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: Use photons:
Think of the radiation as photons. Each photon of energy $\epsilon$ has momentum $\epsilon/c$. For a total energy $E$ the total momentum is $E/c$.

Step 2: Reflection:
A perfectly reflecting surface sends every photon straight back. So each photon changes its momentum from $+p$ to $-p$.

Step 3: Impulse on the surface:
The surface receives the momentum change, $2p$ per photon. Summed over all photons, \[ \Delta p_{surface} = 2 \times \frac{E}{c} \]

Step 4: Compare:
For an absorbing surface the value would be $E/c$. The reflecting case doubles this, so it is $2E/c$, the first option.

Final Answer:
The transferred momentum is $2E/c$. \[ \boxed{\frac{2E}{c}} \]