A light with an energy flux of $25 W/cm^2$ falls on a non-reflecting surface at normal incidence. If the surface has an area of $50\, cm^2$ and the total energy falling on the surface is $10 × 10^5 J$. The total duration of time of the incidence of light is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 800 s
Given:
Energy flux $= 25 \, \text{W/cm}^2$
Area $= 50 \, \text{cm}^2$
Total energy $= 10 \times 10^5 \, \text{J}$
Power incident on the surface is:
$P = \text{Energy flux} \times \text{Area} = 25 \times 50 = 1250 \, \text{W}$
Using the formula:
$\text{Time} = \frac{\text{Total energy}}{\text{Power}}$
$t = \frac{10 \times 10^5}{1250}$
$t = \frac{10^6}{1250} = 800 \, \text{seconds}$