Average force exerted on a non-reflecting sur | vedclass

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(B) For a non-reflecting surface,the radiation pressure $P$ is given by $P = \frac{I}{c}$,where $I$ is the intensity and $c$ is the speed of light.Sin

Vedclass · Accepted Answer

$0.02$

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(B) For a non-reflecting surface,the radiation pressure $P$ is given by $P = \frac{I}{c}$,where $I$ is the intensity and $c$ is the speed of light.Since $P = \frac{F}{A}$,we have $\frac{F}{A} = \frac{I}{c}$.Given intensity $I = 360 \,W/cm^2 = 360 	imes 10^4 \,W/m^2 = 3.6 	imes 10^6 \,W/m^2$.Given force $F = 2.4 	imes 10^{-4} \,N$ and speed of light $c = 3 	imes 10^8 \,m/s$.Substituting the values: $\frac{2.4 	imes 10^{-4}}{A} = \frac{3.6 	imes 10^6}{3 	imes 10^8}$.$\frac{2.4 	imes 10^{-4}}{A} = 1.2 	imes 10^{-2}$.$A = \frac{2.4 	imes 10^{-4}}{1.2 	imes 10^{-2}} = 2 	imes 10^{-2} \,m^2 = 0.02 \,m^2$.

Average force exerted on a non-reflecting surface at normal incidence is $2.4 \times 10^{-4} \,N$. If $360 \,W/cm^2$ is the light energy flux during a span of $1$ hour $30$ minutes,then the area of the surface is: (in $\,m^2$)

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