Light falls on a non-reflecting surface norma | vedclass

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(A) For a non-reflecting (totally absorbing) surface,the radiation pressure $P$ is given by $P = \frac{F}{A} = \frac{I}{c}$,where $I$ is the intensity

Vedclass · Accepted Answer

$20 	imes 10^4 	ext{ Wm}^{-2}$

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(A) For a non-reflecting (totally absorbing) surface,the radiation pressure $P$ is given by $P = \frac{F}{A} = \frac{I}{c}$,where $I$ is the intensity (energy flux) and $c$ is the speed of light.Given:Force $F = 10^{-6} 	ext{ N}$Area $A = 15 	ext{ cm}^2 = 15 	imes 10^{-4} 	ext{ m}^2$Speed of light $c = 3 	imes 10^8 	ext{ ms}^{-1}$Rearranging the formula for intensity $I$:$I = \frac{F \cdot c}{A}$Substituting the values:$I = \frac{10^{-6} 	imes 3 	imes 10^8}{15 	imes 10^{-4}}$$I = \frac{3 	imes 10^2}{15 	imes 10^{-4}}$$I = \frac{3}{15} 	imes 10^6 = 0.2 	imes 10^6 	ext{ Wm}^{-2} = 20 	imes 10^4 	ext{ Wm}^{-2}$.

Light falls on a non-reflecting surface normally. If the average force exerted on a surface with area $15 \text{ cm}^2$ during a $20 \text{ minute}$ time interval is $10^{-6} \text{ N}$,then the energy flux (intensity) of light is (Velocity of light $= 3 \times 10^8 \text{ ms}^{-1}$)

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