A point object 'O' is at the center of curvature of a concave mirror. The mirror starts to move at a speed u, in a direction perpendicular to the principal axis. Then, the initial velocity of the image is

Options

A. 2u, in the direction opposite to theat of mirror's velocity
B. 2u, in the direction same as that of mirror's velocity
C. zero
D. u, in the direction same as that of mirror's velocity

Explanation

To solve the problem of finding the initial velocity of the image when a concave mirror moves with a speed \( u \) in a direction perpendicular to the principal axis, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- A point object \( O \) is located at the center of curvature \( C \) of a concave mirror.
- The image formed by the concave mirror when the object is at the center of curvature is also at the center of curvature, inverted, and of the same size.

2. **Identify the Initial Conditions**:
- The object is at rest, and the mirror is moving upwards with a speed \( u \).

3. **Use the Magnification Formula**:
- The magnification \( m \) of the mirror is given by the formula:
\[ m = -\frac{v}{u} \]
- Here, \( v \) is the image distance, and \( u \) is the object distance. Since both the object and image are at the center of curvature, we have \( v = u \).

4. **Differentiate the Magnification Equation**:
- Differentiate the magnification equation with respect to time \( t \):
\[ \frac{dm}{dt} = -\left(\frac{dv}{dt} \cdot \frac{1}{u} - \frac{v}{u^2} \cdot \frac{du}{dt}\right) \]
- Since \( u \) is constant (the object is stationary), \( \frac{du}{dt} = 0 \).

5. **Relate the Velocities**:
- The height of the object \( H_0 \) and the height of the image \( H_i \) are equal when the object is at the center of curvature. Thus, we can write:
\[ \frac{dH_0}{dt} = -\frac{dH_i}{dt} \]
- Let \( v_o \) be the velocity of the object (which is 0) and \( v_i \) be the velocity of the image.

6. **Express the Velocities**:
- Since the object is stationary, we have:
\[ v_o = 0 \quad \text{(object is at rest)} \]
- The mirror moves with velocity \( u \), so we can express the velocity of the image as:
\[ v_i - u = -(-u) \quad \text{(since the object is at rest)} \]
- Rearranging gives:
\[ v_i = 2u \]

7. **Conclusion**:
- The initial velocity of the image is \( 2u \) in the direction of the mirror's motion.

### Final Answer:
The initial velocity of the image is \( 2u \).