If a 100 watt bulb acts as a point source and has 2.5% efficiency, the intensity of this bulb at 3m is

Options

A. `2.9Wm^-1`
B. `5.8Wm^-1`
C. `1.5Wm^-1`
D. `4.4Wm^-1`

Answer

a

Explanation

In em wave, half of the intensity is
due to electric field vector and half is due to
magnetic field vector. Therefore, intensity due to
electric field vector in e.m. wave is
`=I/2=1/2 in_0 E_(rms)^2 c=1/2xx(0.022W//m^2)`
or `E_(rms)= sqrt((0.022)/(in_(0)C)) =sqrt(0.022/(8.85xx10^(-12)xx3xx10^8))`
`=2.9 V//m`