Questions
The electric field intensity produced by the radiations coming from 100 W bulb at a 3 m distance is E. The electric field intensity produced by the radiations coming from 50 W bulb at the same distance is
detailed solution
Correct option is C
Intensity of electromagnetic waveI=12∈0E02C -------(1)Here E0 = Amplitude of electric field intensity. If ‘P’ is power of source, intensity of electromagnetic radiation at a distance ‘r’from source isI=P4πr2---(2) ; From (1) and (2)12∈0E02C=P4πr2⇒P∝E02 ; E0∝P P2P1=E2E1 ; P2P1=E2E1 50100=12=E2E1 ⇒E2=E12=E2Talk to our academic expert!
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A point source of electromagnetic radiation has an average power output of 800 W. The maximum value of electric field at a distance 3.5 m from source will be
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