Question:
If 2.5 \(\times\) 10\(^{-6}\) N average force is exerted by a light wave on a non-reflecting surface of 30 cm\(^2\) area during 40 minutes of time span, the energy flux of light just before it falls on the surface is ___________ W/cm\(^2\). (Round off to the Nearest Integer)
(Assume complete absorption and normal incidence conditions are there)
If 2.5 \(\times\) 10\(^{-6}\) N average force is exerted by a light wave on a non-reflecting surface of 30 cm\(^2\) area during 40 minutes of time span, the energy flux of light just before it falls on the surface is ___________ W/cm\(^2\). (Round off to the Nearest Integer)
(Assume complete absorption and normal incidence conditions are there)
(Assume complete absorption and normal incidence conditions are there)
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For complete absorption the radiation pressure is \(\frac{I}{c}\), so \(F=\frac{IA}{c}\); the time given is not needed.
Updated On: Nov 4, 2026
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Correct Answer: 25
Solution and Explanation
For a perfectly absorbing surface at normal incidence, the radiation pressure is \(P=\frac{I}{c}\), so the force is \(F=PA=\frac{IA}{c}\).
\(I=\frac{Fc}{A}=\frac{2.5\times10^{-6}\times3\times10^{8}}{30}=\frac{750}{30}=25\ \mathrm{W/cm^2}\).
(The 40-minute duration is not required because the force is an average force.)
Answer: 25
\(I=\frac{Fc}{A}=\frac{2.5\times10^{-6}\times3\times10^{8}}{30}=\frac{750}{30}=25\ \mathrm{W/cm^2}\).
(The 40-minute duration is not required because the force is an average force.)
Answer: 25
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