Q 12-08-077JEE MainJEE Main 2023 (30 Jan, Shift 2)Medium
A point source of $100\ \text{W}$ emits light with $5\%$ efficiency. At a distance of $5\ \text{m}$ from the source, the intensity produced by the electric field component is
Answer: (B) $\dfrac1{40\pi}\ \dfrac{\text{W}}{\text{m}^2}$
Light power $=5\ \text{W}$. Intensity at $5\ \text{m}$: $I=\dfrac{5}{4\pi(25)}=\dfrac{1}{20\pi}\ \text{W m}^{-2}$.
The electric and magnetic fields share the energy equally, so the electric part is $\dfrac{1}{40\pi}\ \text{W m}^{-2}$.
Solution by Sreeraj P, M.Sc Physics