Question:
Light is incident normally on a completely absorbing surface with an energy flux of \(25\ \mathrm{W\,cm^{-2}}\). If the surface has an area of \(25\ \mathrm{cm^2}\), the momentum transferred to the surface in \(40\) min time duration will be :
Light is incident normally on a completely absorbing surface with an energy flux of \(25\ \mathrm{W\,cm^{-2}}\). If the surface has an area of \(25\ \mathrm{cm^2}\), the momentum transferred to the surface in \(40\) min time duration will be :
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For complete absorption, momentum transferred = energy absorbed / c.
Updated On: Nov 5, 2026
- \(6.3\times10^{-4}\ \mathrm{Ns}\)
- \(5.0\times10^{-3}\ \mathrm{Ns}\)
- \(1.4\times10^{-6}\ \mathrm{Ns}\)
- \(3.5\times10^{-6}\ \mathrm{Ns}\)
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The Correct Option is B
Solution and Explanation
Power received: \(P=I A=25\ \mathrm{W/cm^2}\times25\ \mathrm{cm^2}=625\ \mathrm{W}\).
Time: \(t=40\times60=2400\ \mathrm{s}\), so energy \(U=625\times2400=1.5\times10^{6}\ \mathrm{J}\).
For a perfectly absorbing surface, momentum transferred \(p=\dfrac{U}{c}=\dfrac{1.5\times10^{6}}{3\times10^{8}}=5.0\times10^{-3}\ \mathrm{Ns}\).
Answer: (2) \(5.0\times10^{-3}\ \mathrm{Ns}\)
Time: \(t=40\times60=2400\ \mathrm{s}\), so energy \(U=625\times2400=1.5\times10^{6}\ \mathrm{J}\).
For a perfectly absorbing surface, momentum transferred \(p=\dfrac{U}{c}=\dfrac{1.5\times10^{6}}{3\times10^{8}}=5.0\times10^{-3}\ \mathrm{Ns}\).
Answer: (2) \(5.0\times10^{-3}\ \mathrm{Ns}\)
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