Question:
The average force exerted on $5 cm^2$ area of a non-reflecting plate in 10 minutes when light with an energy flux of $20\,W\,cm^{-2}$ incidents normally on it is}
The average force exerted on $5 cm^2$ area of a non-reflecting plate in 10 minutes when light with an energy flux of $20\,W\,cm^{-2}$ incidents normally on it is}
Show Hint
For an absorbing surface,
\[
F=\frac{P}{c}
\]
and for a perfectly reflecting surface,
\[
F=\frac{2P}{c}
\]
Updated On: Jun 17, 2026
- $3.33\times10^{-7}N$
- $6.66\times10^{-4}N$
- $2.5\times10^{-7}N$
- $5.55\times10^{-10}N$
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The Correct Option is A
Solution and Explanation
Concept:
For a perfectly absorbing surface,
\[
P_r=\frac{I}{c}
\]
where \(P_r\) is radiation pressure.
The force exerted is
\[
F=P_rA
\]
Step 1: Calculate the total power incident.
\[ I=20\,Wcm^{-2} \] \[ A=5\,cm^2 \] \[ Power=IA \] \[ Power=20\times5=100W \]
Step 2: Calculate radiation force.
\[ F=\frac{Power}{c} \] \[ F=\frac{100}{3\times10^8} \] \[ F=3.33\times10^{-7}N \] \[ \boxed{3.33\times10^{-7}N} \]
Step 1: Calculate the total power incident.
\[ I=20\,Wcm^{-2} \] \[ A=5\,cm^2 \] \[ Power=IA \] \[ Power=20\times5=100W \]
Step 2: Calculate radiation force.
\[ F=\frac{Power}{c} \] \[ F=\frac{100}{3\times10^8} \] \[ F=3.33\times10^{-7}N \] \[ \boxed{3.33\times10^{-7}N} \]
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