Light falls on a non-reflecting surface normally. If the average force exerted on a surface with area $15…
Light falls on a non-reflecting surface normally. If the average force exerted on a surface with area $15 \mathrm{~cm}^2$ during 20 minute time interval is $10^{-6} \mathrm{~N}$, then energy flux of light is
(Velocity of light $=3 \times 10^8 \mathrm{~ms}^{-1}$ )
$20 \times 10^4 \mathrm{Wm}^{-2}$
$15 \times 10^4 \mathrm{Wm}^{-2}$
$25 \times 10^4 \mathrm{Wm}^{-2}$
$10 \times 10^4 \mathrm{Wm}^{-2}$
Solution
Radiation pressure on a totally absorbing (non reflecting) surface is
Pressure $=\frac{F}{A}=\frac{I}{c}$...(i)
Where, $I=$ Intensity,$c=$ speed of light
it is given that,
$\begin{aligned} & F=10^{-6} \mathrm{~N} \\ & A=15 \mathrm{~cm}^2=15 \times 10^{-4} \mathrm{~m}^2\end{aligned}$
Time for which radiation falls on surface
continuously, $t=20 \mathrm{~min}$
$=20 \times 60 \mathrm{~s}=1200 \mathrm{~s}$
Speed of light, $c=3 \times 10^8 \mathrm{~m} / \mathrm{s}$
From Eq. (i) we have,
$I=\frac{F \times C}{A}$
$\Rightarrow \quad I=\frac{10^{-6} \times 3 \times 10^8}{15 \times 10^{-4}}$
$\begin{aligned} & =2 \times 10^5 \mathrm{~W} / \mathrm{m}^2 \\ & =20 \times 10^4 \mathrm{~W} / \mathrm{m}^2\end{aligned}$