Test: Reflection by Spherical Mirror - Class 10 MCQ

Consider a plane-mirror and a fixed incident ray of light reflecting from the surface at an angle θi. Before the mirror has rotated, the angle of incidence is θ as is the angle of reflection. If the mirror is rotated through an angle φ the normal is rotated by an angle &phi and thus the angle of incidence increases to θ+φ. Therefore, the angle of reflection must also increase by φ to θ+φ The difference between the final angle of reflection and the initial angle of reflection and is 2φ.
Hence, For a fixed incident ray, the angle of the reflection is twice the angle through which the mirror has rotated: thetaf= 2φ

An incident ray which passes through the focus of the mirror is reflected parallel to the principal axis. An incident ray which passes through the centre of curvature of the mirror is reflected back along its own path (since it is normally incident on the mirror).

angle of incidence= angle of reflection

• First Law of Reflection: This law states that the angle of incidence is equal to the angle of reflection. This law is valid for spherical mirrors as well.


• Second Law of Reflection: This law states that the incident ray, the reflected ray, and the normal to the surface of the mirror all lie in the same plane. This law is also valid for spherical mirrors.


• Conclusion: Both laws of reflection are valid for spherical mirrors. Therefore, the correct answer is option A: Both laws of reflection are valid.

Image formed by convex mirror:

• Convex mirrors always form virtual images.
• These images are erect, meaning they are not inverted.
• Virtual images are formed when the reflected light rays diverge and appear to meet at a point behind the mirror.
• These images cannot be projected on a screen as they are not real.

• Therefore, the nature of the image formed by a convex mirror is always virtual and erect.

 

The focal length (f) and radius of curvature (R) of a spherical mirror are related by the equation:
R = 2f
Given:
f = 12 cm
Calculate the radius of curvature:
R = 2 × 12 cm
R = 24 cm
Thus, the correct answer is:C: 24 cm

The distance between mirror and the screen will give the focal length of the mirror as the mirror focuses the light on the screen.

The image formed by a convex mirror is always virtual and erect irrespective of the position of the object.

A ray which is parallel to the principal axis after reflection from  mirror will always pass through principal focus. The distance between  principal focus and pole is known as focal length and is half the centre of curvature of the mirror.

• Objective: The correct answer is Option B:
• Option A: (a) is correct because the pole is the centre of the reflecting surface itself, not the centre of curvature.
• Option C Principal Axis - Line passing through the pole and the centre of curvature. This is because the principal axis is indeed defined as the straight line passing through the pole and the centre of curvature of a spherical mirror.
• Option D: (d) is incorrect because the radius of curvature is the distance between the centre of curvature and the mirror, not the distance between the principal focus and the mirror. Thus, the correctly matched pair is Option B: (b).

The position of the image of a point object formed by a spherical mirror is determined by the intersection of at least two reflected rays. These rays help locate the image of the object, showcasing the principles of reflection in optics.

- When an object is placed between the focus (F) and the center of curvature (C) of a concave mirror:
- The image formed is on the opposite side of the mirror.
- It is real because it can be projected on a screen.
- The image is inverted relative to the object.
- It is enlarged, meaning it appears bigger than the object itself.
- Therefore, option C ("Real, inverted and enlarged") is the correct answer.

To find the height of the image formed by a concave mirror, we use the mirror formula and magnification formula:
Mirror formula: 1/f = 1/v + 1/u
Magnification (m) = -v/u = h'/h
Given:
Object height (h) = 5 cm
Object distance (u) = -20 cm (object distance is negative in the mirror formula)
Focal length (f) = -10 cm (focal length is negative for concave mirrors)
Step-by-step solution:
Using 1/f = 1/v + 1/u:
1/(-10) = 1/v + 1/(-20)
-1/10 = 1/v - 1/20
-1/10 + 1/20 = 1/v
-2/20 + 1/20 = 1/v
-1/20 = 1/v
v = -20 cm
Calculate magnification (m):
m = -v/u = -(-20)/(-20) = 1
Image height (h') = m * object height (h):
h' = 1 * 5 cm = 5 cm
Since the image is formed on the same side as the object (inverted), the height is negative:
h' = -5 cm
Therefore, the correct answer is:
C: -15 cm

Use the mirror formula:
1/v + 1/u = 1/f
Here, u = -10 cm (object distance, negative for concave mirror) and f = -5 cm (focal length, negative for concave mirror)
Calculate v (image distance):
1/v = 1/f - 1/u = -1/5 + 1/10 = -1/10
v = -10 cm
Use magnification formula:
m = -v/u = -(-10)/(-10) = -1
Size of image = m × object height = -1 × 2 cm = -2 cm