Question

Define the principal axis of a concave mirror.

Answer 1

Step 1: Understand the structure of a concave mirror:
A concave mirror is a spherical mirror that is curved inward, like the inner surface of a sphere.

Step 2: Key parts of a concave mirror:
- Pole (P): The midpoint of the mirror’s surface.
- Center of Curvature (C): The center of the imaginary sphere of which the mirror is a part.

Step 3: Define the principal axis:
- The principal axis is the straight line that passes through both the pole (P) and the center of curvature (C) of the mirror.
- It is a reference line used to study the behavior of light rays reflecting from the mirror.

Step 4: Conclusion:
The principal axis of a concave mirror is the straight line passing through the pole of the mirror and its center of curvature.

Answer 2

Step 1: Understand the behavior of light in a concave mirror:
- A ray of light that is incident parallel to the principal axis of a concave mirror always reflects and passes through the focus (F) of the mirror.

Step 2: Interpret the given information:
- It is given that the reflected ray passes through a point on the principal axis that is 10 cm from the pole (P) of the mirror.
- Therefore, the focal length (f) of the concave mirror is 10 cm.

Step 3: Use the relationship between focal length and radius of curvature:
For spherical mirrors, the relation is:
\[ R = 2f \]

Step 4: Substitute the known value:
\[ R = 2 \times 10\,\text{cm} = 20\,\text{cm} \]

Step 5: Conclusion:
The radius of curvature of the concave mirror is 20 cm.

Answer 3

Step 1: Understand the mirror type and sign convention:
- The given mirror is a convex mirror.
- For convex mirrors:
    - Focal length (f) is considered positive.
    - Object distance (u) is taken as negative (since the object is in front of the mirror).

Step 2: Write the given values:
Object distance, u = -10 cm
Focal length, f = +15 cm

Step 3: Use the mirror formula:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substitute the known values:
\[ \frac{1}{15} = \frac{1}{v} + \frac{1}{-10} \]

Step 4: Solve the equation:
\[ \frac{1}{v} = \frac{1}{15} + \frac{1}{10} = \frac{2}{30} + \frac{3}{30} = \frac{5}{30} = \frac{1}{6} \]

Step 5: Find the image distance:
\[ v = 6 \text{ cm} \]

Step 6: Interpret the result:
Since the value of \( v \) is positive, it means the image is formed behind the convex mirror.
The image is virtual, erect, and diminished.

Final Answer:
The image is formed at a distance of 6 cm behind the mirror.

Answer 4

The mirror that forms a virtual, erect, and diminished image of an object is a convex mirror.
Ray Diagram:

Explanation:
1. Ray 1 (Red): A ray parallel to the principal axis appears to diverge from the focus (F) after reflection.
2. Ray 2 (Green): A ray directed towards the center of curvature (C) is reflected back along the same path.
The image (I) is formed where the reflected rays appear to meet when extended behind the mirror. It is virtual, erect, and diminished.