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:

Question

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) (1/40 π) (W/m2) (B) (1/10 π) (W/m2) (C) (1/20 π) (W/m2) (D) (1/2 π) (W/m2)

Tardigrade Admin · Accepted Answer

I EF =(1/2) × (5/4 π × 52) =(1/40 π) W / m 2

Q. 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:

$\frac{1}{40 π} \frac{W}{m ^{2}}$

$\frac{1}{10 π} \frac{W}{m ^{2}}$

$\frac{1}{20 π} \frac{W}{m ^{2}}$

$\frac{1}{2 π} \frac{W}{m ^{2}}$

$I_{EF} = \frac{1}{2} \times \frac{5}{4 π \times 5 ^{2}}$
$= \frac{1}{40 π} W / m^{2}$

Q. A point source of 100W emits light with 5% efficiency. At a distance of 5m from the source, the intensity produced by the electric field component is:

40π1​m2W​

10π1​m2W​

20π1​m2W​

2π1​m2W​