Question

The radiation of energy \( E \) falls normally on a perfectly reflecting surface. The momentum transferred to the surface is:

• A. \( \frac{E}{c} \)
• B. \( \frac{2E}{c} \)
• C. \( \frac{E}{c^2} \)
• D. \( \frac{2E}{c^2} \)

Hint:

For a perfectly reflecting surface, the momentum change is doubled because the radiation undergoes a complete reversal in direction.

Answer 1

The Correct Option is B

Step 1: Understanding the radiation momentum transfer
The momentum \( p \) of a photon is given by the relation: \[ p = \frac{E}{c} \] where \( E \) is the energy of the radiation and \( c \) is the speed of light.
Step 2: Effect of a perfectly reflecting surface
When radiation falls normally on a perfectly reflecting surface, the total momentum transfer is due to the change in momentum. Since the surface reflects the radiation completely, the momentum change is: \[ \Delta p = p_{\text{final}} - p_{\text{initial}} \] For a perfectly reflecting surface, the photon momentum reverses direction, meaning the change in momentum is: \[ \Delta p = \frac{E}{c} - (-\frac{E}{c}) = \frac{E}{c} + \frac{E}{c} = \frac{2E}{c} \] Thus, the total momentum transferred to the surface is: \[ \frac{2E}{c} \] Final Answer: The correct option is \( \frac{2E}{c} \), which corresponds to option (2).