1.

If a 100 watt bulb acts as a point source and has 2.5% efficiency, the intensity of this bulb at 3m isA. `2.9Wm^-1`B. `5.8Wm^-1`C. `1.5Wm^-1`D. `4.4Wm^-1`

Answer» Correct Answer - a
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`


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