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A flashlight of intensity $9 \mathrm{~W} / \mathrm{cm}^2$ illuminates a perfectly reflective surface of area $300 \mathrm{~cm}^2$. The average force exerted on the surface due to the incident light photons is
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$18 \mu \mathrm{N}$
Given, intensity of flash light, $I=9 \mathrm{~W} / \mathrm{cm}^2$
$=9 \times 10^4 \mathrm{~W} / \mathrm{m}^2$
Area, $A=300 \mathrm{~cm}^2=3 \times 10^{-2} \mathrm{~m}^2$
$\therefore$ Radiation pressure due to incident light photon,
$\begin{aligned} p=\frac{2 I}{c} & =\frac{2 \times 9 \times 10^4}{3 \times 10^8} \\ (\because c & \left.=\text { speed of light }=3.0 \times 10^8 \mathrm{~m} / \mathrm{s}\right) \\ p & =6 \times 10^{-4} \mathrm{~N} / \mathrm{m}^2\end{aligned}$
Average force exerted on the surface,
$\begin{aligned} F & =p \times A \\ & =6 \times 10^{-4} \times 3 \times 10^{-2} \\ & =18 \times 10^{-6} \mathrm{~N}=18 \mu \mathrm{N}\end{aligned}$
$=9 \times 10^4 \mathrm{~W} / \mathrm{m}^2$
Area, $A=300 \mathrm{~cm}^2=3 \times 10^{-2} \mathrm{~m}^2$
$\therefore$ Radiation pressure due to incident light photon,
$\begin{aligned} p=\frac{2 I}{c} & =\frac{2 \times 9 \times 10^4}{3 \times 10^8} \\ (\because c & \left.=\text { speed of light }=3.0 \times 10^8 \mathrm{~m} / \mathrm{s}\right) \\ p & =6 \times 10^{-4} \mathrm{~N} / \mathrm{m}^2\end{aligned}$
Average force exerted on the surface,
$\begin{aligned} F & =p \times A \\ & =6 \times 10^{-4} \times 3 \times 10^{-2} \\ & =18 \times 10^{-6} \mathrm{~N}=18 \mu \mathrm{N}\end{aligned}$
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