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

Intensity of electromagnetic wave

$\mathrm{I}=\left(\frac{1}{2}{\in }_0{{\mathrm{E}}_0}^2\right)\mathrm{C} -------\left(1\right)$

Here E₀ = Amplitude of electric field intensity. If ‘P’ is power of source, intensity of electromagnetic radiation at a distance ‘r’

from source is

$\mathrm{I}=\frac{\mathrm{P}}{4{\mathrm{\pi r}}^2}---\left(2\right) ;$

From (1) and (2)

$$\begin{gathered} \frac{1}{2}{\in }_0{{\mathrm{E}}_0}^2\mathrm{C}=\frac{\mathrm{P}}{4{\mathrm{\pi r}}^2}\Rightarrow \mathrm{P}\propto {{\mathrm{E}}_0}^2 ; {\mathrm{E}}_0\propto \sqrt{\mathrm{P}} \\ \sqrt{\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}}=\frac{{\mathrm{E}}_2}{{\mathrm{E}}_1} ; \\ \sqrt{\frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}}=\frac{{\mathrm{E}}_2}{{\mathrm{E}}_1} \\ \sqrt{\frac{50}{100}}=\frac{1}{\sqrt{2}}=\frac{{\mathrm{E}}_2}{{\mathrm{E}}_1} \\ \Rightarrow {\mathrm{E}}_2=\frac{{\mathrm{E}}_1}{\sqrt{2}}=\frac{\mathrm{E}}{\sqrt{2}} \end{gathered}$$