A point source of 100 W emits light with 5% efficiency. At a distance of 5 m from the source, the intensity produced by the electric field component is: (a) (1)/(40π) W/m² (b) (1)/(10π) W/m² (c) (1)/(20π) W/m² (d) 1/2π W/m²

Answer: (a). Radiated power = 5% of 100 W = 5 W Total intensity I = (P)/(4π r²) = (5)/(4π × 25) = (1)/(20π) W/m² Energy is shared equally by E and B fields, so the electric part is I/2 = (1)/(40π) W/m² So the answer is (a) (1)/(40π) W/m².

A point source of 100 W emits light with 5% efficiency. At a distance… | JEE Main 2023 (30 Jan, Shift 2)

Q14JEE Main 2023 · 30 JAN · SHIFT 2Medium

Energy Density & Intensity

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:

Show answer and solution

Answer: (a)

Radiated power $= 5\%$ of $100\ \text{W} = 5\ \text{W}$
Total intensity $I = \dfrac{P}{4\pi r^2} = \dfrac{5}{4\pi \times 25} = \dfrac{1}{20\pi}\ \text{W/m}^2$
Energy is shared equally by $E$ and $B$ fields, so the electric part is $\dfrac{I}{2} = \dfrac{1}{40\pi}\ \text{W/m}^2$
So the answer is (a) $\dfrac{1}{40\pi}\ \text{W/m}^2$ .

Report a mistake

Energy Density & Intensity question from JEE Main 2023 (30 Jan, Shift 2)

More questions on Energy Density & Intensity