Sunlight falls normally on a surface of area | vedclass

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(D) For complete absorption,the force $F$ exerted by light on a surface is given by $F = \frac{I \cdot A}{c}$,where $I$ is the intensity (energy flux)

Vedclass · Accepted Answer

$0.06 \, W/cm^{2}$

Vedclass · Answer

(D) For complete absorption,the force $F$ exerted by light on a surface is given by $F = \frac{I \cdot A}{c}$,where $I$ is the intensity (energy flux),$A$ is the area,and $c$ is the speed of light $(3 	imes 10^{8} \, m/s)$.Given: $A = 36 \, cm^{2} = 36 	imes 10^{-4} \, m^{2}$,$F = 7.2 	imes 10^{-9} \, N$.Rearranging for $I$: $I = \frac{F \cdot c}{A}$.Substituting the values: $I = \frac{7.2 	imes 10^{-9} 	imes 3 	imes 10^{8}}{36 	imes 10^{-4}}$.$I = \frac{21.6 	imes 10^{-1}}{36 	imes 10^{-4}} = 0.6 	imes 10^{3} \, W/m^{2} = 600 \, W/m^{2}$.To convert to $W/cm^{2}$: $I = \frac{600 \, W}{10^{4} \, cm^{2}} = 0.06 \, W/cm^{2}$.

Sunlight falls normally on a surface of area $36 \, cm^{2}$ and exerts an average force of $7.2 \times 10^{-9} \, N$ within a time period of $20 \, minutes$. Considering a case of complete absorption,the energy flux of incident light is:

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