Question

If a plane electromagnetic wave of intensity $9 \times 10^5\text{ W}\cdot\text{m}^{-2}$ incidents normally on a perfectly absorbing surface of area $2\text{ m}^2$ for a time of $180\text{ s}$, then the average force exerted by the electromagnetic wave on the surface during this time is:

• A. $3\text{ mN}$
• B. $12\text{ mN}$
• C. $6\text{ mN}$
• D. $9\text{ mN}$

Hint:

For electromagnetic waves: \[ F=\frac{IA}{c} \] for a perfectly absorbing surface, while \[ F=\frac{2IA}{c} \] for a perfectly reflecting surface. The given time interval does not affect the force because it cancels during the calculation of momentum transfer per unit time.

Answer 1

The Correct Option is C

Step 1: Recall radiation pressure on an absorber.
An electromagnetic wave carries momentum. On a perfectly absorbing surface the radiation pressure is $P_r = \dfrac{I}{c}$, where $I$ is intensity and $c$ the speed of light.
Step 2: Relate pressure to force.
Force is pressure times area, $F = P_r A = \dfrac{IA}{c}$.
Step 3: List the data.
$I = 9\times 10^5\,Wm^{-2}$, $A = 2\,m^2$, $c = 3\times 10^8\,ms^{-1}$. The exposure time of $180\,s$ does not affect the average force.
Step 4: Substitute.
$F = \dfrac{(9\times 10^5)(2)}{3\times 10^8} = \dfrac{18\times 10^5}{3\times 10^8}$.
Step 5: Simplify.
$F = 6\times 10^{-3}\,N$.
Step 6: Convert and conclude.
$6\times 10^{-3}\,N = 6\,mN$, which is option (3).
\[ \boxed{F = 6\,mN} \]