JEE Main · 2015 · 04 Apr 2015 · Offline · EM Waves

The diode pours out $0.1$W uniformly, so at $r=1$m that power is smeared over a sphere and the intensity is $I=\frac{P}{4\pi r^2}$. For a travelling electromagnetic wave the same intensity equals the time-averaged flux carried by a field of amplitude $E_0$, namely $\frac{1}{2}{\epsilon }_0{E}_0^{2}c$. Equating the two and solving, $E_0=\sqrt{\frac{2P}{4\pi r^{2}c{\epsilon }_0}}=\sqrt{\frac{P}{2\pi r^{2}c{\epsilon }_0}}$. Putting in $P=0.1$W, $r=1$m, $c=3\times 10^8$m/s and ${\epsilon }_0=8.85\times 10^{-12}$F/m gives $E_0\approx 2.45$V/m.