50, W/m^2 energy density of sunlight is norma | vedclass

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(D) The intensity of sunlight is $I = 50\, W/m^2$.For a surface,the radiation pressure $P$ is given by $P = \frac{I}{c}(1 + r)$,where $r$ is the refle

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$20 	imes 10^{-8}\, N$

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(D) The intensity of sunlight is $I = 50\, W/m^2$.For a surface,the radiation pressure $P$ is given by $P = \frac{I}{c}(1 + r)$,where $r$ is the reflection coefficient.Here,$r = 0.25$ (since $25\%$ is reflected) and the absorption coefficient is $a = 0.75$.The pressure exerted is $P = \frac{I}{c} 	imes a + \frac{2I}{c} 	imes r$.$P = \frac{I}{c} (0.75 + 2 	imes 0.25) = \frac{I}{c} (0.75 + 0.50) = 1.25 \frac{I}{c}$.Given $I = 50\, W/m^2$ and $c = 3 	imes 10^8\, m/s$,the pressure is $P = 1.25 	imes \frac{50}{3 	imes 10^8} = \frac{62.5}{3} 	imes 10^{-8} \approx 20.83 	imes 10^{-8}\, N/m^2$.Since the area $A = 1\, m^2$,the force $F = P 	imes A = 20.83 	imes 10^{-8}\, N$.This value is closest to $20 	imes 10^{-8}\, N$.

$50\, W/m^2$ 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 $1\, m^2$ surface area will be close to $(c = 3 \times 10^8\, m/s)$.

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