Question

A plane electromagnetic wave has an intensity of $6\frac{\text{W}}{{\text{m}}^2}$ strikes a small pocket mirror which has an area $40{\text{cm}}^2$ held perpendicular to the approaching wave. Then, what momentum does the wave transfer to the mirror each second?

• A. $1.6\times {10}^{-11}\frac{kgm}{s^2}$
• B. $1.6\times {10}^{-10}\frac{kgm}{s^2}$
• C. $1.6\times {10}^{-9}\frac{kgm}{s^2}$
• D. $1.6\times {10}^{10}\frac{kgm}{s^2}$

Answer 1

The correct option is B $1.6\times {10}^{-10}\frac{kgm}{s^2}$

EM waves transport momentum. The wave is reflected by the mirror. The momentum transferred to the mirror per unit area per second is twice the momentum of the light striking the mirror per unit area per second.

Since, $S=6\frac{\text{W}}{{\text{m}}^2},A=40{\text{cm}}^2=4\times {10}^{-3}{\text{m}}^2$

So, as per the concept after reflection the momentum becomes twice, so the momentum per unit time is,

$\frac{P}{t}=\frac{2SA}{c}$

i.e. $\frac{P}{t}=\frac{2\times 6\times 4\times {10}^{-3}}{3\times {10}^8}=1.6\times {10}^{-10}\frac{kgm}{s^2}$