Question:
Average force exerted on a non-reflecting surface at normal incidence is $2.4\times10^{-4}\ N$. If $360\ W/cm^2$ is the light energy flux during span of 1 hour 30 minutes, Then the area of the surface is:
Average force exerted on a non-reflecting surface at normal incidence is $2.4\times10^{-4}\ N$. If $360\ W/cm^2$ is the light energy flux during span of 1 hour 30 minutes, Then the area of the surface is:
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For total absorption the radiation force is $F=\dfrac{IA}{c}$.
Updated On: Oct 26, 2026
- $20\ m^2$
- $0.2\ m^2$
- $0.1\ m^2$
- $0.02\ m^2$
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The Correct Option is D
Solution and Explanation
For a perfectly absorbing (non-reflecting) surface at normal incidence, the force is $F=\dfrac{IA}{c}$, where $I$ is the intensity (energy flux per unit area per unit time).
$I=360\ W/cm^2=360\times10^4\ W/m^2=3.6\times10^6\ W/m^2$.
$A=\dfrac{Fc}{I}=\dfrac{2.4\times10^{-4}\times3\times10^8}{3.6\times10^6}=0.02\ m^2$.
(The duration of 1 hour 30 minutes is not needed for the average force.) Option (4).
$I=360\ W/cm^2=360\times10^4\ W/m^2=3.6\times10^6\ W/m^2$.
$A=\dfrac{Fc}{I}=\dfrac{2.4\times10^{-4}\times3\times10^8}{3.6\times10^6}=0.02\ m^2$.
(The duration of 1 hour 30 minutes is not needed for the average force.) Option (4).
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