Question

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

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

Hint:

Radiation momentum is \(E/c\). Reflection reverses it, so the change is doubled.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Electromagnetic radiation carries momentum as well as energy. For radiation of energy \(E\), the momentum is \(p = E/c\).

Step 2: Key Formula or Approach:
Momentum transferred equals the change in momentum of the radiation. For total reflection at normal incidence the direction reverses.

Step 3: Detailed Explanation:
Initial momentum of radiation: \(+E/c\). Final momentum after reflection: \(-E/c\). The change is
\[ \Delta p = \frac{E}{c} - \left(-\frac{E}{c}\right) = \frac{2E}{c} \]
By conservation of momentum the surface gains \(2E/c\).

Step 4: Check the options:
\(E/c\) is the result for a perfect absorber, where the radiation only stops. \(E/c^2\) and \(2E/c^2\) have the wrong units for momentum. Only \(2E/c\) fits.

Final Answer:
The surface gets a momentum of \(2E/c\). \[ \boxed{\frac{2E}{c}} \]