The radiation pressure 1 ,m away from a 330 , | vedclass

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Question

(B) The intensity $I$ of radiation at a distance $r$ from a point source of power $P$ is given by $I = \frac{P}{4 \pi r^2}$.Given $P = 330 \,W$ and $r

Vedclass · Accepted Answer

$8.75 	imes 10^{-8} \,Pa$

Vedclass · Answer

(B) The intensity $I$ of radiation at a distance $r$ from a point source of power $P$ is given by $I = \frac{P}{4 \pi r^2}$.Given $P = 330 \,W$ and $r = 1 \,m$,we have:$I = \frac{330}{4 	imes 3.14 	imes 1^2} = \frac{330}{12.56} \approx 26.27 \,W/m^2$.The radiation pressure $P_{rad}$ for a perfectly absorbing surface is given by $P_{rad} = \frac{I}{c}$,where $c = 3 	imes 10^8 \,m/s$ is the speed of light.$P_{rad} = \frac{26.27}{3 	imes 10^8} \approx 8.75 	imes 10^{-8} \,Pa$.

The radiation pressure $1 \,m$ away from a $330 \,W$ electric bulb is

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