
The assertion states that an object placed at the focus of a concave mirror forms a highly enlarged and real image. This is true because rays reflecting from the mirror surface become parallel and meet at infinity, thereby forming a real and enlarged image.
The reason mentions that concave mirrors can form virtual images when the object is placed between the focal point and the mirror. This is also true as placing an object between the focal point and the mirror causes diverging rays which appear to come from behind the mirror, thus forming a virtual and erect image.
However, while both statements are true, the reason is not the correct explanation for the assertion. The formation of a real and enlarged image when the object is at the focus is unrelated to the condition mentioned in the reason.

Key Concept: Properties of Concave Mirror Reflections

b) Image diminishes and inverts
[Solution Description]
As an object moves farther from a concave mirror, the image transitions from being enlarged and upright to inverted and reduced in size. Initially, the image appears larger and inverted but continues to shrink as the distance increases until it becomes diminished relative to the object’s size.

Key Concept: Concave Mirror, Real-World Application, Telescope

b) Focal length is 1 meter; enhances image brightness
[Solution Description]
The focal length $(f)$ of a concave mirror is half of its radius of curvature $(R)$:
$f = \frac{R}{2}$
Here, the radius of curvature $R$ is 2 meters. So,
$f = \frac{2}{2} = 1 \text{ meter}$
In reflecting telescopes, this focal length property allows the concave mirror to gather light over a large area and focus it to form sharp images of distant objects. It is crucial in gathering more light and improving the resolution of distant celestial bodies.