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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` |
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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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