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(C) The intensity $I$ of an electromagnetic wave at a distance $d$ from a point source of power $P$ is given by $I = \frac{P}{4\pi d^2}$.Also,the inte

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$\frac{E}{\sqrt{2}}$

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(C) The intensity $I$ of an electromagnetic wave at a distance $d$ from a point source of power $P$ is given by $I = \frac{P}{4\pi d^2}$.Also,the intensity is related to the peak electric field $E_m$ by the relation $I = \frac{1}{2} \epsilon_0 E_m^2 c$.Equating these,we get $\frac{P}{4\pi d^2} = \frac{1}{2} \epsilon_0 E_m^2 c$.Since $4\pi, d^2, \epsilon_0,$ and $c$ are constant for a fixed distance,we have $E_m^2 \propto P$,which implies $E_m \propto \sqrt{P}$.Given $P_1 = 100\ W$ and $P_2 = 50\ W$,and the corresponding electric fields are $E_1 = E$ and $E_2 = E'$,we have $\frac{E'}{E} = \sqrt{\frac{P_2}{P_1}}$.Substituting the values,$\frac{E'}{E} = \sqrt{\frac{50}{100}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.Therefore,$E' = \frac{E}{\sqrt{2}}$.

For an $EM$ wave,the electric field intensity due to a bulb of $100\ W$ at a distance of $3\ m$ is $E$. What is the electric field intensity produced by radiations coming from a $50\ W$ bulb at the same distance?

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