Question

A plane electromagnetic wave of wave intensity $6\, W / m ^{2}$ strikes a small mirror area $40\, cm ^{2}$, held perpendicular to the approaching wave. The momentum transferred by the wave to the mirror each second will be

• A. $6.4 \times 10^{-7} kg - m / s ^{2}$
• B. $4.8 \times 10^{-8} kg - m / s ^{2}$
• C. $3.2 \times 10^{-9} kg - m / s ^{2}$
• D. $1.6 \times 10^{-10} kg - m / s ^{2}$

Answer 1

Answer: (D)

Momentum transferred in one second
$P=\frac{2 U}{c}=\frac{2 S_{a v} A}{c}$
$=\frac{2 \times 6 \times 40 \times 10^{-4}}{3 \times 10^{8}}$
$=1.6 \times 10^{-10} kg - m / s ^{2}$