The radiation pressure exerted by a 450 W light source on a perfectly reflecting surface placed at 2 m away from it, is :

Question

Aicharya · Accepted Answer

$P_{\mathrm{rad}}=\frac{2 I}{\mathrm{C}}$
Where $\mathrm{I}=$ intensity at surface
$\mathrm{C}=$ Speed of light
$\begin{aligned}
& I=\frac{\text { Power }}{\text { Area }}=\frac{450}{4 \pi \mathrm{r}^2} \\ & =\frac{450}{4 \pi \times 4}=\frac{450}{16 \pi} \\ & P_{\mathrm{rad}}=\frac{2 \times 450}{16 \pi \times 3 \times 10^8}=\frac{150}{8 \pi \times 10^8}
\end{aligned}$
$=5.97 \times 10^{-8} \approx 6 \times 10^{-8}$ Pascals

The radiation pressure exerted by a 450 W light source on a perfectly reflecting surface placed at 2 m away…

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