Question

Electromagnetic wave beam of power 20 mW is incident on a perfectly absorbing body for 300 ns. The total momentum transferred by the beam to the body is equal to

• A. 2 × 10^–17 Ns
• B. 1 × 10^–17 Ns
• C. 3 × 10^–17 Ns
• D. 5 × 10^–17 Ns

Answer 1

The Correct Option is A

Momentum Transfer Calculation

1. Total Energy Incident:

The total energy incident is: \[ E = P \cdot t \] where \( P \) is the power and \( t \) is the time.

2. Initial Momentum:

The initial momentum of the radiation is: \[ p_{\text{initial}} = \frac{P \cdot t}{c} \] where \( c \) is the speed of light (\( c = 3 \times 10^8 \, \text{m/s} \)).

3. Final Momentum:

The final momentum is zero because the radiation is absorbed.

4. Total Momentum Transferred:

The total momentum transferred to the object is: \[ \Delta p = \frac{P \cdot t}{c} \] Substituting the values: \[ P = 20 \times 10^{-3} \, \text{W}, \, t = 300 \times 10^{-9} \, \text{s}, \, c = 3 \times 10^8 \, \text{m/s} \] \[ \Delta p = \frac{20 \times 10^{-3} \cdot 300 \times 10^{-9}}{3 \times 10^8} \] Simplifying: \[ \Delta p = \frac{6 \times 10^{-9}}{3 \times 10^8} = 2 \times 10^{-17} \, \text{Ns} \]

Final Answer:

The correct answer is (A): 2 × 10^–17 Ns.