A point source of light having a power of 1000 W radiates isotropically in all directions. A small detector having an area of 10 cm ^2 is placed at a distance of 5 m from the source, such that the light falls normally on it. The total energy received by the detector in 1 minute is:

Step-by-step solution

The intensity of light at a distance r from a point source is given by: I = P 4 r^2 Substituting the given values: I = 1000 4 (5)^2 = 1000 100 = 10 W m ⁻² The energy received by the detector is: E = I A t Converting the given area and time into SI units: A = 10 cm ^2 = 10 10⁻⁴ m ^2 = 10⁻³ m ^2 t = 1 minute = 60 s Calculating the energy: E = 10 10⁻³ 60 = 0.6 J Answer: 0.6 J