A small plane mirror is rotating at constant frequency of n rotations per second. With what linear velocity (in ms-1) will a light spot move along a spherical screen of radius of curvature of R metres if the mirror is at the centre of curvature of the screen?

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the rotation of the mirror
The mirror is rotating at a frequency of \( n \) rotations per second. This means that in one second, it completes \( n \) full rotations.

**Hint:** Remember that one full rotation corresponds to an angle of \( 2\pi \) radians.

### Step 2: Calculate the angular velocity of the mirror
The angular velocity \( \omega \) of the mirror can be calculated using the formula:
\[ \omega = 2\pi n \quad \text{(in radians per second)} \]
This is because for \( n \) rotations, the total angle covered is \( 2\pi n \) radians.

**Hint:** Angular velocity is the rate of change of angle with respect to time.

### Step 3: Relate the angular velocity of the image to the mirror
When the mirror rotates by an angle \( \phi \), the light spot on the screen moves by an angle \( 2\phi \). Therefore, the angular velocity of the image \( \omega_{\text{image}} \) is twice that of the mirror:
\[ \omega_{\text{image}} = 2\omega = 2(2\pi n) = 4\pi n \]

**Hint:** The relationship between the angles is crucial for finding the angular velocity of the image.

### Step 4: Calculate the linear velocity of the light spot
The linear velocity \( v \) of the light spot on the spherical screen can be calculated using the formula:
\[ v = r \cdot \omega \]
where \( r \) is the radius of curvature of the screen, which is given as \( R \). Thus, substituting the angular velocity of the image:
\[ v = R \cdot (4\pi n) \]

**Hint:** Linear velocity is directly proportional to both the radius and the angular velocity.

### Final Answer
The linear velocity of the light spot moving along the spherical screen is given by:
\[ v = 4\pi n R \quad \text{(in m/s)} \]