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50 \mathrm{~W} / \mathrm{m}^2 energy density of sunlight is normally incident on the surface of a solar panel. Some part of incident energy (25 %) is reflected from the surface and the rest is absorbed. The force exerted on 1 \mathrm{~m}^2 surface area will be close to \mathrm{\left(c=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)}

Option: 1

20 \times 10^{-8} \mathrm{~N}\\


Option: 2

10 \times 10^{-8} \mathrm{Nx}\\


Option: 3

35 \times 10^{-8} \mathrm{~N}\\


Option: 4

15 \times 10^{-8} \mathrm{~N}


Answers (1)

best_answer

Given energy density =50 \mathrm{~W} / \mathrm{m}^2

Here, change in momentum \mathrm{\Delta \mathrm{p}=\mathrm{p}_{\mathrm{f}}-\mathrm{p}_{\mathrm{i}}}

\begin{aligned} & \Delta \mathrm{p}=\left(-\mathrm{p}_{\mathrm{i}} / 4\right)-\mathrm{p}_{\mathrm{i}} \\ \\& \Delta \mathrm{p}=-5 \mathrm{p}_{\mathrm{i}} / 4 \because \mathrm{p}_{\mathrm{i}}=\mathrm{E} / \mathrm{c}=50 \mathrm{~W} / \mathrm{s} /\left(3 \times 10^8\right) \\ \\& \Delta \mathrm{p} / \Delta \mathrm{t}=\mathrm{F}=\left|-5 \mathrm{p}_{\mathrm{i}} / 4\right|=(5 / 4) \times\left(50 /\left(3 \times 10^8\right)=20.8 \times 10^{-8} \mathrm{~N} \approx 20 \times 10^{-8} \mathrm{~N}\right. \end{aligned}

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Rishabh

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