A radiation of energy 'E'falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (C = Velocity of light)

Options

A. `(2E)/(C )`
B. `(2E )/(C^2)`
C. `(E )/(C ^2)`
D. `(E )/( C )`

Explanation

To solve the problem of finding the momentum transferred to a perfectly reflecting surface when radiation of energy 'E' falls normally on it, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Initial Momentum**:
When radiation of energy 'E' falls on a perfectly reflecting surface, it carries momentum. The momentum (p) of electromagnetic radiation can be expressed as:
\[ p = \frac{E}{c} \]
where \(E\) is the energy of the radiation and \(c\) is the speed of light.

2. **Momentum of Incoming Radiation**:
The incoming radiation has an initial momentum directed towards the surface. Therefore, the initial momentum \(p_i\) is:
\[ p_i = \frac{E}{c} \]

3. **Momentum of Reflected Radiation**:
When the radiation reflects off the surface, it reverses direction. The momentum of the reflected radiation \(p_r\) is:
\[ p_r = -\frac{E}{c} \]
(the negative sign indicates the change in direction).

4. **Calculating the Change in Momentum**:
The change in momentum (\(\Delta p\)) is given by the difference between the final momentum and the initial momentum:
\[ \Delta p = p_r - p_i \]
Substituting the values we have:
\[ \Delta p = \left(-\frac{E}{c}\right) - \left(\frac{E}{c}\right) = -\frac{E}{c} - \frac{E}{c} = -\frac{2E}{c} \]

5. **Interpreting the Result**:
The negative sign indicates that the momentum is transferred in the opposite direction of the incoming radiation. However, if we are interested in the magnitude of the momentum transferred to the surface, we can express it as:
\[ \Delta p = \frac{2E}{c} \]

### Final Answer:
The momentum transferred to the perfectly reflecting surface is:
\[ \Delta p = \frac{2E}{c} \]