A light with an energy flux of \( 25\,\mathrm{W/cm^{2}} \) falls on a non-reflecting surface at normal incidence. If the surface has an area of \( 50\,\mathrm{cm^{2}} \) and the total energy falling on the surface is \( 10 \times 10^{5}\,\mathrm{J} \). The total duration of time of the incidence of light is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 800\,\mathrm{s} \)
The energy flux is the energy that crosses unit area per unit time, so the power falling on the surface is the flux times the area, and the total energy is the power times the time.
\( \displaystyle P = \text{flux} \times A = 25\,\mathrm{W\,cm^{-2}} \times 50\,\mathrm{cm^{2}} = 1250\,\mathrm{W} \)
\( \displaystyle E = P\,t \;\Rightarrow\; t = \frac{E}{P} = \frac{10\times10^{5}\,\mathrm{J}}{1250\,\mathrm{W}} = 800\,\mathrm{s} \)