Question

If $2.5\times {10}^{-6}N$ average force is exerted by a light wave on a non-reflecting surface of $30c{m}^2$ area , the energy flux of light just before it falls on the surface is _______ $\raisebox{1ex}{$W$}\!\left/ \!\raisebox{-1ex}{$c{m}^2$}\right.$. (Round off to the nearest integer) (Assume complete absorption and normal incidence conditions are there)

Answer 1

Step 1: Given data:

Radiation force= $2.5\times {10}^{-6}N$

Surface of contact= $30c{m}^2$

Step 2: Formula

$$\begin{gathered} {\mathrm{F}}_{\mathrm{radiation}}=\frac{\mathrm{P}}{\mathrm{c}} \\ \mathrm{Where}\mathrm{F}\mathrm{is}\mathrm{radiation}\mathrm{force},\mathrm{P}\mathrm{is}\mathrm{power},\mathrm{C}\mathrm{is}\mathrm{speed}\mathrm{of}\mathrm{light} \\ \mathrm{P}=\mathrm{I}\times \mathrm{A} \\ \mathrm{where}\mathrm{I}\mathrm{is}\mathrm{The}\mathrm{energy}\mathrm{flux}\mathrm{of}\mathrm{light},\mathrm{A}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{contact}. \end{gathered}$$

Step 3: Calculation

$F=\frac{IA}{C}$, where $F=$Radiation force, $C=3\times {10}^{8}m/s(Thespeedoflight)$, $A=$Surface of contact, and $I=$The energy flux of light.

$I=\frac{F\times C}{A}$

$I=\frac{2.5\times {10}^{-6}}{30c{m}^2}N\times 3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

$I=25\raisebox{1ex}{$W$}\!\left/ \!\raisebox{-1ex}{$c{m}^2$}\right.$

Hence, the answer is $25\raisebox{1ex}{$W$}\!\left/ \!\raisebox{-1ex}{$c{m}^2$}\right.$.