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:
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:
Show 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.
Updated On: Jun 15, 2026
- $3\text{ mN}$
- $12\text{ mN}$
- $6\text{ mN}$
- $9\text{ mN}$
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The Correct Option is C
Solution and Explanation
Concept:
Electromagnetic waves carry both energy and momentum. When an electromagnetic wave falls on a surface, it transfers momentum to the surface and exerts a force known as radiation force.
For a perfectly absorbing surface, the radiation pressure is
\[
P_r=\frac{I}{c}
\]
where
• $I$ = intensity of the electromagnetic wave
• $c$ = speed of light in vacuum
The force exerted on a surface of area $A$ is \[ F=P_rA \] Therefore, \[ F=\frac{IA}{c} \]
Step 1: Write the expression for force For a perfectly absorbing surface, \[ F=\frac{IA}{c} \] Given, \[ I=9\times10^5\text{ W m}^{-2} \] \[ A=2\text{ m}^2 \] \[ c=3\times10^8\text{ m s}^{-1} \]
Step 2: Substitute the values \[ F= \frac{(9\times10^5)(2)} {3\times10^8} \] \[ F= \frac{18\times10^5} {3\times10^8} \] \[ F= 6\times10^{-3}\text{ N} \]
Step 3: Convert into millinewton Since \[ 1\text{ mN}=10^{-3}\text{ N} \] we get \[ F=6\text{ mN} \] Hence, the average force exerted by the electromagnetic wave on the surface is \[ \boxed{6\text{ mN}} \]
• $I$ = intensity of the electromagnetic wave
• $c$ = speed of light in vacuum
The force exerted on a surface of area $A$ is \[ F=P_rA \] Therefore, \[ F=\frac{IA}{c} \]
Step 1: Write the expression for force For a perfectly absorbing surface, \[ F=\frac{IA}{c} \] Given, \[ I=9\times10^5\text{ W m}^{-2} \] \[ A=2\text{ m}^2 \] \[ c=3\times10^8\text{ m s}^{-1} \]
Step 2: Substitute the values \[ F= \frac{(9\times10^5)(2)} {3\times10^8} \] \[ F= \frac{18\times10^5} {3\times10^8} \] \[ F= 6\times10^{-3}\text{ N} \]
Step 3: Convert into millinewton Since \[ 1\text{ mN}=10^{-3}\text{ N} \] we get \[ F=6\text{ mN} \] Hence, the average force exerted by the electromagnetic wave on the surface is \[ \boxed{6\text{ mN}} \]
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