If 2.5 times 10^{-6} N average force is exer | vedclass

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(D) For a non-reflecting surface (complete absorption) with normal incidence,the force $F$ exerted by light is given by $F = \frac{P}{c}$,where $P$ is

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(D) For a non-reflecting surface (complete absorption) with normal incidence,the force $F$ exerted by light is given by $F = \frac{P}{c}$,where $P$ is the power and $c$ is the speed of light.Since power $P = I 	imes A$,where $I$ is the intensity (energy flux) and $A$ is the area,the force equation becomes $F = \frac{I A}{c}$.Rearranging for intensity $I$,we get $I = \frac{F c}{A}$.Given values: $F = 2.5 	imes 10^{-6} \ N$,$A = 30 \ cm^{2}$,and $c = 3 	imes 10^{8} \ m/s$.Substituting these values: $I = \frac{2.5 	imes 10^{-6} 	imes 3 	imes 10^{8}}{30} = \frac{7.5 	imes 10^{2}}{30} = \frac{750}{30} = 25 \ W/cm^{2}$.The energy flux is $25 \ W/cm^{2}$.

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