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Question: Answered & Verified by Expert
Light with an energy flux of $18 \mathrm{~W} / \mathrm{cm}^2$ falls on a non-reflecting surface at normal incidence. The pressure exerted on the surface is
PhysicsThermal Properties of MatterAIIMSAIIMS 2009
Options:
  • A $2 \mathrm{~N} / \mathrm{m}^2$
  • B $2 \times 10^{-4} \mathrm{~N} / \mathrm{m}^2$
  • C $6 \mathrm{~N} / \mathrm{m}^2$
  • D $6 \times 10^{-4} \mathrm{~N} / \mathrm{m}^2$
Solution:
2308 Upvotes Verified Answer
The correct answer is: $6 \times 10^{-4} \mathrm{~N} / \mathrm{m}^2$
Radiation pressure $P_{\text {rad }}$ due to light falling on a non-reflecting surface at normal incidence is given by
$\begin{aligned} & P_{\text {rad }}=\frac{\text { Energy flux }}{\text { Speed of light }}=\frac{18 \mathrm{~W} / \mathrm{cm}^2}{3 \times 10^8 \mathrm{~m} / \mathrm{s}} \\ & =\frac{18 \times 10^4 \mathrm{~W} / \mathrm{m}^2}{3 \times 10^8 \mathrm{~m} / \mathrm{s}}=6 \times 10^{-4} \mathrm{~N} / \mathrm{m}^2\end{aligned}$

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