Hello students, these questions are from the subject of Physics of Class 10. These questions are from the chapter Reflection and mirror topics. These questions will really help the students who are looking for test covering all the topics. Yes, this test covers all the topics reflection topic. All the Best.
 |
| Solution of J K Online Classes |
Here is the solution to the questions:
C.B.S.E.
Class X (2023 - 24)
Science (Physics)
Reflection - Mirror
Total Marks - 76 Marks
Time Duration: 1 Hour 30 Minutes
Section A
1. Fill in the blanks: (5 Marks)
a. The laws of reflection state that the incident ray, the reflected ray, and the normal to the surface at the point of incidence, all lie in the same plane.
b. In a concave mirror, when the object is located between the pole and the focal point, a virtual and enlarged image is formed.
c. The focus of a spherical mirror is the point on the principal axis where parallel rays of light converge or appear to diverge from.
d. In a convex mirror, the image formed is always upright, virtual, and smaller in size compared to the object.
e. The mirror formula for a concave mirror is given by 1/f = 1/v - 1/u, where 'f' is the focal length, 'v' is the image distance, and 'u' is the object distance.
Section B
2. Short Answer Type Questions: (1 Marks Each)
a. Plane mirrors have a flat reflecting surface, while spherical mirrors have a curved reflecting surface.
b. The different types of spherical mirrors are concave (or converging) mirrors and convex (or diverging) mirrors.
c. The radius of curvature of a spherical mirror is the radius of the sphere of which the mirror is a part. It is denoted by the symbol 'R.'
d. The focal length of a spherical mirror is the distance between the focus and the pole of the mirror. It is denoted by the symbol 'f.'
e. The relationship between the radius of curvature (R) and the focal length (f) of a spherical mirror is given by the formula: f = R/2.
f. Focal length (f) = Radius of curvature (R) / 2
Therefore, f = 30 cm / 2 = 15 cm.
3. Long answer type Questions: (3 marks each)
a. Reflection is the phenomenon of bouncing back of light when it strikes the surface of an object. The laws of reflection are:
- The incident ray, the reflected ray, and the normal to the surface at the point of incidence, all lie in the same plane.
- The angle of incidence is equal to the angle of reflection.
b. Sign conventions used in spherical mirrors are:
- All distances are measured from the pole of the mirror.
- Distances measured in the direction of incident light are taken as positive, while distances measured in the opposite direction are taken as negative.
- Heights above the principal axis are taken as positive, while heights below the principal axis are taken as negative.
c. In a spherical mirror, the pole is the center of the mirror's reflecting surface. The principal axis is an imaginary line passing through the pole and the center of curvature. The focus is the point on the principal axis where parallel rays of light converge or appear to diverge from.
d. A virtual image is an image formed by the apparent intersection of rays of light after reflection or refraction. It cannot be obtained on a screen. Virtual images are formed by diverging (convex) mirrors or when the object is located inside the focal point of a converging (concave) mirror.
e. A virtual image is an image formed by the apparent intersection of rays of light after reflection or refraction. It cannot be obtained on a screen. Virtual images are formed by diverging (convex) mirrors or when the object is located inside the focal point of a converging (concave) mirror.
f. Characteristics of an image formed by a convex mirror:
- It is always virtual.
- It is always erect.
- It is diminished in size.
- It is formed behind the mirror.
g. The mirror formula for a concave mirror is given by: 1/f = 1/v - 1/u, where 'f' is the focal length, 'v' is the image distance, and 'u' is the object distance.
h. Differences between a real image and a virtual image:
- Real Image: Formed by the actual intersection of light rays, can be obtained on a screen, and is always inverted. It is formed by converging mirrors or converging lenses.
- Virtual Image: Formed by the apparent intersection of light rays, cannot be obtained on a screen, and can be either upright or inverted. It is formed by diverging mirrors or diverging lenses.
i. In relation to images formed by mirrors:
- Erect: When the image appears upright.
- Inverted: When the image appears upside-down compared to the object.
- Enlarged: When the image appears larger than the object.
- Diminished: When the image appears smaller than the object.
j. Uses of concave and convex mirrors in daily life:
- Concave mirrors: Used as shaving mirrors, makeup mirrors, and in headlights of vehicles to get a wider field of view.
- Convex mirrors: Used as rear-view mirrors in vehicles, in department stores for surveillance, and in blind spots to provide a wider field of view.
k. A periscope works based on the principle of reflection. It consists of two plane mirrors set at an angle to each other. Light enters through one end of the periscope and undergoes multiple reflections between the mirrors, allowing the viewer to see objects that are not directly visible. Periscopes are commonly used in submarines, military applications, and some optical instruments.
Section C
4. Numerical-Based Questions: (4 marks each)
a. Given: Object distance (u) = -20 cm, Focal length (f) = 15 cm.
Using the mirror formula: 1/f = 1/v - 1/u
Substituting the values: 1/15 = 1/v - 1/-20
Solving the equation, we find v = -12 cm.
The negative sign indicates that the image is formed on the same side as the object, which means it is a virtual image.
b. Given: Object distance (u) = 20 cm, Focal length (f) = -10 cm.
Using the mirror formula: 1/f = 1/v - 1/u
Substituting the values: 1/-10 = 1/v - 1/20
Solving the equation, we find v = -20 cm.
The negative sign indicates that the image is formed on the same side as the object, which means it is a virtual image.
c. Magnification (m) = Height of the image (háµ¢) / Height of the object (hâ‚’)
Given: Object distance (u) = 10 cm, Image distance (v) = -15 cm.
The magnification can be calculated as m = -v / u = -(-15 cm) / 10 cm = 1.5.
The negative sign indicates that the image is inverted compared to the object.
d. Given: Focal length (f) = 20 cm, Object distance (u) = 30 cm.
Using the mirror formula: 1/f = 1/v - 1/u
Substituting the values: 1/20 = 1/v - 1/30
Solving the equation, we find v = -60 cm.
The negative sign indicates that the image is formed on the same side as the object, which means it is a virtual image.
e. Given: Focal length (f) = -15 cm, Object distance (u) = 40 cm.
Using the mirror formula: 1/f = 1/v - 1/u
Substituting the values: 1/-15 = 1/v - 1/40
Solving the equation, we find v = -30 cm.
The negative sign indicates that the image is formed on the same side as the object, which means it is a virtual image.
f. Radius of curvature (R) = 40 cm.
Focal length (f) = R/2 = 40 cm / 2 = 20 cm.
g. Given: Object distance (u) = 25 cm, Image distance (v) = -10 cm.
Using the mirror formula: 1/f = 1/v - 1/u
Substituting the values: 1/f = 1/-10 - 1/25
Solving the equation, we find 1/f = -1/10 - 1/25
Combining the terms, we get 1/f = -3/50
Taking the reciprocal, f = -50/3 cm.
The negative sign indicates that the focal length is negative, indicating a convex mirror.
h. Given: Focal length (f) = 25 cm, Object distance (u) = 10 cm.
Magnification (m) = -v/u
Using the mirror formula: 1/f = 1/v - 1/u
Substituting the values: 1/25 = 1/v - 1/10
Solving the equation, we find v = -20 cm.
The negative sign indicates that the image is formed on the same side as the object, which means it is a virtual image.
Magnification (m) = -v/u = -(-20 cm)/10 cm = 2.
**POST SCRIPT**
We hope you enjoyed solving these questions. These questions will be very helpful from the examination point of view for all students, whether they are looking for just getting pass.
Okay Folks before leaving one last thing: Beware of Cyber-Fraudulent.
 - Solution){kind=link}
0 Comments